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Time Value of Money - Engineering Economics.

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Page 1: Time Value of Money - Engineering Economics.

Engineering EconomicsEngineering Economics

Time Value of MoneyTime Value of Money

http://www.stellman-greene.com 1

Page 2: Time Value of Money - Engineering Economics.

Road MapRoad Map

Interest: The Cost of Money Economic Equivalence Interest Formulas for Single Cash Flows Uneven-Payment Series Equal-Payment Series Dealing with Gradient Series Composite Cash Flows

2

Page 3: Time Value of Money - Engineering Economics.

The Time Value of Money The Time Value of Money

Money itself is a commodity. Like other goods that are bought and sold, i.e. money costs money

Over time money can earn more money, or interest Interest: way to measure the cost of money The way interest operates reflects the fact that money has a

time value A dollar received today has a greater value than a dollar

received at some future time Money has both earning power and purchasing power over

time (i.e., it can be put to work, earning more money for its owner)

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Page 4: Time Value of Money - Engineering Economics.

The Time Value of Money The Time Value of Money

4

$100

$106

$106

$108

$104

Your money Cost of a fridge

Net loss $2

Net gain $2

Page 5: Time Value of Money - Engineering Economics.

Elements of Transactions Involving Interest Elements of Transactions Involving Interest

1. Principal (P): The initial amount of money invested or borrowed in transactions is called the.

2. The interest rate (i): measures the cost or price of money and is expressed as a percentage per period of time.

3. The interest period (n): A period of time determines how frequently interest is calculated. – Yearly (most common), – daily, – monthly, – quarterly, – semiannually

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Page 6: Time Value of Money - Engineering Economics.

Elements of Transactions Involving Interest Elements of Transactions Involving Interest

4. Number of interest periods (N): a specified length of time marks the duration of the transaction.

5. A plan for receipts or disbursements (A) that yields a particular cash flow pattern over a specified length of time. (For example, we might have a series of equal monthly payments that repay a loan.)

6. A future amount of money (F) results from the cumulative effects of the interest rate over a number of interest periods.

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Page 7: Time Value of Money - Engineering Economics.

Methods of Calculating Interest Methods of Calculating Interest

Money can be loaned and repaid in many ways, and equally, money can earn interest in many different ways

The two computational schemes:– Simple interest:

Total earned Interest: I = iPNTotal amount: F = P + iPN = P(1 + iN)

– Compound interest:After 1st period: P + iP = P(1+i)After 2nd period: P(1+i) + P(1+i)i = P(1+i)2

After 3rd period: P(1+i)2 + P(1+i)2i = P(1+i)3

…….

After Nth period: P(1+i)N-1 + P(1+i)N-1i = P(1+i)N

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Page 8: Time Value of Money - Engineering Economics.

Simple vs. Compound InterestSimple vs. Compound Interest

Suppose you deposit $1,000 in a bank savings account that pays interest at a rate of 8% per year. Assume that you don't withdraw the interest earned at the end of each period (year), but instead let it accumulate. (a) How much would you have at the end of year three with simple interest? (b) How much would you have at the end of year three with compound interest?– Simple: F = P(1 + iN) = 1,000(1 + 0.08*3) = 1,240– Compound: F = P(1 + i)N = 1,000(1 + 0.08)3 = 1,259.71

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Page 9: Time Value of Money - Engineering Economics.

Road MapRoad Map

Interest: The Cost of Money Economic Equivalence Interest Formulas for Single Cash Flows Uneven-Payment Series Equal-Payment Series Dealing with Gradient Series Composite Cash Flows

9

Page 10: Time Value of Money - Engineering Economics.

Definition and Simple CalculationsDefinition and Simple Calculations

Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another in the financial marketplace

For example, that we invest $1,000 at 12% annual interest for five years. At the end of the investment period, our sums grow to

F = 1000(1+0.12)5 = $1,762.34 we can say that at 12% interest, $1,000 received now is

equivalent to $1,762.34 received in five years

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Page 11: Time Value of Money - Engineering Economics.

Definition and Simple CalculationsDefinition and Simple Calculations

Suppose you are offered the alternative of receiving either $3,000 at the end of five years or P dollars today. With 8% interest, what value of P would make you indifferent to your choice between P dollars today and the promise of $3,000 at the end of five years?

