Financial Analysis

Accounting & Financial management

Session Financial AnalysisProf. Dr. R.A Khan

IntroductionFinancial Analysis deals with the problems of investing money or capital.

Money has nominal value and real value

Nominal value: Legal value assigned to a particular unit.

Real value: It is defined as the purchasing power and with the passage of time real value reduce.

The Time therefore is an important element in investment decision

Interest Factors for Discrete Compounding, Discrete Cash Flow

Single Payment compound amount

(F/P, i%, N) = (1+i)nSingle payment present worth (Discount)

(P/F, i%, N) = 1/(1+i)nSinking fund

(A/F, i%, N) = i / [(1+i)n 1]Uniform series compounded amount

(F/A, i%, N) = [(1+i)n 1] / i

Interest Factors for Discrete Compounding, Discrete Cash FlowAnnuity factor

(P/A, i%, N) = [(1+i)n-1] / [i(1+i)n]Capital recovery factor

(A/P, i%, N) = [i(1+i)n] / [(1+i)n -1]Uniform gradient P.W factor

(P/G,i%, N) = (1/i) [ {(1+i)n -1/i(1+i)n} {n/(1+i)n}]Uniform gradient annual series

(A/G, i%, N) = (1/i)- [n / (1+i)n 1]

Criteria of Financial Analysis

Net Present ValueInternal Rate of ReturnBenefit Cost Ratio

Example - 1

A person buys a small piece of land for $5000 down and annual payment of $500 for next 6 years from now. What is the present worth of the investment if the interest rate is 8% per year?

Example - 2

A person buys a small piece of land for $5000 down and deferred annual payments of $500 for 6 years starting 3 years from now. What is the present worth of the investment if the interest rate is 8% per year?

Solution

PW = 5000 + 500 (P/A, 8%, 6) (P/F, 8%,2)

= $6981.6

ExamplePakistan Railway is considering a program of computerizing level crossings by installing automatic barriers. They estimate that the cost of computerizing such level crossing is expected to be Rs.5 millions, resulting in cost saving 0.4 million per year. If the life of the system is 15 years and rate of interest is 8% per year, determine the net present value of the project in zero year.

ExerciseMixer machines are manufactured in a factory, if no improvements are made in the factory, the manufacturing cost of machine is $ 2000 from 1 to 10 months, $ 1600 in moth 11th and $1800 in 12th month. If some improvements could be made in the factory to make it more suited for larger production, these improvements would reduce the manufacturing cost of mixer machine to $1500 for first 10 months, 1400 in 11th month and $1650 in 12th month. The improvement cost is expected to be $ 2500 in year zero.Calculate NPV of the project in year zero, if market rate of interest is 2% per month

ExampleA University is considering installing electric valves with automatic timers on some of their sprinkler systems. They estimate that they need 45 valves and timers at a cost of $85 per set. The initial installation cost is expected to be $2000. At the present time there are four employees who are incharg of maintaining lawns, each of whom earns $12000 per year, spend 25% of their time in watering. The present cost of water is $2200 / year. If the automatic system is installed the manpower cost for watering could be reduced by 80% and water charges by 35%. However extra maintenance cost on the automatic system is expected to be $450/ year. If the automatic system is expected to last for 8 years, which system should be used on the basis of present worth analysis. Assume interest rate is 16% per year.

ExerciseA plant superintendent is trying to decide between two excavating machines with the estimates presented below.

Determine which machine should be selected on the basis of PWC analysis, using an interest rate of 15% per annum.Use EUAC analysis

DescriptionMachine- AMachine -BFirst cost $1100018000Annual O&M cost $35003100Salvage value $10002000Life, years69

ExerciseAn investment company is considering building a 25 units apartment complex in a growing town. Because of the long term growth potential of the town, it is felt that the company could average 90% of full occupancy for the complex each year. If the following items are reasonably accurate estimates, what is the minimum monthly rent that should be charged if a 12% MARR/ year is desired? Use the annual worth method. Land investment $ 50,000Building investment cost $ 225,000Study period 25yearsUpkeep expense per month $ 35Property tax and insurance 10% of total initial investment

PW of alternative evaluationMake a present worth comparison of the equal service machines for which the costs are shown in table below, if i = 10% per year.

DescriptionType AType BFirst cost ($)25003500Annual operating cost ($)900700Salvage value200350Life years55

Gradient Factor

Uniform increase or decrease in the amount is known as gradient amount. It is designated by letter G. Gradient factor is used for converting gradient amount into present worth or in annual worth

% GradientAmount increase or decrease by constant %.

Amount {(1+ r / 1+ i)n - 1} / ( r i)}Where

r = % change in amount i = rate of interest n = Total period of change

Example

Calculate equivalent present worth of $35000 now and annual series of $7000 per year for 5 years beginning 1 year from now, which starts to increase annually at 12% thereafter for the next 8 years. Use interest rate 15% per year.

Solution

PW = 35000 + 7000 (P/A,15%, 4) + [7000(1.12/1.15)9 1/ (0.12 - 0.15)](P/F,15%,4)

PW = $83232

Capitalized Cost

Capitalized cost refer to the present worth value of a project that is assumed last forever or permanent or perpetual life project.

Capitalized Cost Calculation

Cash-flow diagramFind PW of nonrecurring amountsFind EUAW of all recurring amountsEUAW / i% to get the capitalized costAdd value obtained in step 2 to the value obtained in step 4

Example

Calculate the capitalized cost of a project that has an initial cost of $150,000 and an additional investment cost of $ 50,000 after 10 years. The annual operating cost will be $5000 for the first four years and $8000 thereafter. In addition there is expected to be major rework cost of $15000 every 13 years. Assume interest rate is 5% per year.

