TIME VALUE OF MONEY

SUMS and WORTHS

PRESENT (P) FUTURE (F) EQUAL-PAYMENT (A)

PRESENT SUM'S FUTURE WORTH P-->F              
FUTURE SUM'S PRESENT WORTH F-->P              
EQUAL-PAYMENT SERIES SUM'S FUTURE WORTH A-->F
FUTURE SUM'S EQUAL-PAYMENT SERIES WORTH F-->A
EQUAL-PAYMENT SERIES SUM'S PRESENT WORTH A-->P
PRESENT SUM'S EQUAL-PAYMENT SERIES WORTH P-->A

BANKERS' JARGON

SINGLE-PAYMENT COMPOUND-AMOUNT FACTOR F/P,i,n      
SINGLE-PAYMENT PRESENT-WORTH FACTOR P/F,i,n        
EQUAL-PAYMENT SERIES COMPOUND-AMOUNT FACTOR F/A,i,n
EQUAL-PAYMENT SERIES SINKING-FUND FACTOR A/F,i,n   
EQUAL-PAYMENT SERIES PRESENT-WORTH FACTOR P/A,i,n  
EQUAL-PAYMENT SERIES CAPITAL-RECOVERY FACTOR A/P,i,n

TIME VALUE OF MONEY

A present SUM of money P or a yearly series of equal sums (payments) A can be considered to have an inflated value or WORTH Fi,n over time n. This inflation in value is often modeled with the use of interest rates i compounded on a yearly basis over n years. Let the variables of the model be:

P - a present sum of money
Fi,n - a future worth at i interest in n years
A - equal year end payments

then relative to time "now" a money flow diagram can be constructed in the following manner:

where P is a present sum invested at time now, A is a year end payment and Fi,n is the future worth of the investment and payment series.

Often it is easier to partition the problem into components such as the future worth of the P investment and the future worth of the equal-payment series A.
Take the P investment first. The money flow diagram is:

Based on compounding rules, the future worth Fi,n of the present sum P is:

Fi,0 = P                                           
Fi,1 = P+Pi = P(1+i)                               
Fi,2 = P(1+i)+P(1+i)i = P(1+i)(1+i)                
Fi,3 = P(1+i)(1+i) + P(1+i)(1+i)i = P(1+i)(1+i)(1+i)

or in an algebraic extension:

Fi,n = P(1+i)ⁿ

In the past, when calculators where not in everyday use, the above expression was reduced to a tabular form by dividing F by P and creating a table of factorsrelative to i and n or COMPOUND INTEREST FACTORS that represented the future worths of a present sum of one dollar. Symbolically the factors where designated as:

F/P,i,n - the future worth of a present dollar sum
(single payment compound-amount factor)

or:

F/P,i,n = (1+i)ⁿ

which in tabular form is:

COMPOUND INTEREST TABLES FOR F/P,i,n

It is very easy to invert the relationship between time now and a future time so that a future SUM can be evaluated as a present WORTH given the compounding of money. The factor for the present worth of a promised sum of one dollar n years in the future is:

P/F,i,n - the present worth of future dollar sum
(single payment present-worth factor)

which is the invert of the F/P factor:

P/F,i,n = 1/(1+i)_n

Now for the future worth of the equal payment series STARTING AT THE END OF THE CURRENT YEAR AND PROCEEDING THROUGH YEAR n. The money flow diagram for this component of the original model is:

Using the formulation for the F/P factor it is possible to calculated the future worth or each payment A as if it were a P investment held for lesser periods of time. Starting at the payment made at the time of F and proceeding back toward one year from time now, the future worth of the payment series is:

Fi,n = A + A(1+i) + A(1+i)2 ..... A(1+i)^n-1

This series can be simplified by first multiplying the series by (1+i)

Fi,n(1+i) = A(1+i) + A(1+i)² + A(1+i)³ .....A(i+i)ⁿ

and then subtracting the first equation form the second so that the remainder is:

iFi,n = -A + A(1+i)ⁿ

which reduces to:

Fi,n= A[(1+i)ⁿ-1]/i

This reduces to the future worth of a series of payments of one dollar sums:

F/A,i,n - the future worth of a equal payment series of one dollar sums
(equal-payment series compound-amount factor)

or:

F/A,i,n = [(1+i)ⁿ-1]/i

It is also possible to create a factor for the equal payment worth of a future sum of one dollar or what is often called a sinking fund (what sum in equal payments can I sink in an investment if I known I will receive a future sum?):

A/F,i,n - the equal payment series worth of a future dollar sum
(equal-payment series sinking-fund factor)

by inverting the F/A,i,n factor or:

A/F,i,n = i/[(1+i)ⁿ-1]

With this information it is possible to determine the present worth of the original money flow diagram by combining the partitioned values of

Fi,n = P(1+i)ⁿ + A[(1+i)ⁿ-1]/i = P[ F/P,i,n ] + A [ F/A,i,n ]
(Note: The future worth as a composite might not have a single value of n.)

In some cases it is convenient to know the present WORTH of a series of equal payments.

Using the algebraic characteristics of the factors:

P/A,i,n = P/F,i,n x F/A,i,n = 1/(1+i)ⁿ x [(1+i)ⁿ-1]/i = [(1+i)ⁿ-1]/[i(1+i)ⁿ]

or:

P/A,i,n - present worth of a equal-payment series of one dollar sums
(equal-payment series present-worth factor)

is equal:

P/A,i,n = [(1+i)ⁿ-1]/[i(1+i)ⁿ]

Another very interesting factor is derived from the inverse of the present worth of a series of equal payments. This factor is the equal payments required to equal a present sum of one dollar or the rate at which capital must be recovered over n years to be equal to a present sum:

A/P,i,n - equal payment series worth of a present dollar sum
(equal-payment series capital-recovery factor)

or:

A/P,i,n = [i(1+i)ⁿ]/[(1+i)ⁿ-1]

Table of Factors

QUESTIONS 7

1. If you invest $1000 dollars at your local bank in a C.D. at 5% compound interest, what is the value of the C.D. at the end of 10 years?

2. Your friend promises to pay you $50,000 dollars in 5 years, what is the present worth of this promise assuming he will pay as promised and the current interest rates are 6%. What if the interest rates rise to 10%.

3. You put $2000 a year in a investment program for the next 20 years starting at the END of the first year through the 20th year. The long term program you are investing in pays 7% interest. What is the expected future worth of this investment at the end of twenty years?

4. You will inherit a fortune of $100,000 in 10 years. Rather than wait ten years, you have decided to borrow against you inheritance in equal payments at the END of each of the next ten years. If the set interest rate is 9%, what is the maximum amount you can borrow in the equal payment series?

5. You have decided to start a small business in which the planned earnings are $20,000 a year at the end of each of the next five years. You have other opportunities that will earn you 20% interest and your are interested in what the WORTH of the business is relative to today.

