Economy

ENGINEERING ECONOMICS

Ayhan Gerçeker

CHEMICAL ENGINEERING ECONOMICSI. INTRODUCTION

II. PLANT DESIGN PROCESS

III. GENERAL CONSIDERATIONS

IV. CAPITAL INVESTMENT

V. PRODUCT COST – ECONOMIC PRODUCTION

VI. TIME VALUE OF MONEY

VII. PROFITABILITY

VIII. ANALYSIS OF ALTERNATIVES

IX. ANALYSIS OF REPLACEMENTS

X. INFLATION

XI. OPTIMIZATION

XII. PROJECT MANAGEMENT

XIII. BASIC ACCOUNTING

XIV. COST CONTROL

I. INTRODUCTION


* ANY PRODUCT DEVELOPED OR PRODUCED WHICH CANNOT BE MARKETED AT A PRICE ABOVE ITS COST IS A FAILURE

* ENGINEERS SHOULD NEVER FORGET THAT ECONOMICS IS THE ONLY GUIDE IN INDUSTRY BECAUSE ALL ENGINEERING PRACTICES REQUIRE THE USE OF LARGE AMOUNT OF CAPITAL; AND CAPITAL OWNERS EXPECT MAXIMUM RETURN ON THEIR CAPITAL INVESTED

THEREFORE

* SELECTION OF ALTERNATIVES TO ALLOCATE CAPITAL

* DETERMINATION OF CAPACITY, PROCESS AND EQUIPMENT

* RUNNING THE PLANT

SHOULD BE QUIDED BY ECONOMIC PRINCIPLES IN INDUSTRY

SO ENGINEERS SHOULD KNOW ENGINEERING ECONOMY


THE LITERATURE ON ENGINEERING ECONOMY STARTED BY WELLINGTON’S ‘THE ECONOMIC THEORY OF LOCATION OF RAILWAYS’ IN 1887

IN 1920 : GOLDMAN - ‘FINANCIAL PLANNING’

FISH - ‘ENGINEERING ECONOMICS’

IN 1930 : E.L.GRANT (USUALLY CALLED THE FATHER OF ENGINEERING ECONOMY) PUBLISHED ‘PRINCIPLES OF ENGINEERING ECONOMY’

IN INDUSTRY AN ENGINEER MAY ENCOUNTER QUESTIONS LIKE:

* WITH LIMITED CAPITAL AVAILABLE, WHICH INVESTMENT ALTERNATIVE SHOULD BE SELECTED?

* WHICH ONE OF SEVERAL COMPETING ENGINEERING DESIGNS SHOULD BE SELECTED?

* SHOULD THE MACHINE (OR PROCESS) NOW IN USE BE REPLACED WITH A NEW ONE?

ENGINEERING ECONOMICS* WOULD IT BE PREFERABLE TO PURSUE A SAFER COURSE OF

ACTION OR TO FOLLOW A RISKIER ONE THAT OFFERS HIGHER POTENTIAL RETURNS?

* HOW MANY UNITS OF PRODUCT HAVE TO BE SOLD BEFORE A PROFIT CAN BE MADE?

* SHOULD A PLANT PRODUCE A PART IN ITS OWN FACILITY, KNOWING THAT INVESTMENT WILL BE NEEDED, OR SHOULD THE PLANT SUBCONTRACT (MAKE OR BUY) ?

ENGINEERS SHOULD BE ABLE FIND THE CORRECT ANSWERS TO THESE QUESTIONS

THE CORE OF ENGINEERING ECONOMICS IS DECISION MAKING BASED ON COMPARISONS OF THE WORTH OF ALTERNATIVE COURSES OF ACTION WITH RECPECT TO THEIR COSTS

IN THIS COURSE WE WILL STUDY THE ECONOMIC PRINCIPLES THAT IS USED FOR DESIGNING PLANTS OR PROCESSES IN INDUSTRY AND GUIDE THE ENGINEERS IN THEIR DECISIONS.


EFFICIENCY

THE OBJECTIVE OF ENGINEERING IS TO GET THE GREATEST END RESULT PER UNIT OF RESOURCE EXPENDITUTE.

ENGINEER MUST BE CONCERNED WITH TWO LEVELS OF EFFICIENCY. ON THE FIRST LEVEL IS PHYSICAL EFFICIENCY EXPRESSED AS OUTPUT DIVIDED BY INPUTS OF SUCH PHYSICAL UNITS AS BTU’S, KILOWATTS OR KG.

PHYSICAL EFFICIENCY = OUTPUT / INPUT

THIS KIND OF EFFICIENCY IS ALWAYS LESS THAN UNITY.

ON THE SECOND LEVEL ARE ECONOMIC EFFICIENCIES. THESE ARE EXPRESSED IN TERMS OF ECONOMIC UNITS SUCH AS MONEY.

ECONOMIC EFFICIENCY = WORTH / COST

ECONOMIC EFFICIENCIES CAN EXCEED UNITY AND MUST DO SO FOR ECONOMIC PROJECTS TO BE SUCCESSFUL.

II. PLANT DESIGN PROCESS

PLANT DESIGN PROCESS

DETERMINATION OF THE PROCESS AND CAPACITIES, SIZES AND KINDS OF PROCESS EQUIPMENT TO BE USED BEFORE THE CONSTRUCTION AND OPERATION OF A PLANT IS CALLED PROCESS DESIGN

DEVELOPMENT OF A NEW PLANT OR PROCESS FROM CONCEPT EVALUATION TO PROFITABLE REALITY IS OFTEN AN ENORMOUSLY COMPLEX PROBLEM. A PLANT DESIGN PROJECT MOVES TO COMPLETION THROUGH A SERIES OF STAGES:

1. INSPECTION OF THE BASIC IDEA AND INITIAL ANALYSIS

2. MARKET SURVEY AND PRELIMINARY EVALUATION

3. PROCESS DEVELOPMENT (MAY INCLUDE LAB EXPERIMENT OR PILOT PLANT)

4. PRELIMINARY DESIGN (INCLUDING THE NECESSARY INFORMATION TO ESTIMATE THE INVESTMENT AND PRODUCTION COST WITH A NARROW ERROR MARGIN)

5. FEASIBILITY STUDY AND ECONOMIC EVALUATION

PLANT DESIGN PROCESS 6. DECISION 7. DETAILED ENGINEERING DESIGN 8. PROCUREMENT 9. ERECTION10. START-UP AND TRIAL RUNS11. PRODUCTION

AFTER EACH PHASE OF THE DESIGN, AN EVALUATION SHOULD BE MADE AND ENGINEERS SHOULD BE REALISTIC AND PRACTICAL; SHOULD ELIMINATE UNPROFITABLE VENTURES BEFORE THE DESIGN PROJECT APPROACHES ITS FINAL STAGES. COST ESTIMATES SHOULD BE MADE THROUGHOUT ALL THE EARLY STAGES USING VARIOUS COST ESTIMATION METHODS COVERING BOTH INVESTMENT COSTS AND PRODUCTION COSTS. DESIGN ENGINEERS SHOULD KEEP IN MIND THAT THE AIM OF INVESTING MONEY IS TO HAVE A RETURN COVERING THE TIME VALUE OF MONEY AND REASONABLE AFTER-TAX PROFIT.


WHEN THERE ARE SEVERAL ALTERNATIVE METHODS WHICH CAN BE USED FOR OBTAINING THE SAME RESULTS, THE PREFERRED METHOD SHOULD BE THE ONE INVOLVING THE LEAST TOTAL COSTS. THIS IS CALLED OPTIMUM DESIGN.

AN IMPORTANT STAGE IN THE DESIGN PROCESS IS THE FEASIBILITY STUDY WHICH LEADS TO FINAL DECISION.

FEASIBILITY STUDY SHOULD CONTAIN:

1. MARKET SURVEY (PRESENT AND FUTURE DEMAND AND SUPPLY, NEW USES, PRICE RANGE OF PRODUCTS, CONSUMER CHARACTERISTICS)

2. COMPETITION (PRODUCTS,PROCESSES AND PRICES OF COMPETITORS)

3. PRODUCTION METHOD (PRELIMINARY DESIGN)

4. PLANT LOCATION AND LAYOUT

5. TECHNOLOGY AND PATENT SITUATIONS


6. ORGANIZATION AND LABOR REQUIREMENTS

7. RAW MATERIALS (COST AND AVAILABILITY)

8. WASTE DISPOSAL AND SAFETY/SECURITY CONSIDERATIONS

9. STORING, PACKAGING, SHIPPING REQUIREMENTS

10. FACILITIES AND EQUIPMENT THAT SHOULD BE BUILT OR PURCHASED ( INCLUDING LAND, BUILDINGS, UTILITIES, SERVICES, ADMINISTRATIVE FACILITIES, PRODUCTION AND TEST EQUIPMENT)

11. SALES METHODS AND REQUIREMENTS

12. INVESTMENT COST

13. PRODUCTION COST

14. FINANCIAL ANALYSIS

15. PROFIT AND RETURN ON INVESTMENT

16. IMPLEMENTATION PLAN

17. EXECUTIVE SUMMARY

III. GENERAL CONSIDERATIONS

GENERAL CONSIDERATIONS1. PLANT LOCATION THE PLANT SHOULD BE LOCATED WHERE THE MINIMUM TOTAL

COST OF PRODUCTION AND DISTRIBUTION CAN BE OBTAINED. FIRST GENARAL LOCATION THEN EXACT SITE SHOULD BE DETERMINED. YOU SHOULD CONSIDER:

* RAW MATERIALS ( AVAILABILITY, PURITY, PRICE, TRANSPORTATION EXPENSE)

* MARKET ANALYSIS * POWER AND FUEL (AVAILIBILITY, PRICE) * LAND PRICE, SOIL STRUCTURE, EXPANSION FACTOR * CLIMATE, NATURAL EVENTS (FLOOD,EARTQUAKE), WATER

SUPPLY * TRANSPORTATION FACILITIES * WASTE DISPOSAL SITUATION * LABOR (AVAILABILITY, COST, COMPETITION) * TAX BENEFITS * COMMUNITY FACTOR (LIVING CONDITIONS, SOCIAL FACILITIES)

GENERAL CONSIDERATIONS

2. PLANT LAYOUT BESIDES THE BASIC PROCESS, YOU SHOULD CONSIDER: * MINIMUM MATERIAL FLOW * OPERATIONAL CONVENIENCE * UTILITIES, SERVICES, ADMINISTRATIVE FACILITIES * HEALTH, SAFETY, FIRE PROTECTION * WASTE DISPOSAL * STORAGE * TRANSPORTATION * SECURITY


3. INSTRUMENTATION A ‘QUALITY PLAN’ SHOULD BE MADE; APPLYING QUALITY

ASSURANCE, MINIMIZING THE CONTROLS. THE LABORATORIES AND KIND OF EQUIPMENT TO BE USED FOR

QUALITY ASSURANCE SHOULD BE DECIDED BEFORE FINAL DESIGN

USE BUILT-IN-TESTERS WHERE POSSIBLE

4. MAINTENANCE MAINTENANCE REQUIREMENTS AND PLANS SHOULD BE

CONSIDERED BEFORE LAYOUT AND COST CALCULATION TOTAL PRODUCTIVE MAINTENANCE (TPM) SHOULD BE APPLIED


5. UTILITIES

ELECTRICITY, FUEL AND WATER ARE MAJOR COST ITEMS IN MOST CHEMICAL INDUSTRIES. THE SUPPLY METHODS, ALTERNATIVES AND NECESSARY FACILITIES (PIPING, STORAGE TANKS,GENERATORS) SHOULD BE CONSIDERED DURING DESIGN

6. STRUCTURAL DESIGN

FOUNDATION DESIGN IS ESPECIALLY IMPORTANT FOR HEAVY AND VIBRATING EQUIPMENT. FLOORS SHOULD BE DESIGNED TO BE RESISTANT TO CHEMICALS, EASY TO CLEAN AND SAFE. CORROSIVE EFFECTS SHOULD BE CONSIDERED WHEN CHOOSING CONSTRUCTION MATERIAL

7. STORAGE

STORAGE SPACE, PROPER EQUIPMENT FOR GASES AND LIQUIDS AND SAFETY IS IMPORTANT

GENERAL CONSIDERATIONS8. MATERIALS HANDLING IMPORTANT COST IN CHEMICAL INDUSTRY. PUMPS, BLOWERS,

PIPES, DUCTS ARE USED FOR LIQUIDS AND GASES; ELEVATORS, TRUCKS AND PNEUMATIC SYSTEMS ARE USED FOR SOLIDS. PROPER EQUIPMENT SHOULD BE CHOSEN CONSIDERING PROPERTIES OF MATERIALS, TYPE AND DISTANCE OF MOVEMENT AND COST

9. WASTE DISPOSAL - AIR POLLUTION IMPORTANT COST IN CHEMICAL INDUSTRY. TRY TO ELIMINATE

AT SOURCE OR RECOVER BEFORE WASTE TREATMENT10. HEALTH AND SAFETY CAREFULL ANALYSIS SHOULD BE MADE DURING THE DESIGN TO

PREVENT HAZARDOUS ENVIRONMENT OR FIRE AND ACCIDENT POTENTIALS

11. PATENTS A NEW DESIGN SHOULD BE EXAMINED TO MAKE CERTAIN NO

PATENT INFRINGEMENTS ARE INVOLVED

IV. CAPITAL INVESTMENT

CASH FLOW

TAX NET PROFIT + DEPRECIATION

GROSS PROFIT

SALES OPERATIONS EXPENSES

WORKING CAPITAL

FIXED CAPITAL

TOTAL CAPITAL

INVESTMENT

CAPITAL SOURCE

REPAYMENTS AND CAPITAL

DISTIBUTED SINK BONDS

PROFIT LOANS

OTHERS

CASH FLOW

THE BOX “CAPITAL SOURCE…” IS THE COMPANY TREASURY SERVING AS RESERVOIR AND SOURCE OF NECESSARY FUNDS. THE INPUTS OF THIS BOX ARE CAPITAL PAID BY THE STOCKHOLDERS, BONDS SOLD, LOANS FROM BANKS, NET PROFIT + DEPRECIATION COMING FROM OPERATIONS AND OTHERS ( SALES OF ASSETS, PROFIT OF OTHER OPERATIONS…). THE OUT GOING MONEY IS THE TOTAL INVESTMENT, BACK PAYMENTS OF BORROWED MONEY AND PROFIT DISTRIBUTED.

THE BOX “OPERATIONS” IS THE PRODUCTION, ENGINEERING, MARKETING … ACTIVITIES OF THE COMPANY. INPUT IS WORKING CAPITAL AND SALES. OUT GOING MONEY IS THE EXPENSES (ALL COSTS EXCEPT DEPRECIATION) AND GROSS PROFIT. TAX IS PAID FROM THE GROSS PROFIT AND THE REST (NET PROFIT + DEPRECIATION) GOES TO THE TREASURY.

CASH POSITION

Cumulative

cash position, $ Land,salvage,working

capital recovery

Cumulative cash position

Construction

period

-2 -1 0 1 2 3 4 5 6 7 8 9 10 Years

FC Investment

Book value of investment

WC Investment

CASH POSITION

ASSUMING THAT THE CONSTRUCTION STARTS TWO YEARS BEFORE THE PRODUCTION, CASH POSITION GOES DOWN TO NEGATIVE VALUES AS WE SPEND FOR THE FIXED INVESTMENT. AT TIME ZERO WE HAVE THE WORKING CAPITAL AS A NEGATIVE VALUE.

AFTER TIME ZERO, THE CASH POSITION STARTS GOING UP. CUMULATIVE CASH POSITION = NET PROFIT AFTER TAX + DEPRECIATION – TOTAL CAPITAL INVESTMENT. DEPENDING ON THE PROFIT, CUMULATIVE CASH POSITION GOES TO POSITIVE VALUES AFTER SEVERAL YEARS.

LAND, SALVAGE VALUE OF THE PLANT AND THE WORKING CAPITAL ARE RECOVERED AT THE END OF THE PROJECT.

THE DEPRECIATION IS 10 YEAR STRAIGHT- LINE DEPRECIATION.

COST

A PLANT DESIGN MUST PRESENT A PROCESS THAT IS CAPABLE OF OPERATING UNDER CONDITIONS WHICH YIELD A PROFIT. SINCE PROFIT EQUALS INCOME MINUS COSTS, ENGINEERS SHOULD BE AWARE OF ALL TYPES OF COSTS INVOLVED IN THE INDUSTRY.

IN AN ANALYSIS OF COSTS IN INDUSTRIAL PROCESSES,

* CAPITAL INVESTMENT COST AND

* PRODUCTION COST

MUST BE TAKEN INTO CONSIDERATION

CAPITAL INVESTMENT

BEFORE AN INDUSTRIAL PLANT CAN BE PUT INTO OPERATION, A LARGE AMOUNT OF MONEY MUST BE SUPPLIED TO PURCHASE AND INSTALL THE NECESSARY MACHINERY, EQUIPMENT AND BUILDINGS. AN ADDITIONAL AMOUNT OF MONEY IS REQUIRED TO START THE PRODUCTION. THE SUM OF ALL THESE AMOUNTS IS CALLED TOTAL CAPITAL INVESTMENT.

TOTAL CAPITAL INVESTMENT HAS TWO COMPONENTS:

* FIXED CAPITAL INVESTMENT

COVERS LAND, BUILDINGS, UTILITIES, PROCESS AND CONTROL EQUIPMENT AND INSTALLATION

* WORKING CAPITAL

COVERS THE EXPENSES NECESSARY FOR THE OPERATION OF THE PLANT

CAPITAL INVESTMENTBREAKDOWN OF FIXED CAPITAL INVESTMENT DIRECT COSTS 1. PURCHASED EQUIPMENT 2. PURCHASED EQUIPMENT INSTALLATION 3. INSTRUMENTATION AND CONTROLS 4. PIPING 5. INSULATION 6. ELECTRICAL 7. BUILDINGS 8. YARD IMPROVEMENTS 9. UTILITIES 10. SERVICE FACILITIES 11. LAND INDIRECT COSTS 12. ENGINEERING AND SUPERVISION 13. CONTRACTOR’S FEE 14. CONTINGENCY 15. OTHER COSTS (START - UP, DISTRIBUTION, R& D, LICENSE)

CAPITAL INVESTMENT

AN ESTIMATE OF THE CAPITAL INVESTMENT FOR A PROCESS MAY VARY FROM A PREDESIGN ESTIMATE BASED ON LITTLE INFORMATION (+/- 30%) TO A DETAILED ESTIMATE PREPARED FROM COMPLETE DRAWINGS AND SPECIFICATIONS (+/- 5%). PREDESIGN COST ESTIMATES ARE EXTREMELY IMPORTANT FOR DETERMINING IF A PROPOSED PROJECT SHOULD BE GIVEN FURTHER CONSIDERATION AND TO COMPARE ALTERNATIVE DESIGNS.

EQUIPMENT COST ESTIMATES ARE USUALLY THE FIRST STEP FOR MORE DETAILED PLANT COST ESTIMATES. ONCE THE EQUIPMENT COST IS DETERMINED, THE OTHER COST ITEMS OF THE PLANT CAN BE ESTIMATED USING PERCENTAGES AND PREVIOUS DATA.

CAPITAL INVESTMENT

PLANT COST ESTIMATION ITEMS:

1. PURCHASED EQUIPMENT COST (PEC) ESTIMATION

EQUIPMENT COST IS USUALLY 20-30% OF THE TOTAL PLANT COST IN CHEMICAL INDUSTRIES AND IS EITHER FOUND BY QUOTATIONS OR BY USING CHARTS.

A. MANUFACTURERS’ QUOTATIONS

ACCURATE AND PREFERABLE METHOD; BESIDES PRICE, USEFULL TECHNICAL INFORMATION CAN BE OBTAINED BUT NEEDS TIME AND AUTHORIZATION AND DESIGN DETAILES MAY BE REQUIRED BY EQUIPMENT MANUFACTURERS.

B. LITERATURE - ESTIMATING CHARTS

* PRICE CHARTS FOR DIFFERENT PROCESS EQUIPMENT IS AVAILABLE IN LITERATURE AS PRICE vs CAPACITY. THESE ARE ROUGH COST ESTIMATES

CAPITAL INVESTMENT

EQUIPMENT COST DATA ARE CORRELATED ON LOG-LOG PLOTS AS FUNCTIONS OF EQUIPMENT SIZE PARAMETER (HEAT TRANSFER AREA FOR HEAT EXCHANGERS, WORKING CAPACITY FOR MIXERS…).

OVER A LIMITED RANGE OF SIZES THE LOG-LOG CURVE CAN BE APPROXIMATED BY A STRAIGHT LINE, WHICH IS EQUIVALENT TO THE EQUATION

C2 / C1 = (S2 / S1)n

WHERE C IS COST, S IS SIZE AND n IS SLOPE ON THE LOG-LOG PLOT. THE VALUE OF n IS FREQUENTLY AROUND 0.6 0R 0.7 FOR MANY TYPES OF EQUIPMENT.

CAPITAL INVESTMENTIF WE ASSUME THAT n = 0.6 THAN THE RELATION BECOMES:

C2 / C1 = (S2 / S1)0.6 C2 = C1 (S2 / S1)0.6

WHICH IS REFERRED AS SIX-TENTHS-RULE.

EXAMPLE: USE THE SIX-TENTHS-RULE TO ESTIMATE THE % INCREASE IN PURCHASED COST WHEN THE CAPACITY OF A PIECE OF EQUIPMENT IS DOUBLED.

C2 = C1 (S2 / S1)0.6 C2 = C1 (2 / 1)0.6 C2 = 1.52 C1

THIS SIMPLE EXAMPLE SHOWS THAT WHEN YOU DOUBLE THE CAPACITY, THE COSTS OF EQUIPMENT INCREASES ONLY 52%. SO WE CAN STATE : THE LARGER THE EQUIPMENT, THE LOWER THE COST OF EQUIPMENT PER UNIT OF CAPACITY.

WHEN ESTIMATING THE COST OF A SINGLE EQUIPMENT IT IS SAFER TO USE SPECIFIC COST EXPONENTS; THE SIX-TENTH-RULE MAY BE USED MORE RELIABLY FOR A TOTAL PROCESS.

CAPITAL INVESTMENT

ANY COST DATA REFER TO A PARTICULAR YEAR AND IT IS NECESSARY TO KNOW THIS YEAR BEFORE THE DATA ARE USED. SINCE COSTS CHANGE WITH TIME, PUBLISHED COST DATA MUST BE CORRECTED TO THE PRESENT YEAR. THIS IS ACCOMPLISHED WITH COST INDICES USING THE RELATION

Cost in year m = Index value for year m

Cost in year n Index value for year n

EQUIPMENT COST INDEX IS PUBLISHED MONTHLY IN Chemical Engineering. INDEX VALUES ARE GIVEN FOR VARIOUS CATEGORIES OF EQUIPMENT (HEAT EXCHANGERS, PROCESS MACHINERY, PIPES …), CONSTRUCTION LABOR, BUILDINGS, ENGINEERING AS WELL AS COMPOSITE PLANT COST INDEX (CEPCI).

CAPITAL INVESTMENT

EXAMPLE: THE CAPITAL COST OF A 30,000 TONS/YEAR ISOPROPONAL PLANT IN 1986 WAS ESTIMATED TO BE $7 MILLION. ESTIMATE THE CAPITAL COST OF A NEW PLANT WITH A PRODUCTION RATE OF 50,000 TONS/YEAR IN 2004.

COST IN 2004 = (COST IN 1986) (CAPACITY COR.) (INFLATION COR.)

THE CEPCI IS 318 FOR YEAR 1986 AND 442 FOR YEAR 2004

COST IN 2004 = $7,000,000 ( 50/30)0.6 (442/318) = $13,223,000

EXAMPLE :ASSUME A PUMP (4 hp) WAS BOUGHT FOR $1000 EIGHT YEARS AGO AND YOU WILL BUY A NEW PUMP (6 hp) NOW. EXPONENT FOR SIZE OF PUMPS = 0.61 COST INDEX NOW : 580 COST INDEX 8 YEARS AGO: 520

COST OF THE NEW PUMP = (1000)(580/520) (6/4)0.61 = $1428

CAPITAL INVESTMENTTHE DATA IN LITERATURE IS USUALLY FOR EQUIPMENT MADE FROM

MOST COMMON MATERIAL AND USED IN AMBIENT PRESSURE. THE EFFECT OF BETTER MATERIAL AND HIGHER PRESSURES ON THE COST OF THE EQUIPMENT SHOULD BE FOUND AND NECESSARY CORRECTIONS SHOULD BE MADE.

THE COSTS FOUND THROUGH QUOTATIONS OR LITERATURE ARE USUALLY THE PRICE OF THE EQUIPMENT. ALL ADDITIONAL COSTS NECESSARY TO BRING THE EQUIPMENT TO THE FACTORY SITE SHOULD BE ADDED.

FOB OR EX-WORKS PRICE FREIGHT CHARGES (DEPENDS ON DISTANCE – VEHICLE) INSURANCE ANY OTHER CHARGES ( SPECIAL PACKING, DOCUMENT,

TRAINING…) TOTAL CIF(ANKARA) PRICE

CAPITAL INVESTMENT

2. INSTALLATION COST OF EQUIPMENT THE INSTALLATION OF EQUIPMENT INVOLVES COSTS FOR

LABOR, FOUNDATIONS,SUPPORTS, PLATFORMS, CONSTRUCTION EXPENSES AND OTHER FACTORS DIRECTLY RELATED TO THE ERECTION OF PURCHASED EQUIPMENT. IT IS ESTIMATED TO VARY FROM 30 TO 45% OF THE PEC

3. INSTRUMENTATION AND CONTROL TOTAL INSTRUMENTATION COST DEPENDS ON THE AMOUNT OF

CONTROL REQUIRED (QUALITY PLAN OF THE COMPANY) AND MAY AMOUNT TO 6-30 % OF THE PEC

CAPITAL INVESTMENT

4. PIPING

ONE OF THE IMPORTANT COSTS IN CHEMICAL INDUSTRIES. IT COVERS VALVES, FITTINGS, PIPES, SUPPORTS AND LABOR OF PIPING USED FOR RAW MATERIALS, INTERMEDIATE PRODUCTS, PRODUCTS, STEAM, WASTE, AIR. IT IS 15 – 70 % OF PEC.

5. INSULATION COST

WHEN HIGH OR LOW TEMPERATURES ARE INVOLVED, INSULATION COST IS IMPORTANT. INCREASING ENERGY COSTS MAKE INSULATION MORE IMPORTANT. IT IS 2 – 8% OF PEC.

6. ELECTRICAL INSTALLATIONS

POWER WIRING, LIGHTING, TRANSFORMATION, INSTRUMENT- CONTROL WIRING ARE THE MAIN COSTS, ADDING UP TO 10 – 15%OF PEC

7. BUILDINGS

40 –70% OF PEC INCLUDING ADMINISTRATIVE SERVICES, LABS, WAREHOUSES, SOCIAL BUILDINGS, MAINTENANCE FACILITIES

CAPITAL INVESTMENT8. YARD IMPROVEMENTS LANDSCAPING, FENCING, ROADS, SIDEWALKS, PARKING.

ABOUT 10 – 15% OF PEC9. UTILITIES INCLUDE GENERAL SERVICES REQUIRED TO OPERATE THE

PLANT SUCH AS WATER, FUEL, STEAM, ELECTRICITY, FIRE PROTECTION, COMPRESSED AIR, EMERGENCY GENERATOR, TELEPHONE AND WASTE DISPOSAL. 30 – 60% OF PEC

10. SERVICE FACILITIES OFFICE FURNITURE, CAFETERIA EQUIPMENT, SAFETY AND

MEDICAL EQUIPMENT, HOUSE KEEPING EQUIPMENT, MATERIAL HANDLING AND PACKAGING EQUIPMENT. 20 – 40% OF PEC

11. LAND 4 – 8% OF PEC12. ENGINEERING AND SUPERVISION COSTS FOR CONSTRUCTION DESIGN, DRAFTING, PURCHASING,

ENGINEERING AND SUPERVISION. ABOUT 35% OF PEC

CAPITAL INVESTMENT

13. CONSTRUCTION EXPENCES

THE COST OF TEMPORARY FACILITIES, ROADS, OFFICES USED DURING CONSTRUCTION, CONSTRUCTION TOOLS, SECURITY EXPENCES. UP TO 30% OF PEC

14. CONTRACTORS FEE

VARIES BETWEEN 2 – 6% OF FIXED CAPITAL INVESTMENT

15. CONTINGENCY

UNPREDICTABLE COSTS. 5 – 10% OF FIXED CAPITAL INVESTMENT

16. OTHER COSTS

COSTS FOR PLANT START-UP (TRIALS, CHANGES), OFF-SITE FACILITIES LIKE DISTRIBUTION CENTERS AND OFFICES, RESEARCH AND DEVELOPMENT DEPARTMENTS, LICENSE FEES ARE THE OTHER COSTS THAT SHOULD BE CONSIDERED

CAPITAL INVESTMENT

METHODS FOR ESTIMATING FIXED CAPITAL INVESTMENT

THERE ARE DIFFERENT METHODS THAT CAN BE USED.

IN DECREASING DETAIL, PREPARATION TIME AND ACCURACY THEY CAN BE SUMMARIZED AS:

METHOD A – DETAILED ITEM ESTIMATE

ALL ITEMS ARE DESIGNED AND DETERMINED ACCURATELY; SITE SURVEYS ARE MADE, QUOTATIONS FOR EQUIPMENT AND SERVICES ARE OBTAINED, ALL ITEMS ARE SPECIFIED FOR PRICING INCLUDING LABOR. VERY TIME CONSUMING AND EXPENSIVE WORK. USUALLY DONE AT THE FINAL STAGE OR BY THE CONTRACTORS FOR QUOTING. +/- 5% ACCURACY CAN BE OBTAINED

CAPITAL INVESTMENT

METHOD B: MODULE COSTING TECHNIQUE

FIND THE BASE CASE COST (C0P) OF EACH EQUIPMENT FROM

LITERATURE. BASE CASE MEANS, EQUIPMENT MADE OF THE MOST COMMON MATERIAL (USUALLY CARBON STEEL) AND OPERATING AT NEAR AMBIENT PRESSURE.

FIND THE BARE MODULE FACTOR (F0BM) WHICH IS THE

MULTIPLICATION FACTOR TO ACCOUNT FOR OTHER ASSOCIATED COSTS (PIPING, INSULATION, FOUNDATIONS, SUPPORTS, INSTRUMENTATION AND ELECTRICAL, LABOR FOR INSTALLATION, TRANSPORTATION, ENGINEERING AND PROJECT MANAGEMENT).

