Basic Methods for Making Econmy Studies

Chapter 5By:

Gabrielle Plasos

Neil Z. Salvaloza

Alexis Atupan

Basic Methods for Making Economy Studies

By: Gabrielle Plasos

Objectives:

To demonstrate the mechanics of calculations for six basic methods for making economy studies.

To briefly describe the underlying assumptions and interrelationships of those methods

Six different basic methods for economy studies

Annual Worth (A.W.) Present Worth (P.W.) Future Worth (F.W.) Internal Rate of Return (I.R.R.) External Rate of Return (E.R.R.) Explicit Reinvestment Rate of Return

(E.R.R.R.)

Six different basic methods for economy studies

The 1st three methods convert all financial happenings into equivalent worth at some point in time using an interest rate equal to the cost of capital, or the minimum attractive rate of return (M.A.R.R.).

The last three methods are different ways to calculate a rate of profit or savings (annual return as a percent investment) so that, in turn, be compared against the M.A.R.R.

The annual worth method

- the term annual worth means a uniform annual series of net cash flows for certain period of time that is equivalent in amount to particular schedule of cash inflows (receipts or savings) and cash outflows (disbursements or opportunity costs) under consideration.

- If net annual worth ≥ 0 the project is economically justified; otherwise, it is not.

- If disbursements only are considered, the criterion is usually expressed as annual worth-cost (A.W.-C.), or annual cost (A.C.)

Calculation of capital recovery cost

Capital recovery cost (C.R.) – the equivalent uniform annual cost of the capital invested for a project. It covers two items:

Depreciation (loss of value in the asset). Interest (minimum required profit) on invested

capital.


Example: Consider a machine or other asset that will cost $10,000, last 5

years, and have a salvage value of $2,000. Further, the interest on invented capital, i, is 10%.

There are several formulas by which capital recovery cost may be calculated. One formula is just to subtract the annual equivalent of the investment minus the annual equivalent value of the salvage value.

C.R. = P(A/P, i %, N) – F(A/F, i %, N)Where P = investment at the beginning of life

F = salvage value at the end of life N = life of project

C.R. = $10,000(A/P, 10%, 5) - $2,000(A/F, 10%, 5) = $10,000(0.2638) – $2,000(0.1638) = $2,310

Table 5-1Calculation of equivalent cost of interest and capital recovery cost assuming straight line depreciation

Year

Value of investment at the beginning of the year

Interest on Beginning-of-Year investment at 10% Present worth of interest at 10%

1 $10,000 $1,000 $1,000(P/F, 10%, 1) = $ 909

2 8,400 840 840(P/F, 10%, 2) = 694

3 6,800 680 680(P/F, 10%, 3) = 511

4 5,200 520 520(P/F, 10%, 4) = 355

5 3,600 360 360(P/F, 10%, 5) = 224

Total $2,693

Annual equivalent interest = $2,693(A/P, 10%, 5) = $710

Total C.R. cost = annual depreciation + annual equivalent interest= ($10,000 - $2,000)/5 + $710 = $2,310


Another way is to add the annual sinking fund depreciation charge (deposit) to the interest on original investment (minimum required profit).C.R. = (P – F)(A/F, i%, N) + P(i%)C.R. = ($10,000 - $2,000)(A/F, 10%, 5) + $10,000(10%)

= $8,000(0.1638) + $10,000(0.10) = $2310Another way again is:C.R. = (P – F)(A/P, i%, N) F(i%)

= $8,000(0.2638) + $2,000(0.10) = $2,310Lastly, is this approximation formula called the straight line depreciation plus average profit method.C.R. ≈ (P – F)/N + (P – F)[(N + 1)/2N](i) + F(i)

≈ ($10,000 - $2,000)/5 + ($10,000 - $2,000)(0.6)(10%) + $2,000(10%)

≈ $2480


Example 5-1An investment of $10,000 can be made in a project that will produce a uniform annual revenue of $5,310 for 5 years and then have a salvage value of $2,000. Annual disbursements will be $3,000 each year for operation and maintenance costs. The company is willing to accept any project that will earn 10% or more, before income taxes, on all invested capital. Show whether this is a desirable investment using the annual worth method.