F = P(1+i)N or

As you may have already guessed, at a lower interest rate, P must be higher in order to be equivalent to the future

amount. For example, at i = 4%, P = $2,466.

11

042,2$)08.01( 5

000,3

i)(1 NFp

Page 12: Time Value of Money - Engineering Economics.

Equivalence CalculationsEquivalence Calculations

To compare the value of alternative cash flows– present time: yields the present worth– Some point in the future: yields their future worth

Consider the cash flow series given in Figure below. Compute the equivalent lump sum amount at n = 3 at 10% annual interest.

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Page 13: Time Value of Money - Engineering Economics.

Given: The cash flows given in Figure 2.6, and i = 10%. Find: V3 (or equivalent worth at n = 3). We find the equivalent worth at n = 3 in two steps. Step 1: Find the equivalent lump-sum payment of the first

four payments at n = 3:100(1 + 0.10)3+ 80(1 + 0.10)2+ 120(1 + 0.10)1 + 150 = 511.90

Step 2: Find the equivalent lump-sum payment of the remaining two payments at n = 3:

200(1 + 0.10)-1 + 100(1 + 0.10)-2 = 264.46 Step 3: Find V3. the total equivalent value:

V3 = 511.90 + 264.46 = 776.36

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Page 14: Time Value of Money - Engineering Economics.

Road MapRoad Map

Interest: The Cost of Money Economic Equivalence Interest Formulas for Single Cash Flows Uneven-Payment Series Equal-Payment Series Dealing with Gradient Series Composite Cash Flows

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Page 15: Time Value of Money - Engineering Economics.

Compound-Amount FactorCompound-Amount Factor

To calculate the future worth: F = P(1 + i)N . (1 + i)N is called compound-amount factor or single-payment

compound-amount factor– express that factor in a functional notation as (F/P. i, N),

which is read as "Find F, given i, and N”F = P(F/P, i, N)

– process of finding F is often called the compounding process

To find F in any of three ways:– Use calculator: F = P(1 + i)N – Use interest table: F = P(F/P. i, N)– Using a computer: F = FV(i,N,0,P).

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Page 16: Time Value of Money - Engineering Economics.

Present-Worth Factor Present-Worth Factor

The present worth:

Factor 1/(1+i)N single-payment present-worth factor and is designated (P/F, i, N)

16

]i)(1 N

1F[p

Page 17: Time Value of Money - Engineering Economics.

Solving for Time and Interest RatesSolving for Time and Interest Rates

Suppose you buy a share of stock for $10 and sell it for $20; your profit is thus $10. If that happens within a year, your rate of return is an impressive 100% ($10/$10 = 1). If it takes five years, what would be the rate of return on your investment?

Given: P = $10, F = $20, and N = 5. Find: i

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Page 18: Time Value of Money - Engineering Economics.

Solving for Time and Interest RatesSolving for Time and Interest Rates

You have just purchased 100 shares of General Electric stock at $30 per share. You will sell the stock when its market price doubles. If you expect the stock price to increase 12% per year, how long do you expect to wait before selling the stock

Given: P = $3,000, F = $6,000, and i = 12% per year. Find: N (years).

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Page 19: Time Value of Money - Engineering Economics.

Road MapRoad Map

Interest: The Cost of Money Economic Equivalence Interest Formulas for Single Cash Flows Uneven-Payment Series Equal-Payment Series Dealing with Gradient Series Composite Cash Flows

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Page 20: Time Value of Money - Engineering Economics.

Uneven-Payment SeriesUneven-Payment Series

A common cash flow transaction involves a series of disbursements or receipts

When there is no clear pattern over the series: the transaction is called an uneven cash flow series

The present worth of any uneven stream of payments can be found by calculating the present worth of each individual payment and summing the results.

Once the present worth is found, we can make other equivalence calculations

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Page 21: Time Value of Money - Engineering Economics.