NOW YOU CAN CALCULATE BARE MODULE COST AT BASE CONDITIONS FOR EACH EQUIPMENT:

C0BM = C0

P F0BM

CAPITAL INVESTMENT

FOR EQUIPMENT MADE FROM OTHER MATERIALS OF COSTRUCTION ( STAINLESS STEEL, ALUMINUM, TITANIUM …) AND/ OR OPERATING AT NONAMBIENT PRESSURE, YOU SHOULD MAKE THE NECESSARY CORRECTIONS.

FIND FBM WHICH IS THE MULTIPLICATION FACTOR TO ACCOUNT FOR ASSOCIATED COSTS PLUS THE SPECIFIC MATERIALS OF CONSTRUCTION AND OPERATING PRESSURE. THIS IS FOUND FROM LITERATURE. THE BARE MODULE EQUIPMENT COST CAN NOW BE CALCULATED:

CBM = C0P FBM

CAPITAL INVESTMENT

EXAMPLE: FIND THE BARE MODULE COST OF A FLOATING-HEAD SHELL-AND-TUBE HEAT EXCHANGER WITH A HEAT TRANSFER AREA OF 100 m2. THE OPERATING PRESSURE IS 100 barg AND MATERIAL OF CONTRUCTION IS STAINLESS STEEL.

C0P = 250 $/ m2 x 100 m2 = $ 25,000 ; YEAR 2001 (Turton, Appendix A)

FBM = B1 + B2FMFP = 1.63 + 1.66 (2.73) (1.383) = 7.9

ALL THE CONSTANTS ARE FOUND FROM Turton, Appendix A

CBM = C0P x FBM = 25,000 x 7.9 = $197,500

197,500 x 468 / 397 = $232,821 AT YEAR 2005, FEBRUARY

CAPITAL INVESTMENT WHEN FINDING THE TOTAL COST OF AN ALTERATION OR A NEW

PLANT YOU SHOULD CONSIDER THE ADDITIONAL COSTS.

TOTAL MODULE COST(CTM) : THE COST OF MAKING EXPANSIONS OR ALTERATIONS TO AN EXISTING PLANT. YOU SHOULD ADD CONTRACTORS FEE AND CONTINCENCY (18%):

CTM = 1.18 CBM, i

WHERE CBM IS THE BARE MODULE COST OF EACH EQUIPMENT USED IN THE ALTERATION AND i = 1 TO n.

GRASS ROOTS COST(CGR) : THE COST OF MAKING A NEW FACILITY. YOU SHOULD ADD LAND, YARD IMPROVEMENT, UTULITIES, ADMINISTRATIVE AND SERVICE FACILITIES AND BUILDINGS. THESE ARE GENERALLY UNEFFECTED BY MATERIAL OR PRESSURE AND IS TAKEN AS 50% OF C0

BM

CGR = CTM + 0.50 C0BM, i

WHERE C0BM IS THE BARE MODULE COST OF EACH EQUIPMENT AT

BASE CONDITIONS USED IN THE NEW PLANT AND i = 1 TO n.

CAPITAL INVESTMENT METHOD C – PERCENTAGE OF PURCHASED EQUIPMENT COST DETERMINE THE PURCHASED EQUIPMENT COST. ALL OTHER

ITEMS ARE THEN ESTIMATED AS PERCENTAGE OF PEC. THERE IS DATA AVAILABLE FOR AVERAGE PERCENTAGE VALUES FOR DIFFERENT TYPES OF INDUSTRIES

METHOD D – ‘LANG’ FACTORS FOR APPROXIMATION MULTIPLY PEC BY A FACTOR TO OBTAIN TOTAL INVESTMENT

COST (USUALLY 4 – 6). GREATER ACCURACY IS OBTAINED BY USING MORE THAN ONE FACTOR, DIFFERENT FACTORS FOR DIFFERENT KINDS OF EQUIPMENT (4 FOR PUMPS, 3.5 FOR HEAT EXCHANGERS)

METHOD E – POWER FACTOR THE CAPITAL INVESTMENT OF A NEW PLANT CAN BE FOUND BY

GETTING THE EXPONENTIAL POWER OF A SIMILAR PREVIOUS PLANT INVESTMENT. Cn = C (R)X

R: CAPACITY RATIO, C: CAPITAL INVESTMENT FOR X THERE IS DATA FOR DIFFERENT INDUSTRIES;

IN AVARAGE IT IS 0.6- 0.8

nnn

CAPITAL INVESTMENT

METHOD F – COST PER UNIT OF CAPACITY

THERE IS DATA GIVING THE FIXED CAPITAL INVESTMENT REQUIRED PER ANNUAL TON OF A PRODUCT

METHOD G – TURNOVER RATIO

FIXED CAPITAL INVESTMENT =

ANNUAL SALES / TURNOVER RATIO

TURNOVER RATIO IS BETWEEN 0.4 – 3 IN GENERAL. FOR CHEMICAL INDUSTRIES IT IS ABOUT 0.5

CAPITAL INVESTMENT

WORKING CAPITAL

EVERY PLANT HAS A REQUIREMENT FOR A CERTAIN AMOUNT OF CAPITAL TO BE AVAILABLE TO PAY THE BILLS AND SUSTAIN THE OPERATION BEFORE THE PRODUCT IS SOLD AND PAYMENT IS RECEIVED. IT CONSISTS OF

* RAW MATERIALS AND SUPPLIES CARRIED IN STOCK

* SEMIFINISHED PRODUCTS IN THE PROCESS

* FINISHED PRODUCTS NOT SOLD YET

* ACCOUNTS RECEIVABLE

* CASH REQUIRED FOR WAGES,TAXES AND ALL OTHER EXPENSES

FOR RAW MATERIALS IN STOCK, 1 MONTH SUPPLY IS USUAL. SEMIFINISHED AND FINISHED PRODUCTS VALUE CAN BE CALCULATED AS EQUAL TO TOTAL PRODUCTION COST FOR 1 MONTH PRODUCTION. ACCOUNTS RECEIVABLE IS ESTIMATED AS 1 MONTH COST OF PRODUCTION IF 30-DAY PAYMENT PERIOD IS USED, 2 MONTHS OR MORE FOR LONGER PAYMENT CONDITIONS

CAPITAL INVESTMENT

FOR MOST CHEMICAL PLANTS WORKING CAPITAL IS ABOUT 10 – 20% OF TOTAL CAPITAL INVESTMENT. MORE ACCURATE CALCULATIONS CAN BE MADE BY ITEMIZING THE COST COMPONENTS.

A FRACTION OF THE YEARLY MANUFACTURING COST IS A MUCH MORE RELEVANT METHOD OF ESTIMATING WORKING CAPITAL AND IT MAY TYPICALLY BE 10 – 35% OF THE YEARLY OPERATING COSTS.

‘JUST IN TIME’ OPERATING METHODS AND GOOD PLANNING MAY BE EFFECTIVE IN DECREASING THE WORKING CAPITAL

V. PRODUCT COST – ECONOMIC PRODUCTION

PRODUCT COSTCONSISTS OF MANUFACTURING COSTS AND GENERAL EXPENSES. PRODUCT COST SHOULD BE ESTIMATED DURING DESIGN PHASE IN

ORDER TO CALCULATE RETURN ON INVESTMENT AND PROFITABILITY AND DECIDE ON ALTERNATIVES. DURING THE PRODUCTION PHASE, PRODUCT COST SHOULD BE CALCULATED AND CONTROLLED CONTINUOUSLY SINCE COST IS ONE OF THE MAIN FACTORS DETERMINING THE SUCCESS OF AN INDUSTRY.

PRODUCT COST MANUFACTURING COST 1. RAW MATERIALS 2. OPERATION AND SUPERVISION LABOR 3. UTILITIES 4. MAINTENANCE AND REPAIR 5. OPERATING SUPPLIES, LABORATORY CHARGES 6. PATENTS AND ROYALTIES

PRODUCT COST

7. PLANT OVERHEAD COSTS

8. RENT

9. INSURANCE

10. PROPERTY TAXES AND OTHER DUTIES

11. DEPRECIATION

GENERAL EXPENSES

12. ADMINISTRATIVE EXPENSES

13. DISTRIBUTION AND MARKETING COSTS

14. RESEARCH AND DEVELOPMENT COSTS

15. FINANCING COSTS

OTHER

16. CONTINGENCIES

PRODUCT COSTPRODUCT COST ITEMS

1. RAW MATERIALS: AMOUNT OF MATERIALS USED IN THE PLANT SHOULD BE FOUND FROM PROCESS FLOW DIAGRAMS. WHEN CONVERTING HOURLY RATES TO YEARLY CONSUMPTION, THE FRACTION OF TIME THAT THE PLANT IS OPERATING IN A YEAR MUST BE KNOWN (CALLED STREAM FACTOR,SF, WHICH IS ‘DAYS OF OPERATION / 365’ AND IS MOSTLY LESS THEN 0.9). YOU SHOULD ALSO CONSIDER SCRAPS, LOSES AND INEFFICIENCIES.

BESIDES THE MAIN MATERIALS, TREATING AGENTS, CATALYSTS AND FILTER AIDS SHOULD BE CALCULATED.

PRICE FOR MATERIALS CAN BE FOUND FROM LITERATURE OR QUOTATIONS. YOU SHOULD ADD ALL EXPENSES TO THE FACTORY: INSURANCE, TRANSPORTATION AND OTHER COSTS.

SINCE THE LARGEST OPERATING COST IS NEARLY ALWAYS THE COST OF RAW MATERIALS IN CHEMICAL PLANTS, IT IS IMPORTANT TO FIND CORRECT MATERIAL COSTS IN ORDER TO MAKE REALISTIC EVALUATIONS.

PRODUCT COST

2. OPERATION AND SUPERVISION LABOR

NUMBER OF PEOPLE (SKILLED AND UNSKILLED) REQUIRED FOR MANUFACTURING SHOULD BE CALCULATED.

LABOR REQUIREMENT FOR DIFFERENT PROCESSES CAN BE FOUND FROM LITERATURE. WHEN CALCULATING TOTAL NUMBER OF OPERATORS YOU SHOULD KNOW NUMBER OF SHIFTS AND WORKING HOURS. MOST CHEMICAL PLANTS WORK 24 HOURS A DAY AND 365 DAYS A YEAR.

ONE WORKER: 48 WEEKS/YEAR x 5 DAYS/WEEK x 8 HOURS/DAY

: 1920 HOURS/ YEAR

365 DAYS/YEAR x 24 HOURS/DAY = 8760 HOURS/YEAR

8760 / 1920 = 4.56

THAT MEANS FOR SUCH A PLANT 4.56 OPERATORS SHOULD BE HIRED FOR EACH OPERATOR NEEDED.

PRODUCT COST

BESIDES THE OPERATING LABOR YOU NEED SUPPORT AND SUPERVISORY STAFF (TECHNICIANS, ENGINEERS) IN THE MANUFACTURING PROCESS. THE AMOUNT OF SUCH PEOPLE FOR EACH SHIFT SHOULD BE FOUND FROM LITERATURE OR PREVIOUS EXPERIENCES.

TO ESTIMATE THE COST FOR LABOR, AVERAGE WAGE RATES ARE REQUIRED. THE AVERAGE NET WAGES MAY SHOW SIGNIFICANT VARIATIONS DEPENDING ON PLANT LOCATION AND TYPE OF WORK. BUT WHEN CALCULATING THE COSTS YOU SHOULD ALWAYS CONSIDER ADDITIONAL EXPENSES ON THE NET WAGE.

COST TO COMPANY = (NET WAGE)+(TAX)+(SOCIAL SECURITY)+(OTHERS)

IN AVERAGE YOU CAN USE THE FOLLOWING FIGURES:

COST TO COMPANY = (NET WAGE)(1+ 0.20 + 0.40 + 0.10)

COST TO COMPANY = (NET WAGE)(1.70)

PRODUCT COST

3.UTILITIES: THE COST OF ELECTRICITY, FUEL, WATER, STEAM, COMPRESSED AIR AND WASTE TREATMENT IN CHEMICAL INDUSTRIES IS A CONSIDERABLE PERCENTAGE OF THE PRODUCT COST.

MOST OF THE UTILITIES COST IS INFLUENCED BY THE COST OF FUEL. IN TURKEY, COAL AND NATURAL GAS ARE THE CHEAPEST, BUT COAL HAS HANDLING AND ENVIRONMENTAL PROBLEMS AND NATURAL GAS MAY NOT BE AVALIABLE. LPG AND FUEL OIL ARE THE OTHER (MORE EXPENSIVE) ALTERNATIVES.

THERE ARE LOCAL RATES FOR FUEL, ELECTRICITY AND WATER WHICH CAN BE FOUND FROM RELATED ORGANIZATIONS. THE COSTS FOR STEAM, COMPRESSED AIR, COOLING WATER AND WASTE TREATMENT CAN BE CALCULATED OR FOUND FROM LITERATURE.

PRODUCT COST4. MAINTENANCE AND REPAIRS: DEPENDS ON THE QUALITY OF

EQUIPMENT INVESTED AT THE BEGINNING AND INCREASES AS THE PLANT GETS OLDER. 2 – 10% OF FIXED CAPITAL COST

5. OPERATING SUPPLIES, LABORATORY CHARGES: MATERIAL WHICH IS NOT LISTED SEPARATELY BUT NEEDED FOR PRODUCTION (LUBRICANTS, TEST CHEMICALS…) ARE OPERATING SUPPLIES. LABORATORY CHARGES INCLUDE QUALITY CONTROL FUNCTIONS FOR INCOMING INSPECTION, PRODUCTION CONTROL AND FINAL PRODUCT CONTROL. CAN BE ESTIMATED AS 10 – 20% OF OPERATING LABOR

6. PATENTS AND ROYALTIES: EITHER TO PAY A SET AMOUNT OF PATENT RIGHTS OR A ROYALTY ON THE QUANTITY PRODUCED OR SOLD MAY BE REQUIRED. SHOULD BE CALCULATED CASE BY CASE.

7. PLANT OVERHEAD COSTS: INCLUDE COSTS FOR MEDICAL CARE, GENERAL ENGINEERING, SAFETY, SECURITY, CAFETERIA, CLEANING, STORAGE, PACKAGING ETC. CAN BE CALCULATED ITEM BY ITEM IF DATA IS AVAILABLE OR ESTIMATED AS 10 – 15% OF TOTAL PRODUCT COST

PRODUCT COST

8. RENT: IF THERE IS RENTED LAND OR BUILDINGS9. INSURANCE: THE FACILITIES AND ALL MATERIAL AND PRODUCTS

IN THE INVENTORY SHOULD BE INSURED. AVARAGE %1 ANNUALY OF INSURED VALUE

10. PROPERTY TAXES AND OTHER DUTIES: SHOULD BE CALCULATED LOCALLY

11. DEPRECIATION: MEANS OF DISTRIBUTING THE ORIGINAL EXPENSE FOR A PHYSICAL ASSET OVER THE PERIOD DURING WHICH THE ASSET IS IN USE.

IN COMPUTING INCOME TAX ON THE PROFIT, GOVERNMENT ALLOWS A DEDUCTION FOR A FRACTION OF THE INITIAL COST OF THE PLANT AS A ‘HYPOTHETICAL EXPENSE’ TO BE SUBTRACTED FROM THE GROSS PROFIT. THIS DEDUCTION, CALLED DEPRECIATION, MAY BE CONSIDERED AS A FUND TO ALLOW EVENTUAL REPLACEMENT OF THE PLANT.

YOU CAN DEPRECIATE ALL FIXED INVESTMENT EXCEPT LAND

PRODUCT COST

GOVERNMENTS PUBLISH LISTS OF ALLOWABLE DEPRECIATION RATES FOR DIFFERENT EQUIPMENT AND OTHER ITEMS. ACCOUNTING DEPARTMENTS KEEP SEPARATE DEPRECIATION RECORDS ON EACH ITEM. DEPRECIATION CREDIT STOPS WHEN THE TOTAL GOVERNMENT- ALLOWED USEFULL LIFE PERIOD HAS BEEN USED UP, EVEN IF THE EQUIPMENT IS STILL IN SERVICE.

IN TURKEY, DEPRECIATION PERIODS ARE 40-50 YEARS FOR BUILDINGS; 5-10 YEARS FOR MACHINERY, EQUIPMENT AND FURNITURE; 10-20 YEARS FOR UTILITIES; LESS THAN 5 YEARS FOR COMPUTERS AND SPECIAL PUMPS, FILTERS.

METHODS OF DEPRECIATION:

* STRAIGHT LINE DEPRECIATION

ORIGINAL COST – SALVAGE VALUE YEARLY DEPRECIATION =

EQUIPMENT LIFE

SAME PERCENTAGE EVERY YEAR. SALVAGE VALUE IS USUALLY 0.

PRODUCT COST * DOUBLE DECLINING BALANCE DEPRECIATION 2 X (ORIGINAL COST – PREVIOUS DEPRE. – SALVAGE VALUE) EQUIPMENT LIFE * SUM – OF – YEARS DIGIT DEPRECIATION THE NUMBERS FOR EACH YEAR OF THE PLANT LIFE ARE ADDED (FOR

5 YEAR LIFE: 1+2+3+4+5 = 15) AND DEPRECIATION FOR THE FIRST YEAR IS 5/15 OF THE ORIGINAL COST; SECOND YEAR 4/15 …

EXAMPLE: MACHINE PRICE $1000, LIFE 5 YEARS, SALVAGE 0 YEARS STR. LINE DOUBLE-DEC SUM OF DIGITS 1 $200 $400 $333.33 2 200 240 266.67 3 200 144 200.00 4 200 108 133.32 5 200 108 66.67 $1000 $1000 $1000

PRODUCT COST

IN DECLINING BALANCE METHOD, YOU CANNOT DEPRECIATE COMPLETELY IF YOU CONTINUE WITH THE METHOD TO THE END. THEREFORE IT IS ALLOWABLE TO DEPRECIATE AN ASSET OVER THE EARLY PORTION OF ITS LIFE USING DECLINING BALANCE AND THEN SWITCHING TO STRAIGHT LINE DEPRECIATION. IF YOU DO SO, BALANCE WILL BE DIVIDED TO REMAINING YEARS. (YOU CANNOT CHANGE THE METHOD IF YOU START WITH STRAIGHT LINE)

IN TURKEY EITHER STRAIGHT LINE (NORMAL AMORTİSMAN) OR DOUBLE DECLINING BALANCE (AZALAN BAKİYELER) CAN BE USED. STRAIGHT LINE IS EASY TO APPLY. DECLINING BALANCE HAS FINANCIAL ADVANTAGES DUE TO HIGHER DEPRECIATIONS IN THE EARLY-LIFE YEARS. DIFFERENT DEPRECIATION METHODS DO NOT CHANGE THE COMPANY’S OVERALL TAX OBLIGATION; THEY CAUSE THE ANNUAL TAX OBLIGATION TO VARY.

PRODUCT COST

12. ADMINISTRATIVE EXPENSES: WAGES FOR MANAGERS, OFFICE PERSONNEL, ADMINISTRATORS, SECRETARIES, ACCOUNTANTS, COMPUTER OPERATORS AND EXPENSES FOR OFFICE SUPPLIES, COMMUNICATIONS. CAN BE ESTIMATED USING LOCAL RATES.

13. DISTRIBUTION AND MARKETING COSTS: WAGES, SUPPLIES AND OTHER EXPENSES OF SALES OFFICES, TRAVELLING EXPENSES, COMMISSIONS, SHIPPING EXPENSES, ADVERTIZING EXPENSES. MAY BE UP TO 20% OF PRODUCT COST.

14. RESEARCH AND DEVELOPMENT COSTS: ALL EXPENSES RELATED TO DEVELOPING NEW METHODS AND PRODUCTS. INDUSTRY AVERAGE IS 5 – 10% OF PRODUCT COST.

15. FINANCING COSTS: THE INTERESTS PAID FOR THE BARROWED MONEY. DEPENDS ON THE COMPANY CAPITAL AND THE REQUIREMENT FOR THE WORKING CAPITAL.

16. CONTINGENCIES: 1 – 5% CONTINGENCY FACTOR SHOULD BE ADDED.

PRODUCT COST

THE PRODUCT SALES REVENUE (INCOME) MINUS THE TOTAL PRODUCT COST GIVES THE GROSS PROFIT:

SALES REVENUE – TOTAL PRODUCT COST = GROSS PROFIT

THE INCOME TAX IS CALCULATED ON THE GROSS PROFIT, AS A CERTAIN PERCENTAGE (t) OF IT:

INCOME TAX (T) = GROSS PROFIT x t / 100

NET PROFIT IS GROSS PROFIT MINUS THE INCOME TAX:

NET PROFIT = GROSS PROFIT – T = GROSS PROFIT (1 – t /100)

CASH FLOW IS NET PROFIT PLUS DEPRECIATION:

CASH FLOW = NET PROFIT + DEPRECIATION

PRODUCT COSTEXAMPLE :A COMPANY WAS FORMED TO PRODUCE HOUSEHOLD CLEANING

CHEMICALS. THIS COMPANY BOUGHT LAND FOR $220,000, HAD A $900,00 FACTORY BUILDING ERECTED, AND INSTALLED $700,000 WORTH OF CHEMICAL AND PACKAGING EQUIPMENT. THE SALES INCOME FOR THE FIRST YEAR WAS $450,000. SUPPLIES AND ALL OPERATING EXPENSES, EXCLUDING THE DEPRECIATION, WERE $100,000. ALLOWABLE DEPRECIATION PERIODS ARE 5 YEARS FOR EQUIPMENT, 50 YEARS FOR BUILDINGS. FIND THE TAX, NET PROFIT AND CASH FLOW FOR THE FIRST YEAR IF TAX RATE IS %20,USING

a) STRAIGHT LINE b) DOUBLE DECLINING BALANCE DEPRECIATION

a) DEP: (900,000).02 + (700,000) .2 = 158,000 TAX : (450,000 – 100,000 – 158,000) .20 = 38,400 NET PROFIT = (450,000 – 100,000 – 158,000 – 38,400) = 153,600 CASH FLOW = 158,000 + 153,600 = 311,600

b) DEP: 2 x (900,000).02 + 2 x (700,000) .2 = 316,000 TAX: (450,000 – 100,000 – 316,000) .20 = 6,800 NET PROFIT = (450,000 – 100,000 – 316,000 – 6,800) = 27,200 CASH FLOW = 316,000 + 27,200 = 343,200

PRODUCT COST

DEPLETION

A WASTING OR DEPLETING ASSET IS A NATURAL RESOURCE IN WHICH NATURE IS INCAPABLE OF REPLACING THE MINERAL THAT IS EXTRACTED (LIKE COAL MINE OR OIL WELL).

SINCE THE CAPITAL INVESTED IN A DEPLETING ASSET IS CONSUMED AS THE MINERAL IS EXTRACTED, IT IS NECESSARY TO ADJUST THE ACCOUNTING RECORDS OF THE FIRM TO REFLECT THE LOSS.

THE DEPLETION ALLOWANCE FOR A GIVEN YEAR IS DETERMINED BY MULTIPLYING THE GROSS INCOME FROM OPERATIONS OF THAT YEAR BY A SPECIFIED PERCENTAGE SET BY GOVERMENTAL REGULATIONS. SO FOUND DEPLETION ALLOWANCE CANNOT EXCEED %50 OF THE PROPERTY’S TAXABLE INCOME COMPUTED WITHOUT THE DEPLETION DEDUCTION.

PRODUCT COST

EXAMPLE:

AN OIL WELL YIELDED A GROSS ANNUAL INCOME OF $300,000. THE DEDUCTION OF ALL BUSINESS EXPENSES EXCEPT DEPLETION REDUCED THIS AMOUNT TO $125,000. FIND THE DEPLETION AMOUNT IF DEPLETION RATE ON GROSS INCOME IS %22 AND RESTRICTED TO %50 OF THE INCOME AS COMPUTED PRIOR TO DEPLETION.

(300,000)(.22) = 66,000

UPPER LIMIT : (125,000)(.50) = 62,500

SO DEPLETION IS $62,500

GROSS PROFIT = 125,000 – 62,500 = 62,500

PRODUCT COST

EXAMPLE :

A COAL MINE HAS A GROSS INCOME OF $250,000 FOR THE YEAR. MINING EXPENSES EQUAL $190,000. IF COAL MINES HAVE %10 DEPLETION ALLOWANCE AND RESTRICTED TO %50 OF THE INCOME AS COMPUTED PRIOR TO DEPLETION, CALCULATE THE DEPLETION.

(250,000) .10 = 25,000

UPPER LIMIT:

250,000 – 190,000 = 60,000

(60,000) .50 = 30,000

ALLOWABLE DEPLETION REDUCTION IS 25,000

GROSS PROFIT = 60,000 – 25,000 = 35,000

PRODUCT COST

SOME COST ITEMS ARE FIXED, THAT IS, THEY DO NOT VARY WITH THE LEVEL OF OUTPUT. COSTS LIKE PROPERTY TAX, PLANT INSURANCE, DEPRECIATION AND RENT ARE FIXED COSTS (F). SOME OF THE PLANT OVERHEAD COSTS AND ADMINISTRATIVE COSTS ARE ALSO FIXED. OTHER COSTS THAT CHANGE WITH THE LEVEL OF OUTPUT (RAW MATERIALS, LABOR ..) ARE CALLED VARIABLE COSTS (VC).

TOTAL COST C = F + VC

MARGINAL COST (MC) IS THE INCREMENT OR ADDITION TO COST THAT RESULTS FROM PRODUCING ONE MORE UNIT OF OUTPUT.

AVARAGE COSTS (AVERAGE TOTAL COST, AVERAGE VC..) ARE THE COSTS DIVIDED BY THE OUTPUT.

MARGINAL COST CAN BE EITHER ABOVE OR BELOW AVERAGE COST

PRODUCT COST

OUTPUT FIXED VARIABLE TOTAL COST AVERAGE MARGINAL

COST COST COST COST

0 ……… 100 …….. 0 …….. 100 ………. - …… -

1 ………. 100 …….. 10 …….. 110 ……… 110 …… 10

2 ……… 100 …….. 19 …….. 119 ……… 59.5 ……. 9

3 ……… 100 …….. 25 …….. 125 ……… 41.7 ……. 6

4 ……… 100 …….. 32 …….. 132 ……… 33 ……. 7

5 ……… 100 ……. 40 …….. 140 ……… 28 ……. 8

6 ……… 100 ……. 49 …….. 149 ……… 24.8 ……. 9

7 ……… 100 ……. 60 …….. 160 ………. 22.9 ……. 11

8 ………. 100 ……. 73 ……… 173 ……… 21.6 ……. 13

9 ………. 100 ……. 88 ……… 188 ……… 20.9 ……. 15

10 ………. 100 …….. 108 ……… 208 ……… 20.8 ……. 20

PRODUCT COST

ECONOMIES OF SCALE

IF AVERAGE COST FALLS AS OUTPUT INCREASES, THE FIRM IS SAID TO HAVE ECONOMIES OF SCALE (OR INCREASING RETURNS TO SCALE)

IF AVERAGE COST DOES NOT VARY WITH OUTPUT, THE FIRM HAS CONSTANT RETURNS TO SCALE

IF AVERAGE COST RISES WITH OUTPUT, THE FIRM IS SAID TO HAVE DISECONOMIES OF SCALE (OR DECREASING RETURNS TO SCALE)

PRODUCT COST

REASONS OF AVERAGE COST DECREASE : * FIXED COSTS ARE FIXED * SPECIALIZATION SO MORE EFFICIENT USE OF LABOR * CHEAPER RAW MATERIAL (QUANTITY DISCOUNTS)

AS LONG AS MARGINAL COST IS BELOW AVERAGE COST, ECONOMIES OF SCALE EXIST; IF MARGINAL COST EXCEEDS AVERAGE COST THERE ARE DISECONOMIES OF SCALE.

IF WE DEFINE AC/MC = S; ECONOMIES OF SCALE EXIST IF S>1; CONSTANT RETURN OF SCALES EXIST IF S=1; DISECONOMIES OF SCALE EXIST IF S<1.

PRODUCT COST

NUMBER OF PRODUCTION PLANTS

WHEN DECIDING ON THE NUMBER OF PLANTS, PRODUCTION COST IS AN IMPORTANT FACTOR, BUT NOT THE ONLY ONE. EVEN IF YOU HAVE ECONOMIES OF SCALE IN PRODUCTION, HAVING TWO PLANTS WITH LOWER CAPACITIES MAY BE MORE PROFITABLE DUE TO TRANSPORTATION COSTS. SO COST OF RAW MATERIAL TRANSPORTATION AND PRODUCTS TRANSPORTATION SHOULD ALSO BE CONSIDERED AND THE DECISION SHOULD BE ACCORDING TO THE TOTAL COST.

OPPURTUNITY COST

AN ACTION’S OPPURTUNITY COST IS THE VALUE OF THE BEST FORGONE ALTERNATIVE USE OF THE RESOURCES EMPLOYED IN THAT ACTION. IF A FIRM OWNS THE BUILDING IT OCCUPIES AND IF THE BUILDING COULD BE RENTED FOR $1000 / MONTH, THAN THE FIRM SHOULD COUNT THAT AMOUNT (OPPURTUNITY COST) AS ITS COST OF OCCUPYING THE BUILDING.

PRODUCT COSTECONOMIES OF SCOPE WHEN IT IS CHEAPER TO PRODUCE TWO PRODUCTS TOGETHER RATHER THAN

SEPARATELY, THERE IS AN ECONOMY OF SCOPE. FIRMS OFTEN PRODUCE MANY PRODUCTS TO GAIN ECONOMIES OF SCOPE IN MARKETING AND DISTRIBUTION.

CONSIDER THE PRODUCTION OF q1 UNITS OF PRODUCT 1 AND q2 UNITS OF PRODUCT 2.

THE COST OF PRODUCING EACH SEPARATELY IS C (q1 , 0) + C (0 , q2); THE COST OF PRODUCING THEM TOGETHER IS C (q1 , q2).

ECONOMIES OF SCOPE (SC) IS MEASURED AS:

C (q1 , 0) + C (0 , q2) – C (q1 , q2) SC =

C (q1 , q2) SC MEASURES THE RELATIVE INCREASE IN COST THAT WOULD RESULT IF THE

PRODUCTS WERE PRODUCED SEPARATELY. IF SC IS POSITIVE, IT IS CHEAPER TO PRODUCE THE PRODUCTS TOGETHER.