6. I have loaned $50,000 to my neighbor. What is the payments required at the end of the next 5 years to payoff the loan at 6% interest.

CASE STUDY 8 - Investment Study

You invest money in a bank by first depositing $100 and then $10 at the end of

the next 10 years. What is the future sum of money (or future worth) your bank account will have if the compound interest rate is 10 percent. The money flow diagram is:

The diagram can be partitioned into a future worth of a present sum of $100 and a future worth of an equal payment series of $10 and totaled in the format of the compound interest factors:

F.10 = $100[ F/P,.10,10 ] + $10 [ F/A,.10,10 ]

The factors can be either found in the interest tables or calculated:

F/P,.10,10 = (1+.10)¹⁰ = 2.59374

F/A,.10,10 = [(1+.10)¹⁰-1]/.10 = 15.9374

and substituted into the original formulas for a future worth of:

F.10 = $100(2.59374) + $10(15.9374) = $418.748

You are going to invest in a project that is estimated to return $1000 sum at the end of five years. The accepted interest rate for investments is .05. What is the maximum worth you are willing to invest in equal payments at the end of each of the next five years?

The money flow diagram for your investment scheme is:

Given the requirement to find a series worth for a future sum, the algebraic solution to the problem is:

A.05,5 = F [ A/F,.05,5 ] = $1000(.1810) = $181.00

This would indicate that you are willing to invest up to $181 in the project each year.

ANSWERS 7

1. If you invest $1000 dollars at your local bank in a C.D. at 5% compound interest, what is the value of the C.D. at the end of 10 years?

The single payment compound-amount factor is F/P,.05,10 = 1.629

so that 1000(F/P,.05,10) = 1000(1.629)= $1629.00 = F,.05,10

2. Your friend promises to pay you $50,000 dollars in 5 years, what is the worth of this promise assuming he will pay as promised and the current interest rates are 6%. What if the interest rates rise to 10%.

The single payment present-worth factors are P/F,.05,5 = 0.7835 and

P/F,.10,5 = .6209

The present worth at  5% is then 50,000(.7835) = $39,175

The present worth at 10% is then 50,000(.6209) = $31,045

3. You put $2000 a year in a investment program for the next 20 years starting at the END of the first year through the 20th year. The long term program you are investing in pays 7% interest. What is the expected future worth of this investment at the end of twenty years?

The equal-payment series compound-amount factor is F/A,.07,20 = 40.996

The future worth of the series is 2000(40.996) = $81992

4. You will inherit a fortune of $100,000 in 10 years. Rather than wait ten years, you have decided to borrow against you inheritance in equal payments at the END of each of the next ten years. If the set interest rate is 9%, what is the maximum amount you can borrow in the equal payment series?

The equal-payment series sinking fund factor is A/F,.09,10 = .0658

The equal payment series worth is then 100,000(.0658) = $6,580 which will be paid in ten installments at the END of the next ten years.

5. You have decided to start a small business in which the planned earnings are $20,000 a year at the end of each of the next five years. You have other opportunities that will earn you 20% interest and your are interested in what the WORTH of the business is relative to today's earning potential.

The equal-payment series present-worth factor is P/A,.20,5 = 2.9906

The present worth of the adventure is 20,000(2.9906) = $59,812

6. I have loaned $50,000 to my neighbor. What is the payments required at the end of the next 5 years to payoff the loan at 6% interest.

The equal-payment series capital-recovery factor is A/P,.06,5 = .2374

so that the year end payments are 50,000(.2374) = $11,870

EQUIVALENCE

A. VARIABLE INTEREST RATE    
B. VARIABLE NUMBER OF PERIODS

MEASURES OF WORTH

A. SAMPLE PROBLEM
B. PRESENT EQUIVALENT WORTH (PE)
C. FUTURE EQUIVALENT WORTH (FE)
D. ANNUAL EQUIVALENT WORTH (AE)

1. SINGLE LIFE-CYCLE  
2. MULTIPLE LIFE-CYCLES

E. RATE-OF-RETURN (i)           
F. PAYOUT (n)                   
G. PAYBACK (accounting n)       

EQUIVALENCE

Two monetary sums can be equated given that they have the same worth at a given point in time. This is true under the restrictions of the time value of money only if for a given point in time and at a given interest rate the two worths are the same. The factors that determine equivalence are:

1. The amount of the sums.
2. The time of occurrence of the sums.
3. The interest rate.

In algebraic form, two sums P and F are equivalent if:

FEi = P ( F/P,i,n) = Fi,n for some given combination of i and n

(Note: The worth of a sum F at time n is FE.)

As an example, under what conditions are $50 and $100 dollars equivalent? First assume that n is equal to 10 years then:

100 = FEi,10 = 50(F/P,i,10)

F/P,i,10 = 2

From the tables if i = .07 then ( F/P,.07,10 ) = 1.967 and if i = .08 then ( F/P,.08.10 ) = 2.159 so that at equivalence:

i = .07 + .01[(2.000-1.967)/(2.159-1.967] = .07172 interest

as determined with linear interpolation from the tables and:

100 = 50( F/P,.07172,10)

Another way to approach the problem is to use the algebraic formulation for the single payment compound amount factor such that:

100 = 50(1+i)¹⁰

i = 2(1/10)-1 = .07177

Now assume that i=.05 then:

F/P,.05,n = 2.0

From the tables if n = 14 then ( F/P,.05,14 ) = 1.980 and if n = 15 then ( F/P,.05,15 ) = 2.079 so that at equivalence:

n = 14 +1[(2.000-1.980)/(2.079-1.980)] = 14.20202 years

as determined with linear interpolation from the tables and:

100 = 50 (F/P,.05,14.202)

Again, taking the algebraic formulation using the single payment compound amount factor:

100 = 50(1+.05)ⁿ

ln(2) = nln(1.05)

n = 14.2066 years

(Note: The algebraic solution to the problem is not always available when dealing with equal-payment series. In this cases the tabular solution is the only available method to find equivalence.)

QUESTIONS 8

1. Given that you have invested $10,000 at an interest rate of 10%, what would be the number of years to reach equivalence if you were to invest $2000 per year instead.

2. Prince Charming wants to borrow money from his inheritance trust at $20,000 worth per year for 10 years. If the trust will be equal to a sum of $500,000 at the end of 10 years, at what effective interest rate will the fund and the payment series be equivalent.

3. You are given the option to receive your $1,000,000 lottery winnings as 20 year end payments of $50,000 or 240 month end payments of $4,166.67. If the current interest rates are 5% per year, what is the equivalent monthly interest rate? What is the present equivalent worth of each payment series? Are the payment series equivalent?

ANSWERS 8

1. Given that you have invested $10,000 at an interest rate of 10%, what would be the number of years to reach equivalence if you were to invest $2000 per year instead.