Present Values of an Uneven Series by Present Values of an Uneven Series by Decomposition into Single PaymentsDecomposition into Single Payments

Wilson Technology, a growing machine shop, wishes to set aside money now to invest over the next four years in automating its customer service department. The company can earn 10% on a lump sum deposited now, and it wishes to withdraw the money in the following increments:

Year 1: $25,000 to purchase a computer and database software designed for customer service use;

Year 2: $3,000 to purchase additional hardware to accommodate anticipated growth in use of the system;

Year 3: No expenses: and Year 4: $5,000 to purchase software upgrades. How much money must be deposited now in order to cover the

anticipated payments over the next four years?

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Page 22: Time Value of Money - Engineering Economics.

Present Values of an Uneven Series by Present Values of an Uneven Series by Decomposition into Single PaymentsDecomposition into Single Payments

Equivalent to: the present value of each single cash flow of the future expense stream of ($25,000, $3,000, $0, $5,000)

We sum the individual present values as follows: P = $25,000(P/F. 10%. I ) + $3,000(P/F, 10%. 2) + $5.000(P/F, 10%,4) = $28,622

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Page 23: Time Value of Money - Engineering Economics.

Present Values of an Uneven Series by Present Values of an Uneven Series by Decomposition into Single PaymentsDecomposition into Single Payments

Recheck the calculation:1. let's calculate the balance at the end of each year. 2. If you deposit $28,622 now, it will grow to (1.10)($28,622), or $31,484, at

the end of year one. 3. From this balance, you pay out $25.000. The remaining balance, $6,484,

will again grow to (1.10)($6,484), or $7.132, at the end of year two. 4. Now you make the second payment ($3,000) out of this balance, which

will leave you with only $4,132 at the end of year two.5. Since no payment occurs in year three, the balance will grow to

$(1.10)2($4.132), or $5,000, at the end of year four. 6. The final withdrawal in the amount of $5,000 will deplete the balance

completely

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Page 24: Time Value of Money - Engineering Economics.

Road MapRoad Map

Interest: The Cost of Money Economic Equivalence Interest Formulas for Single Cash Flows Uneven-Payment Series Equal-Payment Series Dealing with Gradient Series Composite Cash Flows

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Page 25: Time Value of Money - Engineering Economics.

Compound-Amount Factor: Find Compound-Amount Factor: Find FF, Given , Given AA, , ii, , and and NN

Find F of a fund to which A amount contributed each period at interest rate of i per period. The contributions are made at the end of each of the N periods.

The bracketed term is called the equal-payment-series compound amount factor, or the uniform-series compound-amount factor;

Factor notation is (F/A, i, N)

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N)i,A(F/A,]i

1Ni)(1A[F

Page 26: Time Value of Money - Engineering Economics.

Compound-Amount Factor: Find Compound-Amount Factor: Find FF, Given , Given AA, , ii, , and and NN

Suppose you make an annual contribution of $5,000 to your savings account at the end of each year for five years. If your savings account earns 6% interest annually, how much can be withdrawn at the end of five years

Given: A = $5.000, N = 5 years. and i = 6% per year. Find: F

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Page 27: Time Value of Money - Engineering Economics.

Compound-Amount Factor: Find Compound-Amount Factor: Find FF, Given , Given AA, , ii, , and and NN

Using the equal-payment-series compound-amount factor, we obtainF = $5,000(F/A, 6%, 5) = $5,000(5.6371) = $28,185.46.

To obtain the future value of the annuity on Excel, we may use the following financial command:F = FV(6%,5,5000,0).

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Page 28: Time Value of Money - Engineering Economics.

Handling Time Shifts in a Uniform SeriesHandling Time Shifts in a Uniform Series

In previous example, the first deposit of the five-deposit series was made at the end of period one, and the remaining four deposits were made at the end of each following period. Suppose that all deposits were made at the beginning of each period instead. How would you compute the balance at the end of period five?

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Page 29: Time Value of Money - Engineering Economics.

Handling Time Shifts in a Uniform SeriesHandling Time Shifts in a Uniform Series

Each payment has been shifted one year earlier; thus, each payment is compounded for one extra year

Previous balance can earn interest for one additional yearF5 = 28,185.46(1.06) = 29,876.59

Excel: FV(6%,5,5000,,1) By adding the $5,000 deposit at period zero to the original

cash flow and subtracting the $5,000 deposit at the end of period five

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Page 30: Time Value of Money - Engineering Economics.