ECONOMIC PRODUCTION

BREAKDOWN OF RETAIL PRICE:

RETAILERS SELLING PRICE TO CUSTOMER 100

NET PROFIT OF RETAILER a

RETAILER’S EXPENSES (RENT, LABOUR,ENERGY, TELEPHONE, TAXES, ETC.) b

WHOLESALER’S PRICE TO RETAILER 100-a-b

NET PROFIT OF WHOLESALER c

WHOLESALER’S EXPENSES (RENT,LABOUR, ENERGY, TELEPHONE, DELIVERY, TAXES, ETC) d

FACTORY PRICE TO WHOLESALER 100-a-b-c-d

THE TOTAL OF a,b,c,d MAY CHANGE BUT NORMALLY IT IS IN THE RANGE OF 30-50, SO FOR A PRODUCT SOLD TO 100 YTL, FACTORY PRICE IS ABOUT 50-70 YTL.

ECONOMIC PRODUCTION

RATE OF PRODUCTION IS A VERY IMPORTANT FACTOR FROM ECONOMY POINT OF VIEW. THE DESIGN CAPACITY MAY NOT BE ACHIEVED BECAUSE OF SEVERAL REASONS:

- SALES DEMAND MAY BE LOW

- DUE TO POOR DESIGN OR BAD/OLD EQUIPMENT THE RATE MAY DECREASE

- DUE TO POOR MAINTENANCE LONG IDLE PERIODS MAY OCCUR

- DUE TO BAD MANAGEMENT, BAD PLANNING OR QUALITY PROBLEMS YIELD MAY DECREASE

THE FIXED COSTS WILL NOT DECREASE AS THE RATE OF PRODUCTION DECREASE OR AS THE PLANT STAYS IDLE. SO THE UNIT COST OF THE PRODUCT WILL INCREASE AND THE PROFIT WILL DECREASE.

ECONOMIC PRODUCTION

THE RELATION BETWEEN SALES, COSTS AND PROFIT CAN BE EXPRESSED MATHEMATICALLY IF FOLLOWING ASSUMPTIONS ARE MADE :

- VARIABLE COSTS ARE DIRECTLY PROPORTIONAL TO PRODUCTION RATE

- FIXED COSTS ARE CONSTANT

- THERE ARE NO FINANCIAL COSTS

- UNIT SELLING PRICE IS CONSTANT

- THERE IS NO INCOME OTHER THAN THE SALES

Z = nS – (nV + F) = n (S – V) – F

Z : GROSS PROFIT, S : NET SALES PRICE / UNIT

n : NUMBER OF UNITS PRODUCED / YEAR

V : VARIABLE COST / UNIT, F : FIXED ANNUAL COST

IF PROFIT TAX IS CONSIDERED : NET PROFIT Y = Z ( 1 – t)

ECONOMIC PRODUCTION

WHEN THE GROSS INCOME (SALES) EQUALS THE TOTAL COST OF THE SALES, THE PROFIT IS ZERO :

Z = 0 = nS – (nV + F) = n (S – V) – F

nS = nV + F AND n = F / (S – V)

THIS CAPACITY IS CALLED ‘BREAK – EVEN POINT’. ABOVE THIS

POINT A PROFIT RESULTS. BREAK – EVEN POINT IS SAME

REGARDLESS OF WHETHER OR NOT PROFIT TAX IS INCLUDED.

WHEN A PLANT OPERATES BELOW THE BREAK – EVEN POINT, THAT

IS AT A LOSS, IT DOES NOT MEAN THE PLANT SHOULD BE SHUT

DOWN, BECAUSE THE FIXED COSTS WOULD HAVE TO BE PAID IN

ANY EVENT. THE SHUT DOWN POINT OCCURS WHEN THE ANNUAL

LOSS IS EQUALS OR EXCEEDS THE VALUE OF THE FIXED COSTS.

ECONOMIC PRODUCTION

EXAMPLE:

THE ANNUAL FIXED COSTS FOR A PLANT ARE $100,000, AND THE TOTAL VARIABLE COSTS ARE $140,000 PER YEAR AT 70% CAPACITY WITH SALES INCOME OF $280,000. WHAT IS THE BREAK-EVEN POINT IN UNITS OF PRODUCTION IF THE SELLING PRICE PER UNIT IS $40?

280,000 / 40 = 7,000 UNITS SOLD AT 70% CAPACITY

140,000 / 7,000 = $20 / UNIT VARIABLE COST

n = F / (S-V) = 100,000 / 20 = 5,000 UNITS AT BREAK-EVEN

ECONOMIC PRODUCTION

EXAMPLE:

IF THE RATIO OF VARIABLE COSTS TO SALES PRICE IS 0.5 AND THE FIXED COSTS ARE $100,000 FOR A PRODUCT SELLING AT $40 PER TON, WHAT IS THE COST FOR UNIT OF PRODUCT a) AT MAXIMUM CAPACITY OF 10,000 TONS AND b) AT $200,000 TOTAL SALES?

a) V / S = 0.5 V = 0.5 x 40 = $20

CTOTAL ={ ( 10,000 TONS x $20 / TON ) + $100,000 } / 10,000 TON

= $ 30 / TON

b) $200,000 / $40 /TON = 5,000 TONS SOLD

CTOTAL ={ ( 5,000 TONS x $20 / TON ) + $100,000 } / 5,000 TON

= $ 40 / TON

ECONOMIC PRODUCTION

EXAMPLE: ANNUAL FIXED COSTS ARE $100,000; VARIABLE COSTS ARE $20 PER TON; DESIGN CAPACITY IS 10,000 TONS/YEAR; SELLING PRICE IS $40 PER TON. WHAT IS THE BREAK – EVEN POINT CAPACITY?

n = F / (S – V) = 100,000 / (40 – 20) = 5,000 TONS

‘ DUMPING’ OCCURS WHEN A COMPANY SELLS A PORTION OF HIS PRODUCTION AT ONE SALE PRICE S1 AND THE REMAINING AT A LOWER PRICE S2. HE OBTAINS GREATER TOTAL PROFIT BY RUNNING HIS PLANT AT HIGHER CAPACITY TO OBTAIIN LOWER UNIT COSTS THAN WOULD RESULT FROM PRODUCING AT A LESSER CAPACITY. IF n1 IS THE NUMBER OF UNITS SOLD AT PRICE S1, AND n2 AT PRICE S2, THE GROSS PROFIT IS:

Z = n1S1 + n2S2 – (n1V + n2V + F)

ECONOMIC PRODUCTION

$, x1000

TONS / YEAR, x1000

0 2 4 6 8 10

400

300

200

100

Net Sales, nS

Total Cost

Variable Costs

Fixed Costs

chart

Break-even Point

ECONOMIC PRODUCTION

THE RELATION BETWEEN PRODUCTION RATE – COST – SALES CAN

BE USED FOR EVALUATION OF NEW PLANTS, EXISTING PLANTS

OR NEW PRODUCTS. BREAK – EVEN POINT IS AN IMPORTANT

INFORMATION TO JUDGE ON THE RISK OF A NEW PLANT OR A

NEW PRODUCT. IF WE DEFINE ‘RATE OF RETURN’ (ROI) AS THE

RATIO OF YEARLY NET PROFIT TO TOTAL INVESTMENT :

ROI = NET ANNUAL PROFIT / TOTAL INVESTMENT

= n [ S – (V + F / n)] (1 – t) / P

FROM THIS RELATION WE CAN CALCULATE THE NUMBER OF UNITS

THAT SHOULD BE SOLD YEARLY FOR A CERTAIN TOTAL

INVESTMENT (P) TO ACHIEVE THE REQUIRED ROI

ECONOMIC PRODUCTION EXAMPLE:A PLANT IS DESIGNED TO PRODUCE 1,200 TONS/YEAR OF CHEMICAL

X WITH A TOTAL INVESTMENT OF $10,000,000. THE FIXED COSTS PER YEAR ARE $2,000,000 AND VARIABLE COSTS ARE $4/kg. THE INCOME TAX RATE IS 30% AND DESIRED ROI IS 0.20. WHAT SHOULD BE THE YEARLY SALES IF SELLING PRICE IS a) $10/kg, b) $12/kg

ROI = n [ S – (V + F/n)] (1 – t) / P

a) 0.20 = n [10 – (4 + 2,000,000/n) ] 0.70 / 10,000,000 2,000,000 = (10n – 4n – 2,000,000) 0.70 4,857,142 = 6n n = 809,524 kg/year

b) 0.20 = n [12 – (4 + 2,000,000/n) ] 0.70 / 10,000,000 2,000,000 = (12n – 4n – 2,000,000) 0.70 4,857,142 = 8n n = 607,143 kg/year

ECONOMIC PRODUCTIONBY LOWERING OUR UNIT SELLING PRICE, WE CAN INCREASE SALES.

IN THAT CASE WE HAVE A NEW BREAK- EVEN CURVE. HERE WE HAVE TWO POSSIBLE REVENUE LINES. THE TOTAL PROFIT AT POINT B MAY BE LARGER OR SMALLER THAN THE PROFIT AT POINT A. BY DISCUSSING ON THESE CURVES, MANAGEMENT MAY DECIDE ON THE OPTIMUM SELLING PRICE MAXIMIZING THE PROFIT

revenue at

$ low price TC revenue at

high price

FC

a b QUANTITY

ECONOMIC PRODUCTIONEXAMPLE:THE FIXED COSTS OF MAKING A PRODUCT ARE $2000 AND THE

VARIABLE COSTS ARE $10 PER KG. THE PRODUCT IS SOLD FOR $12 PER KG AND 2000 KG IS SOLD.

IF THE SELLING PRICE IS REDUCED TO $11 PER KG, HOW MUCH THE PRODUCT SHOULD BE SOLD BEFORE THE REDUCED SELLING PRICE INCREASES PROFIT?

TCA = 2,000 + 20,000 = 22,000

RA = 12 x 2,000 = 24,000

PA = 2,000

TCB = 2,000 + 10B

RB = 11B

PB = 11B – 2,000 – 10B > 2,000 B > 4000 KG

ECONOMIC PRODUCTION

BREAK- EVEN CURVES MAY ALSO BE USED TO DECIDE ON THE LEVEL OF INVESTMENT. IN MOST OF THE CASES, HIGHER INVESTMENT COSTS (FC), LOWER THE VARIABLE COSTS. SO DEPENDING ON THE SALES VOLUME, MANAGEMENT DECIDES ON THE INVESTMENT TYPE (MANUAL VS. AUTOMATIC) TO MAXIMIZE THE PROFIT.

$ revenue

TC for low FC

TC for high FC

high FC

low FC

a b c QUANTITY

ECONOMIC PRODUCTIONEXAMPLE:WITH FIXED COSTS OF $2,000, VARIABLE COSTS OF $10 A UNIT AND A

SELLING PRICE OF $12 A UNIT, 2000 OF A PRODUCT ARE SOLD. THE COMPANY IS CONSIDERING IMPROVING ITS MACHINERY. THIS WILL

RAISE FIXED COSTS TO $4,000 BUT REDUCE VARIABLE COSTS TO $8 A UNIT. IF THE SELLING PRICE IS REDUCED TO $10, TO WHAT FIGURE THE SALES BE INCREASED TO JUSTIFY THE EXTRA INVESTMENT ON TOOLING?

TCA = 2,000 + 20,000 = 22,000

RA = 12 x 2,000 = 24,000

PA = 2,000

TCB = 4,000 + 8B

RB = 10B

PB = 10B – 4,000 – 8B > 2,000 B > 3000

ECONOMIC PRODUCTION

NON LINEAR BREAKEVEN

IN SOME CASES, COSTS AND SALES REVENUES MAY BE NONLINEAR. SECOND OR THIRD SHIFTS OR OVERTIME PAYMENTS MAY HAVE INCREASING EFFECT ON THE UNIT VARIABLE COSTS. TO INCREASE SALES QUANTITY, MORE ADVERTISING OR PROMOTIONS MAY BE REQUIRED WHICH WILL ALSO INCREASE THE UNIT COSTS. THE REVENUE MAY ALSO BE NONLINEAR DUE TO LOWER SELLING PRICE REQUIREMENT FOR INCREASING THE SALES QUANTITY.

IN NONLINEAR CASES WE HAVE TWO BREAKEVEN POINTS AND BETWEEN THESE POINTS WE HAVE THE PROFIT AREA. IN THESE CASES WE HAVE THE MAXIMUM PROFIT AT A OPTIMUM POINT.

ECONOMIC PRODUCTION

be2

sales

$ max. profit

tc

be1

vc

fc

output

ECONOMIC PRODUCTION

ASSET – INTENSIVE INDUSTRIES

AUTOMATED PRODUCTION PLANTS ARE LIKE THIS; THEY HAVE HIGH FIXED COSTS AND RELATIVELY LOW VARIABLE COSTS

THE TOTAL-COST LINE HAS A LOW SLOPE SO WHEN SALES ARE LOW IT IS HARD TO LOWER THE TOTAL COST. IN THESE INDUSTRIES, PROFIT – LOSS IS MORE SENSITIVE TO QUANTITY

TR

$ TC

FC

VC

Q

ECONOMIC PRODUCTION

LABOR – INTENSIVE INDUSTRIES

THESE ARE LESS AUTOMATED PLANTS, WHERE MOST OPERATIONS ARE MANUAL; THEY HAVE RELATIVELY HIGH VARIABLE COSTS AND LESS FIXED COSTS

IN THESE PLANTS, TOTAL COST CAN BE MORE EASILY CONTROLLED SO THE PROFIT - LOSS IS LESS SENSITIVE TO QUANTITY

TC

$ TR VC

FC

Q

CAPITAL INVESTMENT - PROBLEM

• THE TOTAL CAPITAL INVESTMENT FOR A CHEMICAL PLANT IS $1 MILLION AND THE WORKING CAPITAL IS $100,000. IF THE PLANT CAN PRODUCE AN AVERAGE OF 8000 KG OF FINAL PRODUCT PER DAY DURING A 365-DAY YEAR, WHAT SELLING PRICE IN DOLLARS PER KG OF PRODUCT WOULD BE NECESSARY TO GIVE A TURNOVER RATIO OF 1.0? (ASSUME ALL PRODUCTION IS SOLD)

FIXED CAPITAL INVESTMENT = 1,000,000 – 100,000 = $900,000

TURNOVER RATIO = GROSS ANNUAL SALES / FIXED CAPITAL INV.

GROSS ANNUAL SALES = 900,000 x 1.0

GROSS ANNUAL SALES = SELLING PRICE x ANNUAL PRODUCTION

$900,000 = SELLING PRICE x (8000 KG/DAY)(365 DAYS/YEAR)

SELLING PRICE = $0.308

PRODUCT COST - PROBLEM

• A PIECE OF EQUIPMENT ORIGINALLY COSTING $40,000 WAS PUT INTO USE 12 YEARS AGO. AT THE TIME THE EQUIPMENT WAS PUT INTO USE, THE SERVICE LIFE WAS ESTIMATED TO BE 20 YEARS AND THE SALVAGE VALUE WAS ASSUMED TO BE ZERO. ON THIS BASES, A STRAIGHT-LINE DEPRECIATION FUND WAS SET UP. THE EQUIPMENT CAN NOW BE SOLD FOR $10,000 AND A MORE ADVANCED MODEL CAN BE INSTALLED FOR $55,000. ASSUMING THE DEPRECIATION FUND IS AVAILABLE FOR USE, HOW MUCH NEW CAPITAL MUST BE SUPPLIED TO MAKE THE PURCHASE?

YEARLY DEPRECIATION = 40,000 / 20 = 2,000

TOTAL DEPRECIATION FOR 12 YEARS = 12 x 2,000 = $24,000

55,000 – 24,000 – 10,000 = $21,000 REQUIRED

PRODUCT COST - PROBLEM

• A COMPANY HAS A TOTAL INCOME OF $1 MILLION/YEAR AND ALL EXPENSES EXCEPT DEPRECIATION ARE $600,000/YEAR. THE COMPOSITE AMOUNT OF ALL DEPRECIABLE ITEMS HAS A VALUE OF $850,000, OVERALL SERVICE LIFE OF 20 YEARS AND SALVAGE VALUE OF $50,000. THE INCOME TAX RATE IS 35%. WHAT WOULD BE THE REDUCTION IN INCOME TAX CHARGES FOR THE FIRST YEAR OF OPERATION IF DOUBLE DECLINING BALANCE METHOD WERE USED FOR DEPRECIATION INSTEAD OF STRAIGHT-LINE METHOD?

STARIGHT-LINE: d = (850,000 – 50,000) / 20 = $40,000/YR

TAX : T = (1,000,000 – 600,000 – 40,000)(0.35) = $126,000/YR

DOUBLE DECLINING: d = 2 (850,000 – 50,000) / 20 = $80,000/ FIRST YR

TAX : T = (1,000,000 – 600,000 – 80,000)(0.35) = $112,000/ FIRST YR

REDUCTION IN TAX FOR FIRST YEAR = 126,000 – 112,000 = $14,000

ECONOMIC PRODUCTION - PROBLEM

*A COMPANY HAS FIXED COSTS OF $100,000 WITH VARIABLE COSTS EQUAL TO 50% OF NET SALES AND IS PLANNING TO INCREASE ITS PRESENT CAPACITY OF $400,000 SALES BY 30% WITH A 20% INCREASE IN FIXED COSTS. THE PROFIT TAX RATE IS 38%.

a) WHAT NEW SALES DOLLARS ARE REQUIRED TO OBTAIN THE SAME GROSS PROFIT AS THE PRESENT PLANT OPERATION?

b) WHAT WOULD BE THE NET PROFIT IF THE ENLARGED PLANT IS OPERATED AT FULL CAPACITY?

c) WHAT WOULD BE THE NET PROFIT FOR THE ENLARGED PLANT IF SALES REMAINED THE SAME AS AT PRESENT?

a) Z = 400,000 – 100,000 – 0.5(400,000) = $100,000

Z* = Z = 100,000 = S* – 120,000 – 0.5S* S* = $440,000

b) Y* = [1.3(400,000) – 120,000 – 0.5 (1.3)(400,000)] 0.62 = $86,800

c) Y* = [400,000 – 120,000 – 0.5(400,000)] 0.62 = $49,600


* A COMPANY DESIGNED A PLANT TO PRODUCE A CHEMICAL TO BE

SOLD AT $50,000 PER TON. THE EXPECTED PROFIT AT DESIGN

CAPACITY IS $10,000 PER TON. THE TOTAL CAPITAL INVESTMENT

REQUIRED FOR THE PLANT IS $800,000. THE WORKING CAPITAL

REQUIREMENT IS 15% OF THE TOTAL CAPITAL INVESTMENT. THE

EXPECTED SERVICE LIFE IS 10 YEARS AND STRAIGHT LINE

DEPRECIATION IS USED.

DURING THE TESTING PERIOD THE PLANT WAS PUT INTO TRIAL

RUNS OF OPERATION AT 60% OF ITS DESIGN CAPACITY AND IT

WAS FOUND OUT THAT VARIABLE COSTS ARE $15,000 PER TON

AND ANNUAL FIXED COSTS OTHER THAN DEPRECIATION IS

$200,000.

WHAT IS THE BREAK – EVEN POINT PRODUCTION RATE AS PERCENT

OF THE DESIGN CAPACITY OF THE PLANT?

ECONOMIC PRODUCTION - PROBLEM TCI : $800,000 WCI = .15 x TCI = $120,000 FIXED CAPITAL INVESTMENT (FCI) = $680,000 DEPRECIATION PER YEAR = 680,000 / 10 = 68,000

THE TOTAL PROFIT AT DESIGN CAPACITY, A :PROFIT = CAPACITY x PROFIT PER TON = A x 10,000 = A x PRICE – [ DEPRECIATION + FC + (A x VC) ] 10,000A = 50,000A – ( 68,000 + 200,000 + 15,000A) A = 10.72 TONS PER YEAR

LET B = BREAK – EVEN CAPACITY WHERE SALES = COSTS 50,000B = 68,000 + 200,000 + 15,000B B = 7.66 TONS PER YEAR B / A = 7.66 / 10.72 = .715


A PLANT MAKING 4,000 TONS PER YEAR OF PRODUCT AND SELLING AT $0.8 PER kg HAS ANNUAL VARIABLE COSTS OF $2,000,000 AT 100% CAPACITY AND ANNUAL FIXED COSTS OF $700,000. WHAT IS THE FIXED COSTS PER kg AT THE BREAK – EVEN POINT ? IF THE SELLING PRICE IS INCREASED BY 10%, WHAT IS THE DOLLAR INCREASE IN NET PROFIT AT FULL CAPACITY IF THE INCOME TAX RATE IS 20% OF GROSS PROFIT?

V = 2,000,000 / 4,000,000 = $0.5 PER kg

AT BREAK – EVEN : n = 700,000 / (0.8 – 0.5) = 2,333,333

FC PER kg = 700,000 / 2,333,333 = $0.30 PER kg

ORIGINAL NET PROFIT = [(0.8)(4,000,000) – 2,700,000](0.80) = $400,000

NEW NET PROFIT = [(0.8)(1.1)(4,000,000) – 2,700,000](0.80) = $656,000

INCREASE IN NET PROFIT = 656,000 – 400,000 = $256,000

ECONOMIC PRODUCTION-PROBLEM

A COMPANY WITH 50,000 kg/YEAR CAPACITY HAS A TURNOVER RATIO OF 0.5. TOTAL VARIABLE COST OF THE COMPANY AT FULL CAPACITY 5,000,000 TL/YEAR, AND ANNUAL FIXED COST IS 30% OF THE FIXED CAPITAL INVESTMENT. AT FULL CAPACITY YEARLY GROSS PROFIT IS 2,000,000 TL. COMPANY MAKES A NEW INVESTMENT INCREASING THE FIXED CAPITAL INVESTMENT 20% RESULTING WITH 10% INCREASE IN CAPACITY AND SALES. THE YEARLY VARIABLE COST AT THE NEW FULL CAPACITY IS 4,400,000 TL AND ANNUAL FIXED COST IS AGAIN 30% OF THE NEW FIXED CAPITAL INVESTMENT. WHAT IS THE NEW GROSS PROFIT AT THE NEW CAPACITY?

TR = SALES/FCI SALES = 0.5FCI, V = 100 TL, F = 0.3FCI

PROFIT = 2,000,000 = 0.5FCI – 100 x 50,000 – 0.3FCI FCI = 35,000,000

NEW PROFIT = 35,000,000 x 0.5 x 1.1 – 4,400,000 – 35,000,000 x 0.3 x 1.2

= 2,250,000

VI. TIME VALUE OF MONEY

TIME VALUE OF MONEYINTEREST REPRESENTS THE EARNING POWER OF MONEY. IT IS THE

PREMIUM PAID TO COMPENSATE A LENDER FOR THE LOSS OF USE OF THE LOANED MONEY, THE RISK OF NONREPAYMENT AND THE ADMINISTRATIVE COST OF MAKING A LOAN. ON THE OTHER HAND A BORROWER PAYS INTEREST CHARGES FOR THE OPPURTUNITY TO DO SOMETHING NOW THAT OTHERWISE WOULD HAVE TO BE DELAYED OR WOULD NEVER BE DONE.

THE AMOUNT OF INTEREST DEPENDS UPON THE SCARCITY OF MONEY AT THE TIME OF THE LOAN AND WHAT ALTERNATIVE INVESTMENTS MIGHT HAVE YIELDED. IT ALSO DEPENDS UPON THE RISK THE LENDER FEELS THAT HE IS TAKING THAT THE MONEY MIGHT NOT BE REPAID OR WHAT SECURITY MAY BE PLEDGED TO THE LENDER THAT HAS AN EQUAL OR SOMEWHAT GREATER VALUE.

THE ECONOMIC GAIN THROUGH THE USE OF MONEY IS WHAT GIVES MONEY ITS TIME VALUE. BECAUSE MONEY CAN EARN AT AN INTEREST RATE THROUGH ITS INVESTMENT FOR A PERIOD OF TIME, A DOLLAR RECEIVED AT SOME FUTURE DATE IS NOT WORTH AS MUCH AS A DOLLAR IN HAND AT PRESENT. THIS LEADS TO THE CONCEPT OF TIME VALUE OF MONEY.

TIME VALUE OF MONEY

A DOLLAR IN HAND NOW IS WORTH MORE THAN A DOLLAR RECEIVED n YEARS FROM NOW SINCE HAVING A DOLLAR NOW PROVIDES THE OPPURTUNITY FOR INVESTING THAT DOLLAR FOR n YEARS MORE THAN THE DOLLAR TO BE RECEIVED n YEARS HENCE. THIS OPPURTUNITY WILL EARN A RETURN SO THAT AFTER n YEARS THE ORIGINAL DOLLAR PLUS ITS INTEREST WILL BE A LARGER AMOUNT THAN THE ONE DOLLAR RECEIVED AT THAT TIME.

THE TIME VALUE OF MONEY IS LIMITED TO THE FACT THAT MONEY HAS AN EARNING POWER. THE EFFECT OF INFLATION IS SEPARATE.

SINCE ENGINEERING PROJECTS REQUIRE THE INVESTMENT OF MONEY, IT IS IMPORTANT THAT THE TIME VALUE OF THE MONEY USED BE PROPERLY REFLECTED IN THE EVALUATION OF THE PROJECTS.

TIME VALUE OF MONEY

THE AMOUNT OF CAPITAL ON WHICH INTEREST IS PAID IS DESIGNATED AS THE PRINCIPAL (P), AND THE AMOUNT OF INTEREST EARNED BY A UNIT OF PRINCIPAL IN A UNIT OF TIME AS THE RATE OF INTEREST (i). THE TIME UNIT IS USUALLY TAKEN AS ONE YEAR. FOR EXAMPLE, IF $100 WERE THE COMPENSATION DEMANDED FOR GIVING SOMEONE THE USE OF $1000 FOR A PERIOD OF ONE YEAR, THE PRINCIPAL WOULD BE $1000 AND THE RATE OF INTEREST WOULD BE 100 / 1000 = 0.1 OR 10% / YEAR.

TYPES OF INTEREST:

1. SIMPLE INTEREST

THE SIMPLEST FORM OF INTEREST REQUIRES COMPENSATION PAYMENT AT A CONSTANT INTEREST RATE BASED ONLY ON THE ORIGINAL PRINCIPAL. THE AMOUNT OF SIMPLE INTEREST I DURING n INTEREST PERIOD IS:

I = (P)(i)(n)

TIME VALUE OF MONEY

THE PRINCIPLE MUST BE REPAID EVENTUALLY; THEREFORE THE ENTIRE AMOUNT S OF PRINCIPLE + INTEREST DUE AFTER n INTEREST PERIOD IS:

S = P + I = P + (P)(i)(n) = P(1 + in)

IF THE INTEREST RATE IS EXPRESSED ON THE REGULAR YEARLY BASIS AND d REPRESENTS THE NUMBER OF DAYS IN AN INTEREST PERIOD:

ORDINARY SIMPLE INTEREST = P(i)(d/360)

EXACT SIMPLE INTEREST = P(i)(d/365)

TIME VALUE OF MONEY2. COMPOUND INTEREST INTEREST HAS AN IMPORTANT TIME VALUE AND WHENEVER THE

INTEREST IS PAID THE RECEIVER CAN IMMEDIATELY PUT THE INTEREST TO WORK AND EARN ADDITIONAL INTEREST. THIS IS CALLED ‘COMPOUND INTEREST’ WHICH ASSUMES THAT THE INTEREST RECEIVED IS NOT WITHDRAWN BUT ADDED TO THE PRINCIPLE AND INTEREST IS RECEIVED UPON THIS ENLARGED PRINCIPLE DURING THE FOLLOWING PERIOD. THE COMPOUND AMOUNT (S) DUE AFTER DISCREET NUMBER OF INTEREST PERIODS CAN BE DETERMINED AS FOLLOWS:

YEAR P AT THE START INTEREST EARNED S AT THE END 1 P P(i) P + P(i) = P(1 + i) 2 P(1+i) P(1+i)(i) P(1+i)+P(1+i)(i) = P(1+i)2

3 P(1+i)2 P(1+i)2(i) P(1+i)2+P(1+i)2(i) = P(1+i)3 n P(1+i)n-1 P(1+i)n-1(i) P(1+i)n

TIME VALUE OF MONEY

SO IN COMPOUND INTEREST :

S = P(1+i)n

EXAMPLE: IF P = $1000 AND i = 10%

n COMPOUND SIMPLE

1 S1 = 1000(1+0.1) = 1100 ; = 1000(1+0.1) = 1100

2 S2 = 1000(1+0.1)2 = 1210 ; = 1000(1+0.2) = 1200

3 S3 = 1000(1+0.1)3 = 1331 ; = 1000(1+0.3) = 1300

7 S7 = 1000(1+0.1)7 = 1949 ; = 1000(1+0.7) = 1700

COMPOUND INTEREST IS THE MOST USED TYPE OF INTEREST AND IF NOTHING ELSE IS INDICATED WE SHALL UNDERSTAND COMPOUND INTEREST COMPOUNDED YEARLY

TIME VALUE OF MONEY

NOMINAL AND EFFECTIVE INTEREST RATES: THERE ARE CASES WHERE TIME UNITS OTHER THAN ONE YEAR ARE USED. CONSIDER AN EXAMPLE IN WHICH THE INTEREST RATE IS 3% PER PERIOD AND INTEREST IS COMPOUNDED AT HALF-YEAR PERIODS. A RATE OF THIS TYPE WOULD BE REFERRED TO AS ‘6% COMPOUNDED SEMIANNUALLY’. INTEREST RATES STATED IN THIS FORM ARE KNOWN AS ‘NOMINAL INTEREST RATES’. THE ACTUAL ANNUAL RETURN ON THE PRINCIPAL WOULD NOT BE EXACTLY 6% BUT WOULD BE SOMEWHAT LARGER BECAUSE OF THE COMPOUNDING EFFECT AT THE END OF THE SEMIANNUAL PERIOD. NOMINAL INTEREST RATES SHOULD ALWAYS INCLUDE A QUALIFYING STATEMENT INDICATING THE COMPOUNDING PERIOD (IF NO STATEMENT, WE WILL ASSUME ONE YEAR).

P = $100,NOMINAL INTEREST RATE 6% COMPOUNDED YEARLY:

S1 = $106

P= $100,NOMINAL INTEREST RATE 6% COMPOUNDED SEMIANNUALLY

S1 = $100(1.03)(1.03) = $106.09

TIME VALUE OF MONEYSOMETIMES IT IS DESIRABLE TO EXPRESS THE EXACT INTEREST RATE

BASED ON THE ORIGINAL PRINCIPAL AND THE TIME UNIT OF ONE YEAR. A RATE OF THIS TYPE IS KNOWN AS THE ‘EFFECTIVE INTEREST RATE’ (6.09% IN THE ABOVE CASE). THE ONLY TIME THAT THE NOMINAL AND EFFECTIVE INTEREST RATES ARE EQUAL IS WHEN THE INTEREST IS COMPOUNDED ANNUALLY.