The cash flow diagram for your investment is:

In algebraic form this reduces to

If we can solve for n or the number of years we have the answer.

5(.1)(1.1)ⁿ = (1.1)ⁿ -1

(1.1)ⁿ = 2

nln(1.1) = ln(2)

n = 7.2725

2. Prince charming wants to borrow money from his inheritance trust at $20,000 worth per year for 10 years. If the trust will be equal to a sum of $500,000 at the end of 10 years, at what effective interest rate will the fund and the payment series be equivalent.

The cash flow diagram is:

or algebraically:

(1+i)¹⁰ - 25i - 1 = 0

This a bit hard to crack. Try the tables where we desire (A/F,i,10)=.04 but (A/F,.15,10)=.0493, (A/F,.20,10)=.0385 so that:

i = .15 + .05[(.0493-.0400)/(.0493-.0385)] = .19306 interest

3. You are given the option to receive your $1,000,000 lottery winnings as 20 year end payments of $50,000 or 240 month end payments of $4,166.67. If the current interest rates are 5% per year, what is the equivalent monthly interest rate? What is the present worth of each payment series?

First start with the equivalent interest rate. Set the future worth of one dollar compounded yearly equal to the future worth of one dollar compounded monthly.

(1+.05) = (1+i)12, i = (1.05)(1/12)-1 = .004075 effective monthly interest

The yearly payment series present equivalent worths are then:

PE.05 = 50,000(P/A,.05,20) = 50,000(12.4622) = $623,110.00

The monthly payment series has a factor not in the book so algebraically:

The big difference arises because almost $50,000 is received in the monthly payments before the first yearly payments is received such that:

which is not equivalent.

Are the to series equivalent? No.

MEASURES OF WORTH

ACCOUNTANT                   ENGINEER      

CASH FLOW       ----->     PRESENT EQUIVALENT WORTH
PAY-BACK PERIOD ----->    PAYOUT PERIOD           

ALTERNATIVE EVALUATION

A. PRESENT EQUIVALENT WORTHS COMPARISONS (PE)
B. FUTURE EQUIVALENT WORTHS COMPARISONS (FE)
C. ANNUAL EQUIVALENT WORTHS COMPARISONS (AE)
D. RATE OF RETURN COMPARISONS                

1. INTERNAL RATES-OF-RETURN (i)
2. RATE-OF-RETURN EVALUATION (i)
   (break-even rate of return)

E. PAYOUT EVALUATION (break-even period n)  

BREAK-EVEN ANALYSIS

A. CUMULATIVE PROFIT EVALUATION (cost-profit-volume)
B. ANNUAL EQUIVALENT WORTH EVALUATION (N)           

MEASURES OF WORTH

The most common MEASURE OF WORTH is equivalent worth. This is usually in the form of PRESENT equivalent worth (PEi), ANNUAL equivalent worth (AEi) or FUTURE equivalent worth (FEi) for a given i. Other measures of worth is equivalent RATE-OF-RETURN (i) or PAYOUT time (n).

Present equivalent worth PEi

This measure assumes that a reasonable interest value i has been established for the time value of money. With i, it is possible to determine the worth of an alternatives at a given "time now". This present equivalent worth establishes the metric of worth of the alternative.

As an example, say that i=.05 and that the anticipated six year budget for a project is:

Equipment purchase                              $35
Annual operating expense per year for 6 years    10
Equipment salvage after 6 years                  10
Annual income per year for 6 years               15

The money flow diagram for the project is:

Under the accounting view of the project the cash flow for the project over the next six years is:

The accounting view does not measure the worth of the project. A common method is to find the worth of all sums at a given point in time or a "time now"
(Note: "Time now" does not have to be the first day of the project or the current moment. In some cases the start day of the production phase of a project is considered "time now".)

For our project, say that time now is when the equipment is purchased then:

PE.05 = -35 + 10( P/F,.05,6) - 10( P/A,.05.6) + 15( P/A,.05,6)

PE.05 = -35 + 10(.7462) - 10(5.0757) + 15(5.0757) = -$2.16   

or the project's present worth is a loss of -$2.16 at the date of equipment purchase.

Another time now might have been two years earlier during the preliminary design phase of the project. In this case the "present" worth is:

PE.05 = -35( P/F,.05,2 ) + 10( P/F,.05,8 ) - [10( P/A,.05,6 )]( P/F,.05,2)
+ [15( P/A,.05,6 )](P/F,.05,2)

PE.05 = -35(.9070) + 10(.6768) - 10(5.0757)(.9070) +15(5.0757)(.9070) = -$1.96

or

-$2.16( P/F.,05,2) = -$2.16(.9070) = -$1.96

Future equivalent worth FEi

The use of future equivalent worth is the same as determining the worth of an alternative at a future date much as the worth of a bank account at a future date. If for the example project the future worth is set at the end of year six assuming "time now" is time 0 then:

FE.05 = -35( F/P,.05,6) + 10 - 10( F/A,.05,6 ) + 15( F/A,.05,6 )

FE.05 = -35(1.340) +10 - 10(6.802) + 15(6.802) = -$2.89 at time 6

This can be returned to "time now" with:

PE.05 = -2.89( P/F,.05,6) = -2.89(.7462) = -$2.16

Equal-payment series or annual equivalent worth AEi

Many times the durations of alternatives are different. Some lasting for a few years and then repeating, will others lasting for many years. This makes determining a comparable present worth measure difficult. One solution to this problem is to consider the annual equivalent worth of a project from "time now" to the end of the project. Taking the original example with a six year life and a "time now" at the purchase of equipment then:

AE.05 = - 35( A/P,.05,6 ) + 10( A/F,.05,6) - 10 + 15    

AE.05 = -35(.1970) + 10(.1470) - 10 - 15 = -$.43 per year

This can be done in another way by taking the present equivalent worth of the project and converting it to the annual equivalent worth:

AE.05 = -$2.16( A/P,.05,6) = -2.16(.1970) = -$.43 per year

Now say that the six year alternative is made into a 12 year alternative by simply repeating the project. The money flow diagram might be simplified as:

which has an annual equivalent worth of:

AE.05 = [-2.16 - 2.16( P/F,.05,6 )]( A/P,.05,12)     

AE.05 = [ -2.16 -2.16(0.7462)](.1128) = -$.43 per year

or the same as the six year project. In other words, if an alternative is assumed to repeat in cycles forever in time, then the annual equivalent worth for one cycle is the annual equivalent worth forever.

Internal rate-of-return

Most alternatives are intended to make money. That is, there is a stream of expenses and a stream of incomes over time. It is hoped that the income stream is greater than the expense stream so that the alternative's present worth Pi is greater than zero. Now assume that the expense stream is borrowed on interest and that the added interest is just high enough to make the present equivalent worth zero. This interest rate is the internal rate-of-return that DISCOUNTS the alternative cash flows to zero worth.