Sinking-Fund Factor: Find A, Given F, Sinking-Fund Factor: Find A, Given F, i, and Ni, and N

The term within the brackets is called the equal-payment-series sinking-fund factor, or just sinking-fund factor

Referred to with the notation (A/F, i, N)

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N)i,F(A/F,]1Ni)(1

1F[A

Page 31: Time Value of Money - Engineering Economics.

College Savings Plan: Find A, Given College Savings Plan: Find A, Given F, N, and iF, N, and i

You want to set up a college savings plan for your daughter. She is currently 10 years old and will go to college at age 18. You assume that when she starts college, she will need at least $100,000 in the bank. How much do you need to save each year in order to have the necessary funds if the current rate of interest is 7%? Assume that end-of-year payments are made.

Using the sinking-fund factors, we obtainA = $100,00O(A/F, 7%, 8) = $9,746.78.

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Page 32: Time Value of Money - Engineering Economics.

Capital-Recovery Factor (Annuity Factor): Find Capital-Recovery Factor (Annuity Factor): Find AA, , Given Given PP, , i i and and NN

The portion within the brackets: the equal-payment-series capital-recovery factor, or simply capital-recovery factor,

which is designated (A/P, i, N). In finance, A/P factor is referred to as the annuity factor.

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N)i,P(A/P,]1Ni)(1

Ni)i(1P[A

Page 33: Time Value of Money - Engineering Economics.

Paying Off an Educational Loan: Find A, Given Paying Off an Educational Loan: Find A, Given P, P, i, and Ni, and N

You borrowed $21,061.82 to finance the educational expenses for your senior year of college. The loan will be paid off over five years. The loan carries an interest rate of 6% per year and is to be repaid in equal annual installments over the next five years. Assume that the money was borrowed at the beginning of your senior year and that the first installment will be due a year later. Compute the amount of the annual installments.

Given: P = $21,061.82, i = 6% per year, and N = 5 years. Find: A.

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Page 34: Time Value of Money - Engineering Economics.

Paying Off an Educational Loan: Find A, Given Paying Off an Educational Loan: Find A, Given P, P, i, and Ni, and N

Using the capital-recovery factor, we obtainA = $21,061.82(A/P, 6%, 5) = $21,061.82(0.2374) = $5,000.

The Excel solution using annuity function commands is as follows:

A = PMT(i,N,P) = PMT(6%,5, 21,061.82)

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Page 35: Time Value of Money - Engineering Economics.

Deferred Loan RepaymentDeferred Loan Repayment

Suppose in previous Example that you had wanted to negotiate with the bank to defer the first loan installment until the end of year two (but still desire to make five equal installments at 6% interest). If the bank wishes to earn the same profit as in previous Example, what should be the annual installment

Given: P = $21,061.82, i = 6% per year, and N = 5 years, but the first payment occurs at the end of year two.

Find: A.

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Page 36: Time Value of Money - Engineering Economics.

Deferred Loan RepaymentDeferred Loan Repayment

In deferring one year, the bank will add the interest accrued during the first year to the principal, i.e. we need to find the equivalent worth of $21,061.82 at the end of year 1, P':

P’ = 21,061.82(F/P,6%,1) = 22,325.53

Thus, you are borrowing $22,325.53 for five years. To retire the loan with five equal installments, the deferred equal annual payment, A', will be

A = 22,325.53(A/P, 6%, 5) = 22,325.53 (0.2374) = 5,300

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Page 37: Time Value of Money - Engineering Economics.

Present-Worth Factor: Find Present-Worth Factor: Find PP, Given , Given AA, , ii, and , and N N

The bracketed term is referred to as the equal-payment-series present-worth factor

and is designated (P/A, i, N)

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N)i,A(P/A,]Ni)i(1

1Ni)(1A[P

Page 38: Time Value of Money - Engineering Economics.

Present-Worth Factor: Find Present-Worth Factor: Find PP, Given , Given AA, , ii, and , and N N

Lottery problem: a couple gave up the installment plan of $7.92 million a year for 25 years to receive a cash lump sum of $104 million. If the couple could invest its money at 8% annual interest, did it make the right decision? What is the lump-sum amount that would make the couple indifferent to each payment plan?