LET r BE THE NOMINAL INTEREST RATE UNDER CONDITIONS WHERE THERE ARE m CONVERSIONS OR INTEREST PERIODS PER YEAR. THEN THE INTEREST RATE BASED ON THE LENGTH OF ONE INTEREST PERIOD IS r / m .

SINCE S = P (1+i)n

Safter 1 year = P (1+r/m)m

IF WE DESIGNATE EFFECTIVE INTEREST RATE AS ieff

S1 = P (1+ieff) = P (1+r/m)m

ieff = (1+r/m)m - 1

TIME VALUE OF MONEYEXAMPLE : P = $1000, MONTHLY INTEREST RATE OF 2% * TOTAL AMOUNT PRINCIPAL + SIMPLE INTEREST DUE AFTER 2

YEARS IF NO INTERMEDIATE PAYMENTS ARE MADE S = P(1+in) = $1000 ( 1+ 0.02 X 24) = $1480 a) TOTAL AMOUNT PRINCIPAL + COMPOUNDED INTEREST AFTER 2

YEARS IF NO INTERMEDIATE PAYMENTS ARE MADE S = P (1+I)n = $1000 (1.02)24 = $1608

b) NOMINAL INTEREST RATE WHEN THE INTEREST IS COMPOUNDED MONTHLY

r = 0.02 X 12 = 0.24 OR 24% COMPOUNDED MONTHLY

c) EFFECTIVE INTEREST RATE WHEN THE INTEREST IS COMPOUNDED MONTHLY

ieff = (1+r/m)m – 1 = (1+0.24/12)12 – 1 = 0.268 OR 26.8%

TIME VALUE OF MONEY

3. CONTINUOUS INTEREST

FOR INTEREST ACCUMULATION YOU CAN TAKE SHORTER TIME INTERVALS THAN ONE YEAR LIKE ONE MONTH, ONE DAY, ONE HOUR OR EVEN SHORTER; AND THE EXTREME CASE IS WHEN THE TIME INTERVAL BECOMES INFINITESIMALLY SMALL SO THAT THE INTEREST IS COMPOUNDED CONTINUOUSLY.

r : NOMINAL INTEREST RATE ; m : INTEREST PERIODS / YEAR

Sn = Plim m →∞ (1+r/m)mn = Plimm→∞ (1+r/m)m/r x rn

SINCE limm→∞(1+r/m)m/r = e Sn = Pern

TO USE ieff WHICH IS (1+r/m)m – 1 = (1+r/m)m/r x r- 1

SO ieff = er – 1 AND Sn = P(ieff +1)n

TIME VALUE OF MONEY

EXAMPLE : IF r = 0.2 (NOMINAL INTEREST RATE OF 20%)

a) TOTAL AMOUNT TO WHICH $1 OF INITIAL PRINCIPAL WOULD ACCUMULATE AFTER ONE YEAR IF COMPOUNDED DAILY

S = P (1+r/m)m = 1(1 + 0.2/365)365 = $ 1.2213

b) TOTAL AMOUNT TO WHICH $1 OF INITIAL PRINCIPAL WOULD ACCUMULATE AFTER ONE YEAR WITH CONTINUOUS COMPOUNDING

S = Pern = 1 (e)0.2 = $ 1.2214

c) ieff FOR CONTINUOUS COMPOUNDING

ieff = er – 1 = 1.2214 – 1 = 0.2214 OR 22.14%

TIME VALUE OF MONEY

PRESENT WORTH AND DISCOUNT

THE PRESENT WORTH (OR PRESENT VALUE) OF A FUTURE AMOUNT IS THE PRESENT PRINCIPLE WHICH MUST BE DEPOSITED AT A GIVEN INTEREST RATE TO YIELD THE DESIRED AMOUNT AT SOME FUTURE DATE.

SINCE S = P (1 + i)n

P = S / (1 + i)n FOR COMPOUND INTEREST

AND SINCE S = Pern

P = S / ern FOR CONTINUOUS INTEREST

THE DISCOUNT = FUTURE VALUE – PRESENT VALUE = S – P

TIME VALUE OF MONEY

EXAMPLE : A BOND HAS A MATURITY VALUE OF $1000 AND IS PAYING DISCRETE COMPOUND INTEREST AT AN EFFECTIVE ANNUAL RATE OF 3%. DETERMINE THE FOLLOWING AT A TIME 4 YEARS BEFORE THE BOND REACHES MATURITY VALUE:

a) PRESENT WORTH

P = S/ (1 + i)n = 1000 / (1 + 0.03)4 = $888

b) DISCOUNT

DISCOUNT = 1000 – 888 = $112

c) DISCRETE COMPOUND RATE OF EFFECTIVE INTEREST WHICH WILL BE RECEIVED BY A PURCHASER IF THE BOND WAS OBTAINED FOR $700

P = S / (1+i)n ; 700 = 1000 / (1+i)4 i = (1000/700)1/4 – 1 = 0.0933

d) REPEAT (a) FOR THE CASE WHERE THE NOMINAL BOND INTEREST IS 3% COMPOUNDED CONTINUOUSLY

P = S / ern = 1000 / e(0.03)x4 = $887

TIME VALUE OF MONEYANNUITIES : AN ANNUITY IS A SERIES OF EQUAL PAYMENTS

OCCURING AT EQUAL TIME INTERVALS. AN ‘ANNUITY TERM’ IS THE TIME FROM THE BEGINNING OF THE FIRST PAYMENT PERIOD TO THE END OF THE LAST PAYMENT PERIOD. THE ‘AMOUNT OF AN ANNUITY’ IS THE SUM OF ALL THE PAYMENTS PLUS INTEREST.

LET R REPRESENT THE UNIFORM PERIODIC PAYMENTS MADE DURING n DISCRETE PERIODS IN AN ORDINARY ANNUITY. THE INTEREST RATE BASED ON THE PAYMENT PERIOD IS i, AND S IS THE AMOUNT OF ANNUITY. THE FIRST PAYMENT R IS MADE AT THE END OF THE FIRST PERIOD AND WILL BEAR INTEREST FOR n – 1 PERIODS. THUS AT THE END OF THE ANNUITY TERM, THIS FIRST PAYMENT WILL HAVE ACCUMULATED TO AN AMOUNT OF R(1 + i)n-1. THE SECOND PAYMENT WILL BE R(1 + i)n-2 AND SO ON.

S = R(1+i)n-1+ R(1+i)n-2 + …. + R(1+i) + RMULTIPLY BOTH SIDES BY (1+i) S + Si = R(1+i)n + R(1+i)n-1…+R(1+i)SUBTRACT THE FIRST EQUATION FROM THIS ONESi = R (1+i)n – R S = R { (1+i)n - 1} / i

TIME VALUE OF MONEY

FOR DISCRETE CASH FLOW AND CONTINUOUS COMPOUNDING:

SINCE S = Rern ; THE FIRST PAYMENT WILL BE Rer(n-1) AT THE END, SECOND WILL BE Rer(n-2) AND SO ON.

S = Rer(n-1) + Rer(n-2) + …. + Rer + R

MULTIPLY BOTH SIDES BY er

Ser = Rern + Rern-1 + …. + Rer

SUBTRACT THE FIRST EQUATION FROM THIS

Ser – S = Rern – R

S (er – 1) = R (ern – 1)

S = R { ( ern – 1) / ( er – 1) }

TIME VALUE OF MONEY

FOR CONTINUOUS CASH FLOW AND INTEREST COMPOUNDING, LET r REPRESENT NOMINAL INTEREST RATE WITH m INTEREST PERIODS PER YEAR SO THAT i = r / m AND THE TOTAL NUMBER OF INTEREST PERIODS IN n YEARS IS mn. LET Ř REPRESENT THE TOTAL OF ALL ORDINARY ANNUITY PAYMENTS OCCURING UNIFORMLY THROUGHOUT THE YEAR SO THAT Ř / m IS THE UNIFORM ANNUITY PAYMENT AT THE END OF EACH PERIOD:

S = Ř / m [ { (1 + r/m)mrn/r – 1 } / r/m ]

= Ř { (ern – 1) /r }

(THE SYMBOLS S, R REPRESENT DISCRETE LUMPSUM PAYMENTS. A BAR ABOVE THE SYMBOL, SUCH AS Š, Ř MEANS THAT THE PAYMENTS ARE MADE CONTINUOUSLY THROUGHOUT THE TIME PERIOD.)

TIME VALUE OF MONEYPRESENT WORTH OF ANNUITY: SINCE P = S / (1 + i)n ; P = R [ { (1 + i)n – 1 } / i(1 + i)n ] FOR CONTINUOUS CASH FLOW AND INTEREST COMPOUNDING : P = Ř { (ern – 1) / rern }ANNUITY DUE : IF PAYMENTS ARE MADE AT THE BEGINNING OF

EACH PERIOD INSTEAD OF END OF THE PERIODSDEFERRED ANNUITY : IF THE FIRST PAYMENT IS DUE AFTER A

DEFINITE NUMBER OF YEARS

EXAMPLE : YOU WANT TO ACCUMULATE $10,000 IN 10 YEARS BY MAKING EQUAL INSTALMENTS EACH YEAR, BEGINNING AT THE END OF THE FIRST YEAR ; DETERMINE THE YEARLY INSTALMENTS

a) IF i = 0.06 ANNUALLY R = S [ i / { (1 + i)n – 1 } ] = $10,000 [ 0.06 / { (1.06)10 - 1} ] = $759 / YEAR b) IF ANNUAL INTEREST OF 0.06 COMPOUNDED CONTINUOUSLY

WITH CONTINUOUS CASH FLOW Ř = S { r / (ern – 1) } = $10,000 { 0.06 / (e0.6 – 1) } = $730 / YEAR

TIME VALUE OF MONEYSUMMARY OF BASIC INTEREST RELATIONS

• SIMPLE INTEREST : S = P (1 + in) n: NUMBER OF PERIODS

i : INTEREST PER PERIOD • COMPOUND INTEREST: S = P (1 + i)n

• CONTINUOUS COMPOUNDING:

S = Pern = P (ieff + 1)n r:NOMINAL INTEREST RATE

• ANNUITIES WITH DISCRETE PAYMENTS, R :

S = R [ {(1 + i)n – 1} / i ] R:UNIFORM PERIODIC PAYMENTS

P = R [ {(1 + i)n – 1} / i (1 + i)n ] MADE AT EACH n DISCRETE

PERIOD

• ANNUITIES WITH DISCRETE PAYMENTS,R, CONTINUOUS COMPOUNDING:

S = R { (ern – 1) / (er – 1) }

P = R { (ern – 1) / ern(er – 1) }• ANNUITIES WITH CONTINUOUS CASH FLOW, Ř, AND CONTINUOUS

COMPOUNDING: S = Ř { (ern – 1) / r Ř: TOTAL ANNUITY PAYMENT

P = Ř { (ern – 1) / rern PER YEAR

TIME VALUE OF MONEY

WHAT HAPPENS WHEN THE INVESTMENT IS CONSIDERED TO BE PERMENANT, THAT IS n ?

WHEN n , (1 + i)n = SINCE S = P (1 + i)n S = FOR ANY P

P = S / (1 + i)n P = 0 FOR ANY S

SINCE R = P [{i (1 + i)n} / { (1 + i)n – 1 }]

{i (1 + i)n} / { (1 + i)n – 1 } = [ i / { (1 + i)n – 1} ] + i

WHEN n = 0 + i

R = Pi

TIME VALUE OF MONEYCAPITILIZED COST WE MAY DESIRE TO DETERMINE A TOTAL COST FOR A PIECE OF EQUIPMENT

UNDER CONDITIONS WHICH PERMIT THE EQUIPMENT TO BE REPLACED PERPATUALLY WITHOUT CONSIDERING INFLATION OR PRICE CHANGES.

IF REPLACEMENT AT THE END OF n YEARS COSTS CR , THEN WE SHOULD HAVE A PRINCIPAL (P) IN THE BEGINNING WHICH WILL ACCUMULATE TO S IN n YEARS, SUCH THAT WHEN YOU SPEND CR, YOU STILL HAVE P IN HAND TO ACCUMULATE AGAIN TO S IN n YEARS AND SO ON.

S = P + CR

SINCE S = P(1 + i)n = P + CR

(1 + i)n = 1 + CR / P

P = CR / {(1 + i)n – 1 }

TIME VALUE OF MONEY

THE CAPITILIZED COST (K) IS DEFINED AS THE ORIGINAL COST OF THE EQUIPMENT (CV) PLUS P

K = CV + CR /{ (1 + i)n – 1 } IF CV = CR K = CV [ 1 + 1/{ (1 +i)n – 1 }]

ENGINEERS USE CAPITILIZED COST PRINCIPALLY FOR COMPARING ALTERNATIVES

EXAMPLE : A NEW EQUIPMENT COSTS $12,000 AND WILL HAVE A SCRAP VALUE OF $2000 AT THE END OF ITS USEFULL LIFE OF 10 YEARS. WHAT IS THE CAPITILIZED COST IF INTEREST IS 0.06 ?

K = CV + CR /{ (1 + i)n – 1} = 12,000 + 10,000 / (1.0610 - 1 ) = $24,650

THAT MEANS OUT OF THIS AMOUNT $12,000 WILL BE USED FOR BUYING THE EQUIPMENT IN THE BEGINNING; REMAINING $12,650 WILL ACCUMULATE TO $22,650 IN 10 YEARS, OUT OF WHICH $10,000 WILL BE USED TOGETHER WITH THE SCRAP VALUE TO RENEW THE EQUIPMENT; AND THIS WILL GO ON.

TIME VALUE OF MONEY

EXAMPLE : IF A REACTOR IS MADE FROM MILD STEEL IT COSTS $5000 AND ITS USEFULL LIFE IS 3 YEARS. IF IT IS MADE FROM STAINLESS STEEL IT COSTS $15,000. WHAT SHOULD BE THE USEFULL LIFE PERIOD OF THE STAINLESS STEEL REACTOR TO HAVE EQUAL CAPITILIZED COST IF SCRAP VALUES ARE ZERO AND MONEY IS WORTH 6% COMPOUNDED ANNUALLY ?

FOR THE MILD STEEL REACTOR :

K = CV + CR /{ (1 + i)n – 1} = 5000 + 5000 / (1.063 – 1) = $31,176

FOR THE STAINLESS STEEL TO HAVE EQUAL K

31,176 = 15,000 + 15,000 / (1.06n – 1)

1.06n = 1.9273

n = 11.3

THAT MEANS, IF THE LIFE IS > 11.3 YEARS, CHOOSE STAINLESS STEEL FOR MAKING THE REACTOR

TIME VALUE OF MONEY

WHEN THERE IS A PAYMENT AT THE END OF EACH YEAR (LIKE MAINTANENCE COST), WE SHOULD ADD IT TO THE ‘CAPITILIZED COST’ FORMULA:

P = R [ (1 + i )n – 1 / i (1 + i )n ]

DIVIDE BY (1 + i )n

P = R [ 1 – 1/ (1 + i )n / i ]

WHEN n GOES TO ∞

P = R / i

CAPITILIZED COST FORMULA BECOMES:

K = R / i + CV + CR / (1 + i )n – 1 WHERE R IS THE YEARLY EXPENSE

TIME VALUE - PROBLEM

• AT WHAT RATE WILL $65.07 YIELD $8.75 IN SIMPLE INTEREST IN 3 YEARS 6 MONTHS?

S = P(1 + in) 73.82 = 65.07 (1 + i x 3.5) i = 3.84%

• FIND THE COMPOUND AMOUNT OF $100 FOR 4 YEARS AT 6% COMPOUNDED ANNUALLY.

S = P(1 + i)n = 100(1.06)4 = $126.25

• ACCUMULATE A PRINCIPLE OF $1000 FOR 5 YEARS 9 MONTHS AT A NOMINAL RATE OF 12% COMPOUNDED MONTHLY. HOW MUCH INTEREST IS EARNED?

S = 1000(1 + 0.12/12)69 = $1986.9 ; I = $986.9

TIME VALUE - PROBLEM• AT WHAT ANNUAL INTEREST RATE WILL $1000 INVESTED TODAY

BE WORTH $2000 IN 9 YEARS ? P = 1000 ; S = 2000 ; n = 9 S = P(1 + i)n

2000 / 1000 = (1 + i)9 21/9 = 1 + i i = 0.08

• A LOAN OF $1000 IS MADE TODAY UNDER AN AGREEMENT THAT $1400 WILL BE RECEIVED IN PAYMENT SOMETIME IN FUTURE. WHEN SHOULD THE $1400 BE RECEIVED IF THE LOAN IS TO EARN INTEREST AT A RATE OF NOMINAL YEARLY 8% COMPOUNDED QUARTERLY?

P = 1000 ; S = 1400 ; r = 0.08 ; m = 4

ieff = (1 + r/m)m – 1 = (1 + 0.08/4)4 – 1 = 0.0824

S = P(1 + ieff)n 1400 = 1000(1.0824)n

1.4 = (1.0824)n ln 1.4 = nln1.0824 n = 4.25

TIME VALUE – PROBLEM • NOW IS JUNE 30, 2002. 3 PAYMENTS, EACH OF $500, ARE TO BE

RECEIVED EVERY 2 YEARS FROM NOW AND DEPOSITED IN A BANK WHERE THEY WILL EARN INTEREST AT 7% PER YEAR. HOW LARGE WILL THE BANK ACCOUNT BE ON JUNE 30, 2010?

2004 $500 ; 2006 $500 ; 2008 $500 S = P (1 + i)n = 500 { (1.07)6 + (1.07)4 + (1.07)2 } = $1978.3

• A PERSONAL LOAN OF $1000 IS MADE FOR A PERIOD OF 18 MONTHS AT A INTEREST RATE OF 1.5% PER MONTH ON THE UNPAID BALANCE. IF THE ENTIRE AMOUNT IS PAID AS A LUMP SUM AT THE END OF 18 MONTHS, DETERMINE

a) THE EFFECTIVE ANNUAL INTEREST RATE b) THE TOTAL AMOUNT OF INTEREST PAID a) SINCE r/m = 0.015

ieff = (1 + r/m)m – 1 = (1 + 0.015)12 – 1 = .1956 b) S = P (1 + i)n = 1000 ( 1 + 0.015)18 = $1307.4 ; I = $307.4

TIME VALUE - PROBLEM• FIND THE COMPOUND AMOUNT OF $500 AT 6% FOR 4, 8, 12 YEARS. S = P (1 + i)n = 500 (1.06)4,8,12 = $631.24 ; $796.92 ; $1006.10

• A LOAN OF $200 IS MADE FOR A PERIOD OF 13 MONTHS AT A SIMPLE INTEREST RATE OF 10%. WHAT FUTURE AMOUNT IS DUE AT THE END OF THE LOAN PERIOD?

S = P(1 + in) = 200 { 1 + 0.10(1 + 1/12) } = $221.67

• A CREDIT PLAN CHARGES INTEREST AT A RATE OF NOMINAL YEARLY 18% COMPOUNDED MONTHLY. WHAT IS THE EFFECTIVE INTEREST RATE ?

ieff = (1 + r/m)m – 1 = (1 + 0.18/12)12 – 1 = 0.1956 OR 19.56%

• HOW MUCH WOULD A PERSON HAVE HAD TO INVEST 1 YEAR AGO TO HAVE $2500 AVAILABLE TODAY WHEN NOMINAL YEARLY INTEREST r = 0.12 COMPOUNDED MONTHLY ?

S = P(1 + r/m)m 2500 = P (1 + 0.12/12)12 P = $2218.6


• HOW MANY YEARS DOES IT TAKE FOR A MONEY TO DOUBLE IF a) r = 0.08 COMPOUNDED ANNUALLY?

S = P(1 + i)n 2P = P(1.08)n n = 9 YEARS

b) r = 0.08 WITH CONTINUOUS INTEREST?

S = Pern 2P = Pe0.08n n = 8.66 YEARS

• AT WHAT NOMINAL INTEREST RATE WILL A MONEY DOUBLE USING CONTINUOUS COMPOUNDING IN 5 YEARS?

S = Pern 2P = Pe5r r = 0.138 OR 13.8%

• A BUILDING WAS PURCHASED 10 YEARS AGO FOR $50,000 AND HAS RECENTLY BEEN SOLD FOR $120,000. DISREGARDING ANY TAXES, DETERMINE THE RATE OF INTEREST OBTAINED ON THE INITIAL INVESTMENT.

S = P(1 + i)n 120,000 = 50,000 (1 + i)10 i = 0.0915 OR 9.15%


• A CONSTRUCTION FIRM CAN LEASE A CRANE REQUIRED IN A PROJECT FOR 3 YEARS FOR $180,000 PAYABLE NOW, WITH MAINTENANCE INCLUDED. THE ALTERNATIVE IS TO BUY A CRANE FOR $240,000 AND SELL IT AT THE END OF 3 YEARS FOR $100,000. ANNUAL MAINTENANCE COSTS FOR THIS ALTERNATIVE IS $5000 THE FIRST 2 YEARS AND $10,000 THE THIRD YEAR (PAYABLE AT THE END OF EACH YEAR). WHICH ALTERNATIVE IS GOOD IF i = 0.07?

FIRST ALTERNATIVE : P1 = $180,000

SECOND ALTERNATIVE : P2 = $240,000 – PS + PM

PS = S / (1 + i)n = 100,000 / (1.07)3 = $81,630

PM = 5000 / (1.07) + 5000 / (1.07)2 + 10,000 / (1.07)3 = $17,203

P2 = 240,000 – 81,630 + 17,203 = $175,573 2nd ALT. IS BETTER

IF i = 0.10 THAN PS = $75,131 ; PM = $16,191

P2 = 240,000 – 75,131 + 16,191 = $181,060 1st ALT. IS BETTER


• ASSUME THAT YOU SOLD A PROPERTY TODAY FOR $2421 WHICH YOU HAD PURCHASED 4 YEARS AGO WITH $2000 WITHDRAWN FROM YOUR SAVING ACCOUNT. DURING THE FOUR YEAR PERIOD YOUR SAVINGS WOULD HAVE EARNED 6% PER YEAR. COMPARE THE NOMINAL INTEREST RATE RECEIVED FROM YOUR PROPERTY PURCHASE.

S = P(1 + i)n 2421 = 2000 (1 + i)4 ln 1.2105 = 4 ln (1+i)

i = 0.0489 < 0.06

TIME VALUE - PROBLEM• WHAT ANNUAL YEAR- END PAYMENT MUST BE MADE EACH YEAR TO

HAVE $20,000 AVAILABLE 5 YEARS FROM NOW IF THE COMPOUND ANNUAL INTEREST RATE IS 6% ?

S = R [{(1 + i)n – 1} / i ] 20,000 = R [{(1.06)5 – 1} / 0.06] R = $3,548

• IF YOU DEPOSIT $10,000 TODAY WHAT EQUAL AMOUNTS CAN YOU WITHDRAW AT THE END OF EACH QUARTER FOR THE NEXT 4 YEARS WHEN THE NOMINAL INTEREST RATE IS 10%, COMPOUNDED QUARTERLY ?

n = 4 x 4 = 16 ; i = .10 / 4 = 0.025 P = R [{(1 + i)n - 1} / i (1 + i)n ] 10,000 = R [{ (1.025)16 – 1} / 0.025 (1.025)16] R = $765.74

• A LOAN OF $5000 IS SCHEDULED TO BE PAID IN EQUAL MONTHLY INSTALLMENTS OVER 2.5 YEARS. THE NOMINAL INTEREST RATE IS 6% COMPOUNDED MONTHLY. HOW LARGE IS EACH PAYMENT ?

n = 2.5 x 12 = 30 i = .06 / 12 = 0.005 P = R [{(1 + i)n – 1} / i (1 + i)n ] 5000 = R [{(1.005)30 – 1} / 0.005 (1.005)30] R = 179.89


• THERE IS A TRAINING PROGRAM WHICH WILL COST $12,000 IN THE BEGINNING AND $4000 PER YEAR. IT IS ESTIMATED TO PRODUCE SAVINGS OF $7000 EACH YEAR FOR 5 YEARS. SHOULD WE START THE PROGRAM IF i = 0.06 ?

ANNUAL SAVING = 7000 – 4000 = 3000 P = R [{1 + i)n – 1} / i(1 + i)n ] P = 3000 [{1.06)5 – 1} / 0.06 (1.06)5 ] = $12,636

SINCE PRESENT WORTH OF SAVINGS IS > $12,000 , START .

• YOU CAN BUY A NEW CAR FOR $12,000, PAYING $2000 DOWN AND THE DEALER WILL FINANCE THE REMAINDER AT A NOMINAL ANNUAL RATE OF 6%, COMPOUNDED MONTHLY FOR 5 YEARS.

DETERMINE THE AMOUNT OF YOUR MONTHLY PAYMENTS

n = 5 x 12 = 60 ; i = 0.06 / 12 = 0.005 ; P = R[{(1 + i)n – 1} / i(1 + i)n ]

10,000 = R[{(1.005)60 – 1} / 0.005(1.005)60] R = $193.33

TIME VALUE - PROBLEM• THE AMOUNT OF $1200 PER YEAR IS TO BE PAID INTO AN ACCOUNT OVER

EACH OF THE NEXT 5 YEARS. USING A NOMINAL INTEREST RATE OF 12% PER YEAR, DETERMINE THE TOTAL AMOUNT THAT THE ACCOUNT WILL CONTAIN AT THE END OF THE FIFTH YEAR UNDER THE FOLLOWING CONDITIONS :

a) DEPOSITS MADE AT THE BEGINNING OF EACH YEAR WITH SIMPLE INTEREST

S = P(1 + ni) = 1200{ (1+5x.12)+(1+4x.12)+(1+3x.12)+(1+2x.12)+(1+.12) } = $8160

b) DEPOSITS MADE AT THE END OF EACH YEAR WITH INTEREST COMPOUNDED ANNUALLY

S = R [{ (1+i)n – 1} / i ] = 1200 [{ (1.12)5 – 1} / .12] = $7623.42

c) DEPOSITS MADE AT THE END OF EACH MONTH ($100) WITH INTEREST COMPOUNDED MONTHLY

S = R [{ (1+i)n – 1} / i ] = 100 [{ (1+.12/12)60 - 1} / (.12/12) ] = $8166.97

d) DEPOSITS MADE AT THE END OF EACH YEAR WITH INTEREST COMPOUNDED MONTHLY

ieff = (1 + r/m)m – 1 = (1 + .12/12)12 – 1 = 0.1268

S = R [{ (1+i)n – 1} / i ] = 1200 [{ (1.1268)5 – 1} / .1268] = $7727.08


• A MANUFACTURING FIRM IN A FOREIGN COUNTRY HAS AGREED TO PAY $25,000 IN ROYALTIES AT THE END OF EACH YEAR FOR THE NEXT 5 YEARS. IF THE PAYMENTS ARE LEFT IN THE FOREIGN COUNTRY, INTEREST ON THE RETAINED FUNDS WILL BE PAID AT AN ANNUAL RATE OF 15%.

a) WHAT TOTAL AMOUNT WILL BE AVAILABLE IN 5 YEARS?

S = R [{(1 + i)n – 1} / i ] = 25,000 [{ (1.15)5 – 1} / 0.15] = $168,560

b) HOW LARGE WOULD THE UNIFORM ANNUAL PAYMENTS HAVE TO BE IF THE PATENT OWNERS INSISTED THAT A MINIMUM $175,000 BE ACCUMULATED BY THE END OF 5 YEARS ?

175,000 = R [{(1.15)5 – 1} / 0.15]

R = $25,955


• A COMPANY 3 YEARS AGO BORROWED $40,000 TO PAY FOR A NEW MACHINE TOOL, AGREEING TO REPAY THE LOAN IN 100 MONTHLY PAYMENTS AT AN ANNUAL NOMINAL INTEREST RATE OF 12% COMPOUNDED MONTHLY. THE COMPANY NOW WANTS TO PAY OFF THE LOAN. HOW MUCH WOULD THIS PAYMENT BE?

i = 0.12 / 12 = 0.01 MONTHLY ; n (total) = 100

P = R [{(1 + i)n - 1} / i(1 + i)n]

40,000 = R [{(1.01)100 – 1} / 0.01(1.01)100] R = $634.63

n (up to now) = 3 x 12 = 36

S = R [{(1 + i)n – 1} / i ] = 634.63 [{(1.01)36 – 1} / 0.01]

= $27,337.88 (today value of what was paid)

ieff = (1 + r/m)m – 1 = (1 + 0.12 / 12)12 = 0.126825

S = P (1 + ieff)n = 40,000 (1.126825)3

= $57,230.75 (today value of 40,000)

57,230.75 – 27,337.88 = $29,892.87 SHOULD BE PAID

TIME VALUE - PROBLEM• COMPARE THE COST TO PAY OFF A $3000 LOAN IN 1 YEAR WITH

12 EQUAL PAYMENTS WHEN INTEREST IS 12% COMPOUNDED MONTHLY AS OPPOSED TO MAKING A SINGLE PAYMENT WHEN THE EFFECTIVE INTEREST RATE IS 12%

a) AS OUT OF POCKET AMOUNT P = R [{ (1 + i)n – 1} / i(1 + i)n]

3000 = R [{(1+.12/12)12 – 1} / .12/12(1+.12/12)12] R = 266.67 12 x 266.67 = $3200 when paid in installments S = 3000 (1 + i)n = 3000 (1.12) = $3360 when paid as single payment 3200 – 3360 = - $160 b) AS S AMOUNT S = R [{(1 + i)n – 1} / i] = 266.67 [{(1.01)12 - 1} / 0.01] = $3382 3382 – 3360 = $22 c) WHICH IS ADVANTAGEOUS AND WHY? single payment is advantageous because :

ieff = (1 + r/m)m – 1 = (1 + 0.01)12 = 0.1268 which is > 0.12


• AT THE END OF EACH YEAR A SINGLE PAYMENT OF $1766 IS DEPOSITED IN AN ACCOUNT THAT EARNS 6% COMPOUNDED CONTINUOUSLY. WHAT IS THE AMOUNT IN THE ACCOUNT AFTER 5 YEARS?

S = R {(ern – 1) / (er – 1)} = 1766 {(e5x0.06 – 1) / (e0.06 – 1)} = $9992

• THE EXPECTED LIFE OF A NEW MACHINE IS 5 YEARS. ANNUAL CASH FLOW DUE TO THIS MACHINE IS $27,000. ASSUMING CONTINUOUS CASH FLOW AND CONTINUOUS COMPOUNDING, WHAT IS THE AMOUNT THAT CAN INITIALLY BE PAID TO THIS MACHINE IF WE WANT TO EARN 15% ON THE INVESTMENT ?