Annual Worth

Annual Revenue $5,310 Annual Disbursements

-$3,000

C.R. cost -$2,310 Total -$5,310Net A.W. $ 0

Solution:

Present worth method

the present worth method for economy is based on the concept of equivalent worth of all cash flows relative to some base or beginning point in time called the present. All cash inflows and outflows are discounted back at an interest rate that is generally the M.A.R.R.

If the net present worth ≥0, the project is economically justified.

If only outflows (disbursements) are considered, the method is characterized by negative-valued present worth amounts expressed as present worth-cost (P.W.-C.).

Present worth method

Example 5-2

Considering the same project as in example 5-1, show whether it is justified using the P.W. method.

Present worth

Annual revenue: $5,310(P/A, 10%, 5 $20,000

Salvage value: $2,000(P/F, 10%, 5) $ 1,245

Investment -$10,000

Annual disbursements: $3,000(P/A, 10%, 5)

-$11,370

Total -$21,370

Net P.W. $ 0

Four different methods for economy studies

By: Neil Z. Salvaloza

Four different methods for economy studies

The future worth method(FW)The internal rate of return method

(I.R.R.)The external rate of return

method(ERR)Explicit reinvestments rate of return

method(ERRR)

The future worth method (F.W.)

It is just exactly comparable to the present worth method except that all cash inflows and outflows are compounded forward to a reference point in time called the future


Example – An investment of 10,000 can be made in a project

that will produce a uniform annual revenue of 5,310 for 5 years and then have a salvage value of 2,000. annual disbursements will be 3,000 each year for operation and maintenance costs. The company is willing to accept any project that will earn 10% or more, before income taxes, on all invested capital.


Solution future

worth

Annual revenue: 5310(F/A, 10%, 5) 32,420

salvage value 2,000

Investment : 10,000(F/P, 10%, 5) -16,105

Annual disbursements: 3,000(F/A, 10%, 5) -18,315

total -34,420

NET F.W. 0

The internal rate of return method (I.R.R.)

the rate of return promised by an investment project over its useful life.

The most general and widely used rate of method– Investors method– Discounted cash flow method– Receipts versus disbursements method– Profitability index


N N

∑Rk(P/F,i%,k) = ∑Dk(P/F,i%,k) k=0 k=0

Rk = net receipts for kth year

Dk = net disbursements for kth year

N = project life for a maximum number of years for study


Using the example above:

I’ = 5%: -10,000 + 2,3140(4.3295) + 2,000(.7835) = + 1,568

I’ = 25%: -10,000 + 2,3140(2.6893) + 2,000(.3277) = - 3,132

(25%-5%)/(1,568-(-3,132)) = (I’% - 5%)/(1,568-0)

= 5% + 6.7%

= 11.7%

Selecting trial rates of return when using the I.R.R. method

To reduce the number of trials required to obtain an acceptably accurate answer

Cash inflow:annual receipts: 5,310 x 5 26,550

salvage 2,000

total28,550

Selecting trial rates of return when using the I.R.R. method

Cash outflow:Annual disbursements: 3,000 x 5 - 15,000

Investment - 10,000

Net cash inflow (profit) 13,550

Average profit per year = 13,550/5=710

Average investment = (10,000+2,000)/2 =6,000

Average profit per year/ Average investment =

710/ 6,000= 11.8%

The external rate of return method (E.R.R.)

Involves the assumption that all recovered funds can be reinvested at some specified rate of return

– Two advantages over I.R.R Can be solved directly rather than by trial and error Not subject to the possibility of multiple rates of return


N N

∑Dk(P/F,e%,k)(F/P,i’%,N) = ∑Rk(F/P,e%,N-k)k=0 k=0

Dk= net outflow(excess of disbursements over receipts) for kth year

Rk = net inflow (excess of receipts over disbursements) for kth year

N = project life for maximum number of years for studye = external reinvestment rate


– Consider the same example

M.A.R.R. = 10%

10,000(F/P,i’%,5) = (5,310-3,000)(F/a,10’%,5) + 2,000

= 2,310(6.105) + 2,000

(F/P,i’%,5) = 1.61

Explicit reinvestments rate of return method (E.R.R.R)

Is a computationally easy means of calculating a rate of return when there is a single lump sum investment and uniform cash savings or returns at the end of each period throughout the life , N, of the investment project