As an example, take the present equivalent worth for the example project and set it to zero:

PEi = -35 + 10( P/F,i,6) + (15-10)( P/A,i,6) = 0

Now find the internal rate-of-return that balances the equation. This is done with trial and error. First try i=.05 or:

( P/F,.05,6 ) = .7462 ( P/A,.05,6 ) = 5.075 

PE.05 = -35 + 10(.7462) + 5(5.0757) = -$2.16

This places to high a premium on the "borrowed" money, Try i=.04 or:

( P/F,.04,6 )=1/(1.04)⁶=.7903, ( P/A,.04,6 )=[(1.04)⁶-1)]/(.04(1.04)⁶)=5.2421

PE.04 = -35 + 10(.7903) + 5(5.2421) = -$.89

Try i=.03 or:

( P/F,.03,6 )=1/(1.03)⁶=.8374, ( P/A,.03,6 )=[(1.03)⁶-1)]/(.03(1.03)⁶)=5.4171

PE.03 = -35 + 10(.8374) + 5(5.4171) = $.46

This brackets the zero value, so using linear interpolation, an internal rate-of-return is estimated as:

Payout evaluation

Many times it is important to know at what point in time n an alternative can be terminated leaving a present equivalent worth PEi,n equal zero. For the example project let us assume that the project is terminated at the time of purchase so that the purchase price is immediately compensated with the salvage value:

PE._05,0 = -35 + 10 = -$25

It is apparent that the one year payout time for the project will give the project a negative worth. The next year will generate an income stream that is the difference between the yearly expense and income. Now for a project of 1 year.

PE.05,1 = -35 + 10( P/F,.05,1 ) + 5( P/A,.05,1)

PE.05,1 = -35 + 10(.9524) + 5(.9524) = -$20.71

This process can be continued until a positive present equivalent worth is found. Now for a project of 2 - 7 years:

PE.05,2 = -35 + 10(.9070) + 5(1.8594) =-$16.63

PE.05,3 = -35 + 10(.8638) + 5(2.7233) =-$12.75

PE.05,4 = -35 + 10(.8227) + 5(3.5460) = -$9.04

PE.05,5 = -35 + 10(.7835) + 5(4.3295) = -$5.52

PE.05,6 = -$2.16

PE.05,7 = -35 + 10(.7107) + 5(5.7864) = $1.04

interpolation determines the final value:

The example project was terminated too early in the previous examples to have a positive equivalent WORTH even though the accounting cash flow was positive. The time the project should have run is for 6.675 years or more to have a positive worth.

Pay-back Evaluation

As a warning to those working with accountants, many times "payout" is calculated with an interest rate of 0%. This form of payout is referred to as the PAYBACK and does not reflect the time value of money. Using the above example:

QUESTIONS 9

1. I drive my cars until they become museum pieces. The car I now drive was purchased at the end of 1974 for $8000 and I am planning on selling it at the end of 1994 for $25. The maintenance cost for the first ten years (year end) were $100 dollars per year and for the next ten years were $500 dollars per year. The revenue generated by my driving was $2000 per year and the time value of my money is 5%. Given these costs and times, what is the worth of the car relative to the end of 1974? What is the worth of the car relative to the end of 1994? What is the worth of the car as equal payments over the twenty years? What was the internal rate of return for my driving over the past twenty years? What was the payout time for the car?

ANSWER 9

1. The cash flow diagram for the car is:

Using the diagram the present equivalent worth of the car in 1974 is:

PE.05
= -8000-100(P/A,.05,10)- 500(P/A,.05.10)(P/F,.05,10)+2000(P/A,.05,20)+25(P/A,.05,20)
= -8000-100(7.7217)-500(7.7217)(.6139)+2000(12.4622)+25(.3769)
= $13791.48

This converts to an equivalent worth in 1994 of:

FE.05 = 13,791.48(F/P,.05,20) = 13,791.48(2.653) = $36,588.80

or as an annual equivalent series:

AE.05,20 = 13,791.48(A/P,.05,20) = 13,791.48(.0803) = $1,107.46

The effective rate-of-return for the car is bracketed by 20% and 25% where:

-8000-100(4.1925)-500(4.1925)(.1615)+2000(4.8696)+25(.0261) =  $982.05
-8000-100(3.5705)-500(3.5705)(.1074)+2000(3.9539)+25(.0115) = -$640.69

and:

The payout period is bracketed by n=4 and n=5

-8000-100(3.5460)+2000(3.5460)+25(.8227)= -$1242.03
-8000-100(4.3295)+2000(4.3295)+25(.7835)=   $245.63

ALTERNATIVE EVALUATION

The comparison of mutually exclusive alternatives can be based on present equivalent worth, future equivalent worth, or an equal-payment series or annual equivalent worth all relative to a "time now". It should be noted that these methods entail setting a "time now" date which can be either the present time or a milestone date set within a given alternative's schedule.

Present Equivalent Worth Comparisons (determine PE)

A simple example of PRESENT EQUIVALENT WORTH comparison would be the comparison between a series of ten $10 payments starting after two years and a lump sum payment of $80. Assuming that the interest rate is 10%, then the equal-payments series as an equivalent present worth is:

PE.10 = 10( P/A,.10,10 )( P/F,.10,2 ) = 10(6.1446)(.8265) = $50.79

compared to the present worth of $80.

(Note: Remember that an equal-payment series is considered to be paid out at the end of the year. That would mean that a payment series starting after two years delay would actually have the first payment falling on the end of the third year.)

Future equivalent worth comparisons (determine FE)

The same example for a FUTURE EQUIVALENT WORTH comparison assuming a "time now" date of year 12 for the equal-payment series:

FE.10 = 10 ( F/A,.10,10) = 10(15.937) = $159.37

and for the lump sum:

FE.10 = 80 ( F/P,.10,12) = 80(3.138) = $251.04

again results in the lump sum as the best alternative.

Present Equivalent Worth of a life-cycle (determine PE at milestone)

Another method is to establish a fixed MILESTONE as the "time now" for each alternative. As a example, say you have two alternative projects with the following money flow diagram.

Now assume that the proposed production phases for the alternative projects are marked on the diagrams by the time 0, then the "time now" equivalent worth for the first alternative is:

PE.10=-75 - 75( F/P,.10,2) + 50( P/A,.10,6) =-75-75(1.210)+50(4.3553) = $127.015

and the second alternative is:

PE.10 = -80 - 80( F/P,.10,3 ) - 80( F/P,.10,5 ) + 60( P/A,.10,6 )
PE.10 = -80 - 80(1.331) - 80(1.611) + 60(4.3553) = $73.227

indicating that the first alternative is best.