Ř = 27,000 ; ieff = er – 1 0.15 = er – 1 r = 0.1398

P = Ř {(ern – 1) / rern} = 27,000 {(e5x0.1398 – 1) / 0.1398e5x0.1398}

= $97,130


• USING THE CAPITILIZED COST PROCEDURE COMPARE THE MERITS OF BUYING A PUMP THAT COSTS $2000, NEEDS REPLACEMENT EVERY 5 YEARS AND REQUIRES $300 / YEAR AS MAINTENANCE EXPENSE WITH ANOTHER ONE THAT COSTS $4000 BUT WOULD HAVE A 10 YEAR LIFE AND ONLY NEEDS $100 / YEAR FOR MAINTENANCE. ASSUME 10% INTEREST RATE PER YEAR.

K1 = R / i + CV + CR / {(1 + i)n – 1}

= 300 / .1 + 2000 + 2000 / {(1.1)5 – 1} = $8,276

K2 = R / i + CV + CR / {(1 + i)n – 1}

= 100 / .1 + 4000 + 4000 / {(1.1)10 – 1} = $7,510

SECOND ALTERNATIVE IS BETTER

VII. PROFITABILITY

PROFITABILITY

BEFORE CAPITAL IS INVESTED IN A PROJECT, IT IS NECESSARY TO KNOW HOW MUCH PROFIT CAN BE OBTAINED AND WHETHER OR NOT IT MIGHT BE MORE ADVANTAGEOUS TO INVEST THE CAPITAL IN ANOTHER FORM OF ENTERPRICE. THUS, DETERMINATION OF PROFIT IS A MAJOR GOAL OF AN ECONOMIC ANALYSIS.

PROFIT CAN BE CALCULATED WITH SOME ASSUMPTIONS ABOUT FUTURE (DEMAND, PRICE, AMOUNT OF PRODUCTION), SO IT IS NOT AN UNFAILING VALUE AND CAN ONLY SERVE AS A GUIDE.

PROFIT ALONE CANNOT BE USED FOR DETERMINING IF AN INVESTMENT SHOULD BE MADE. SUPPOSE TWO INVESTMENTS ARE UNDER CONSIDERATION : ONE REQUIRES $100,000 OF CAPITAL AND WILL YIELD A PROFIT OF $10,000/ YEAR; THE OTHER REQUIRES $1 MILLION OF CAPITAL AND WILL YIELD $50,000/ YEAR. THE SECOND GIVES A GREATER YEARLY PROFIT, BUT THE RATE OF RETURN IS ONLY 5% WHILE THE RATE OF RETURN OF THE FIRST IS 10%.

PROFITABILITY

THE MOST COMMONLY USED METHODS FOR PROFITABILITY EVALUATION ARE:

1. RATE OF RETURN

2. PAYOUT PERIOD

3. NET RETURN

4. DISCOUNTED CASH FLOW

5. NET PRESENT WORTH

6. EQUIVALENT ANNUAL COST – ANNUAL WORTH

EACH OF THESE METHODS HAVE ITS ADVANTAGES AND DISADVANTAGES AND SINCE NO SINGLE METHOD IS BEST FOR ALL SITUATIONS THE ENGINEER SHOULD KNOW THEM ALL AND CHOOSE THE BEST SUITING ONE.

PROFITABILITY

1. RATE OF RETURN ON INVESTMENT

THE YEARLY PROFIT DIVIDED BY THE TOTAL INITIAL INVESTMENT (FIXED + WORKING CAPITAL) REPRESENTS THE RETURN ON INVESTMENT.

PROFITS AND THEREFORE RATE OF RETURNS MAY BE EXPRESSED ON THE BEFORE-TAX OR AFTER-TAX BASIS AND THIS SHOULD BE INDICATED.

ROI = NET PROFIT PER YEAR / TOTAL INVESTMENT

RATE OF RETURN MAY BE COMPARED FOR ALTERNATIVE INVESTMENTS, INCLUDING PUTTING THE MONEY IN THE BANK, AND GIVES A FIRST APPROXIMATION OF HOW ATTRACTIVE THE NEW PROJECT MAY BE. IF THE PROFIT VARIES FROM YEAR TO YEAR, AN AVERAGE MAY BE ASSUMED.

IN RATE OF RETURN CALCULATION, WE DO NOT CONSIDER THE TIME VALUE OF MONEY.

PROFITABILITYEXAMPLE : A PROPOSED MANUFACTURING PLANT REQUIRES AN

INITIAL FIXED CAPITAL OF $900,000 AND A WORKING CAPITAL OF $100,000. IT IS ESTIMATED THAT ANNUAL INCOME WILL BE $800,000 AND ANNUAL EXPENSES (INCLUDING DEPRECIATION) WILL BE $520,000 BEFORE INCOME TAX. INCOME TAX IS 20%. FIND % RETURN ON INVESTMENT BEFORE AND AFTER TAX.

800,000 – 520,000 = $280,000 / YEAR PROFIT BEFORE TAX 280,000 / 1,000,000 = 0.28 28% ROI BEFORE TAX (280,000)(0.80) / 1,000,000 = .224 22.4% ROI AFTER TAX

THE ‘MINIMUM ACCEPTABLE RATE OF RETURN’ (MARR) IS THE RATE SET BY AN ORGANIZATION TO DESIGNATE THE LOWEST LEVEL OF RETURN THAT MAKES AN INVESTMENT ACCEPTABLE. IT IS A DEVICE DESIGNED TO MAKE THE BEST POSSIBLE USE OF MONEY AND APPLIED FOR EVALUATING NEW PROJECTS, COST REDUCTION PROPOSALS, RESEARCH AND DEVELOPMENT PROGRAMS.

PROFITABILITY

2. PAYOUT PERIOD (PAYBACK METHOD)

PAYOUT PERIOD OR TIME IS THE MINIMUM LENGTH OF TIME THEORETICALLY NECESSARY TO RECOVER THE ORIGINAL FIXED CAPITAL INVESTMENT IN THE FORM OF CASH FLOW TO THE PROJECT (NET PROFIT + DEPRECIATION).

DEPRECIABLE FIXED CAPITAL INVESTMENT

PAYOUT PERIOD =

AV. PROFIT/YEAR + AV. DEPRECIATION/YEAR

INTEREST IS NEGLECTED AND ONLY DEPRECIABLE FIXED INVESTMENT IS CONSIDERED. IF TIME VALUE OF MONEY IS ALSO CONSIDERED, THAN WE HAVE ‘PAYOUT PERIOD INCLUDING INTEREST’. IN THIS CASE THE ANNUAL CASH FLOWS ARE DISCOUNTED.

PROFITABILITYEXAMPLE: A COMPANY PLANS AN INVESTMENT OF $300,000 TO

MANUFACTURE A NEW PRODUCT. ALLOWABLE DEPRECIATION IS 10 YEARS, ANNUAL NET PROFIT IS $45,000. STRAIGHT- LINE DEPRECIATION WILL BE USED. WHAT IS THE PAYOUT PERIOD a) IF TIME VALUE OF MONEY IS NOT CONSIDERED? b) IF TIME VALUE IS CONSIDERED AND i IS 8%?

ANNUAL CASH FLOW = NET PROFIT + DEPRECIATION = 45,000 + 30,000 = 75,000 a) PAYOUT PERIOD = 300,000 / 75,000 = 4 YEARS

b) P = R [ {(1 + i)n – 1} / i (1 + i)n 300,000 = 75,000 {(1.08)n – 1} / 0.08 (1.08)n

4 x 0.08 (1.08)n = (1.08)n – 1 0.68(1.08)n = 1 n = log 1.47 / log 1.08

PAYOUT PERIOD n = 5 YEARS

PROFITABILITY

THE PAYOUT PERIOD METHOD IS USED WIDELY TO RATE RELATIVELY SMALL INVESTMENT PROPOSALS IN PRODUCTION DEPARTMENTS. IT MAY LEAD TO INCORRECT CONCLUSIONS SINCE IT DOES NOT RECOGNIZE THE CASH FLOW OCCURING AFTER THE PAYOUT PERIOD. FOR EXAMPLE AN INVESTMENT OF $1000 WITH LIFE OF 1 YEAR AND HAS A NET RETURN OF $1000 WILL YIELD A PAYOUT PERIOD OF 1 YEAR. ANOTHER INVESTMENT OF $1000 PROMISES TO RETURN $250/YEAR DURING ITS ECONOMIC LIFE TEN YEARS. THIS WILL YIELD A PAYOUT PERIOD OF 4 YEARS. IF YOU CONSIDER ONLY PAYOUT PERIODS, YOU SHALL CHOOSE THE FIRST ALTERNATIVE WHICH ACCUALLY EARNS NOTHING.

PROFITABILITY

3. NET RETURN

NET RETURN IS THE AMOUNT OF CASH FLOW OVER AND ABOVE THAT REQUIRED TO MEET THE MINIMUM ACCEPTABLE RATE OF RETURN AND RECOVER THE TOTAL CAPITAL INVESTMENT. THIS IS CALCULATED BY SUBTRACTING THE TOTAL AMOUNT EARNED AT THE MINIMUM RATE OF RETURN AND THE TOTAL CAPITAL INVESTMENT FROM THE TOTAL CASH FLOW. IN THIS CALCULATION WE NEGLECT THE TIME VALUE OF MONEY.

TOTAL CASH FLOW: TOTAL NET PROFIT + TOTAL DEPRECIATION +SALVAGE VALUE + RECOVERED WORKING CAPITAL

INVESTMENT + EARNING OF INVESTMENT: TOTAL CAPITAL INVESTMENT (FIXED + WORKING) + MARR MULTIPLIED BY TOTAL CAPITAL INVESTMENT FOR ALL YEARS

PROFITABILITY

Rn = (NP,j + dj) + S + WC – TC – (MARR)(N)(TC)

WHERE NP,j IS NET PROFIT, dj IS DEPRECIATION FOR YEAR j, S IS THE SALVAGE VALUE

IN CASES WHERE dj + S + WC = TC

Rn = NP,j - (MARR)(N)(TC)

IF WE HAVE AN AVERAGE PROFIT (PAVE), THEN

Rn,AVE = PAVE – (MARR)(TC)

N

j 1

N

j 1

PROFITABILITY

EXAMPLE: A COMPANY PLANS TO START A NEW PRODUCT WHICH REQUIRES $24 MILLION OF NEW MACHINERY AND $4 MILLION WORKING CAPITAL. ALL FIXED COSTS EXCEPT DEPRECIATION IS $1 MILLION/YEAR, VARIABLE COSTS AT FULL CAPACITY IS $5 MILLION/YEAR. DEPRECIATION IS BY DOUBLE-DECLINING IN 5 YEARS, INCOME TAX IS 35%. THE PRODUCTION RATE AT 100% CAPACITY IS 2 x 106 kg/YEAR. IN THE FIRST YEAR CAPACITY IS USED 50%, IN THE SECOND YEAR 90% AND AFTER THE SECOND YEAR 100%. MARR IS 30%.

CALCULATE THE SALES PRICE (p) REQUIRED TO ACHIEVE MARR IN TEN YEARS BY NET RETURN METHOD.

USING THE TABLE : PAVE = 1/10 (18.8p – 81)(1 - 0.35) 106

= (1.222p – 5.265) 106

Rn = 0 = (1.222p) 106 – (5.265) 106 – (0.30)(28) 106 p = $11.18

PROFITABILITY

YEAR

1 2 3 4 5 6 7 8 9 10 SUM

-----------------------------------------------------------------------------------

A. PERCENT OF

OPERATING TIME 50 90 100 100 100 100 100 100 100 100

B. PRODUCT RATE,

106 kg/yr 1 1.8 2 2 2 2 2 2 2 2 18.8

C. ALL VARIABLE

COSTS, $ 106/yr 2.5 4.5 5 5 5 5 5 5 5 5 47

D. ALL FIXED COSTS

(EXCEPT DEP.) $ 106/yr 1 1 1 1 1 1 1 1 1 1 10

E. DEPRECIATION,

$ 106/yr 9.6 5.76 3.456 2.592 2.592 0 0 0 0 0 24

F. TOTAL PRODUCT

COST(C+D+E) $ 106/yr 13.1 11.26 9.456 8.592 8.592 6 6 6 6 6 81

PROFITABILITY4. NET PRESENT WORTH THE NET PRESENT WORTH OF A PROJECT IS THE DIFFERENCE

BETWEEN THE PRESENT VALUE OF THE ANNUAL CASH FLOWS AND THE INITIAL REQUIRED INVESTMENT ,

OR MORE GENERALLY:

P = PRESENT WORTH OF BENEFITS – PRESENT WORTH OF COSTS

YOU CAN USE PRESENT WORTH TO

A. COMPARE ALTERNATIVES (ALLWAYS USING SAME NUMBER OF YEARS AND CONSTANT i); LARGER P IS BETTER

B. DECIDE IF A PROJECT IS FEASIBLE OR NOT (TAKING i = MARR); IF P ≥ 0, YOUR PROJECT IS EQUAL OR BETTER THAN MARR,

IF P < 0, LESS THAN MARR

PROFITABILITY

EXAMPLE : THERE ARE TWO ALTERNATIVE MACHINES YOU CAN BUY TO YOUR FACTORY, A AND B. A COSTS $9000, B COSTS $14,500. THE NET CASH FLOWS ARE:

1.YEAR 2.YEAR 3.YEAR MACHINE A : $ 4,500 4,500 4,500 MACHINE B : $ 6,000 6,000 8,000 FIND PRESENT WORTH IF i = 0.8 AND SALVAGE VALUE IS ZERO

PW(A) = (4500) [{(1+0.08)3 - 1} / 0.08 (1+0.08)3] – 9000 = $2594

PW(B) = (6000) [{(1+0.08)2 - 1} / 0.08 (1+0.08)2] + 8000 / (1+0.08)3 – 14,500 = $2550

MACHINE A IS BETTER

PROFITABILITY

EXAMPLE : FIXED CAPITAL INVESTMENT IS $100,000 AND WORKING CAPITAL INVESTMENT IS $10,000, SALVAGE IS $10,000 AT THE END OF 5 YEAR LIFE; FIND PW IF i = 0.15 AND AFTER TAX CASH FLOW AT THE END OF EACH YEAR IS AS FOLLOWS:

YEAR 1 2 3 4 5

$ 30,000 31,000 36,000 40,000 43,000

PWINCOME = 30,000 / (1.15) + 31,000 / (1.15)2 + 36,000 / (1.15)3

+ 40,000 / (1.15)4 + 43,000 / (1.15)5 + (10,000+10,000) / (1.15)5

= $127,327

NET PRESENT WORTH IS 127,327 – 110,000 = $17,327

PROFITABILITYEXAMPLE: A REACTOR TO BE USED FOR OXIDATION OF PARAFFIN VAX TO

MAKE FATTY ACIDS IS REQUIRED. TWO ALTERNATIVES ARE SUGGESTED: REACTOR A MADE OF ORDINARY STEEL COSTS $12,000 INSTALLED AND HAS A LIFE OF TWO YEARS WITH A JUNK VALUE OF $200, WHEREAS REACTOR B, MADE OF STEEL BUT GLASS LINED, COSTS $28,000 INSTALLED AND HAS A LIFE OF 6 YEARS WITH A SALVAGE VALUE OF $800 FOR FITTINGS. LABOR AND OTHER OPERATING COSTS FOR BOTH REACTORS ARE THE SAME, BUT REACTOR A IS EXPECTED TO REQUIRE ABOUT $400 FOR MAINTENANCE DURING THE SECOND YEAR, WHEREAS REACTOR B WILL REQUIRE $100 MAINTENANCE DURING THE THIRD YEAR AND $300 DURING THE FIFTH YEAR. IT IS EXPECTED THAT REACTOR B WILL GIVE A SLIGHTLY BETTER YIELD AND A BETTER QUALITY PRODUCT. THE VALUE OF THESE IMPROVEMENTS IS ESTIMATED AT $400 PER YEAR. IF MONEY IS WORTH 10%, WHICH INSTALLATION SHOWS THE LOWER EQUIVALENT CAPITAL REQUIREMENT AT THE PRESENT TIME?

PA = –12,000 –12,200 / 1.12 – 12,200 / 1.14 – 200 / 1.16 = – 30,529

PB = –28,000 + 400 / 1.1 + 400 / 1.12 + 300 / 1.13 + 400 / 1.14 + 100 / 1.15 + 1400 / 1.16

= – 26,015 REACTOR B HAS LOWER REQUIREMENT

PROFITABILITY

• A COMPANY IS CONSIDERING A NEW INVESTMENT WHICH COSTS 12,000,000 TL AS TCI. THE LAND COST IS 2,000,000 TL AND REST OF THE FCI COST IS 8,000,000 TL. THE YEARLY EXPECTED CASH FLOW IS 2,500,000 TL.

IF MARR IS 20% AND LIFE OF THE PLANT IS 15 YEARS, IS THIS INVESTMENT FEASIBLE?

WC = 12,000,000 – 10,000,000 = 2,000,000

P = 2,500,000 [ (1.215 – 1) / .20 (1.215)] + 4,000,000 / 1.215 - 12,000,000

P = - 51,696 TL

NOT FEASIBLE

PROFITABILITY

EXAMPLE: IN AN EXISTING PLANT A NEW CHEMICAL WILL BE PRODUCED WITH A RATE OF 100,000kg/YEAR. THE ESTIMATED LIFE OF THIS PROJECT IS 8 YEARS. NECESSARY INVESTMENT IS 800,000 TL WORTH MACHINERY WHICH WILL BE DEPRECIATED BY STRAIGHT- LINE IN 8 YEARS. YEARLY PRODUCTION COST IS ESTIMATED TO BE 1,200,000 TL EXCEPT DEPRECIATION AND WORKING CAPITAL IS 20% OF THE TOTAL YEARLY PRODUCTION COST. IF TAX RATE IS 20%, AND i = 6%, WHAT SHOULD BE THE SELLING PRICE OF THE CHEMICAL IF MANAGEMENT REQUIRES A PRESENT WORTH OF MINIMUM 1,000,000 TL?

WC = (1,200,000 + 100,000) .20 = 260,000

P=1,000,000 = -800,000 -260,000 + 260,000/1.068 +(CF) (1.068 -1)/.06(1.068 )

CF (CASH FLOW) = 120,245 = NET PROFIT + DEPRECIATION

120,245 = (100,000p – 1,300,000)0.80 + 100,000

p = 13.25 TL

PROFITABILITY WORK SHEET FOR NET PRESENT WORTH

YEARS 0 1st 2nd 3rd 4th 5th 1. FIXED CAPITAL INVESTMENT 2. WORKING CAPITAL 3. TOTAL CAPITAL INVESTMENT (1+2) 4.OPERATING RATE (% OF CAPACITY) 5. ANNUAL INCOME (SALES) 6. ANNUAL MANUFACTURING COST 7. DEPRECIATION 8. ANNUAL GENERAL EXPENSES 9. TOTAL PRODUCT COST (6 + 7 + 8)10. ANNUAL GROSS PROFIT (5 - 9)11. INCOME TAX12. ANNUAL NET PROFIT (10 – 11) 13.ANNUAL OPERATING CASH FLOW (12 + 7)14.TOTAL ANNUAL CASH FLOW (13 + 3)

PROFITABILITY

15.NET PRESENT WORTH

ENTER EXPENDITURES AS NEGATIVE, INCOMES AS POSITIVE

ASSUME ALL INVESTMENT IS MADE IN YEAR 0 AND PRODUCTION STARTS IN THE BEGINNING OF YEAR 1; WORKING CAPITAL AND SALVAGE VALUE IS RECOVERED AT THE END AND SINCE THEY ARE LUMP-SUM, FINITE CASH FLOW SHOULD BE APPLIED IN ALL CASES.

NET PRESENT WORTH (LINE 15) IS CALCULATED BY DECIDING ON r OR i AND USING THE BASIC FORMULAS:

a) P = S / ern b) P = S / (1 + i)n

PROFITABILITY5. DISCOUNTED CASH FLOW RATE OF RETURN BASED ON DISDOUNTED CASH FLOW IS ALSO

CALLED PROFITABILITY INDEX, TRUE RATE OF RETURN OR INTERNAL RATE OF RETURN. THIS METHOD TAKES INTO ACCOUNT THE TIME VALUE OF MONEY. A TRIAL AND ERROR PROCEDURE IS USED TO ESTABLISH A RATE OF RETURN WHICH CAN BE APPLIED TO YEARLY CASH FLOW SO THAT THE ORIGINAL INVESTMENT IS REDUCED TO ZERO (OR TO SALVAGE AND LAND VALUE PLUS WORKING CAPITAL INVESTMENT) DURING THE PROJECT LIFE.

EXAMPLE : FIXED CAPITAL INVESTMENT IS $100,000 AND WORKING CAPITAL INVESTMENT IS $10,000, SALVAGE IS $10,000 AT THE END OF 5 YEAR LIFE; WHAT IS THE INTERNAL RATE OF RETURN IF PREDICTED AFTER-TAX CASH FLOW AT THE END OF EACH YEAR IS AS FOLLOWS:

YEAR 1 2 3 4 5 $ 30,000 31,000 36,000 40,000 43,000 S = (30,000)(1+i)4 + (31,000)(1+i)3 + (36,000)(1+i)2 + (40,000)(1+i) + 43,000 =

(110,000)(1+i)5 – 10,000 – 10,000 BY TRIAL AND ERROR i = 0.207

PROFITABILITY

USE OF CONTINUOUS INTEREST COMPOUNDING :

WHEN CONTINUOUS INTEREST IS USED WE SHOULD REPLACE (1 + i)n WITH ern (r IS NOMINAL INTEREST)

EXAMPLE : DETERMINE THE DISCOUNTED CASH FLOW RATE OF RETURN (i.e. PROFITABILITY INDEX) FOR THE FOLLOWING PROJECT:

ONE YEAR PRIOR TO START - UP, LAND IS PURCHASED FOR $200,000. DURING THE YEAR PRIOR TO START - UP INVESTMENT IS MADE WITH UNIFORM CONTINUOUS SPENDING OF $600,000. A WORKING CAPITAL OF $200,000 IS NEEDED. ESTIMATED LIFE IS 10 YEARS, SALVAGE VALUE IS $100,000. ANNUAL CASH FLOW (NET PROFIT + DEPRECIATION) WILL BE $310,000 AFTER TAX AND IS FLOWING UNIFORMLY. USE CONTINUOUS INTEREST COMPOUNDING AND CONTINUOUS CASH FLOW.

PROFITABILITY

LAND VALUE AT TIME 0 S = Pern = 200,000 er

FIXED INVESTMENT AT TIME 0 S = {Ř (ern – 1) / r} = 600,000 {(er – 1) / r}

TOTAL CASH POSITION AT TIME 0 :

CP0 = 200,000 er + 600,000 {(er – 1) / r} + 200,000

AT THE END OF THE PROJECT WE HAVE SALVAGE + WORKING CAPITAL + LAND WHICH IS $500,000

PWS = 500,000 / e10r

PRESENT WORTH OF CASH FLOW :

PWC = Ř {(ern – 1) / rern} = 310,000 {(e10r – 1) / re10r}

WE SHOULD FIND THE r WHICH DECREASES THE NET CASH POSITION TO 0 AT THE END OF 10 YEARS : CP0 = PWS + PWC

200,000 er + 600,000 {(er – 1) / r} + 200,000 =

500,000 / e10r + 310,000 { (e10r – 1) / re10r}

BY TRIAL AND ERROR r = 0.26 ; SO PROFITABILITY INDEX IS 26%

PROFITABILITY6. EQUIVALENT ANNUAL COST – ANNUAL WORTH EQUIVALENT ANNUAL COST (EAC) IS CONVERTING ALL COSTS TO

AN EQUIVALENT SERIES OF UNIFORM END-OF-YEAR PAYMENTS. IF WE ARE COMPARING INCOMES, WE CONVERT INCOMES SIMILARLY TO OBTAIN EQUIVALENT ANNUAL WORTH (EAW).

EXAMPLE: A MACHINE COSTS $1,600, HAS 5 YEARS LIFE. ANNUAL OPERATING COSTS IS $500 PER YEAR AND i = 0.08. WHAT IS EAC a) IF SALVAGE VALUE IS 0 , b) IF SALVAGE VALUE IS $300 ?

a) R = P [ i (1 + i)n / {(1 + i)n – 1}] = 1,600 [0.08 x 1.085 / (1.085 – 1)] = 400 EAC = 400 + 500 = $900 b) R = S [ i / {(1 + i)n – 1}] = 300 [0.08 /( 1.085 – 1)] = 51 EAC = 400 – 51 + 500 = $849

PROFITABILITYEXAMPLE:EQUIPMENT CAN BE PURCHASED FOR $10,000 AND LAST FOR N

YEARS (WITH NO SALVAGE VALUE), OR IT CAN BE LEASED FOR 2,200 A YEAR. IF INTEREST IS 12% PER YEAR, WHAT IS THE MINIMUM N TO JUSTIFY PURCHASING?

a) PRESENT WORTH METHOD P = R [ (1 + i)n – 1 / i (1 + i)n ] 10,000 = 2,200 [ (1.12)N – 1 / .12 (1.12)N ] .545 = (1.12)N – 1 / (1.12)N

(1.12)N = 2.2 N ln 1.12 = ln 2.2 N = 7

b) EAC METHOD R = P [ i (1 + i)n / (1 + i)n – 1] 2,200 = 10,000 [.12 (1.12)N / (1.12)N – 1] N = 7

PROFITABILITY

EXAMPLE: A PACKAGING MACHINE WITH ECONOMIC LIFE OF 10 YEARS COSTS $20,000 WITH 0 SALVAGE VALUE. THIS MACHINE HAS OPERATING AND MAINTENANCE COSTS OF $10 PER HOUR. IT CAN PACKAGE 100 UNITS PER HOUR. THE CURRENT METHOD IS BY HAND LABOR COSTING $12 PER HOUR TO PACKAGE AN AVARAGE OF 100 UNITS IN 1.5 HOURS. HOW MANY PACKAGES SHOULD BE MADE PER YEAR TO JUSTIFY THE PURCHASE OF THIS MACHINE IF i = 6%?

R = P { i (1 + i)n / (1 + i)n – 1} = 20,000 { 0.06 (1.06)10 / (1.06)10 – 1} = 2,719

YEARLY COST WITH MACHINE = $2,719 + ($10/HR)(1/100 HR/PACK)(X PACK/YEAR) = 2719 + 0.1X

YEARLY COST WITH CURRENT METHOD =

($12/HR)(1.5/100 HR/PACK)(X PACK/YEAR) = 0.18X

2719 + 0.1X = 0.18X

X = 33,988 PACKS / YEAR

PROFITABILITY

EXAMPLE: A FURNACE INSTALLATION COSTING $12,000 WITH OPERATING COSTS OF $4,800 PER YEAR AND A 10-YEAR LIFE IS OFFERED BY ONE SUPPLIER; A SECOND ALTERNATIVE GUARANTEES TO PROVIDE THE SAME SERVICE WITH $1,000 A YEAR LOWER OPERATING COSTS AT A COST OF $25,000. SALVAGE VALUE ON BOTH FURNACES IS ESTIMATED AT $1,000. WHAT INCREASE IN SERVICE LIFE WOULD BE REQUIRED FOR THE SECOND PROPOSAL TO WARRANT ITS SELECTION IF i = 8%?

R = P { i (1 + i)n / (1 + i)n – 1} AND R = S { i / (1 + i)n – 1 }

R = 12,000 { 0.08 (1.08)10 / (1.08)10 – 1} - 1,000 { 0.08 / (1.08)10 – 1 } = 1,719

1,719 + 4,800 = 25,000 { 0.08 (1.08)n / (1.08)n – 1} – 1,000 { 0.08 / (1.08)n – 1 } + 3800

2719 { (1.08)n – 1 } = 25,000 (0.08) (1.08)n – 1000 (0.08)

(1.08)n = 3.67

n = 16.88

PROFITABILITY

RISK ANALYSIS AND ACCEPTABLE RETURNS:

IN INDUSTRIAL OPERATIONS THERE IS ALWAYS A DEGREE OF UNCERTANITY AND CALCULATED RETURN DEPENDS ON SOME ASSUMPTIONS ABOUT THE FUTURE. SO THERE IS ALWAYS A RISK FACTOR. A SIMPLE METHOD IS TO MAKE THE CALCULATIONS FOR WORST, AVERAGE AND BEST CASES. A BETTER METHOD IS TO MAKE RISK ANALYSIS USING PROPER SOFTWARES. YOU SHOULD FIRST DECIDE ON THE RANGE OF VALUES FOR EACH FACTOR (TOTAL INVESTMENT, PLANT LIFE, ANNUAL CASH FLOW…) AND THE LIKELIHOOD OF OCCURANCE OF EACH VALUE. COMPUTER THAN CALCULATES THE PROBABILITY OF ALL RETURNS.

DUE TO THE UNCERTANITIES, IN AVERAGE 20 – 30% RETURN BEFORE TAX (NOT CONSIDERING THE EFFECT OF INFLATION) WHICH MEANS 14 – 20% RETURN AFTER TAX IS REQUIRED FOR INVESTING IN INDUSTRY.

VIII. ANALYSIS OF ALTERNATIVES

ALTERNATIVES

AN ‘ALTERNATIVE’ IN ENGINEERING ECONOMICS IS AN INVESTMENT POSSIBILITY. IT IS A SINGLE UNDERTAKING WITH A DISTINGUISHABLE CASH FLOW.

IN INDUSTRIAL OPERATIONS, IT IS OFTEN POSSIBLE TO PRODUCE EQUIVALENT PRODUCTS IN DIFFERENT WAYS. ALTHOUGH THE PHYSICAL RESULTS MAY BE APPROXIMATELY THE SAME, THE CAPITAL REQUIRED AND THE EXPENSES INVOLVED CAN VARY CONSIDERABLY DEPENDING ON THE METHOD CHOSEN. SIMILARLY, ALTERNATIVE METHODS INVOLVING VARYING CAPITAL AND EXPENSES CAN OFTEN BE USED TO CARRY OUT OTHER TYPES OF BUSINESS VENTURES. IT MAY BE NECESSARY, THEREFORE, NOT ONLY TO DECIDE IF A GIVEN BUSINESS VENTURE WOULD BE PROFITABLE, BUT ALSO TO DECIDE WHICH OF SEVERAL POSSIBLE METHODS WOULD BE THE MOST DESIRABLE. DECISION USUALLY HAS LONG-TERM CONSEQUENCES SO IT IS IMPORTANT FOR THE ENGINEERS TO BE COMPETENT ECONOMIC ANALYSTS.