E.R.R.R. = (R-D-P[P-F][A/F,e,N])/P

– R = uniform annual receipts or savings– D = uniform annual costs or disbursements– P = investment– F = salvage value– e = reinvestment rate


M.A.R.R. = 10%– Solution

annual revenue 5,310

annual disbursements -3,000

depreciation: (10,000-2000)(A/F,10%,5) -1,310

TOTAL -4,310

Net annual profit 1,000

ERRR = (net annual profit)/(investment)

=1,000/10,000 = 10%

Summary comparison of economy study methods

You may wonder why there are six different methods for economy studies have been presented, when any one of these methods will give a valid answer for the reinvestment assumption in that method.– The best answer is that it preferences differ

between analyst and decision makers and also that often one method is easier to use than the other method because of the particular cash flow patterns

An undesirable economy study method

By: Alexis Atupan

An undesirable economy study method

=) used as supplemental information in conjunction with a correct economy study method

Payout period method =) also called payback period method =)determines the number of year that will have to elapse in order for the

invested capitral to be recovered out of the net incoming cash flow(determination must be made on an after-cash basis)

=) does not consider possible earnings from reinvested capital that is recovered during the payout period

=) does not take into consideration the economic life of the physical asset

*if revenue and cost are uniform each year, the determination form is: Number of years for payment = (P-F)/(R-D) *if the salvage value is ignored, the formula is Number of years for payment = P/(R-D) Where: R-D represents the net annual cash flow

Economy study of a new venture using various methods

Example 5-7 involving a single, initial investment, and uniform revenue and cost data

Mr. Brown has an opportunity to purchase an apartment house that has just been completed. It is located in the suburban area in a medium-size city that contains a fair amount of industrial plants. It is also within a walking distance of a quite large university. The apartment house is in the process of being rented and now is more than 80% occupied, with prospect of being fully rented within a few weeks. The purchase price would be $85,000, of which $10,000 represents the value of the land. The building is consists of 10 four-room apartments, a small apartment for the caretaker, and garage space for 11 automobiles. Form a study of similar apartment buildings, Mr. Brown estimates that each apartment can be rented for $110 per month, with at least 95% occupancy at all times. Heat and water are included in the rental. The operating costs are estimated to be as follows:


Caretaker $175 per month plus his apartment

Fuel $400 per year Water $150 per year Maintenance and repair equal to 1 month’s rental on each

rental unit per year Taxes $4 per $100 of assessed value ;

assessed value will be approximately 30% of costs of building and land

Insurance 0.5% of first cost building, per year

Agent’s commission 2 ½ of gross rental revenue


At present Mr. Brown's capital is invested in bonds that yield approximately 5% before income taxes and he feels that the project of this should earn at least 7% before income taxes. E estimates that the economic life of the apartment house will be at least 40 years and that the $10,000 for the land will be the only salvage value at the end of that time..


The revenue each year can be computed as 10 x $110 x 12 x 0.95 = $12,540. The out pocket costs (disbursement) each year can be computed as: Caretaker : $175 x 12 $2,100 Fuel 400 Water 150 Maintenance and repair:10 x $110 1,100 Taxes; ($85,000/$100)x 0.3 x $4 1,020 Insurance: $75,000 x 0.005 375 agent's commission: $12,550 x 0.025 314

total $5,495 hence the net cash inflow (revenue minus disbursement) each year is $12,540 -

$5,495 = $7,081.


Show whether Mr. Brown should purchase the apartment house using the I.R.R. Method.

Solution:

the present worth can be written

-$85,000 + $7,081 (P/A,i', 40) + $10,000 (P/F,i', 40) =0

at i' = 5%: - $85,000 + $7,081 (17.1591) + $10,000 (0.1420) ~= $38,000

at i' = 10%: - $85,000 + $7,081 (9.7791) + $10,000 (0.0221)~= -$15,500

Since we are seeking the I' at which the present worth equation is zero, by linear interpolation we find that

i'= I.R.R. ~= 5% + [$38,000/($38,000 + $15,500)](10% - 5%) ~= 8.6%

since 8.6% is greater than 7% the project is apparently worthy of investment.


Example 5-8

Analyze the same investment project as in Example 5-7 using E.R.R.R. Method. Assume that he expects to reinvest accumulated depreciation funds in bonds earning 5%.