Equal-payment series or annual equivalent worth comparisons (determine AE)

Comparisons can also be based on an ANNUAL EQUIVALENT WORTH comparison of alternatives that can be cycled. As an example, take two projects:

where one has a four year life and the other has a three year life. If the objective is to cycle the alternative's lives to provide a continuous series of lives then the annual equivalent worth of the first alternative is:

AE.10 =-150(A/P,.10,4)+50+50(A/F,.10,4)=-150(.3155)+50+50(.2155)=$13.45 per year

and the second alternative:

AE.10 =-150(A/P,.10.3)+50+70(A/F,.10,3)=-150(.4021)+50+70(.3021)=$10.83 per year

making the first alternative the better choice.

Rate-of-Return Evaluation (determine IROR)

A possible method for evaluating alternatives is to compare the internal rates- of-return of each alternative. If the internal rate-of-return for one alternative is higher than the internal rate-of-return for another, then the higher rated alternative MIGHT appear best. Unfortunately this does not take into consideration the EQUIVALENT WORTH of the two alternatives (Consider a high rate-of-return on a small worth but lower rate-of-return on a large worth.)

A better approach is to find the interest rate where both alternative have equal present equivalent worth. As the interest rate rises, one alternative will generally have a decreasing equivalent worth while the other will have an increasing worth. The interest rate that equates the equivalent worths is the BREAK-EVEN INTEREST RATE that is found by setting the present equivalent worths of both alternatives equal and solving for the variable i.

As an example, take two alternatives:

where for the first alternative the present equivalent worth is:

PEi = -100 + 40( P/A,i,4 )

and for the second alternative the present equivalent worth is:

PEi = -110 + 80( P/A,i,2 )

When the two present equivalent worths are equated then:

-100 + 40( P/A,i,4 ) = -110 + 80( P/A,i,2 )
PEA-B,i = -100 +40( P/A,i,4 ) - (-110 + 80( P/A,i,2 )

Using trial and error the interest rate i that satisfies the equations is found as:

i=.05   10 + 40(3.5460) - 80(1.859) =  3.088 = PEA-B
i=.06   10 + 40(3.4651) - 80(1.833) =  1.932       
i=.07   10 + 40(3.3872) - 80(1.808) =  0.848       
i=.08   10 + 40(3.3121) - 80(1.783) = -0.172       

This rate-of-return is the interest rate at which the two projects have equal equivalent worth. If the first alternative is observed, it can be seen that the delay in income makes it very sensitive to increasing interest rates. From this observations, if the current interest rates are below .07831 then the first alternative is the best. If the interest rates are above .07831, the second alternative is best.

(Note: Most authors consider rate-of-return evaluation as a BREAK-EVEN or STAND-OFF method where the two alternative break-even in present equivalent worth. Your book has grouped the methods in the PE,FE,AE,i,n approaches.)

Multiple Mutually Exclusive Alternatives using Rate-of-return

There is a method for combining the internal rate of return and the break-even rate-of-return. Say that a series of MUTUALLY EXCLUSIVE alternative have greater and greater capital investment so that they can be ranked by present worths of INVESTMENTS based on a minimum accepted rate-of-return. The question is whether to invest in the cheapest investment, or NOTHING, or increase the investment and choose a higher ranked investment, or not.

This can be accomplished by first ranking the alternatives in order of present equivalent worth of INVESTMENT for a minimum acceptable rate of return and then setting up a table with the rate-of-returns i that satisfy the present equivalent worths relations shown in each box. To determine which investment to make, simply determine a minimum acceptable rate of return (or effective rate) and follow the table starting at NOTHING and going down the column until the highest rate-of-return is found. If the highest rate-of-return in the column is lower than acceptable, stop at NOTHING. If the greatest rate-of-return in the first column is higher than the minimum acceptable rate, then go to the column corresponding to the alternative with the highest interest rate. Now go down that column until the highest rate-of-return is again found. If the highest rate of return is lower than acceptable, stop at the current alternative. If not, continue to the next alternative columns until the best alternative is found.

As an example, say that the rate-of-returns are as listed:

and the minimum acceptable rate is 5%. Follow the NOTHING column and choose alternative 2 at 15%. This is still greater than the minimum acceptable rate. Next go to the alternative 2 column and choose the added investment of alternative 4 at 2%. This added investment earns less than the minimum acceptable rate and is not attractive. Therefore stay with alternative 2.

Payout Evaluation (determine n)

Alternative projects may not have a fixed duration over which to determine a present equivalent worth. Since there is no n, the only approach is to determine a point in time n where the present equivalent worths of the two alternatives become equal. As an example, take two alternatives where the first alternative has a purchase price of $100, a salvage value of $50, and an annual income of $10. The interest rate is 10% but the project life is unknown.

The second alternative has a purchase price of $150, a salvage value of $50, and an annual income of $20.

For each alternative the present equivalent worth in terms of n is:

PE.10,n = -100 + 50( P/F,.10,n ) + 10( P/A,.10,n )
PE.10,n = -150 + 50( P/F,.10,n ) + 20( P/A,.10,n )

Equating the present equivalent worths results in:

-150+50( P/F,.10,n )+20( P/A,.10,n ) = -100+50( P/F,.10,n )+10( P/A,.10,n )

and:

10( P/A,.10,n ) = 50

which reduces to:

( P/A,.10,n ) = 5

Using interpolation:

( P/A,.10,7 ) = 4.8684
( P/A,.10,8 ) = 5.3349

The project duration for which both alternatives have the same present equivalent worth is 7.282 years. If the duration is less than 7.282 years then the first alternative will recover its initial investment faster and have the greater present worth.

(Note: Payout evaluation is considered a break-even method by most authors were the two alternatives break-even in present equivalent worth.)

CASE STUDY 9 - Career Counseling

You are to evaluate the potential of four career paths. To simplify the calculations, the careers will be limited to ten years.

1. The first career is as an engineer. The engineer must first complete four years of college at $10,000 per year before starting six years of earnings at$35,000 per year with expenses and taxes of $12,000 per year.
2. The second career is as a doctor. The doctor must first complete four years of undergraduate studies at $10,000 per year. Then he/she must complete four years of graduate school at $20,000 per year before starting two years of earnings at $200,000 per year with expenses and taxes of $80,000 per year.
3. The third career is as a gambler. The gambler "invests" in the lottery $3000 dollars per year before winning the $100,000 jack pot in the ninth year. After winning the jack pot, $40,000 in taxes are due in the 10th year. To maintain the proper image, the gambler must spend $2000 dollars each year for expenses.
4. The fourth career is as a bank robber. The bank robber robs a bank at time now for $100,000 which he/she deposits in a Swiss bank account. To satisfy the IRS the robber pays $40,0000 of income tax on his "take" at the end of the first year. The robber starts his ten year prison sentence immediately after robbing the bank during which time he/she makes license plates for $1000 per year and has $50 in yearly cigarette expenses.