ALTERNATIVESANALYSIS START WITH THE IDENTIFICATION OF ALTERNATIVES.

AFTER PREPARING ENOUGH DATA FOR EACH ALTERNATIVE, COMPARISON IS MADE.

HOW MANY ALTERNATIVES SHOULD WE HAVE? YOU CAN NEVER DECIDE IF YOU CONTINUE SEARCHING FOR A STILL- BETTER OPTION. ON THE OTHER SIDE, ACTING ON THE FIRST OPTION THAT COMES TO MIND MAY RESULT WITH A WRONG DECISION. AIM SHOULD BE TO DEVELOP A SET OF ALTERNATIVES LARGE ENOUGH TO INCLUDE THE BEST POSSIBLE SOLUTION.

WE CAN CLASSIFY ALTERNATIVES INTO DEPENDENT AND INDEPENDENT CATEGORIES. AN INDEPENDENT ALTERNATIVE IS NOT AFFECTED BY THE SELECTION OF ANOTHER ALTERNATIVE. EACH PROPOSAL IS EVALUATED ON ITS MERIT AND IS APPROVED IF IT MEETS THE CRETERIA OF ACCEPTABILITY. COMPARISONS OF INDEPENDENT INVESTMENT PROPOSALS ARE DESIGNED TO DETERMINE WHICH PROPOSALS SATISFY A MINIMUM LEVEL OF ECONOMIC VALUE. ALL THOSE THAT SURPASS THE MINIMUM LEVEL MAY BE IMPLEMENTED AS LONG AS SUFFICIENT CAPITAL IS AVAILABLE.

ALTERNATIVES

DEPENDENT ALTERNATIVES ARISE WHEN ALTERNATIVES ARE RELATED IN A WAY THAT INFLUENCES THE SELECTION PROCESS. THERE ARE TWO CLASSIFICATIONS FOR DEPENDENT ALTERNATIVES : MUTUALLY EXCLUSIVE DEPENDENT AND CONTINGENCY DEPENDENT. ALTERNATIVES ARE MUTUALLY EXCLUSIVE WHEN THE SELECTION OF ONE ELIMINATES THE OPPURTUNITY TO ACCEPT ANY OF THE OTHERS. OPERATIONAL PROBLEMS NORMALLY FIT INTO THIS CATEGORY, BECAUSE A SINGLE COURSE OF ACTION IS SOUGHT TO SOLVE A PARTICULAR, OFTEN URGENT, PROBLEM.

THE CONTINGENCY DEPENDENT ALTERNATIVES ARISE WHEN INDIVIDUAL INVESTMENT OPPURTUNITIES ARE LINKED TO OTHER ALTERNATIVES. THEN THE ACCEPTANCE OF ONE ALTERNATIVE DEPENDS ON THE SIMULTANEOUS ACCEPTANCE OF ONE OR MORE RELATED ALTERNATIVES. THE PURCHASE OF NEW COMPUTER TERMINALS IS CONTINGENT ON THE PURCHASE OF HARDWARE INCREASING THE CAPACITY OF THE MAIN COMPUTER.

ALTERNATIVES

1. ANALYSIS OF MUTUALLY EXCLUSIVE ALTERNATIVES

WE CAN USE DIFFERENT METHODS FOR EVALUATION:

A. PRESENT WORTH METHOD

RELIABLE, EASY

YOU CAN USE IT IN ALL CASES

WHEN ECONOMIC LIFE OF EQUIPMENT IS DIFFERENT, YOU HAVE TO FIND THE COMMON NUMBER OF YEARS, AND IN SOME CASES IT CAN BE A LONG CALCULATION.

B. EQUIVALENT ANNUAL COST (WORTH) METHOD

RELIABLE, EASY

MAY BE USEFULL WHEN ECONOMIC LIFES OF ALTERNATIVES ARE DIFFERENT

ALTERNATIVES

C. CAPITILIZED COST METHOD

EASY

MAYBE USEFULL WHEN ECONOMIC LIFES OF ALTERNATIVES ARE DIFFERENT

YOU CAN USE THIS METHOD WHEN YOU ARE COMPARING ONLY COST DATA (INCOMES SAME)

D. INTERNAL RATE OF RETURN (IRR) METHOD

MAYBE MISLEADING, SO YOU HAVE TO USE ‘INCREMENTAL IRR’ METHOD

NOT EASY

ALTERNATIVES

EXAMPLE:TWO TYPES OF EQUIPMENT ARE AVAILABLE FOR PERFORMING A

MANUFACTURING OPERATION AND THE COST DATA ARE RECORDED AS: TYPE A TYPE BFIRST COST,$ 88,000 45,000SALVAGE, $ 7,500 4,000ANNUAL MAINTENANCE, $ 4,300 5,200LIFE, YEARS 12 6

DETERMINE THE MORE ECONOMIC TYPE IF i = %8

ALTERNATIVESa. PRESENT WORTH METHOD

WE SELECT A 12- YEAR ANALYSIS PERIOD:

PA = 88,000 + 4,300 [ (1.0812 – 1) / .08 (1.0812 )] - 7,500 / 1.0812 = $ 117,415

PB = 45,000 + 41,000 / 1.086 + 5,200 [ (1.0812 – 1) / .08 (1.0812 )] - 4,000 / 1.0812

= $ 108,432TYPE B IS BETTER

b. EAC METHOD

EACA = 88,000 [.08 (1.0812 ) / (1.0812 – 1) ] + 4,300 - 7,500 [.08 / (1.0812 – 1) ] = 15,582.4

EACB = 45,000 [.08 (1.086 ) / (1.086 – 1) ] + 5,200 – 4,000 [.08 / (1.086 – 1) ] = 14,387.9TYPE B IS BETTER

ALTERNATIVES

EXAMPLE:

COMPARE THE INVESTMENTS HAVING THE FOLLOWING COST DATA ON THE BASIS OF PRESENT WORTH OF COSTS USING i = %10:

TYPE A TYPE B

FIRST COST,$ 46,000 34,000

SALVAGE,$ 4,000 0

ANNUAL MAINTENANCE,$ 3,700 3,200

LIFE, YEARS 7 5

SINCE THE LOWEST COMMON MULTIPLE OF 7 AND 5 IS 35, WE SELECT AN ANALYSIS PERIOD OF 35 YEARS:

PA = 46,000 + 42,000 / 1.17 + 42,000 / 1.114 + 42,000 / 1.121 + 42,000 / 1.128

– 4,000 / 1.135 + 3,700 [ (1.135 – 1) / .1(1.135 ) = $122,720

PB = 34,000 + 34,000 / 1.15 + 34,000 / 1.110 + 34,000 / 1.115 + 34,000 / 1.120

+ 34,000 / 1.125 + 34,000 / 1.130 + 3,200 [ (1.135 – 1) / .1(1.135 ) = $117,378

TYPE B IS BETTER AND PA / PB = 122,720 / 117,378 = 1.046

ALTERNATIVES

SOLVE THE SAME PROBLEM USING CAPITILIZED COST METHOD:

K = R / i + CV + CR / [(1 + i)n – 1]

KA = 3,700 / .1 + 46,000 + 42,000 / (1.17 – 1)

= 127,270

KB = 3,200 / .1 + 34,000 + 34,000 / (1.15 – 1)

= 121,691

TYPE B IS BETTER AND KA / KB = 127,270 / 121,691 = 1.046

WE HAVE THE SAME RESULT BOTH METHODS AND CAPITILIZED COST CALCULATION IS SIMPLER

ALTERNATIVES

IN IRR INCREMANTAL APPROACH, YOU START WITH THE LOWEST INVESTMENT ALTERNATIVE, CALCULATE IRR, COMPARE WITH MARR; IF IRR > MARR, THAT IS AN ACCEPTABLE ALTERNATIVE. NOW CALCULATE IRR FOR THE INCREMENT OF THE NEXT ALTERNATIVE. IF THIS SATISFIES MARR IT IS ACCEPTABLE; IF NOT ELIMINATE. CONTINUE LIKE THIS TILL YOU EVALUATE ALL ALTERNATIVES.

THE BIGGEST INVESTMENT THAT SATISFIES THE MARR IS THE BEST ALTERNATIVE

ALTERNATIVESSTEP 1A: CALCULATE THE IRR FOR STEP 1B: IF THE ALTERNATIVES ARE BASED

THE ALTERNATIVE REQUIRING ONLY ON RELATIVE COST, ASSUME THE THE LEAST INVESTMENT LOWEST COST OPTION IS ACCEPTABLE

STEP 2: COMPARE THE IRR WITH THE MARR TO DETERMINE

WHETHER THE ALTERNATIVE IS ACCEPTABLE STEP 3A: IF IRR < MARR , ELIMINATE STEP 3B: FOR AN ACCEPTABLE THE ALTERNATIVE, CONSIDER ALTERNATIVE, DETERMINE THE NEXT THE NEXT HIGHER INVESTMENT INCREMENT OF INVESTMENT

STEP 4: COMPARE THE IRR FOR THE INCREMENT OF INVESTMENT WITH THE MARR. IF IRR < MARR, ELIMINATE THE ALTERNATIVE. OTHERWISE ALTERNATIVE IS ACCEPTABLE

STEP 5: IF NO HIGHER LEVEL OF INVESTMENT, GO TO STEP 6. OTHERWISE CALCULATE IRR FOR THE TOTAL INCREMENT OF INVESTMENT BETWEEN THE LAST ACCEPTABLE ALTERNATIVE AND NEXT HIGHER LEVEL OF INVESTMENT.GO TO ST4

STEP 6: DECISION. WE HAVE FOUND THE ACCEPTABLE ALTERNATIVE WITH THE GREATEST INVESTMENT ASSUMING AVAILABILITY OF CAPPITAL

ALTERNATIVESEXAMPLE : A B C D INITIAL INVESTMENT : 170,000 260,000 300,000 330,000 ANNUAL CASH FLOW: 44,000 49,000 66,000 68,000 MARR = 10% IRR FOR A : P = R [{ (1 + i)n – 1} / i(1 + i)n] 170,000 = 44,000 [{ (1+i)10 – 1} / i(1+i)10] i = 0.225 by trial and error > MARR; ALTERNATIVE A IS ACCEPTABLE INCREMENT OF ALT. B : 260,000 – 170,000 = 90,000 inc. investment 49,000 – 44,000 = 5,000 inc. annual return ; 10 x 5,000 = 50,000 inc. for 10 years SINCE INC. RETURN FOR 10 YEARS < INC. INVEST., ALT B NOT ACCEPTABLE IRR FOR INCREMENT OF C TO A : 300,000 – 170,000 = (66,000 – 44,000) [{ (1 + i)10 – 1} / i(1 + i)10] i = 0.109 by trial error > MAAR ; ALTERNATIVE C IS ACCEPTABLE INCREMENT OF ALT. D : 330,000 – 300,000 = 30,000 inc. investment 68,000 – 66,000 = 2,000 inc. annual return ; 10 x 2,000 = 20,000 inc. for 10 years SINCE INC. RETURN FOR 10 YEARS < INC. INVEST., ALT D NOT ACCEPTABLE CHOOSE ALTERNATIVE C

ALTERNATIVES

WHEN WE DIRECTLY COMPARE ALTERNATIVES :

A B C D

INVESTMENT : 170,000 260,000 300,000 330,000

ANNUAL CASH FLOW 44,000 49,000 66,000 68,000

IRR (%) : 22.5 13.5 17.7 15.9

IF WE DO NOT USE THE INCREMENTAL METHOD, WE MAY MAKE MISTAKES. IF WE CHOOSE ALTERNATIVE A (SINCE IT HAS THE GREATEST IRR) WE WILL LOOSE MONEY COMPARED WITH CHOOSING ALTERNATIVE C. IF WE CHOOSE ALTERNATIVE D (GREATEST INVESTMENT WITH IRR > MARR) WE WILL AGAIN LOOSE MONEY BY INVESTING AN EXTRA 30,000 WITH LOWER RETURN.

ALTERNATIVES

IF WE USE PRESENT WORTH FOR COMPARISON, WE WILL GET THE SAME (CORRECT) RESULT (MUCH EASIER)

P = R [{ (1 + i)n – 1} / i(1 + i)n] ; i = 0.10

P = R{ (1.110 – 1) / 0.1(1.1)10} = R (6.14475)

ALT. A : PW = - 170,000 + 44,000 x 6.14475 = 100,369

ALT. B : PW = - 260,000 + 49,000 x 6.14475 = 41,092

ALT. C : PW = - 300,000 + 66,000 x 6.14475 = 105,553

ALT. D : PW = - 330,000 + 68,000 x 6.14475 = 87,843

ALTERNATIVE C GIVES THE BIGGEST PRESENT WORTH SO IT SHOULD BE CHOOSEN

ALTERNATIVES2. ANALYSIS OF INDEPENDENT ALTERNATIVES FOR INDEPENDENT ALTERNATIVES THE CAPITAL AVAILABLE MAY

BE THE LIMITING FACTOR. WITH THIS LIMIT IN MIND, THE BEST ALTERNATIVE OR ALTERNATIVES MAY BE CHOSEN AFTER FINDING PW OR IRR VALUES FOR EACH ALTERNTIVE. SINCE THEY ARE NOT EXCLUSIVE, WE CAN CHOOSE MORE THAN ONE.

ONE USEFULL METHOD TO HANDLE INDEPENDENT ALTERNATIVES IS TO CONVERT THEM TO MUTUALLY EXCLUSIVE GROUPS.

EXAMPLE: YOU HAVE THE FOLLOWING 4 INDEPENDENT PROPOSALS PROPOSAL INVESTMENT END OF EACH YEAR IRR% PW CASH FLOW (i = 0.10) 1 2 3 1 $300 $130 130 130 14.4 23.29 2 500 210 210 210 12.5 22.24 3 600 250 250 250 12.0 21.71 4 1500 628 628 628 12.3 61.74 YOU CAN CONVERT THIS TO 16 MUTUALLY EXCLUSIVE ALTERNATIVES

ALTERNATIVESGROUP PROPOSALS INVESTMENT ANNUAL RECEIPT PW 1 2 3 4 FOR 3 YEARS 1 0 0 0 0 0 0 0 2 1 0 0 0 $300 $130 $23.29 3 0 1 0 0 500 210 22.24 4 1 1 0 0 800 340 43.53 5 0 0 1 0 600 250 21.71 6 1 0 1 0 900 380 45.00 7 0 1 1 0 1100 460 43.95 8 1 1 1 0 1400 590 67.24 9 0 0 0 1 1500 628 61.74 10 1 0 0 1 1800 758 85.03 11 0 1 0 1 2000 838 83.98 12 1 1 0 1 2300 968 107.27 13 0 0 1 1 2100 878 83.45 14 1 0 1 1 2400 1008 106.74 15 0 1 1 1 2600 1088 105.69 16 1 1 1 1 2900 1218 128.98

ALTERNATIVES

PRESENT WORTHS CAN BE CALCULATED BY ADDING THE PRESENT WORTHS OF THE GROUP MEMBERS OR BY

R {(1.13 – 1) / 0.1(1.1)3} – PINV

IF THERE IS NO CAPITAL LIMITATION, WE SHOULD CHOOSE GROUP 16 COVERING ALL PROPOSALS SINCE IT HAS THE BIGGEST PW. IF THERE IS A LIMIT OF $2400 WE SHOULD CHOOSE GROUP 12, IF THERE IS A LIMIT OF $1000 WE SHOULD CHOOSE GROUP 6.

IN ANY ANAYLSIS ‘DO NOTHING’ ALTERNATIVE SHOULD BE CONSIDERED. WHEN ALL IRR’S ARE < MARR OR PW’S ARE NEGATIVE, THIS ALTERNATIVE SHOULD BE CHOSEN (IF THE INVESTMENT IS NOT A MUST). THEN THE INVESTOR OR THE COMPANY WILL ‘DO NOTHING’ ABOUT THE PROJECTS BEING CONSIDERED AND THE AVAILABLE FUNDS WILL BE USED IN OTHER PROJECTS OR INVESTMENT TOOLS.

ALTERNATIVESWE MAY HAVE TWO INDEPENDENT SETS OF MUTUALLY EXCLUSIVE

PROPOSALS. THAT IS, PROPOSALS A1 AND A2 ARE MUTUALLY EXCLUSIVE AND B1 AND B2 ARE ALSO MUTUALLY EXCLUSIVE BUT A’S AND B’S ARE INDEPENDENT (TWO KINDS OF COMPUTERS AND TWO KINDS OF FORKLIFTS).

THAN WE HAVE :

GROUP A1 A2 B1 B2

1 0 0 0 0 2 1 0 0 0 3 0 1 0 0 4 0 0 1 0 5 0 0 0 1 6 1 0 1 0 7 1 0 0 1 8 0 1 1 0 9 0 1 0 1

ALTERNATIVES

IT IS QUITE POSSIBLE TO COMPARE A SERIES OF ALTERNATIVE INVESTMENTS BY EACH OF THE DIFFERENT PROFITABILITY METHODS AND FIND DIFFERENT RESULTS DEPENDING ON THE EVALUATION TECHNIQUE USED.

IF THERE IS ANY QUESTION AS TO WHICH METHOD SHOULD BE USED FOR A FINAL DETERMINATION, NET PRESENT WORTH SHOULD BE CHOSEN.

ALTERNATIVESEXAMPLE:INVESTMENTS A ($160,000) AND B ($150,000) HAVE THE CASH FLOWS SHOWN

BELOW. DETERMINE WHICH INVESTMENT SHOULD BE UNDERTAKEN IF THE SOLE CRITERION IS a) THE PAYBACK PERIOD, b) THE PRESENT WORTH.

ALT 1 2 3 4 5 6 YEARS A $100,000 80,000 20,000 10,000 10,000 10,000 B $ 25,000 25,000 25,000 100,000 150,000 80,000

a) PAYBACK FOR A: 2 YEARS, PAYBACK FOR B: 4 YEARS

b) PWA : 100,000 / 1.1 + 80,000 / 1.12 + 20,000/ 1.13 + 10,000/ 1.14 + 10,000/ 1.15 + 10,000 / 1.16 – 160,000 = 30,742

PWB : 25,000 / 1.1 + 25,000 / 1.12 + 25,000/ 1.13 + 100,000/ 1.14 + 150,000/ 1.15 + 80,000 / 1.16 – 150,000 = 118,838

ALTERNATIVE B IS MUCH BETTER, BUT PAYBACK METHOD MAY MISLEAD

ALTERNATIVES

REVENUE : TOTAL INCOME (OR TOTAL SAVINGS)

NET PROFITS : REVENUE – ALL EXPENSES – INCOME TAX

ALL EXPENSES : CASH EXPENSES + DEPRECIATION

INCOME TAX : (REVENUE – ALL EXPENSES) (TAX RATE)

CASH FLOW : NET PROFIT + DEPRECIATION

IX. ANALYSIS OF REPLACEMENTS

REPLACEMENTS

REPLACEMENT : AN ALTERNATIVE CASE IN WHICH FACILITIES ARE CURRENTLY IN EXISTENCE AND IT MAY BE DESIRABLE TO REPLACE THESE FACILITIES WITH DIFFERENT ONES

THE REASONS FOR REPLACEMENTS CAN BE DIVIDED INTO TWO GENERAL CLASSES:

1. AN EXISTING PROPERTY MUST BE REPLACED TO CONTINUE A SATISFACTORY OPERATION BECAUSE

a) PROPERTY IS WORN OUT

b) CAPACITY IS NOT SUFFICIENT

C) ECONOMICALLY NOT FEASIBLE ANYMORE

2. DUE TO TECHNOLOGICAL IMPROVEMENTS, BETTER OR MORE ECONOMICAL EQUIPMENT IS AVAILABLE

REPLACEMENTSFOR THE FIRST CLASS OF REASONS, YOU SHOULD EITHER MAKE

THE REPLACEMENT OR GO OUT OF BUSINESS. SO ECONOMIC ANALYSIS IS MADE BETWEEN THE NEW ALTERNATIVES. FOR THE SECOND CLASS OF REASONS, ANALYSIS IS MADE BETWEEN THE EXISTING FACILITY AND NEW ALTERNATIVE. IN THIS CASE, IN ORDER TO DECIDE, THE OPERATING EXPENSES OF THE PRESENT EQUIPMENT MUST BE COMPARED WITH THOSE THAT WOULD EXIST IF THE CHANGE WERE MADE.

THE DIFFERENCE BETWEEN THE TOTAL COST OF THE NEW PROPERTY AND THE NET RELIZABLE VALUE OF THE EXISTING PROPERTY EQUALS THE NECESSARY INVESTMENT FOR THE REPLACEMENT. IN THESE ANALYSIS, THE NET RELIZABLE VALUE OF AN EXISTING PROPERTY SHOULD BE ASSUMED TO BE THE MARKET VALUE. ALTHOUGH THIS MAY BE LESS THAN THE ACTUAL VALUE OF THE PROPERTY AS FAR AS THE OWNER IS CONCERNED OR LESS THAN THE BOOK VALUE (UNAMORTIZED VALUE), IT STILL REPRESENTS THE AMOUNT OF CAPITAL WHICH CAN BE OBTAINED FROM THE OLD EQUIPMENT. ANY ATTEMP TO ASSIGN A VALUE GREATER THAN THIS TENDS TO FAVOR REPLACEMENTS WHICH ARE UNECONOMICAL.

REPLACEMENTS

SUPPOSE YOU BOUGHT AN EQUIPMENT FOR $100,000 WITH 10 YEARS ECONOMIC LIFE AND DEPRECIATE IT WITH STRAIGHT LINE METHOD. AFTER 8 YEARS IT HAS A BOOK VALUE OF $20,000. BUT IF YOU TRY TO SELL IT, ITS VALUE MAY BE LOWER OR HIGHER THAN $20,000 DEPENDING ON MANY FACTORS. SO FOR REPLACEMENT ANALYSIS YOU SHOULD FORGET $20,000 AND TRY TO FIND OUT THE MARKET VALUE.

DURING REPLACEMENT, ADDITIONAL EXPENSES OCCUR SUCH AS FREIGHT, CONSTRUCTION OF FOUNDATIONS, NEW CONNECTIONS OF WIRING OR PIPING, TEST AND ADJUSTMENT EXPENSES, REMOVAL OF OLD EQUIPMENT, REPLACING OF FLOORS OR OTHER STRUCTURAL ELEMENTS AND THE SHUT- DOWN TIME DURING THE REPLACEMENT. THESE COSTS WHICH MAY BE AS BIG AS THE PRICE OF THE NEW EQUIPMENT OR EVEN BIGGER SHOULD ALLWAYS BE CONSIDERED WHEN ANALYSING THE REPLACEMENT.

REPLACEMENT

REPLACEMENT ANALYSIS:

ALL KINDS OF EQUIPMENT USED IN PLANTS GETS OLD AND SHOULD BE REPLACED ONE DAY. SO THE QUESTION IS NOT IF THE EXISTING EQUIPMENT WILL BE REPLACED, BUT WHEN IT WILL BE REPLACED.

THIS LEADS US TO THE DECISION:

SHALL WE REPLACE THE EXISTING EQUIPMENT NOW, OR SHALL WE KEEP IT FOR ONE OR MORE ADDITIONALYEARS.

THE ECONOMIC EVALUATION MADE FOR THE ABOVE DECISION IS CALLED ‘REPLACEMENT ANALYSIS’

REPLACEMENT

WE CALL THE EXISTING EQUIPMENT ‘DEFENDER’, AND THE BEST AVAILABLE REPLACEMENT EQUIPMENT ‘CHALLENGER’.

1. FIND THE MARGINAL COST DATA OF THE DEFENDER2. FIND THE EAC (EQUIVALENT ANNUAL COST) OF THE

CHALLENGER3. MAINTAIN THE DEFENDER AS LONG AS THE MARGINAL COST OF

OWNERSHIP FOR ONE MORE YEAR IS LESS THAN THE MINIMUM EAC OF THE CHALLENGER. WHEN THE MARGINAL COST OF THE DEFENDER BECOMES GREATER THAN THE MINIMUM EAC OF THE CHALLENGER, THAN REPLACE THE DEFENDER WITH THE CHALLENGER

MARGINAL COST OF AN ASSET FOR ANY YEAR IS THE COST OF KEEPING THE ASSET AND INCLUDE LOSS IN MARKET VALUE AND YEARLY OPERATING AND MAINTENANCE EXPENSES

REPLACEMENT

EXAMPLE: AN ASSET PURCHASED 5 YEARS AGO CAN BE SOLD TODAY FOR $15,000.

OPERATING AND MAINTENANCE EXPENSES FOR THE FIRST YEAR IS $10,000 BUT THESE ARE ESTIMATED TO INCREASE IN THE FUTURE BY $1,500 PER YEAR EACH YEAR. IT IS ESTIMATED THAT THE MARKET VALUE OF THE ASSET WILL DECREASE BY $1,000 PER YEAR.

A NEW PIECE OF EQUIPMENT THAT MAY REPLACE THE EXISTING ONE, WILL COST $25,000 INCLUDING ALL KIND OF REPLACEMENT COSTS. ANNUAL OPERATING AND MAINTENANCE COSTS OF THIS NEW MACHINE IS $7,000 THE FIRST YEAR AND INCREASING AT $1,000 PER YEAR THEREAFTER. USEFUL LIFE OF THIS EQUIPMENT IS 5 YEARS ANDTHE SALVAGE VALUE IS $4,000.

IF MARR = %15, DETERMINE WHEN, IF AT ALL, A REPLACEMENT DECISION

SHOULD BE MADE

REPLACEMENT

MARGINAL COST OF THE DEFENDER1. YEAR: 10,000 + [(15,000)1.15 – (14,000)] = 13,2502. YEAR: 11,500 + [(14,000)1.15 – (13,000)] = 14,6003. YEAR: 13,000 + [(13,000)1.15 – (12,000)] = 15,9504. YEAR: 14,500 + [(12,000)1.15 – (11,000)] = 17,3005. YEAR: 16,000 + [(11,000)1.15 – (10,000)] = 18,650

EAC OF THE CHALLENGER [(25,000)1.155 – 4,000 + 11,000 + (10,000)1.15 + (9,000)1.152 + (8,000)1.153 +

(7,000)1.154] .15 / (1.155 – 1) = 15,606

AT THE BEGINNING OF THE THIRD YEAR YOU SHOULD REPLACE YOUR EQUIPMENT SINCE MARGINAL COST IS BIGGER THAN THE EAC OF THE NEW EQUIPMENT: 15,950 > 15,606

ALTERNATIVES - PROBLEMS

*ALTER INVESTMENT SALVAGE VALUE LIFE ANNUAL NET CASH FLOW

X $11,000 $0 6 YR $2,600

Y 12,000 3000 6 2,500

Z 18,000 0 6 4,000

IF MARR = 0.1, WHICH ALTERNATIVE(S) SHOULD BE SELECTED ?

a) WHEN ALTERNATIVES ARE MUTUALLY EXCLUSIVE

b) WHEN ALTERNATIVES ARE INDEPENDENT

PWX = R[{(1+i)n – 1} / i(1+i)n ] – PINV = 2600[{ (1.1)6 – 1} / 0.1(1.1)6] – 11,000 = 323.68

PWY = R[{(1+i)n – 1} / i(1+i)n ] – PINV + 3000 / 1.16 = 581.60

PWZ = R[{(1+i)n – 1} / i(1+i)n ] – PINV = – 578.96

a) ALTERNATIVE Y IS BEST

b) ALTERNATIVE Y + X IS BEST

ALTERNATIVE PROBLEMS

A MANUFACTURING FIRM HAS A CHOICE BETWEEN TO MACHINES TO PRODUCE A STANDART COMMODITY. IF i = %13.6, WHAT ANNUAL PRODUCTION IS REQUIRED TO JUSTIFY PURCHASE OF MACHINE B?

MACHINE A MACHINE B

FIRST COST,$ 30,000 46,000

SALVAGE,$ 2,000 0

FIXED ANNUAL PRODUCTION COST,$ 4,800 5,000

VARIABLE PRODUCTION COST, $ / UNIT 5 2

LIFE, YEARS 10 6

LET X BE THE NUMBER OF UNITS PRODUCED ANNUALLY.

EACA = 30,000[ .136 (1.136)10 / (1.13610 – 1)] – 2000[.136 / (1.13610 – 1)] + 4,800 + 5X

= 10,356 + 5X

EACB = 46,000[ .136 (1.136)6 / (1.1366 – 1)] + 5,000 + 2X

= 16,700 + 2X

EACA = EACB 10,356 + 5X = 16,700 + 2X X = 2115 UNITS / YEAR

MACHINE B IS ECONOMIC WHEN YEARLY PRODUCTION IS > 2115 UNITS

ALTERNATIVES – PROBLEMS

• A SPECIAL EQUIPMENT IS REQUIRED FOR A R&D PROJECT THAT WILL LAST FOR 3 YEARS. FIRST ALTERNATIVE IS A $50,000 EQUIPMENT WITH A SALVAGE VALUE OF $25,000 AND YEARLY OPERATING COST OF $15,000. A SIMPLER EQUIPMENT WILL COST $30,000 WHICH WILL HAVE A SALVAGE VALUE OF $10,000 AND YEARLY OPERATING COST OF $20,000. FIND THE BETTER ALTERNATIVE USING EQUIVALENT ANNUAL COSTS IF i = 0.1.

RA = P[ i(1 + i)n / {(1 + i)n – 1}]

= (50,000 – 25,000/1.13) { .1x1.13/ (1.13 – 1)} = 12,553

EACA = 12,553 + 15,000 = $ 27,553

RB = (30,000 – 10,000/1.13) { .1x1.13/ (1.13 – 1)} = 9,042

EACA = 9,042 + 20,000 = $ 29,042

ALTERNATIVE A IS BETTER

ALTERNATIVES - PROBLEM• A COMPANY MAY BUY A MACHINE THAT COSTS $10,000 WITH A SALVAGE

VALUE OF $4,000 AT THE END OF 6 YEARS LIFE. THE ANNUAL OPERATING DISBURSEMENTS ARE $5,000 A YEAR FOR THE FIRST 3 YEARS AND $6,000 A YEAR FOR THE LAST 3 YEARS. THERE IS AN AUTOMATIC VERSION WHICH COSTS $20,000 WITH A SALVAGE VALUE OF $6,000 AT THE END OF THE 6th YEAR AND YEARLY OPERATING DISBURSEMENTS OF $3,000 A YEAR. FIND THE BETTER ALTERNATIVE USING EQUIVALENT ANNUAL COSTS IF i = 0.15.