ExampleCDGK has estimated the first cost of new amusement park to be $ 40,000. They expect to improve the park by adding new rides every year for the next 5 years at a cost of $6000 per year. Annual operating cost are expected to be $12000the first year: These will increase by $2000per year until year 5,after that it will remain same. CDGK expect to receive $11000 in income the first year, 14000 the second and amounts increasing by $3000 per year until year 8 after which the income will remain constant. Calculate the capitalized cost of the park if the interest rate is 6% per year.

ExerciseA businessman purchased a building and

insulated the ceiling with 6 inches of foam. This cut the heating bill by $25/ month and the air conditioning cost by $20/ month. Assuming that the winter season is the first 6 months of the year and the summer season is the next 6 months, what was the equivalent amount of his savings after the first 3 years at an interest rate of 1% per month

ExerciseA company purchased a machine for $18000. Its annual maintenance and operation cost was $ 2700. After 4 years from the initial purchase, the company decided to purchase an additional unit for the machine which would make it fully automatic. The additional unit had a first cost of $9100. The cost for operating the machine in fully automatic condition was $1200/ yr. If the company used the machine for 13 years with no salvage value, what was its EUAC at interest rate of 9% per year.

Internal Rate of Return

IRR is the rate of interest that equate present worth of benefit to present worth of cost. The critical value of interest rate at which NPV = 0

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Trial & Error Method for ROR

IRR = A% + [a / (a-b)] (B% - A%)

whereA% : low interest rateB% : High interest ratea : Positive NPVb : Negative NPV

Example

If $5000 is invested now in common stock that is expected to yield $100 per year for 10 years and $7000 at the end of10 years what is the rate of return.

SolutionAssume i= 5%NPV at 5% rate of interest = $69.46Assume i= 6%NPV at 6% rate of interest = -$355.19

IRR = 5% + [ 69.46/ (69.46 +355.19)](6% - 5%)IRR = 5.16%

Amortized Loan

If a loan is to be repaid in equal periodic amounts, it is said to be an amortized loan.

ExampleA firm borrows $10000, from a bank at 6% per year rate of interest on balance amount that is outstanding at the beginning of each year. The loan is to be repaid in three equal payment at the end of each year. Determine the amount of installment and develop loan amortization table.

Solution

Installment Amount: 10000 (A/ P, 6%, 3) = $ 3741.1

Amortization Table

YearBeginning AmountInst.AmountInterest AmountPaid PrincipalBalance Amount1$10000$3741.1$600$3141.1$6850.926850.93741.1411.53329.63529.333529.33741.1211.83529.30.00

Exercise

Set up an amortizing schedule for a $25000 loan to be repaid in equal installments at the end of each of the next five years, assume interest rate is 10% per year.

Benefit Cost Ratio

It is the ratio of equivalent worth of benefit to equivalent worth of cost.B/C = AWB / AWC > 1 Desirable

Modified Approach:B/C = [AB ADB O&M] / CR

ExampleA small flood control dam is expected to have an initial cost of $2.8 million and an annual upkeep cost of $20000. In addition minor reconstruction will be required every five years at a cost of 100,000. As a result of the dam flood damage will be reduced by an average of $120,000 per year Using an interest rate 7% to determine B/C of the dam. Assume the dam is expected to last 25 years.

(b) Assume the dam will be permanent.

ExampleTwo independent sites are under consideration for a business. Which site should be selected on the basis of B/C analysis using an interest rate 12 % per year. Assume that the financial details are as follows

Site A (000)Site B (000)Initial CostRs.100,000Rs.150,000Annual maintenanceRs.800Rs.1000Annual IncomeRs.25000Rs.60000Salvage valueRs. 20000Rs.30000Life, years2025

Mutually Exclusive AlternativesThe steps for the incremental B/C analysis are summarized below.

Alternatives arrange in ascending order according to their annual costCompare lower cost alternative to higher cost alternative only if that lower cost alternative is justifiedChoose that alternative which requires heavy cost for which funds are available and for which incremental B/C is justified

ExampleFour different building locations have been suggested of which only one will be selected. Cost and annual cash flow information are given in table below. If the MARR is 10%, use B/C analysis to select the most economically best location.

LocationA (000)B (000)C (000)D (000)Buildg Cost ($)200275190350Annual cash-inflow 223519.542Life, years30303030

ExerciseThe Corps of Engineer intendsto construct an earthen damon the river Indus. Six differentsites are suggested, and theenvironmental impact havebeen approved.The construction cost andannual benefits are tabulated below.If a MARR of 6% per year for publicprojects is used and dam life is longenough to be considered infinite foranalysis purposes, select the bestlocation from an economicperspective.

siteCost,$(miln)A. Inc ($)A6350000B8420000C3125000D10400000E5350000F11700000

ExampleTwo road routes are under consideration for a new inter state highway segment. The northerly route N would be located about 5 km from central business district and would require longer travel distance and time by local commuter traffic. The southerly route S would pass directly through the down town area , and although its construction cost would be higher, it would reduce the travel time and distance for local commuters. Assume that the costs for the routes are as follows..

Cont..

If the roads are assumed to last 30 years with no salvage value, which route should be accepted on the basis of B/C analysis using an interest rate of 5% per year?

Route N, $(000)RouteS,$(000)Initial cost10,00015,000Maintenance cost per year3555Road user cost per year450200

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