Based on the career descriptions, determine the present equivalent worths, future equivalent worths at the ten year point, and annual equivalent worths of each ten year career assuming the interest rate is 6%.

What would appear to be the best career choice based on the measures of worth?

At what time n will the doctor and the engineer have equal present equivalent worths assuming that their pay continues at the current rate?

What is the rate of return for which the doctor and the engineer have equal present equivalent worths? What career is favored by high interest rates?

The first career is the engineer for which the cash flow diagram is:

The present equivalent worth for the career is:

PE.06 = -10,000(P/A,.06,4) + 23,000(P/A,.06,6)(P/F,.06,4)
        =-10.000(3.4651) + 23,000(4.9173)(.7921) = $54,993.85

This is equal to a future equivalent worth in ten years or:

FE.06 = 54,993.85(F/P,.06,10) = 54,993.85(1.791) = $98,386.53

or an annual equivalent worth of:

AE.06,10 = 54,993.85(A/P,.06,10) = 54,993.85(.1359) = $7465.51

The second career is the doctor for which the cash flow diagram is:

The present equivalent worth of the career is:

PE.06 = -10,000(P/A,.06,4)-20,000(P/A,.06.10)(P/F,.06,4)+120,000(P/A,.06.2)(P/F,.06,8)
      =-10,000(3.4651)-20,000(3.4651)(.7921)+120,000(1.8334)(.6274)= $48,487.91

This is equal to a future equivalent worth in ten years of:

FE.06 = 48,487.91(1.791) = $86,841.84

or an annual equivalent worth of:

AE.06,10 = 48,487.91(.1359) = $6589.50

The third career is the gambler for which the cash flow diagram is:

The present equivalent worth for the career is:

PE.06 = -5000(P/A,.06,9)-42,000(P/F,.06,10)+100,000(P/F,.06,9) =
-5,000(6.8017)-42,000(.5584)+100,000(.5919) = $1728.70

This is equal to a future equivalent worth in ten years of:

FE.06 = 1,728.70(1.791) = $3096.10

or an annual equivalent worth of:

AE.06,10 = 1,728.70(.1359) = $234.93

The fourth career is the bank robber for which the cash flow diagram is:

The present equivalent worth for the career is:

PE.06 = 100,000 - 40,000(P/F,.06,1) + 950(P/A,06,10) =
100,000 - 40,000(.8900) + 950(7.3601) = $71,392.10

This is equal to a future equivalent worth in ten years of:

FE.06 = 71,392.10(1.791) = $127,863.24

or an annual equivalent worth of:

AE.06,10 = 71,392.10(.1359) = $9,702.19

Based on these measures of worth, the career choices appear to be:

The payout evaluation for the engineer and the doctor requires setting the present equivalent worths equal so that:

Engineer

-10,000(3.4651) + 23,000(P/A,.06,n-4)(.7921) =

Doctor

-10,000(3.4651)-20,000(3.4651)(.7921)+120,000(P/A,.06,n-8)(.6274)

which reduces to:

75.288(P/A,.06,n-8)-18.2183(P/A,.06,n-4)-54.8941 = 0

now try n = 10 so that:

75.288(1.8334)-18.2183(4.9173)-54.8941 = -6.44

then n = 11 so that:

75.288(2.6730)-18.2183(5.5824)-54.8941 = 44.64

This has bracketed the zero value and:

The rate-of-return again sets the present equivalent worths equal so that:

Engineer

-10,000(P/A,i,4) + 23,000(P/A,i,6)(P/F,i,4) =

Doctor

-10(P/A,i,4) - 20,000(P/A,i,4)(P/F,i,4) + 120,000(P/A,i,2)(P/F,i,8)

which reduces to:

[23(P/A,i,6)+20(P/A,i,4)](P/F,i,4) - 120(P/A,i,2)(P/F,i,8) = 0

Now try i=.06

[23(4.9173)+20(3.4651)](.7473) - 120(1.8334)(.6274) = 6.44

Now try i=.05

[23(5.0757)+20(3.5460)](.8227) - 120(1.8594)(1.477) = -175.17

This brackets the zero value so:

BREAK-EVEN ANALYSIS

Accountant considers BREAK-EVEN as the cumulative production N where cumulative cash flow is zero or the CUMULATIVE REVENUE equals CUMULATIVE COST. Cost engineers consider Break-even as the production or usage LEVEL N, such as units per year, where the annual present equivalent worth is equal to zero or the annual equivalent worth of two alternatives are equal.

Cumulative Profit Evaluation (determine production quantity N)

One method for finding cumulative production N is the COST-VOLUME-PROFIT analysis. The method calculates the cumulative cash flow of an operation and determines the number of production units N at which the cumulative cash flow reaches zero. As an example, assume that a product sells for $20 a unit. Then as N units are sold the cumulative earnings from sales is:

I = 20N

There is also a FIXED cost of the production which is NOT DEPENDENT ON N and a VARIABLE cost of the production which is dependent on N. For the example, say that plant investment is FC = $200,000 and the cost to produce a unit is VC = $10. The cumulative cost is then:

TC = FC + VC = 200,000 + 10N

Equating the two cash flows:

TC = I

200,000 + 10N = 20N

for a total production quantity of N = 20,000 units.

(Note: This accounting method fails to consider the time value of money.)

Annual Equivalent Worth Evaluation (determine production or usage level N)

Most times in a product-system the number of units N produced affects the annual equivalent worth. When production rises to the LEVEL where the annual equivalent worth of the system is zero, then a break-even is reached.

Another common application of this idea is the buy/lease decision in which the usage in hours or quantity N used per year determines a "break-even" of the annual equivalent worths of leasing and ownership. (The cost to lease is equally offset by the cost to purchase.)

As an example, say a machine can be either leased or purchased. The lease agreement charges $10 for every N hours of usage in a given year. The purchase price of the machine is $5000 with a salvage value of $500 in 10 years. The operating cost of the machine is $2 per N hours of usage in a given year. The annual interest rate is 5 percent. The question is how many hours per year must the machine operate before the lease and the buy break-even in terms of annual equivalent worths.

The first step is to determine how much it cost PER YEAR to operate the purchased machine no matter how many hours it is operated. (The term COST is from accounting. The better term is WORTH since the time value of money is used to determine the actual values.) This is often called FIXED COSTS and for the machine is:

ANNUAL FIXED COST = 5000 ( A/P,.05,10) - 500( A/F,.05,10)

The VARIABLE COSTS is the expense that varies with the number of hours operated PER YEAR or:

ANNUAL VARIABLE COST = 2N

(Note: The number of hours operated is on a yearly basis so the variable cost is also an annual worth.)

The total annual cost for the ownership of the machine is:

ANNUAL COST TO BUY = 5000(.1295) - 500(.0795) + 2N = 607.75 + 2N

The lease annual cost is simply 10N since N is based on yearly hourly usage. If the two costs are equated:

10N = 607.75 + 2N

N = 75.96 hours per year

This would indicate that for less than 75.96 hours per year the lease agreement is better than the buy.