Given: i = 8% per year, A = $7.92 million, and N = 25 years. Find: P..

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Page 39: Time Value of Money - Engineering Economics.

Present-Worth Factor: Find Present-Worth Factor: Find PP, Given , Given AA, , ii, and , and N N

Tabular Solution:P = $7.92(P/A, 8%, 25) = $7.92(10.6748) = $84.54 million.

Excel solution: = PV(8%,25,7.92,,0) = $84.54 million..

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Page 40: Time Value of Money - Engineering Economics.

Start Saving Money as Soon as PossibleStart Saving Money as Soon as Possible

Composite Series that Requires Both (F / P ,i, N) and (F/A, i, N) Factors

Consider the following two savings plans that you consider starting at the age of 21:

Option I: Save $2,000 a year for 10 years. At the end of 10 years, make no further investments, but invest the amount accumulated at the end of 10 years until you reach the age of 65. (Assume that the first deposit will be made when you are 22.)

Option 2: Do nothing for the first 10 years. Start saving $2,000 a year every year thereafter until you reach the age of 65. (Assume that the first deposit will be made when you turn 32.)

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Page 41: Time Value of Money - Engineering Economics.

Start Saving Money as Soon as PossibleStart Saving Money as Soon as Possible

Option I: Compute the final balance in two steps. First, compute the accumulated balance at the end of 10

years (when you are 31).F31 = 2,000(F/A,8%,10) = 28,973

Then use this amount to compute the result of reinvesting the entire balance for another 34 years. F65 = 28,973(F/P,8%,34) = 396,645

Option 2: Since you have only 34 years to invest, the resulting balance will beF65 = 2000(F/A,8%,34) = 317,253

With the early savings plan, you will be able to save 79,391 more.

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Page 42: Time Value of Money - Engineering Economics.

Present Value of PerpetuitiesPresent Value of Perpetuities

A perpetuity is a stream of cash flows that continues forever. A good example is a share of preferred stock that pays a fixed

cash dividend each period (usually a quarter-year) and never matures

You cannot compute the future value of its cash flows, because it is infinite

We know how to calculate the present value for an equal-payment series with the finite stream: – Find P, Given A, i, and N

Take N →∞: P = A / i

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]Ni)i(1

1Ni)(1A[P

Page 43: Time Value of Money - Engineering Economics.

Road MapRoad Map

Interest: The Cost of Money Economic Equivalence Interest Formulas for Single Cash Flows Uneven-Payment Series Equal-Payment Series Dealing with Gradient Series Composite Cash Flows

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Page 44: Time Value of Money - Engineering Economics.

Linear Gradient SeriesLinear Gradient Series

Sometimes cash flows will vary linearly They increase or decrease by a set amount, G,

the gradient amount.

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Page 45: Time Value of Money - Engineering Economics.

Linear Gradient SeriesLinear Gradient Series

Note: each payment is An = (n - 1)G. Note The series begins with a zero cash flow at the

end of period zero. If G > 0: an increasing gradient series. If G < 0: a decreasing gradient series.

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Linear Gradient SeriesLinear Gradient Series

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Page 47: Time Value of Money - Engineering Economics.

Linear Gradient SeriesLinear Gradient Series

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Page 48: Time Value of Money - Engineering Economics.

Linear Gradient SeriesLinear Gradient Series Choosing how to receive your winning a SuperLotto Plus jackpot?! You have a choice:

– the entire jackpot in 26 annual graduated payments– receiving one lump sum that will be less than the announced

jackpot. What wouId these choices come out to for an announced jackpot

of $7 million? Lump-sum cash-value option:

– Receive about 49.14%. or $3.44 million, in one lump sum (less tax withholdings).This cash value is based on average market costs determined by U.S. Treasury zero-coupon bonds with 5.3383% annual yield.

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Page 49: Time Value of Money - Engineering Economics.

Linear Gradient SeriesLinear Gradient Series

Annual-payments option: – Receive the jackpot in 26 graduated annual payments. – Receive $175,000 as the first payment (2.5% of the total

jackpot amount). – The second payment would be $189,000. Over the course of

the next 25 years these payments would gradually increase each year by $7,000 to a final payment of $357,000

If the U.S. Treasury zero-coupon rate is reduced to 4.5% (instead of 5.338%) at the time of winning. what would be the equivalent cash value of the lottery?