RA = P [ i(1 + i)n / {(1 + i)n – 1}]

SINCE $5,000 IS SPEND EVERY YEAR, $1,000 FOR THE LAST 3 YEARS WILL BE CONSIDERED

RA =(10,000 – 4000/1.156 +1000/1.154 +1000/1.155 +1000/1.156){.15x1.156/(1.156-1)}

RA = $2582 EACA = 2582 + 5000 = $7582

RB = (20,000 – 6000/1.156) {.15x1.156/(1.156-1)} = $4599

EACB = 4599 + 3000 = $7599

ALTERNATIVE A IS BETTER


* 5 INVESTMENT PROPOSALS WITH 5 YEARS LIFE ARE AS FOLLOWS :

A B C D E

INVESTMENT $ 30,000 60,000 20,000 40,000 30,000

SALVAGE VALUE $ 0 10,000 0 10,000 5,000

NET ANNUAL CASH FLOW $ 7,500 13,755 5,000 10,000 7,500MARR = 9%

a) WHICH PROPOSAL IS PREFERRED IF ONLY ONE CAN BE SELECTED?

b) HOW MUCH SHOULD BE INVESTED IF THE PROPOSALS ARE INDEPENDENT AND UNLIMITED CAPITAL IS AVAILABLE?

PWA=R[{(1+i)n – 1)} / i(1+i)n] – PINV =7500{(1.095 – 1) / 0.9x1.095} – 30,000 = – 827.66

PWB = 13,755 {(1.095 – 1) / 0.9x1.095} +(10,000 / 1.095) – 60,000 = 0

PWC = 5,000{(1.095 – 1) / 0.9x1.095} – 20,000 = – 552

PWD = 10,000{(1.095 – 1) / 0.9x1.095} +(10,000 / 1.095) – 40,000 = 5,395.31

PWE = 7,500{(1.095 – 1) / 0.9x1.095} +( 5,000 / 1.095) – 30,000 = 2,421.66

a) ALTERNATIVE D IS SELECTED SINCE IT HAS THE BIGGEST PW

b) ALT B + ALT D + ALT E = $130,000


* A COMPANY PLANS TO BUY A FACTORY BUILDING FOR A TEN YEAR PROJECT AND SELL IT AT THE END. THE LOCATION OF THE BUILDING EFFECTS THE COSTS, SO DIFFERENT CASH FLOWS OCCUR. IF MARR = 15%, WHICH BUILDING IS BEST ?

SITE 1 SITE 2 SITE 3

PURCHASE PRICE : $140,000 190,000 220,000

RESALE VALUE : $125,000 155,000 175,000

CASH FLOW : $ 24,000 31,000 41,000

PW1 = 24,000{(1.1510 – 1)/0.15x1.1510} + 125,000/1.1510 – 140,000 = 11,349

PW2 = 31,000{(1.1510 – 1)/0.15x1.1510} + 155,000/1.1510 – 190,000 = 3,896

PW3 = 41,000{(1.1510 – 1)/0.15x1.1510} + 175,000/1.1510 – 220,000 = 29,027

HIGHEST PW IS SITE 3, SO CHOOSE SITE 3


* A FIRM FINDS THAT AIRCONDITIONING IS NECESSARY FOR A ROOM AND AS MONEY SPEND ON INSULATING THE WALLS GETS LARGER, INVESTMENT FOR AIRCONDITIONING GETS SMALLER :

ALTERNATIVES 1 2 3 4

FIRST COST OF INS. $ 35,000 45,000 60,000 80,000

FIRST COST OF AC $ 52,000 45,000 38,000 32,000

ANNUAL POWER COST $ 6,500 5,100 4,100 3,500

INSULATION HAS 20 YEARS, AC HAS 10 YEARS LIFE WITH 0 SALVAGE VALUE.

WHICH ALT. IS BEST IF i = 15%, AND REPLACEMENT COST IS CONSTANT ?

CR (AC) = C / 1.1510 = 0.2472 x C

PW = C(INS) + C(AC) + CR(AC) + R {(1.1520 – 1) / 0.15 x 1.1520}

PW1 = 35,000 + 52,000 + 12,854 + 6,500 {(1.1520 – 1) / 0.15 x 1.1520} = $140,538

PW2 = 45,000 + 45,000 + 11,124 + 5,100 {(1.1520 – 1) / 0.15 x 1.1520} = $133,045

PW3 = 60,000 + 38,000 + 9,394 + 4,100 {(1.1520 – 1) / 0.15 x 1.1520} =$133,056

PW4 = 80,000 + 32,000 + 7,910 + 3,500 {(1.1520 – 1) / 0.15 x 1.1520} =$141,817

CHOOSE THE MINIMUM PW : ALTERNATIVE 2

ALTERNATIVES - PROBLEM* TWO MUTUALLY EXCLUSIVE PROPOSALS HAVE THE FOLLOWING CASH FLOW :

INVESTMENT 1 2 3 4 5YEARS A $ 100,000 60,000 50,000 40,000 30,000 0 B 100,000 60,000 15,000 60,000 40,000 0ALL RECEIPTS FROM PROPOSALS A AND B CAN BE REINVESTED AT 20% FOR

A AND 25% FOR B. THE REINVESTMENT RATES ARE EXPECTED TO CONTINUE THROUGHOUT THE 5-YEAR COMPARISON PERIOD. MARR = 15%

a) WHICH IS BETTER WHEN REINVESTMENT OPPURTUNITY IS IGNORED?

PW = S/(1+i)n

PWA = 60,000/1.15 + 50,000/1.152 + 40,000/1.153 + 30,000/1.154 – 100,000 = 33,434

PWB = 60,000/1.15 + 15,000/1.152 + 60,000/1.153 + 40,000/1.154– 100,000 = 25,837 ALT. A IS BETTER b) WHICH IS BETTER WHEN RECEIPTS ARE INVESTED AT GIVEN RATES ? S = P (1+i)N PW = P(1+i)N / (1+i)n WHERE N : RECEIPT INVESTING TIME

PWA =(60,000x1.24 + 50,000x1.23 + 40,000x1.22 + 30,000x1.2)/1.155- 100,000=51,348

PWB =(60,000x1.254 + 15,000x1.253 + 60,000x1.252 + 40,000x1.25)/1.155- 100,000 =58864

58,864 ALT B IS BETTER

ALTERNATIVES - PROBLEMS* CASH FLOWS FOR 5 ALTERNATIVES ARE BELOW : (i = 0.15) ALT INVESTMENT 1 2 3 4 5 A $15,000 4,500 4,500 4,500 4,500 4,500 B 25,000 12,000 10,000 8,000 6,000 4,000 C 20,000 2,000 4,000 6,000 8,000 10,000 D 30,000 0 0 15,000 15,000 15,000 E 10,000 4,500 4,500 -2,500 4,500 7,500a) IF ALTERNATIVES ARE MUTUALLY EXCLUSIVE , WHICH IS THE BEST ?

PWA = R[ {(1+i)n – 1} / i(1+i)n] – PINV = 4,500 {(1.155 – 1) / 0.15x1.155} – 15,000 = 84

PWB =12,000/1.15+10,000/1.152+8,000/1.153+6,000/1.154+4,000/1.155- 25,000 =3,675

PWC =2,000/1.15+4,000/1.152+6,000/1.153+8,000/1.154+10,000/1.155- 20,000 =-1,754

PWD =15,000/1.153+15,000/1.154+15,000/1.155- 30,000 =- 4,103

PWE =4,500/1.15+4,500/1.152-2,500/1.153+4,500/1.154+7,500/1.155- 10,000 =1,974 ALT B IS THE BESTb) IF INVESTMENT CAPITAL IS LIMITED WITH $30,000 AND THE PROPOSALS

ARE INDEPENDENT, WHICH PROPOSALS SHOULD BE EXPECTED ? ALT B WILL BE EXPECTED


* PROPOSAL A REQUIRES AN INVESTMENT OF $250,000 WHICH WILL BRING AN ANNUAL INCOME OF $100,000 FROM ANNUAL DISBURSEMENTS OF $40,000. THE LIFE OF THE PROJECT IS EXPECTED TO BE 20 YEARS WITH $75,000 SALVAGE AT THAT DATE. PROPOSAL B REQUIRES $175,000 INVESTMENT WITH ANNUAL INCOME OF $90,000 AND DISBURSEMENTS 0F $50,000. AT THE END OF 20 YEARS ITS SALVAGE WILL BE $60,000. THE PROPOSALS ARE MUTUALLY EXCLUSIVE ALTHOUGH THE COMPANY MAY TURN DOWN BOTH IF NEITHER IS ACCEPTABLE. THE MARR IS 20%.

PWA = – 250,000 + 75,000/1.220 + 60,000 [(1.220 – 1)/ (.2 x 1.220)] = $44,131

PWB = – 175,000 + 60,000/1.220 + 40,000 [(1.220 – 1)/ (.2 x 1.220)] = $21,348

BOTH ALTERNATIVES ARE ACCEPTABLE, BUT SINCE THEY ARE EXCLUSIVE, ALTERNATIVE A SHOULD BE CHOOSEN

ALTERNATIVES - PROBLEMS* A DAM COSTING $100,000 TO CONSTRUCT WILL COST $15,000 A

YEAR TO OPERATE AND MAINTAIN. ANOTHER DESIGN COSTING $150,000 TO BUILD WILL COST $10,000 A YEAR TO OPERATE AND MAINTAIN. BOTH INSTALLATIONS ARE FELT TO BE PERMENANT. THE MINIMUM REQUIRED RATE OF RETURN IS 5%.

a) WHICH ALTERNATIVE IS BETTER ? b) AT WHAT TIME THEY ARE EQUAL ?

a) WHEN n P = R / i

PWA = 100,000 + 15,000 / 0.05 = $400,000

PWB = 150,000 + 10,000 / 0.05 = $350,000 ALTERNATIVE B IS BETTER

b) 100,000 + 15,000 [ (1.05n – 1) / 0.05 x 1.05n] = 150,000 + 10,000 [ (1.05n – 1) / 0.05 x 1.05n]

1.05n – 1 = 0.5 x 1.05n 1.05n = 2 n = 14.2 YEARS

ALTERNATIVES - PROBLEM* AN AUTOMATIC MACHINE COSTS $20,000 AND IS EXPECTED TO

HAVE AN ANNUAL OPERATING COST OF $900 AT THE END OF EACH YEAR OF ITS 10-YEAR LIFE. A SEMIAUTOMATIC MACHINE WHICH COSTS $10,000 WILL HAVE ANNUAL OPERATING COSTS OF $2,000 OVER ITS 10-YEAR LIFE. IF SALVAGE VALUES ARE 0, WHICH ONE COSTS LESS a) WHEN i = 0 AND b) WHEN i = 0.10 ?

a) CA = 20,000 + 10 x 900 = $29,000

CS = 10,000 + 10 x 2,000 = $30,000 AUTOMATIC COSTS LESS

b) PW = PINV + R [{(1+i)n – 1} / i(1+i)n]

PWA = 20,000 + 900 {(1.110 – 1) / 0.1x1.110} = $25,529

PWS = 10,000 + 2,000 {(1.110 – 1) / 0.1x1.110} = $22,288 SEMIAUTOMATIC COSTS LESS


A FIRM INSTALLED A FACILITY WITH A FIRST COST OF $80.000, WHICH HAS A LIFE OF 9 YEARS AND A SALVAGE VALUE OF $4,000. YEARLY MAINTENANCE COST IS $6,800. AFTER THE FACILITY HAD BEEN IN OPERATION FOR 5 YEARS AN IMPROVEMENT WAS PROPOSED TO INCREASE ITS LIFE TO 11 YEARS, REDUCE YEARLY MAINTENANCE TO $5,000 FOR THE REMAINING LIFE, AND INCREASE THE SALVAGE VALUE TO $7,500. IF MONEY IS WORTH %10.5, WHAT IS THE MAXIMIM AMOUNT THE FIRM SHOULD EXPEND FOR THIS IMPROVEMENT?

FIND THE ORIGINAL EAC:

EAC1 = 80,000[.105 (1.1O5)9 / (1.1O59 – 1)] – 4,000 [.105 / (1.1O59 – 1)] + 6,800

= 20,681

FIND THE IMPROVED EAC ( LET X BE THE EXPENSE OF IMPROVEMENT):

P = 80,000 + X / 1.1O55 – 7,500 / 1.1O511 + 6,800 [ (1.1O55 – 1) / .105 (1.1O5)5 ]

+ 5,000 [ (1.1O56 – 1) / .105 (1.1O5)6 ] ( 1 / 1.1O55 ) = 115,985 + .607 X

EAC2 = (115,985 + .607 X) [.105 (1.1O5)11 / (1.1O511 – 1)] = 18,268 + .0956 X

EAC1 = EAC2 20,681 = 18,268 + .0956 X X = $25,240


A FIRM IS CONSIDERING BUYING EQUIPMENT THAT WILL REDUCE ANNUAL LABOR COSTS BY $10,500. THE EQUIPMENT COSTS $51,000 AND HAS A SALVAGE VALUE OF $5,000 AND LIFE OF 7 YEARS. ANNUAL MAINTENANCE WILL BE $1,200. IF THE MARR IS %10,IS THE FIRM JUSTIFIED IN PURCHASING THE EQUIPMENT?

• BY PRESENT WORTH

P = – 51,000 + 5,000 / 1.17 + (10,500 – 1,200) [ (1.17 – 1) / .1(1.17) ] = – 3,128

SINCE P IS NEGATIVE, IT IS NOT ECONOMIC

• BY EAC

EAC = 51,000 [ .1(1.17) / (1.17 – 1) ] – 5,000 [ .1 / (1.17 – 1) ] + 1,200 = 11,142

SINCE EAC > ANNUAL SAVING (10,500), IT IS NOT ECONOMIC

• BY IRR

– 51,000 + 5,000 / (1+i)7 + (10,500 – 1200) [ (1+i)7 – 1) / i(1+i)7 ] = 0

BY TRIAL- ERROR i = %8.2 WHICH IS LESS THAN MARR, NOT ECONOMIC

X. INFLATION

INFLATION

INFLATION CAUSES PRICES TO RISE AND DECREASES THE PURCHASING POWER OF MONEY. INFLATION RATES ARE USUALLY MEASURED BY WHOLESALE PRICE INDEX AND CONSUMER PRICE INDEX.

WHEN INFLATION IS MODEST, 2 TO 4 PERCENT PER YEAR, IT IS GENERALLY IGNORED IN ECONOMIC EVALUATIONS OF PROPOSALS. IT IS ARGUED THAT ALL PROPOSALS ARE AFFECTED SIMILARLY BY PRICE CHANGES AND THAT THERE IS TOO LITTLE DIFFERENCE BETWEEN CURRENT AND FUTURE COSTS TO INFLUENCE THE ORDER OF PREFERENCE. THESE ARGUMENTS LOSE SUBSTANCE WHEN INFLATION IS HIGH AND SOME GOODS ESCALATE MORE RAPIDLY THAN OTHERS.

INFLATION

THERE ARE TWO ALTERNATIVE METHODS OF TREATING INFLATION IN FINANCIAL ANALYSIS :

1. CONVERT ALL CASH FLOWS TO CONSTANT- WORTH (REAL OR TODAY’S DOLLAR) AMOUNTS TO ELIMINATE THE EFFECT OF INFLATION. THEN USE THE REGULAR INTEREST RATES. THIS IS APPROPRIATE WHEN ALL COMPONENTS INFLATE AT UNIFORM RATE.

2. EXPRESS THE FUTURE CASH FLOWS IN THEN- CURRENT OR ACTUAL DOLLARS AND USE AN INTEREST RATE THAT INCLUDES INFLATION.

IN ANALYSIS THAT CONSIDERS INFLATION, MARKET INTEREST RATE (if) SHOULD BE USED. if INCLUDES THE COMBINED EFFECT OF THE EARNING VALUE OF CAPITAL AND EXPECTED INFLATION. MARKET INTEREST RATES ARE ALSO REFERRED TO AS ‘COMPOSITE’ OR INFLATION ADJUSTED INTEREST RATES.

INFLATION

THE INTEREST RATE if IS DEFINED AS

if = (1 + i)(1 + f) – 1

WHERE f IS THE AVERAGE INFLATION RATE.

IF i = 0.12 AND f = 0.1 THEN if = (1.12)(1.1) – 1 = 0.232

CONSTANT (REAL) DOLLARS IN ANY FUTURE YEAR N CAN BE INFLATED TO CURRENT DOLLARS BY:

CURRENT DOLLARS = (CONSTANT DOLLARS)(1 + f)N

WHEN f IS DIFFERENT FOR EACH YEAR :

(1+ f)N (1 + f1)(1 + f2) ….. (1 + fN)

INFLATION

EXAMPLE : AN INVESTMENT IS MADE FOR $4000. IT WILL BRING $1500 (REAL DOLLARS) EVERY YEAR FOR 4 YEARS. IF i = 0.12 AND f = 0.06 (CONSTANT FOR 4 YEARS), WHAT IS THE PW ?

a) IN REAL DOLLARS WE DO NOT CONSIDER INFLATION

PW = – 4000 + 1500 [{ (1.12)4 – 1} / 0.12(1.12)4]

= $556

b) WHEN WE FIND CURRENT DOLLARS

N1 : R = 1500 (1.06) = 1590

N2 : R = 1500 (1.06)2 = 1685

N3 : R = 1500 (1.06)3 = 1787

N4 : R = 1500 (1.06)4 = 1894

if = (1.12)(1.06) – 1 = 0.1872

PW = – 4000 + (1590 / 1.1872) + (1685 / 1.18722) + (1787 / 1.18723) +

(1894 / 1.18724) = $556

INFLATIONEXAMPLE : A MACHINE CAN BE PURCHASED FOR $100,000. IT WILL

NO SALVAGE VALUE AT THE END OF 5 YEAR USEFULL LIFE. OPERATING COSTS WILL BE $12,000 PER YEAR, WHILE IT PROVIDES A REVENUE OF $40,000 ANNUALLY. ESTIMATES ARE BASED ON CURRENT ECONOMIC CONDITIONS WITHOUT CONSIDERATION OF INFLATION. EVALUATE THE PROPOSAL ACCORDING TO REAL- DOLLAR DATA AND ACTUAL- DOLLAR CASH FLOW WHEN INFLATION RATE IS 8%. THE MARR IS 10% WITHOUT ANY ADJUSTMENT FOR INFLATION.

a) REAL- DOLLARS : 40,000 – 12,000 = 28,000 / YEAR PW = – 100,000 + 28,000 [(1.15 – 1) / 0.1(1.1)5] = $6142 b) CONVERT $28,000 TO CURRENT DOLLARS FOR f = 0.08

N1 : R = 28,000 x 1.08 = 30,240 N2 : R = 32,659

N3 : R = 35,272 N4 : R = 38,094 N5 : R = 41,141

if = (1.1 x 1.08) – 1 = 0.188 PW = – 100,000 + 30,240/1.188 + 32,659/1.1882 + 35,272/1.1883 + 38,094/1.1884 + 41,141/1.1885 = $6142

INFLATION

OFTEN ONE OR MORE COMPONENTS IN A CASH FLOW STREAM HAVE ESCALATION RATES DIFFERENT FROM THE GENERAL INFLATION RATE.

EXAMPLE : PREVIOUS PROBLEM BUT REVENUE ($40,000) WILL ESCALATE AT 5% WHILE INFLATION IS 8% AND EXPENSES INCREASE AT 8%

REVENUE EXPENSES NET RECEIPT

N1 $42,000 $12,960 $29,040

N2 44,100 13,997 30,103

N3 46,305 15,116 31,189

N4 48,620 16,326 32,294

N5 51,051 17,632 33,419

PW = – 100,000 + 29,040/1.188 + 30,103/1.1882 + 31,189/1.1883 +

32,294/1.1884 + 33,419/1.1885

= – $5,289

XI.OPTIMIZATION

OPTIMIZATION

THE OPTIMUM FOR A PROCESS DESIGN IS THE MOST COST-EFFECTIVE SELECTION, ARRANGEMENT, SEQUENCING OF PROCESSING EQUIPMENT AND OPERATING CONDITIONS FOR THE DESIGN.

HOW OPTIMIZATION IS MADE:

1. OPTIMIZATION CRITERIA IS ESTABLISHED: SELECTION OF AN ECONOMIC CRITERION THAT IS TO BE THE OBJECTIVE FUNCTION (total cost, profit, production rate…)

2. PROBLEM IS DEFINED: EXAMINATION OF PROCESS TO DETERMINE VARIABLES AND CONSTRAINTS THAT EFFECT THE OBJECTIVE FUNCTION AND ESTABLISHMENT OF MATHEMATICAL RELATIONS (temperature, pressure, flow rate, equipment specifications…)

3. PROBLEM IS SOLVED FOR THE OPTIMAL VALUES

OPTIMIZATION

WHEN A DESIGN VARIABLE IS CHANGED, OFTEN SOME COSTS INCREASE AND OTHERS DECREASE. THE TOTAL COST MAY GO THROUGH A MINIMUM AT ONE VALUE OF THAT VARIABLE AND THIS VALUE IS THE OPTIMUM VALUE OF THAT VARIABLE.

EXAMPLE: Optimum Insulation Thickness for a steam pipe

Cost per

Year, $ total cost (fixed+heat loss)

fixed costs

cost of heat loss

optimum insulation thickness

Insulation thickness,cm

OPTIMIZATION

IN PROCESS DESIGN YOU SHOULD OPTIMIZE IN TWO FIELDS:

STRUCTURAL OPTIMIZATION: TO FIND THE BEST FLOWHEET, YOU SHOULD CONSIDER SELECTING EQUIPMENT, ARRANGEMENT AND SEQUENCING OF EQUIPMENT

PARAMETRIC OPTIMIZATION: TO FIND THE BEST CONDITIONS, YOU SHOULD CONSIDER YOU EQUIPMENT DESIGN PARAMETERS AND OPERATING VARIABLES

TYPICALLY, A FEW NEAR OPTIMAL FLOWSHEETS ARE OBTAINED BY ESTABLISHING SEVERAL FEASIBLE FLOWSHEETS AND THEN COMPARING THEIR NET PRESENT WORTHS OR ANNULAIZED COSTS. ONE OR MORE OF THESE IS THEN SUBJECTED TO PARAMETRIC OPTIMIZATION, AND THE MOST ECONOMICALLY ATTRACTIVE ONE IS SELECTED.

OPTIMIZATION

WHEN THE FACTOR TO BE OPTIMIZED (MINIMIZED OR MAXIMIZED), IS AN ANALYTICAL FUNCTION OF A SINGLE VARIABLE, SOLVING THE PROBLEM IS SIMPLE.

EXAMPLE: SUPPOSE WE HAVE A COST FOR AN OPERATION BEING A FUNCTION OF ONE VARIABLE:

C = ax + b + c / x + d

WHERE a, b, c, d ARE CONSTANTS, x IS THE VARIABLE AND C IS THE OBJECTIVE FUNCTION TO BE MINIMIZED.

WE SET THE DERIVATIVE OF C WITH RESPECT TO X EQUAL TO 0:

dC / dx = a – c / x2 = 0

x = (c / a)1/2

TO CHECK IF THE VALUE IS MINIMUM OR MAXIMUM YOU CAN TAKE THE SECOND DERIVATIVE (IF IT IS POSITIVE, THE VALUE IS MINIMUM), OR CALCULATE C FOR A SLIGHTLY DIFFERENT VALUE OF x.

OPTIMIZATION

EXAMPLE:

A BATCH OPERATION IN A PRODUCTION PLANT HAS ANNUAL OPERATION COST EQUAL TO $2,000 Q0.8 WHERE Q IS THE KG OF PRODUCTION FOR BATCH. THE PREPARATION COST FOR BATCH IS $60. WHAT IS THE OPTIMUM BATCH SIZE IF THE YEARLY PRODUCTION IS 50,000 KG.

TOTAL COST C = 2,000 Q0.8 + 60(50,000/Q)

dC / dQ = 0.8(2,000 Q-0.2 ) – 3,000,000 / Q2 = 0

Q1.8 = 1875

Q = 65.8 KG / BATCH

OPTIMIZATION

WHEN TWO OR MORE INDEPENDENT VARIABLES AFFECT THE OBJECTIVE FUNCTION, WE USE A SIMILAR PROCEDURE

EXAMPLE: SUPPOSE YOU HAVE A TOTAL COST TO BE MINIMIZED WHICH IS A FUNCTION OF TWO VARIABLES x AND y

C = ax + b / xy + cy + d

∂C / ∂x = a – b / x2y

∂C / ∂y = c – b / xy2

IF WE SET BOTH EQUATIONS EQUAL TO ZERO AND SOLVE FOR x AND y, WE CAN FIND THE CORRECT VALUES OF BOTH VARIABLES OPTIMIZING THE COST.

OPTIMIZATION

EXAMPLE: FOR AN OPERATION, THE TOTAL COST IS FOUND TO BE:

C = 2.33x + 11,900 / xy + 1.8y + 10

FIND THE x AND y VALUES GIVING THE MINIMUM TOTAL COST.

∂C / ∂x = 2.33 – 11,900 / x2y = 0

∂C / ∂y = 1.86 – 11,900 / xy2 = 0

2.33 x2y = 11,900

1.86 xy2 = 11,900

x ( 11,900 / 2.33x2)2 = 11,900

x = 16, y = 20, C = 121.7

XII. PROJECT MANAGEMENT

PROJECT MANAGEMENT

PROJECT :A MAJOR UNDERTAKING THAT IS UNLIKELY TO BE REPEATED IN EXACTLY THE SAME WAY AT FUTURE

PRODUCTION MANAGEMENT IS REPETITIVE IN NATURE, PROJECT MANAGEMENT IS MORE OF A SINGLE, MAJOR JOB. PROJECT ACTIVITIES HAVE PRECEDENCE RELATIONSHIPS.

IN PROJECT MANAGEMENT WE PLAN AND CONROL:

* BUDGET

* TIME

AIM IS TO MINIMIZE BUDGET AND TIME

PROJECT LIFE CYCLE

CONCEPT

- INITIATE BROAD DISCUSSION OF PROJECT

PROJECT DEFINITION

- DEVELOP PROJECT DESCRIPTION

- DESCRIBE HOW TO ACCOMPLISH THE WORK

- DETERMINE TENTATIVE TIMING

- IDENTIFY BROAD BUDGET, PERSONNEL, RESOURCE REQUIREMENTS

PLANNING

- DEVELOP DETAILED PLANS IDENTIFYING TASKS, TIMING, BUDGETS AND RESOURCES

- CREATE ORGANIZATION TO MANAGE THE PROJECT

PROJECT LIFE CYCLE

PRELIMINARY STUDIES

- VALIDATE THE ASSUMPTIONS MADE IN THE PROJECT PLAN BY DATA COLLECTION, LITERATURE SEARCH

PERFORMANCE

- EXECUTE THE PROCECT PLAN AND PERFORM WORK

- USE PROJECT CONTROL TOOLS AND TECHNIQUES

POSTCOMPLETION

- CONFIRM PROJECT RESULTS

- REASSIGN PERSONNEL AND EQUIPMENT

- DOCUMENT PROJECT FILES FOR FUTURE REFERENCE

GANTT CHART

ONE OF THE OLDEST TECHNIQUES USED IS GANNT CHART. IT SHOWS PLANNED ACTIVITIES VERSUS ACTUAL ACCOMPLISMENTS ON THE SAME TIME SCALE.

IT IS NOT SUITED FOR MANAGEMENT OF LARGE PROJECTS. INTERRELATIONS OF ACTIVITIES ARE NOT SHOWN. IT IS DIFFICULT TO SEE HOW PROJECT IS EFFECTED IF YOU DELAY AN ACTIVITY. RESCHEDULING, WHEN THERE ARE DEVIATIONS, IS HARD TO MAKE.

GANTT CHART

ACTIVITY 16 IS EXACTLY ON SCHEDULE. ACT 12 IS 2 WEEKS AHEAD OF SCHEDULE, ACT 18B IS 1/2 WEEK BEHIND SCHEDULE

THE GANTT CHART MUST BE UPDATED VERY FREQUENTLY TO REFLECT THE CHANGING SITUATION SINCE CERTAIN ACTIVITIES WILL TAKE LONGER OR SHORTER THEN EXPECTED

AC

TIV

ITY

START OPERATION

END OPERATION

PLANNED PROCESS TIME

ACTUAL TIME EXPANDED

NOW

8 12 12A

16

18 18B 22

0 5 10 15WEEKS

PROJECT PLANNING NETWORKS

TO OVERCOME SOME SHORTCOMINGS OF GANTT CHART, CPM AND PERT METHODS WERE DEVELOPED, IN LATE 1950 s. CRITICAL PATH METHOD (CPM) WAS DEVELOPED BY DUPONT COMPANY DURING PLANNING FOR MAINTENANCE OF A CHEMICAL PLANT.

PERFORMANCE EVALUATION AND REVIEW TECHNIQUE (PERT) WAS DEVELOPED DURING PLANNING POLARIS PROJECT.

USING CPM, DUPONT DECREASED THE DOWNTIME OF ITS PLANT FOR MAINTENANCE FROM 125 TO 78 HOURS.

BY PERT, THE ORIGINAL PROGRAM OF POLARIS DEVELOPMENT PROJECT WAS DECREASED BY 2 YEARS.

PERT AND CPM HAVE SIMILARITIES. THEY BOTH ALLOW US TO DETERMINE THE MOST CRITICAL ACTIVITIES.

CRITICAL PATH METHOD (CPM)

CPM UTILIZE A NETWORK REPRESANTATION TO SHOW EACH TASK TO BE PERFORMED, ITS PREDECESSOR AND ITS SUCCESSORS.

CPM REQUIRES THAT WE BEGIN BY IDENTIFYING ALL PROJECT ACTIVITIES, PRESEDENCE RELATIONSHIPS AND TIMES.

ACTIVITY PREDECESSOR DURATION

A None 8

B None 20

C None 33

D A 18

E A 20

ACTIVITY PREDECESSOR DURATION

F B 9

G C 10

H D 8

I E,F 4


20 30

3

29 39

6

3929

3020

F, 9

3929

3020

0 0

1

20 30

3

33 33

4

26 35

5

29 39

6

43 43

7

90

178

A, 8

100

B, 20 3020

F, 9 4333

3929

I, 4

3526

178

D, 18

00

3333

C, 33

198

3928

E, 20

3526

4334

H,8

33

33

43

43

G, 10

8 17

2

LATEST ALLOWABLE EVENT OCCURANCE TIME

LATEST ALLOWABLE ACTIVITY START TIME

EARLIEST EVENT OCCURANCE TIME

EARLIEST TIME ACTIVITY CAN START

ACTIVITY DURATION

NODE NUMBER

EARLIEST TIME ACTIVITY ENDS

LATEST ALLOWABLE ACTIVITY FINISH TIME


FORWARD PASS : CALCULATE THE EARLIEST TIME AN ACTIVITY CAN START.

BACKWARD PASS : CALCULATE LATEST TIME AT WHICH EVENTS CAN BE COMPLETED.

TOTAL ACTIVITY SLACK : LATEST ALLOWABLE OCCURANCE TIME OF AN ACTIVITY'S SUCCESSOR EVENT, LESS THE EARLIEST FINISH TIME OF THE ACTIVITY. TELLS AMOUNT OF DELAY WE CAN HAVE WITHOUT AFFECTING THE EARLIEST START OF AN ACTIVITY ON CP.

CRITICAL PATH : SEQUENCE OF ACTIVITIES HAVING NO SLACK.


OFFERS A SYSTEMATIC PROCEDURE FOR SELECTING THE CRITICAL PATH.

AMOUNT OF SLACK OR FREE TIME ON NONCRITICAL PATHS MAY BE DETERMINED. THIS PERMITS US TO TRADE OFF MANPOWER AND EQUIPMENT RESOURCES FROM NONCRITICAL ACTIVITIES TO CONCENTRATE ON AND SHORTEN THE CRITICAL PATH.

MANAGEMENT FOCUS MORE ATTENTION TO SMALL PERCENTAGE OF CRITICAL ACTIVITIES.

EFFECTS OF CHANGES CAN BE SEEN EASILY.

PERT

PROGRAM EVALUATION AND REVIEW TECHNIQUE (PERT) IS SIMILAR TO CPM. DIFFERENCE IS CPM REQUIRES SINGLE ESTIMATE FOR EACH ACTIVITY, PERT REQUIRES 3 : MOST LIKELY TIME, MINIMUM TIME, MAXIMUM TIME.

PERT IS USED WHERE GREAT DEAL OF UNCERTANITY EXISTS FOR DURATIONS OF ACTIVITIES, LIKE RESEARCH PROGRAMS OR APPLICATION OF NEW TECHNOLOGIES.

to : OPTIMISTIC TIME, tm : MOST LIKELY TIME,tp : PESSIMISTIC TIME, te : EXPECTED TIME, (AVERAGE)

to + 4 tm + tp te =

6 tp - to 2

VARIANCE OF DISTRIBUTION = Vt = 6

( )

PERT

EXAMPLE : ACTIVITY t o t m t p t e V t

C 21 33 45 33 16G 7 9 17 10 2.778

SINCE THE CRITICAL PATH IS C AND G

TE = te = 33 + 10 = 43

Vt = Vt = 16 + 2.788 =18.778

STANDARD DEVIATION : St = Vt

St = 18.778 = 4.333

THE PROJECT MOST PROBABLY WILL FINISH LATEST

43 + 4 = 47 DAYS

C. G

C. G

TIME-COST TRADE-OFFS

IN CPM AND PERT WE USE NORMAL WORKING CONDITIONS SUCH AS REGULAR WORK DAY AND STANDARD EQUIPMENT. ALL JOBS CAN BE SHORTENED BY USING NEW MACHINES, EXTRA MAN-POWER, OVER TIME WORK ETC. THESE INCREASE THE DIRECT COSTS. BUT INDIRECT COST LIKE PENALTIES FOR TIME OVERRUNS, EARLY COMPLETION REWARDS MAY HAVE POSITIVE EFFECT TO COST BY SHORTENING PROJECT DURATION.

WE CALCULATE DIRECT COSTS FOR EACH ACTIVITY FOR NORMAL TIME AND CRASH TIME (MINIMUM TIME) AND INDIRECT OUTCOME AS TOTAL TIME DECREASE

DURATION (TOTAL) : 43 42 41 40 39 38 37 36 35 34

REWARD (1000$) : - 10 18 25 31 36 40 43 45 46

TIME-COST TRADE-OFFS

ACTIVITY A B C D E F G H INORMAL TIME 8 20 33 18 20 9 10 8 4CRASH TIME 6 16 21 14 17 7 7 8 3NORMAL COST 10 22 30 20 6 4 9 3 6CRASH COST 15 38 42 26 7.5 8 18 3 9.5SLOPE (Cost/Day) 2.5 4 1 1.5 .5 2 3 - 3.5

PROJECT DURATION ACTIVITY CHANGE COST REWARD RESULT

43 - 110 - - 42 C: 33 TO 32 111 10 +9 41 C: 32 TO 31 112 18 +16 40 C: 31 TO 30 113 25 +22

KEEP LOWERING ACTIVITY TIMES ON CP AS LONG AS TOTAL COST DECREASES TILL TO CRASH TIMES. THERE MAY BE A NEW CP.

x 1

00

0 $

WBS

WORK BREAKDOWN STRUCTURE (WBS) THE PURPOSE OF WBS IS TO SUB-DIVIDE THE SCOPE

OF WORK INTO MANAGEABLE WORK PACKAGES WHICH CAN BE ESTIMATED, PLANNED AND ASSIGNED TO A RESPONSIBLE PERSON OR DEPARTMENT FOR COMPLETION

IT IS A NICE TOOL USED FREQUENTLY IN PROJECT MANAGEMENT BREAKING COMPLEXITY INTO SIMPLE, MANAGEABLE COMPONENTS

BY USING WBS, ESTIMATING THE DURATION AND THE COST OF THE PROJECT AND CONTROLLING IS EASIER

THERE ARE MANY METHODS (SYSTEM BREAKDOWN, PRODUCT BREAKDOWN, CONTRACTOR BREAKDOWN..) OF SUBDIVIDING THE WORK AND THE BEST METHOD IS THE ONE THAT WORKS WELL FOR THAT PROJECT

WBS

AN EXAMPLE FOR WBS

HOUSE

FOUNDATIONS WALLS / ROOF PIPING SEWERAGE WIRING APPLIANCES

CIVIL PLUMBING ELECTRICAL

DECISION TREE ANALYSIS

THE DECISION TREE TECHNIQUE FACILITATES PROJECT EVALUATION BY ENABLING THE FIRM TO WRITE DOWN THE POSSIBLE FUTURE EVENTS AS WELL AS THEIR MONETARY OUTCOMES, IN A SYSTEMATIC MANNER.

IN MAKING THE DECISION TREE WE START WITH DECISION NODE. THE DECISION ALTERNATIVES ARE REPRESENTED AS BRANCHES FROM THE DECISION NODE. GENERALLY, TRAVERSING EACH BRANCH ON THE DECISION TREE WILL BRING SOME REWARD, POSITIVE OR NEGATIVE, TO THE DECISION MAKER. THIS REWARD WILL BE CASH FLOW IN INDUSTRY.

alternative 1

alternative 2

alternative 3


FOR EACH ALTERNATIVE SHOULD PREDICT AN OUTCOME FOR GOOD, MODERATE AND BAD CONDITIONS. THAN, FOR EACH CONDITION YOU SHOULD ESTIMATE THE PROBABILITY OF OCCURANCE.

WITH THIS INFORMATION, YOU CAN CALCULATE EMV, EXPECTED MONETARY VALUE, BY MULTIPLYING THE PRESENT VALUES OF EACH CONDITION BY THE PROBABILITY OF THAT CONDITION AND SUMMING THESE FOR EACH ALTERNATIVE. THE ALTERNATIVE WITH THE LARGEST EMV IS THE BEST ALTERNATIVE.


EXAMPLE: YOUR COMPANY IS DISCUSSING AN EXPANSION PROJECT. THERE ARE 3 ALTERNATIVES; LARGE EXPANSION, SMALL EXPANSION AND NO EXPANSION. IN THE LARGE EXPANSION THE NEW MACHINERY INVESTMENT IS $140,000, WORKING CAPITAL IS $20,000, THE SALVAGE VALUE AFTER 5 YEARS IS ASSUMED TO BE 0. IN THIS ALTERNATIVE YEARLY INCREASE IN EXPENSES WILL BE $20,000. IN THE SMALL EXPANSION THE INVESTMENT REQUIRED IS $60,000, WORKING CAPITAL IS $10,000 AND NO SALVAGE VALUE AFTER 5 YEARS. THE INCREASE IN THE YEARLY EXPENSES IS $10,000. THE ESTIMATED INCREASE IN THE SALES REVENUES AND THE PROBABILITIES ARE AS FOLLOWS:

LARGE EXPANSION SMALL EXPANSION

REV. INCREASE PROBABILITY REV. INCREASE PROBABILITY

GOOD $100,000 0.25 $50,000 0.25

MODERATE $ 80,000 0.60 $35,000 0.60

POOR $ 50,000 0.15 $24,000 0.15

FIND THE BEST ALTERNATIVE IF TAX RATE IS 30%, MAAR 15% USING STRAIGHT LINE DEPRECIATION.


LARGE EXPANSION:

1 2 3 4 5

REVENUE INCREASE 100,000 100,000 100,000 100,000 100,000

EXPENSE INCREASE 20,000 20,000 20,000 20,000 20,000

DEPRECIATION 28,000 28,000 28,000 28,000 28,000

TAX 15,600 15,600 15,600 15,600 15,600

NET INCOME 36,400 36,400 36,400 36,400 36,400

WOR. CAP. - - - - 20,000

CASH FLOW 64,400 64,400 64,400 64,400 84,400

PVGOOD = 64,400 [ {(1.15)4 – 1} / .15(1.15)4] + 84,400 /(1.15)5 – 160,000 = 65,852

PVMOD = 18,917

PVPOOR = – 51,486


SMALL EXPANSION:

1 2 3 4 5

REVENUE INCREASE 50,000 50,000 50,000 50,000 50,000

EXPENSE INCREASE 10,000 10,000 10,000 10,000 10,000

DEPRECIATION 12,000 12,000 12,000 12,000 12,000

TAX 8,400 8,400 8,400 8,400 8,400

NET INCOME 19,600 19,600 19,600 19,600 19,600

WOR. CAP. - - - - 10,000

CASH FLOW 31,600 31,600 31,600 31,600 41,600

PVGOOD = 31,600 [ {(1.15)4 – 1} / .15(1.15)4] + 41,600 /(1.15)5 – 70,000 = 40,915

PVMOD = 5,428 PVPOOR = – 20,101

EMVLARGE EXPANSION= (65,852)(0.25) + (18,915)(0.60) – (51,486)(0.15) = 20,089

EMVSMALL EXPANSION= (40,915)(0.25) + (5,428)(0.60 – (20,101)/0.15) = 10,471

XIII. BASIC ACCOUNTING

ACCOUNTING

FINANCE DEPARTMENTS HAVE TWO PRIMARY FUNCTIONS IN COMPANIES:

1. RECORDING, MONITORING AND CONTROLLING OF FINANCIALCONSEQUENCES OF PAST AND CURRENT OPERATIONS.

2. ACQUIRING FUNDS TO MEET CURRENT AND FUTURE NEEDS.

THE FIRST ACTIVITY IS CALLED ACCOUNTING.

ALTHOUGH IT IS NOT NECESSARY TO KNOW THE DETAILS INVOLVED IN ACCOUNTING, A KNOWLEDGE OF BASIC PRINCIPLES IS AN INVALUABLE AID TO THE ENGINEERS.

ACCOUNTING

BUSINESS TRANSACTIONS

JOURNAL

LEDGER

BALANCE SHEET INCOME STATEMENT

JOURNAL IS A BOOK IN WHICH THE ORIGINAL RECORD OF TRANSACTIONS ARE LISTED DAILY.

LEDGER IS A GROUP OF ACCOUNTS GIVING CONDENSED AND CLASSIFIED INFORMATION FROM THE JOURNAL

ACCOUNTING

BALANCE SHEET AND INCOME STATEMENT ARE TWO KEY DOCUMENTS OF FINANCIAL ACCOUNTING.

BALANCE SHEET : SHOWS THE WEALTH IN MONETORY (DOLLARS, TL) OR OTHER FORMS (BUILDING, LAND, ACCOUNTS RECEIVABLE) AVAILABLE TO THE OWNERS AND OBLIGATIONS DUE TO OWNERS AT A GIVEN TIME.

INCOME STATEMENT : SHOWS THE FLOW OF WEALTH, THAT IS RECEIPT OR DISBURSEMENT OF WEALTH OCCURING BETWEEN TWO POINTS IN TIME.

ACCOUNTING

ASSETS ARE CASH ON HAND, ACCOUNTS RECEIVABLE, INVENTORIES, INVESTMENTS, LAND, MACHINERY ETC. TO WHICH ENTITY HAS PRIMARY CLAIM. THE RIGHT HAND SIDE CONTAINS OUTSIDERS' (LENDERS + OWNERS) CLAIMS ON THE ASSETS OF THE ENTITY.

BALANCE SHEETAS OF 1 APRIL 2004

BALANCE SHEETAS OF 31 MAY 2004

INCOME STATEMENT

FOR THE PERIOD 1 APRIL - 31 MAY 2004

BALANCE SHEET :

THE FORMAT OF BALANCE SHEET CONSIST OF A T , WITH THE ASSETS ON LEFT HAND SIDE, LIABILITIES AND OWNERSHIP AT RIGHT

ASSETS LIABILITIESOWNERSHIP

ACCOUNTING

ASSETS = LIABILITIES + EQUITY

ASSETS : CURRENT ASSETS CONSISTS OF CASH OR CLAIMS ON OUTSIDERS THAT CAN BE CONVERTED TO CASH IN

LESS THAN ONE YEAR. FIXED ASSETS ARE THINGS LIKE LAND, BUILDING, MACHINERY.

LIABILITIES : CURRENT LIABILITIES ARE ACCOUNTS PAYABLE AND SHORT TERM CREDITS THAT ARE EXPECTED TO BE

DISCHARGED IN LESS THAN A YEAR. LONG TERM LIABILITIES ARE CREDITS OR BONDS NOT DUE WITHIN THE NEXT FISCAL YEAR.

EQUITY : CAPITAL, RETAINED INCOME.

ACCOUNTING

ASSUME COMPANY X FORMED IN JANUARY 1, 2004 WITH 5 MILLION DOLLAR PAID CAPITAL.

CASH : . . . . . . . . $ 5.000.000

TOTAL ASSETS : $ 5.000.000

BALANCE SHEET, COMPANY XAS OF JANUARY 1, 2004

CAPITAL : . . . . . . . . . $ 5.000.000

TOTAL EQUITIES . . . $ 5.000.000

ASSUME FOLLOWING HAPPENED IN JANUARY FOR COMPANY X :

1. A BUILDING WAS BOUGHT FOR $ 7 MILLION, $ 2 MILLION BY CASH, $ 5 MILLION BY BANK CREDIT.

2. EQUIPMENT WORTH $ 300.000 ACQUIRED BY CASH.

3. RAW MATERIAL WORTH $ 200.000 BOUGHT BY CREDIT.

ACCOUNTING

CASH : $ 2.700.000INVENTORIES : 200.000

CURRENT ASSETS : $ 2.900.000

EQUIPMENT : 300.000BUILDING : 7.000.000

FIXED ASSETS : $ 7.300.000


ACCOUNT PAYABLE : $ 200.000CREDIT : 5.000.000

LIABILITES : $ 5.200.000

CAPITAL : 5.000.000

EQUITIES : $ 5.000.000

TOTAL LIABILITES : $ 10.200.000AND EQUITIES

BALANCE SHEET, COMPANY XAS OF JANUARY 31, 2004

ACCOUNTING

INCOME STATEMENT : TRANSACTIONS (REVENUES AND EXPENSES) IN A COMPANY DURING A PERIOD. IT IS ALSO CALLED ‘PROFIT AND LOSS STATEMENT’.

REVENUE: CAN BE FROM THE SALE OF PRODUCTS OR SERVICES, OR FROM OTHER SOURCES. OTHER SOURCES INCLUDE THE SALE OF OLD, USED EQUIPMENT, THE SALE OF SCRAP, PROFIT FROM OTHER FIRMS AND INTEREST GAINED.

EXPENSES: BOTH CASH (PURCHASES OF RAW MATERIAL, LABOR COSTS, ADMINISTRATIVE COSTS ETC.) AND NONCASH (DEPRECIATION) EXPENSES ARE SHOWN.

TAX: INCOME TAX, CERTAIN PERCENT OF PROFIT

DIVIDENDS: PAYMENT TO THE SHAREHOLDERS

RETAINED INCOME: STAYS IN THE COMPANY AS A FUND

ACCOUNTING

INCOME STATEMENT

REVENUE …………………….A

SALES, OTHER INCOME

EXPENSES ……………………B

OPERATING EXPENSES, ADM.COSTS, DEPRECIATION

INCOME BEFORE TAX………C = (A-B)

TAX………………………………D

NET INCOME OR NET PROFIT AFTER TAX…E = (C-D)

DIVIDENDS……………………..F (CERTAIN % OF E)

RETAINED INCOME…………..E-F

ACCOUNTING

ASSUME FOLLOWING HAPPENED IN COMPANY X IN FEBRUARY: 1. RECEIPT FROM SALES : $ 380.000 2. RAW MATERIAL USED : $ 114.000

3. SALARIES PAID : $ 208.000 4. INTEREST ON MORTGAGE : $ 40.000

INCOME STATEMENT (1-28 FEBRUARY 2004)

REVENUE :SALES …………………………………………………………………………………………... $ 380.000

EXPENSES :COST OF GOODS SOLD : $ 114.000SALARIES : 208.000INTEREST : 40.000

$ 362.000

INCOME $ 18.000

ACCOUNTINGSO CASH IS : 2.700.000 + [ 380.000 - ( 208.000 + 40.000 ) ] = $ 2.832.000

INVENTORIES IS : 200.000 - 114.000 = $ 86.000NEW BALANCE SHEET IS :

BALANCE SHEET, COMPANY XAS OF FEBRUARY 28, 2004

CASH : $ 2.832.000INVENTORIES : 86.000

CURRENT ASSETS : $ 2.918.000

EQUIPMENT : 300.000BUILDING : 7.000.000

FIXED ASSETS : $ 7.300.000


ACCOUNTS PAYABLE : $ 200.000CREDIT : 5.000.000

LIABILITES : $ 5.200.000

CAPITAL : 5.000.000RETAINED INCOME : 18.000

EQUITIES : $ 5.018.000

TOTAL LIAB.+ EQU. : $ 10.218.000

FINANCIAL RATIOS

SOME OF THE RELATIONSHIPS FROM THE BALANCE SHEET AND INCOME STATEMENT WERE JUDGED TO HAVE PARTICULAR SIGNIFICANCE AS INDICATORS OF FINANCIAL HEALTH AND GOOD MANAGEMENT. THEY ARE KNOWN AS FINANCIAL RATIOS.

MOSTLY USED RATIOS ARE AS FOLLOWS :

LIQUIDITY RATIOS :

CURRENT ASSETSCURRENT RATIO :

CURRENT LIABILITES

CURRENT ASSETS - INVENTORIESQUICK RATIO (ACID TEST) :

CURRENT LIABILITIES

FINANCIAL RATIOS PROFITABILITY RATIO

NET PROFITON SALES :

SALES

PROFIT AFTER TAXESON EQUITY :

EQUITY

ACTIVITY RATIOSSALES

ASSETS TURNOVER :TOTAL ASSETS

SALESINVENTORY TURNOVER :

INVENTORY

LEVERAGE RATIOSTOTAL DEBT

DEBT TO TOTAL ASSETS :TOTAL ASSETS

FINANCIAL MANAGEMENT

SHORT TERM FINANCING :

- TRADE CREDITS : THE USE OF MATERIALS WITHOUT IMMEDIATELY PAYING FOR THEM

- NOTES : WRITTEN PROMISE TO PAY LATER

- BANK CREDITS : BORROWING FROM BANKS AND PAYING AN INTEREST

- FACTORING : TO SELL THE ACCOUNTS RECEIVABLE AT A DISCOUNT IN EXCHANGE FOR IMMEDIATE MONEY


LONG TERM FINANCING :

- LONG TERM BANK CREDITS

- STOCKS : SELLING STOCKS. OWNERS GET SHARE FROM PROFIT

- BONDS : YOU SELL BONDS AND AGREE TO PAY A CHARGE ANNUALLY OR BUY WITH A HIGHER PRICE AT THE MATURITY DATE

- PROFIT INVESTED IN CURRENT ASSETS

- SALES OF FIXED ASSETS


BORROWED FUNDS: COMPANIES MAY REQUIRE OUTSIDE FUNDS FOR THEIR INVESTMENTS OR PRODUCTION EXPENSES. THEY CAN SELL BONDS OR GET CREDITS FROM BANKS OR OTHER ORGANIZATIONS. THEY PAY INTEREST FOR THESE LOANS.

THE BORROWED FUNDS ARE NOT CONSIDERED AS INCOME (SO ARE NOT TAXED); WHEN THEY ARE REPAID, THIS IS NOT CONSIDERED AS AN EXPENSE (NOT INCLUDED IN YOUR COSTS). BUT THE INTERESTS PAID ARE CONSIDERED AS EXPENSE. SO THE REAL COST OF BORROWING FUNDS CAN BE FOUND BY DEDUCTING THE TAX AMOUNT.


EXAMPLE:

IF YOU PAY 30% INCOME TAX, WHAT IS THE COST OF BORROWING $1,000,000 WITH 10% INTEREST RATE?

INTEREST/ YEAR = 0.1 x 1,000,000 = $100,000 / YEAR

YOU CAN INCLUDE THIS $100,000 INTO YOUR EXPENSES, SO YOUR TAX WILL DECREASE

0.30 x 100,000 = $30,000 / YEAR

COST OF THE FUND = 100,000 – 30,000 = $70,000 / YEAR

OR 7%

FINANCIAL MANAGEMENTLOAN REPAYMENT: THE BORROWED FUNDS CAN BE REPAID BY

DIFFERENT METHODS. THE MOST COMMON TYPE IS BY CONSTANT PERIODIC PAYMENTS. EACH PAYMENT COVERS THE INTEREST DUE ( INTEREST FOR THE REMAINING PORTION OF THE PRINCIPLE) AND SOME OF THE REMAINING PRINCIPLE. THE TOTAL PERIODIC PAYMENT IS CONSTANT; BUT SINCE THE PRINCIPLE BALANCE DECREASES, THE INTEREST PORTION OF EACH PAYMENT IS SMALLER THAN THE PREVIOUS ONE AND THE PRINCIPLE PORTION OF EACH PAYMENT IS LARGER THAN THE PREVIOUS ONE.

L = Ij + pj WHERE L IS THE CONSTANT PAYMENT EACH PERIOD; Ij IS THE jth PERIOD INTEREST PAYMENT; pj IS THE jth PERIOD PRINCIPLE PAYMENT.

YOU CAN DRIVE AND USE THE FOLLOWING FORMULA TO

CALCULATE THE CONSTANT PERIODIC REPAYMENT OF LOANS

L = P0 [ 1 + i 1

N (1 + i) j-1 ] / 1

N (1 + i) j-1

WHERE P0 IS THE INITIAL AMOUNT OF LOAN.


EXAMPLE:

A LOAN OF $100,000 AT AN INTEREST RATE OF 10% PER YEAR IS MADE FOR A REPAYMENT PERIOD OF 10 YEARS. DETERMINE THE CONSTANT PAYMENT PER PERIOD FOR ANNUAL, END-OF-THE-YEAR PAYMENTS.

L = P0 [ 1 + i 1

N (1 + i) j-1 ] / 1

N (1 + i) j-1

= 100,000 [ 1+ 0.1 1

N (1.1) j-1 ] / 1

N (1.1) j-1

= 100,000 [ 1 + 1.59374] / 15.9374

= $16,274.54 per year

XIV. COST CONTROL

COST CONTROLCONTROLLING THE COSTS IS ONE OF THE MOST IMPORTANT

FUNCTIONS IN ANY COMPANY. SINCE THE SELLING PRICE IS LIMITED BY THE MARKET, THE

COMPETING POWER AND THE PROFIT MARGIN OF A COMPANY DEPENDS ON THE PRODUCT COST.

INCREASING GLOBAL COMPETITION DEMANDS EFFECTIVE METHODS FOR IMPROVED OPERATIONAL PERFORMANCE AND EFFICIENCY AIMING LOWER COSTS. THE PRODUCT COSTS ARE BECOMING INCREASINGLY CRITICAL, AND IN SOME CASES, MAY MAKE THE DIFFERENCE BETWEEN SURVIVAL AND EXTINCTION.

COST CONTROL IS THE JOB OF EVERYONE IN THE ORGANIZATION AND IT CAN BE DONE EFFECTIVELLY BY GOOD ENGINEERING AND MANAGEMENT PRACTICES.

COST CONTROL SHOULD BE DONE IN TWO STEPS: 1. CONTROLLING THE COSTS TO AVOID ANY INCREASE 2. REDUCING THE COSTS

COST CONTROL

1. CONTROLLING THE COSTS TO AVOID ANY INCREASE

A MANAGEMENT INFORMATION SYSTEM (MIS) IS REQUIRED TO MONITOR ALL COST ITEMS IN THE COMPANY. COST STANDARDS FOR ALL ITEMS SHOULD BE SET UP AND CONSTANTLY UPDATED. PURCHASING AND RELATED COSTS FOR ALL MATERIAL, LABOR COSTS, RAW MATERIAL USAGE, PRODUCTION EFFICIENCY, UTILITY COSTS, ALL OVERHEAD COSTS SHOULD HAVE STANDARDS AND SHOULD BE RECORDED. THE COMPARISON OF ACTUAL COSTS WITH STANDARDS WILL GIVE YOU THE MEASUREMENT OF PERFORMANCE AND ALARM YOU WHEN GOING IN THE WRONG DIRECTION.

VARIANCES (OR VARIATIONS FROM STANDARD) AND THE COST PERFORMANCE INDEX (OR PERCENT EFFECTIVENESS; RATIO OF ACTUAL TO STANDARD) CONSTITUDE THE MOST VALUABLE TOOLS OF COST MANAGEMENT.

COST CONTROL

A GOOD MANAGEMENT PRACTICE FOR COST CONTROL IS TO DECENTRALIZE THE SYSTEM SO DEPARTMENTS HAVE THEIR OWN COST DATA AND ARE RESPONSIBLE FOR THE COST PERFORMANCE. THIS WILL BRING A CONSCIOUSNESS OF PROFIT RESPONSIBILITY AND WILL RESULT WITH THE CONTROL OF MINOR COST ELEMENTS EFFECTIVELY.

THE RULE IS ‘THE BEST PLACE TO CONTROL COSTS IS AS CLOSE TO THE SOURCE AS POSSIBLE’. SO DEFINING COST CENTERS AS SMALL AS POSSIBLE AND GIVING THE RESPONSIBILITY TO A SMALL GROUP OF MEN WILL IMPROVE THE COST CONTROL SYSTEM CONSIDERABLY.

COST CONTROL

THERE ARE 3 MAJOR TYPES OF STANDARDS THAT CAN BE USED FOR COST CONTROL SYSTEMS:

a) BUDGET STANDARD:

MOST COMMONLY USED STANDARD; BASED ON THE PAST HISTORY. DETERMINED BY MATHEMATICAL AND STATISTICAL AVERAGES

b) ENGINEERING STANDARD:

DETERMINED BY ENGINEERING STUDIES AND BENCHMARKING

c) COST REDUCTION STANDARD:

DETERMINED BY ENGINEERING STUDIES INVOLVING IMPROVEMENT OR CHANGE OF PROCESS, MATERIAL OR EQUIPMENT

COST CONTROL

2. REDUCING THE COSTS

THE PRIMARY GOAL OF ALL COMPANIES IS TO INCREASE PROFIT WHICH IS POSSIBLE BY COST REDUCTION. THERE ARE THREE MAIN METHODS USED IN INDUSTRY FOR THIS PURPOSE:

- DAILY INCREMENTAL IMPROVEMENTS

USING THE EXISTING TECHOLOGY AND FACILITIES, SMALL IMPROVEMENTS IN ALL PROCESSES CAN BE MADE TO REDUCE COSTS

- TECHNOLOGICAL CHANGES

REDUCING THE COSTS BY CHANGING THE PROCESS OR EQUIPMENT

- PRODUCT CHANGES

DESIGNING NEW PRODUCTS WHICH CAN BE PRODUCED WITH LESS COST

COST CONTROL

THERE ARE MANY POSSIBILITIES OF COST REDUCTION. BASIC PRINCIPLE IS TO INCREASE EFFICIENCY BY DECREASING THE INPUTS (MATERIAL, LABOR, OVERHEAD COSTS) WITHOUT DECREASING THE QUALITY OF YOUR PRODUCTS.

IT IS HARD TO NAME ALL POSSIBILITIES BUT SOME FREQUENTLY USED METHODS ARE:

- BUY MATERIAL CHEAPER - DECREASE THE COSTS OF PURCHASING, TRANSPORTATION - DECREASE YOUR INVENTORY (INVENTORY IS MONEY KEPT WITH

ZERO INTEREST; IDEAL IS ZERO INVENTORY) - IMPROVE YIELD SO USE LESS MATERIAL - SUBSTITUDE LESS EXPENSIVE MATERIAL - DECREASE THE LEAD-TIMES - DECREASE LOSS AND SCRAP (TRY TO WORK ZERO-DEFECT) - IMPROVE QUALITY ASSURANCE SYSTEMS, INCREASE TRAINING

COST CONTROL

- IMPROVE ALL PROCESSES - WORK WITH MINIMUM NUMBER OF PEOPLE IN ALL KIND OF JOBS - USE AUTOMATION - MINIMIZE SUPERVISION AND CONTROL ACTIVITIES - MINIMIZE MATERIAL FLOW - MINIMIZE PAPER WORK AND BUREAUCRACY - ELIMINATE UNUSED BUSINESS REPORTS AND DUPLICATIONS - MINIMIZE SET-UP PERIODS, DECREASE CYCLE TIME - INCREASE RATES OF FLOW IN PROCESSES - ANALYZE JOBS, FIND BEST METHODS - MINIMIZE UTULITY COSTS (AMOUNT AND UNIT COSTS) - CHECK AND MINIMIZE ALL OVERHEAD COSTS - MINIMIZE DISTRIBUTION COSTS - MINIMIZE BREAK-DOWNS BY EFFECTIVE MAINTENANCE - DO ‘MAKE-OR-BUY’ ANALYSIS REGULARLY - LOOK FOR IDLE TIMES OF PEOPLE AND FILL THEM

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