QUESTIONS 10

1. You have $150,000 in your bank account earning 2%. Your husband/wife has abetter ideas for the money and is looking for a new house. The three mutually exclusive choices on the market are the "Bungalow" for $50,000, the "Ranch" for $100,000, and the "Estate" for $150,000. Based on your budget forecast the Bungalow will cost $500 per year to maintain with a sales value of $60,000 in ten years; the Ranch will cost $1000 per year to maintain with a sales value in of $150,000 in ten years; and the Estate will cost $1500 per year to maintain with a sales value of $200,000 in ten years. Your husband/wife has pointed out that your money is only earning $32,848.50 in the bank over the next ten years and that you could make $35,000 on the purchase of the Estate over the next ten years. Use your knowledge of alternative evaluation to select a course of action.

2. McCann Farms Inc. has a cow-calf operation in Appanoose County, Ia. In 1989 the Corporation invested $50,000 in renovating the properties and $20,000 in rebuilding the 80 head cow herd. We sell our calves at $500 a piece and we invest in the cow/calf about $200 a year. Under these conditions, when will our accountant think we have produced enough calves to have "broken-even"? If we expect a life for our facilities and heard of 20 years, what is the production level of calves per year where we will have an annual equivalent worth of zero at 10% interest? What is the annual equivalent worth of our operation if we produce 78 calves per year?

3. I am considering the option of leasing a car rather than buying a car. The motor company is willing to lease me a car at a flat rate of $400 per month for five years. To purchase a new car will cost $18,000.00, 12 cents a mile for upkeep (gas and oil cost the same for both options), and $500 in salvage at the end of 10 years. If the effective interest rate is .5% per month, how many miles per month must I drive before the lease and buy are a standoff? What if I drive the purchased car for 5 years and salvage the vehicle for $5000? What if I drive the purchase car for 20 years and salvage the vehicle at $25? What does all this imply?

4. Another way to look at the lease option is to determine the number of MONTHS that a car must be leased before a lease/buy decision is a breakeven proposition. Say that I drive 1100 miles per month. The value of the car depreciates at a rate of 1% per month so that the salvage value for any given month is known. If the other data in problem three applies, determine the number of months before breakeven and the option that is best when the time period is LESS than breakeven.

ANSWERS 10

1. You have $150,000 in your bank account earning 2%. Your husband/wife has abetter ideas for the money and is looking for a new house. The three mutually exclusive choices on the market are the "Bungalow" for $50,000, the "Ranch" for $100,000, and the "Estate" for $150,000. Based on your budget forecast the Bungalow will cost $500 per year to maintain with a sales value of $60,000 in ten years; the Ranch will cost $1000 per year to maintain with a sales value in of $150,000 in ten years; and the Estate will cost $1500 per year to maintain with a sales value of $200,000 in ten years. Your husband/wife has pointed out that your money is only earning $32,848.50 in the bank over the next ten years and that you could make $35,000 on the purchase of the Estate over the next ten years. Use your knowledge of alternative evaluation to select a course of action.

Your husband/wife has probably "cashflowed" the houses and found that the Bungalow(B) returns $5,000, the Ranch(R) returns $40,000, and the Estate (E) returns $35,000 while the bank account earns only $32,848.50 over the next ten years.

Your too smart for this trick, so you consider the present worth of each house relative to the interest rate provided by the bank or 2%.

PEB =  -$50,000 -  $500(8.9824) +  $60,000(.82035) = -$5,270.14
PER = -$100,000 - $1000(8.9824) + $150,000(.82035) = $14,070.26
PEE = -$150,000 - $1500(8.9824) + $200,000(.82035) =    $596.61

The question now is whether to invest in the Ranch and leave the rest of the money in the bank at 2% or to invest in the Estate the whole $150,000. This can be resolved using the rate-of-return approach. First set up comparison table in order of present worth of invested capital at 2% where:

iB.02 =   50,000 +  500(8.9824 =  $54,491.20
iR.02 = 100,000 + 1000(8.9824) = $108,982.40
iE.02 = 150,000 + 1500(8.9824) = $163,473.60

and:

The values in the table are found by calculating the internal rates-of-return for the first column such that for the Bungalow:

PEB,i   = -50,000 - 500(P/A,i,10) + 60,000(P/F,i,10) =         0
PEB,.01 = -50,000 - 500(9.4711)   + 60,000(.9053)    = -$418.24
PEB,.00 = -50,000 - 500(10.000)   + 60,000(1.000)    = $5000.00

and the Ranch:

PER,i   = -100,000 - 1,000(P/A,i,10) + 150,000(P/F,i,10) =        0.00
PER,.04 = -100,000 - 1,000(8.1108)   + 150,000(.6756)    = -$6,775.93
PER,.03 = -100,000 - 1,000(8.5303)   + 150,000(.7441)    =   $3,083.48

and the Estate:

PEE,i   = -150,000 - 1,500(P/A,i,10) + 200,000(P/F,i,10) =         0.00
PEE,.03 = -150,000 - 1,500(8.5303)   + 200,000(.7441)    = -$13,977.03
PEE,.02 = -150,000 - 1.500(8.9824)   + 200,000(.8203)    =      $596.61

The step for the investment gap between Bungalow and Ranch is then:

PEB,i = PER,i
-50,000-500(P/A,i,10)+60,000(P/F,i,10)=-100,000-1,000(P/A,i,10)+150,000(P/F,i,10)

i=.05 50,000+500(7.7217)-90,000(.6139) = -$1390.15
i=.06 50,000+500(7.3601)-90,000(.5584) = $3424.05

between Bungalow and Estate:

PEB,i = PEE,i
-50,000-500(P/A,i,10)+60,000(P/F,i,10)=-150,000-1,500(P/A,i,10)+200,000(P/F,i,10)

i = .02 100,000 + 1000(8.9824) - 140,000(.8203) = -$5866.74
i = .03 100,000 + 1000(8.5303) - 140,000(.7440) =  $4357.40

and between Ranch and Estate:

PER,i = PEE,i
-100,000-1,000(P/A,i,10)+150,000(P/F,i,10)=-150,000-1,500(P/A,i,10)+200,000(P/F,i,10)

i=-.01 50,000+500(10.5727)-50,000(1.1057)= $0.0025

The network for the rates-of-return in order of increasing investment present worths is then:

resulting in the best path for the investment as Bank -> Ranch and only a negative return for the extra investment cost of the Estate. The best decision is to invest in the Ranch and leave the rest of the money in the bank at 2%.

(Note: You can also point the vectors in the direction of the highest present worth of the two alternatives on each vector when the evaluated interest rate is LESS than on the vector. This diagram will give the highest present equivalent worth satisfying the minimum acceptable interest rate.)