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Page 50: Time Value of Money - Engineering Economics.

Linear Gradient SeriesLinear Gradient Series

Given: A1 = $175,000, A2 = $189,000, G = $7,000 (from payment periods 3 to 26), i = 4.5% per year, and N = 26 years

Find: P

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Page 51: Time Value of Money - Engineering Economics.

Linear Gradient SeriesLinear Gradient Series Since the linear gradient series starts at period 2 for this

example (i.e., is delayed by one period), we can calculate the present value in two steps: First compute the value at N = 1 and then extend it to N = 0

P = [175,00 + 189,000(P/A,4.5%,25) + 7,000(P/G,4.5%,25)](P/F,4.5%,1)

P = 3,818,363

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Page 52: Time Value of Money - Engineering Economics.

Geometric Gradient SeriesGeometric Gradient Series

Cash flow varies by some fixed rate expressed as percentage

Many engineering economic problems, particularly those relating to construction costs or maintenance costs, involve cash flows that increase or decrease over time by a constant percentage (geometric)

That process is called compound growth. Price changes caused by inflation are a good example

of such a geometric series.

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Page 53: Time Value of Money - Engineering Economics.

Geometric Gradient SeriesGeometric Gradient Series

An = A1(1 + g)n-1 n = 1, 2, 3, 4, … N If g > 0 the series will increase if g < 0 the series will decrease

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Page 54: Time Value of Money - Engineering Economics.

Geometric Gradient SeriesGeometric Gradient Series

The present worth Pn of any cash flow An at an interest rate i Pn = An(1 + i)-n = A1(1 + g)n-1(1 + i)-n

The present amount for the entire series

P = ∑ Pn = ∑ A1(1 + g)n-1(1 + i)-n

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N

n=1

N

n=1

Page 55: Time Value of Money - Engineering Economics.

Geometric Gradient SeriesGeometric Gradient Series

P= A1(P/A1,g,I,N) (P/A1,g,I,N): geometric-gradient-series present-worth factor

Special case i=g: P = [A1/(1+i)]N

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Page 56: Time Value of Money - Engineering Economics.

Required Cost-of-Living Required Cost-of-Living Adjustment CalculationAdjustment Calculation

Suppose that your retirement benefits during your first year of retirement are $50.000. Assume that this amount is just enough to meet your cost of living during the first year. However, your cost of living is expected to increase at an annual rate of 5%, due to inflation. Suppose you do not expect to receive any cost-of-living adjustment in your retirement pension. Then. some of your future cost of living has to come from your savings other than retirement pension. If your savings account earns 7% interest a year, how much should you set aside in order to meet this future increase in cost of living over 25 years

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Page 57: Time Value of Money - Engineering Economics.

Required Cost-of-Living Required Cost-of-Living Adjustment CalculationAdjustment Calculation

Given: A1 = 50,000 g = 5%, i = 7%, N = 25 years Find: P

- Find the equivalent amount of total benefits paid over 25 years Pincome

- Find the equivalent amount of total cost of living with inflation Pcost-of-living

- The required additional savings to meet the future increase in cost of living will be

ΔP = Pcost-of-living - Pincome

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Page 58: Time Value of Money - Engineering Economics.

Required Cost-of-Living Required Cost-of-Living Adjustment CalculationAdjustment Calculation

- Find the equivalent amount of total benefits paid over 25 years:

Pincome = 50,000(P/A,7%, 25)

Pincome = 582,679

- Find the equivalent amount of total cost of living with inflation Pcost-of-living =

Pcost-of-living = 940,696

- The required additional savings to meet the future increase in cost of living will be

ΔP = 940,696 – 582,679 = 358,017

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Home Work 02Home Work 02

Do end chapter problems: 2.6, 2.10, 2.20, 2.26, 2.31, 2.32, 2.37, 2.40, 2.41, 2.42

Each problem is worth 10 points. Total points of homework 02 = 100 points Any extra problem worth 10 bonus points Due at 11:50am thứ ba 20/07/2010 at class.

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