Solution:

Annual revenue $12,550

Annual costs:

Out - of – pocket $5,495

Depreciation:

$75,000(A/F,5%,40)= $75,000(0.0083) 622

total $ 6,081

Net annual profit $ 6,469

E.R.R.R. = $6,469 / $85,000 = 7.6%

Since 7.6% is greater than 7% he would probably invest the apartment house.


Example 5-9

Analyze the same investment project as in Ex. 5-7 using the annual worth method.

Solution:

Annual revenue $12,550

Annual costs:

Out -of – pocket $5,495

Depreciation:

$75,000(A/F,7%,40)= $75,000(0.0050) 375

Minimum required profit :$85,000(0.07) 5,950

total $11,784

Since the annual revenue of $12,550 exceeds the total annual costs of $11,784, the investment would be justified.


Example 5-10

Analyze the same investment project as in Ex. 5-7 using the present worth method.

Solution:

P.W. Of inflow:

Revenue: $12,550(p/a,7%,40) = $12,550(13.3317) $167,500

Salvage of land: $10,000(P/F,7%,40) = $10,000(0.0668) 668

total$168,168

P.W. Of outflow:

Investment $85,000

Out – of – pocket : $5,495(P/A,7%,40) 72,700

total$157,700

Since $168,168 > $157,700, the project is once again shown to be worthy of investment.

Discussion of decision criteria to supplement economy study or studies

In order to arrive at a decision on the project for which economy studies were made in Ex.'s 5-7 through 5-10 Mr. Brown would have to consider a number of factors, first item undoubtedly should br the revenue estimate. Some question to be answered:

>Is the assumed rental rate reasonable, particularly in relationship to similar apartments?

>Is the assumed occupancy rate a reasonable one for this type of housing?

>How will possible changes in economic conditions affect the rentals, both as to price and occupancy rate?

An example with important intangibles In many economy studies the final decision is determined primarily by the intangibles that exist in the situation. The following illustrate this type of problem

Example 5-11

The management of a hotel in a city an an inland valley of California was considering the installation of an air-cooling system for all the rooms. This hotel had 150 guest rooms and was considered to be one of three first-class hotels in the city. One of the other hotel had installed such a system the previous year .A bid of $18,000 had been received for installing the system. It was estimated that the cooling system would have to be operated at full capacity for 14 weeks of each year, and at reduced capacity at 6 weeks. Operation costs at full capacity would be $17 per day and, at reduced capacity, $12 per day. The annual maintenance expense was estimated to be$125, and taxes and insurance to be $200. The life of the installation was estimated to be not less than 15 years.


If the cooling system were installed, it was estimated that 90% of the rooms would be rented during the 20 weeks of the hot weather, whereas only 80%, of the rooms could be rented if no air cooling was available. These estimates were based on results of similar hotels in the other cities. The existing average profit on each room that was rented was $2 per day. The owners had capital invested in stocks, paying about 6% before taxes, which could be used to finance the project. Determine if the investment should be made using E.R.R.R method.


Solution:

Annual income:150 x (0.9 – 0.8) x 7 x 20 $4,200

Annual expenses:

Depreciation:418,000(A/F,6%,15) $ 774

Out – of – pocket costs:

Operation:$17 x 7 x 14 1,666

$12 x 7 x 6 504

Maintenance 125

Taxes and insurance 200

total $3,260

Profit $ 931

E.R.R.R = $931 / $18,000 = 5.5%

An example of a proposed investment to reduce costs Example 5-12

A manufacturer of jewelry is contemplating the installation of a system that will recover a large portion of the fine particles of gold and platinum that result from the various manufacturing operations. At the present time a little over $5,000 worth of these metals is being lost per year, and it is anticipated that, because of the growth of the company, this amount will increase by $500 each year for the next ten years. The proposed system, involving a network of exhaust ducts and separators, will recover at least two-thirds of the gold and platinum that otherwise would be lost. The complete installation would cost $14,000.The best estimates for the operating cost of the system. obtained from operations of similar systems, are $1,000 per year for operating expense,$180 per year for maintenance and repairs, and 2% of the first cost annually for taxes and insurance. The company would require the investment to be written off within 10 years. The average earnings of the company, before taxes, have been about 15%.Should the recovery system be installed?

An example of a proposed investment to reduce costs

Solution :

Such an investment is made to reduce some of the operating expenses, in this case the cost of the material used. Thus the saving (income) to be obtained by making an investment is almost entirely within the control of the investors. The company knows exactly what expenses have been. If the efficiency of the proposed equipment is known, the only factors that should affect the saving are the variation of production, operation, and maintenance expenses of the proposed equipment and depreciation expense. In most cases of this type these items are known or may be predicted quite accurately. The company would have a good idea of how its volume would vary. Operation and maintenance expenses can usually be estimated accurately, especially if historical data are available on the proposed equipment. Depreciation cost can be placed on the safe side by using a write-off period shorter than actual physical life.

Investment where income is unknown

Decisions to invest or not to invest capital often must be made when it is impossible to know or evaluate the return. In some cases it is not particularly necessary to know what the income will be. These often occurs when public or governmental improvements are made. The returns are often in a non-monetary form, yielding convenience and satisfaction to the public. When companies spend large sums of money to create customer or employee goodwill, a precise measurement of the return is impossible. Yet such investments must often be made.

Investment where income is unknown

Probably the only rule that can be established for such cases is that generally no more should be invested than is required to give a satisfactory result. Obviously, this rule cannot be followed rigidly. When public roads are built, for example, it may be advisable to spend more than a bare minimum to assure permanence and low maintenance cost. I such cases the rule rule of least annual cost becomes important. The income (or at least a large part of it) resulting from many projects that at first thought do not appear subject to the type of analysis under discussion can be measured if the undertakings are analyzed fully. Whenever this can be done, it should not be neglected. Greater efficiency in the use of capital is bound to result.

An example of an investment to reduce risk

There are cases where we consider making an investment in order to reduce, or eliminate, future costs that may result from various types of risk, usually associated with such events as fires or floods. The basic economic aspects can be illustrated by a simple example. Assume that a man agrees that once each year, for the next 20 years, he will cut a deck of 52 playing cards and that if he cuts an ace he will pay $500 to a second individual. How much could he afford to pay, in one immediate sum,to avoid this obligation and risk, assuming that capital is worth 6% to him?


With four aces in a deck of 52 cards it is apparent that, on the average, an ace will be cut four times out of 52 attempts. In other words , it is probable that an ace will be cut in 1/13 of the attempts. Therefore the expected cost to him in any one year is 1/13 * $500 = $38.46.

With money worth 6%, the present worth of 20 annual payments of $38.46 would be $38.46 * (P/A, 6%, 20) = $38.46 * 11.4699 = $441.13.

Thus, if the probability of occurrence of an event as well as the cost of that event when it occurs is known, a single or annual amount that we can afford to pay to avoid the liability resulting from its occurrence can be calculated.


The foregoing type of calculation forms the basis for all insurance. The annual premium that is paid for insurance represents the average annual loss that might occur, except that it also must include an extra charge to pay the overhead and profit costs of the insurance company.

An application of an economy study aspects of such a situation is illustrated in the following problem.

EXAMPLE 5-13

A company has a warehouse that has no automatic sprinkler system. Annual fire insurance premiums have been $1,200. If a sprinkler system costing $8,000 is installed, the insurance premium will be

reduced by 50%. The system is estimated to have a life of 30 years with the annual costs for taxes and maintenance amounting to $75. Capital is worth 8%.

The company engineer estimates that the actual loss resulting from any fire, including such factors as lost orders and personnel

required to arrange for repairs, would be at least 50% greater than the physical loss covered by insurance. It is assumed that the reduction in

insurance premium is a good measure of the difference in the probable annual physical loss without and with a sprinkler system.

EXAMPLE 5-13

Solution:

With the premium reduction being equal to $600 per year, the probable saving in actual loss due to fire is estimated to be $600 * 1.5= $900. The costs would be

Capital recovery:$8,000(A/P, 8%, 30)=$8,000(0.0888) $710

Taxes and maintenance 75

TOTAL ANNUAL COST $785

Thus the probable annual savings of $900 exceeds the annual cost of $785, so from a long range viewpoint, the sprinkler system would be a good investment.

Basic Methods for Making Economy Studies

QUESTIONS

Basic Methods for Making Econmy Studies

Documents

annual worthcost

annual worth method

cost of capital

term annual worth

economy studies annual

f net annual worth

equivalent uniform annual

annual equivalent value