2. McCann Farms Inc. has a cow-calf operation in Appanoose County, Ia. In 1989 the Corporation invested $50,000 in renovating the properties and $20,000 in rebuilding the 80 head cow herd. We sell our calves at $500 a piece and we invest in the cow/calf about $200 a year. Under these conditions, when will our accountant think we have produced enough calves to have "broken-even"? If we expect a life for our facilities and heard of 20 years, what is the production level of calves per year where we will have an annual equivalent worth of zero at 10% interest? What is the annual equivalent worth of our operation if we produce 78 calves per year?

The fixed costs for the operation are the $50,000 invested in renovating and the $20,000 in the herd. The variable costs are the $200 spent on each cow and calf. The revenue generated is the $500 per calf sold. In the form of the accounts break-even analysis:

FC + VC = Revenue
70,000 + 200N = 500N where N is total number sold
N = 234 calves

Taking into account the time value of money and restructuring our problem so that N is the level of calf production per year, the break-even point is now:

70,000(A/P,.10,20) + 200N = 500N
70,000(.1175) + 200N = 500N
N = 27.41 calves per year

If we produce 78 calves per year then the annual equivalent worth is:

500(78)-200(78)-70,000(.1175) = $15,175.00

3. I am considering the option of leasing a car rather than buying a car. The motor company is willing to lease me a car at a flat rate of $400 per month for five years. To purchase a new car will cost $18,000.00, 12 cents a mile for upkeep (gas and oil are the same for both), and $500 in salvage at the end of 10 years. If the effective interest rate is .5% per month, how many miles per month must I drive before the lease and buy are a standoff? What if I drive the purchased car for 5 years and salvage the vehicle for $5000? What if I drive the purchase car for 20 years and salvage the vehicle at $25? What does all this imply?

The MONTHLY cost to operate the purchased car for 10 years or 120 months can be equated to the lease costs per month such that:

18,000(A/P,.005,120)-500(A/F,.005,120)+.12N = 400
18,000(.011102)-500(.006102)+.12N = 400
N = 1,693.45 miles per month

The MONTHLY cost to operate the purchased car for 5 years or 60 months can be equated to the rental costs per month such that:

18,000(.0193328)-5000(.0143328)+.12N = 400
N = 1030.61 miles per month

The MONTHLY cost to operate the purchased car for 20 years or 240 months can be equated to the rental costs per month such that:

18,000(.0071643)-25(.0021643)+.12N = 400
N = 2289.14 miles per month

This implies that the longer you own your car the more miles you must drive per month to make a lease car of interest. And even if you trade your car in early, you still have to drive more than 10,000 miles per year to make the lease attractive.

4. Another way to look at the lease option is to determine the number of MONTHS that a car must be leased before a lease/buy decision is a breakeven proposition. Say that I drive 1100 miles per month. The value of the car depreciates at a rate of 1% per month so that the salvage value for any given month is known. If the other data in problem three applies, determine the number of months before breakeven and the option that is best when the time period is LESS than breakeven.

Solving this problem requires equating the present worths on day of purchase or lease of each alternative and finding n or number of months required to satisfy the relationship. For the buy option the present worth of the car for any n is negative the purchase price of the car, negative the maintenance costs through month n and plus the depreciated value of the car at month n or:

PEB = -18,000 - .12(1100)(P/A,.005,n) + 18,000(1-.01n)(P/F,.005,n)

The present worth of the lease is simply negative the lease payment through month n (assuming payments made at the end of the month)

PEL = 400(P/A,.005,n)

The two present worths can be equating and restructured to reflect the marginal gain of the buy option over the lease option.

PEB-L = 18,000[(1-.01n)(P/F,.005,n)-1] + 268(P/A,.005,n)

Now for trial and error to find a n that equates the two options or PB-PL=0.

PEB-L = 18,000[(1-.05).9753 - 1] + 4.9256(268) -$2.30

n=6 P/A = 5.8768 P/F = .9705

PEB-L = 18,000[(1-.06).9705 - 1] 5.8768(268) = $1.3716

The resulting solution is then:

It would appear from the calculations that for a period of time less than 5.62 months the PEB-L < 0 so that the lease has the advantage up to 5.62 months and the buy has the advantage after 5.62 months.

DECISION MAKING

DECISION MAKING PROCESS

A. COURSES OF ACTION (i)            
B. FUTURE SITUATION (j)             
C. PREDICT OUTCOMES (ij)            
D. MEASURE OUTCOME VALUES (Vij)    
E. PREDICT PROBABILITIES OF OUTCOMES
F. DETERMINING BEST COURSE OF ACTION

PAYOFF MATRIX

A. CERTAIN FUTURE       
B. DOMINANCE            
C. ASPIRATION LEVELS    
D. MOST PROBABLE OUTCOME
E. EXPECTED VALUE       
F. LAPLACE CRITERION    
G. MAXIMIN (CYA)        
H. MAXIMAX (GO FOR BROKE)
I HURWICZ CRITERION     

DECISION MAKING

Decision making is the selection of a ALTERNATIVE course of action from a number of OUTCOMES of FUTURE SITUATIONS defined in terms of technical performance measures (TPM). For most design and operational decisions, the objective is to optimization the outcome. The major component steps that go into decision making are:

1. Determining alternative courses of action i.
2. Determining future situations j.
3. Predicting outcomes for a given action in a given situation.
4. Measuring outcome values based on TPM.
5. Predicting the likelihood of outcomes pj.
6. Deciding on a course of action i.

These components are often organized as a DECISION EVALUATION MATRIX which appears as:

where:

1. A set of i mutually exclusive alternatives as the rows.
2. A set of j mutually exclusive situations as columns.
3. A set of equivalent worth or value measures Vij for outcomes.
4. A set of probabilities pj for the situations.

The most common form of decision matrix is the PAYOFF MATRIX which is built on the assumption that:

1. All possible future situations are known or > pj = 1.
2. The technical performance measure is economic worth PE.
3. The occurrence of one outcome precludes the occurrence of another.
4. The occurrence of a specific situation is not influenced by the alternative selected (the pi value is the same for all alternatives).
5. The future is uncertain or no pi = 1.

An example of a payoff matrix would be:

Depending on ones view of the future, and willingness to take risks, the payoff matrix can be evaluated in a number of different ways.

Certain Future

If the future situation is known, then the payoff matrix can be reduced to a single column. (This is like having a certain tip at the race track.) If our example matrix is a certainty for the first situation, then:

and the best alternative results in the maximum worth max [10,20,-5,30] = 30 outcome or alternative 4 for $30.

Maximum and Minimum Dominance

In many cases a given alternative will dominate the matrix as either a consistent winner or consistent loser. This dominance can be found by finding the maximum and minimum payoff for each situation and determining if a single alternative is the source of the min or max value. For the example, say the table is changed to be: