TM 661 Engineering Economics for Managers Unit 1 Time Value of Money.

TM 661 Engineering TM 661 Engineering Economics for Economics for

ManagersManagers

Unit 1Time Value of Money

Time Value-of-MoneyTime Value-of-Money

Suppose we invest $100. If the bank pays 5% interest, how much will I have after 1 year?

100

F

0 1 t


Suppose we invest $100. If the bank pays 5% interest, how much will I have after 1 year?

100

F1

0 1 t F1 = 100 + interest= 100 + 100 (.05)= 105


Now suppose I keep that $105 in the bank for another year. Now how much do I have?

100

0 1 2 t F2 = 105 + interest= 105 + 105 (.05)= 110.25

105

F2


In General

P

0 1 2 t

F1= P + interest= P + iP= P(1+i)

F1


In General

P

0 1 2 t

F1= P + interest= P + iP= P(1+i)

F2 = F1 + interest= F1 + iF1

= F1(1+i)

F2


In General

P

0 1 2 t

F1= P(1+i)

F2 = F1 + interest

= F1 + iF1

= F1(1+i)

= P(1+i)(1+i)

= P(1+i)2

F2


In General

P

0 1 2 3 n Fn= P(1+i)n

Fn

ExampleExample

I would like to have $20,000 for my son to go to college in 15 years. How much should I deposit now ifI can earn 10% interest?

Fn= P(1+i)n

P

0 1 2 3 n

20,000

ExampleExample


Recall, Fn= P(1+i)n

P

0 1 2 3 n

20,000

ExampleExample


Recall, Fn= P(1+i)n

Then, P= Fn(1+i)-n

P

0 1 2 3 n

20,000

ExampleExample


Then, P= Fn(1+i)-n

= 20,000(1+.1)-15

= 20,000(0.2394)

= $2,394

P

0 1 2 3 n

20,000

Future Worth Given Future Worth Given Annuity Annuity

Suppose I wish to compute annual installments to save for college.

0

1 2 3 4

. . . .

n

A

F

Annuities Given Annuities Given FutureWorthFutureWorth

F = A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3+...+ A(1+i) + A

0

1 2 3 4

. . . .

n

A

F

Annuities Given Future Annuities Given Future WorthWorth

(1+i)F = A(1+i)n + A(1+i)n-1 + A(1+i)n-2 + . . . + A(1+i)

F = A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3+...+ A(1+i) + A

0

1 2 3 4

. . . .

n

A

F


iF = A(1+i)n + 0 + 0 + . . . - A

(1+i)F = A(1+i)n + A(1+i)n-1 + A(1+i)n-2 + . . . + A(1+i)

F = A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3+...+ A(1+i) + A

0

1 2 3 4

. . . .

n

A

F


iF = A(1+i)n + 0 + 0 + . . . - A

= A[(1+i)n - 1]

(1+i)F = A(1+i)n + A(1+i)n-1 + A(1+i)n-2 + . . . + A(1+i)

F = A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3+...+ A(1+i) + A

0

1 2 3 4

. . . .

n

A

F


iF = A[(1+i)n - 1]0

1 2 3 4

. . . .

n

A

F

F Ai

iAF

Ai n

n

)( , , )

1 1(


0

1 2 3 4

. . . .

n

A

F

A Fi

i n

( )1 1


0

1 2 3 4

. . . .

n

A

F

A Fi

i n

( )1 1

A

20 0001

1 1 1

20 000 0 0315

50

15,(. )

( . )

, ( . )

$629.

Annuities Given Annuities Given Present WorthPresent Worth

P A ikk

nk

1

1( )

A i k

k

n

( )11

, for A1 = A2 = A3 = ... = A

0

1 2 3 4

. . . .

n

A

P


0

1 2 3 4

. . . .

n

A1

F

A2 A3 AnA4

F A ikk

nk

0

1

1( )

A i k

k

n

( )10

1

, for A1 = A2 = A3 = . . . = A

Annuities Given Annuities Given Present Worth Present Worth

Suppose we wish to compute the monthly payment of a car if we borrow $15,000 at 1% per month for 36 months.

0

1 2 3 4

. . . .

n

A

P

Recall:Fn = A

andP = Fn(1 + i) -n


0

1 2 3 4

. . . .

n

A

P

( )1 1 i

i

n


Inverting and solving for A gives

0

1 2 3 4

. . . .

n

A

PP Ai

i i

n

n

( )

( )

1 1

1

A Pi i

iP A P i n

n

n

( )

( )( / , , )

1

1 1

A = 15,000[.01(1.01)36/(1.0136 - 1)]

= $498.22

Car ExampleCar Example

We have an initial loan of $15,000 at a rate of i=1% per month for n=36 months. The monthly loan payment is then

0

1 2 3 4

. . . .

36

A

P

Car Example; Car Example; AlternativeAlternative

We have an initial loan of $15,000 at a rate of i=1% per month for n=36 months. The monthly loan payment is then

A = 15,000(A/P,i,n)=

15,000(A/P,1,36)=

15,000(.0032)=

$498.21

0

1 2 3 4

. . . .

36

A

P

Car Example; TablesCar Example; Tables

Discrete Compoundingi = 1.00%

Single Payment Uniform Series Gradient Series

Compound Present Compound Sinking Present Capital Uniform Presentamount worth amount fund worth recovery series worthfactor factor factor factor factor factor factor factor

To find F To find P To find F To find A To find P To find A To find A To find PGiven P Given F Given A Given F Given A Given P Given G Given G

33 1.3887 0.7201 38.8690 0.0257 27.9897 0.0357 15.0995 422.629134 1.4026 0.7130 40.2577 0.0248 28.7027 0.0348 15.5441 446.157235 1.4166 0.7059 41.6603 0.0240 29.4086 0.0340 15.9871 470.158336 1.4308 0.6989 43.0769 0.0232 30.1075 0.0332 16.4285 494.620737 1.4451 0.6920 44.5076 0.0225 30.7995 0.0325 16.8682 519.532938 1.4595 0.6852 45.9527 0.0218 31.4847 0.0318 17.3063 544.8835

Example; Home Example; Home MortgageMortgage

Example: Suppose we borrow $75,000 for house at 9% for 30 years. Find monthly payment. Assume that the monthly interest rate is 9/12 = 3/4%.

Gradient SeriesGradient Series

Suppose we have an investment decision which is estimated to return $1,000 in the first year, $1,100 in the second, $1,200 in the third, and so on for the 7 year life of the project.

P

1 2 3

0

7

1600

10001100

1200

Gradient Equivalent Gradient Equivalent FlowsFlows

We can replace the cash flow as the equivalent of an annuity and a constant growth

PA

1 2 3

0

n

A A A A

+

PG

1 2 3

0

n

(n-1)G

G 2G

PW = PA + PG

PA = A (P/A, i, n)

PG = 0(1+i)-1 + G(1+i)-2 + 2G(1+i)-3 +...

Gradient DerivationGradient Derivation

PA

1 2 3

0

n

A A A A

+

PG

1 2 3

0

n

(n-1)G

G 2G

Gradient DerivationGradient Derivation(1+i)PG = G[ (1+i)-1 + 2(1+i)-2 + 3(1+i)-3 +4(1+i)-4 ] PG = G[ (1+i)-2+ 2(1+i)-3 + 3(1+i)-4 +4(1+i)-

5 ]

Gradient DerivationGradient Derivation(1+i)PG = G[ (1+i)-1 + 2(1+i)-2 + 3(1+i)-3 +4(1+i)-4 ] PG = G[ (1+i)-2+ 2(1+i)-3 + 3(1+i)-4 +4(1+i)-5 ]

iPG = G[ (1+i)-1 + (1+i)-2 + (1+i)-3 +(1+i)-4 - 4(1+i)-5 ]

ExampleExample

Estimated return of $1,000 in the first year, $1,100 in the second, $1,200 in the third, and so on for the 7 year life of the project.

PG = A(P/A,10,7) +G(P/G,10,7)

= 1000(4.8684) +100(12.7631)

= $6,144.71

P

1 2 3

0

7

1600

10001100

1200

Determine PW of a depreciation scheme which saves $1000 in the first year and declines by $100 per year for the next 10 years if i = 10%

Depreciation Depreciation ExampleExample

PW

1 2 3

0

10

1000 900 800

100

Example; (cont.)Example; (cont.)

PW = 1000(P/A, 10, 10) - 100(P/G, 10, 10)= 1000(6.1446) - 100(22.8913)= $3,855.47

0

PA

1 2 3 10

1000 10001000 1000

+

PG

1 2 3

0

10

900

100 200

Gradient AlternativeGradient Alternative

A = G(A/G, 10, 10)= 100(3.7255)= $372.55

PW = (1000 - 372.55)(P/A, 10, 10)

= 627.45(6.1446)= $3,855.43

0

PA

1 2 3 10

1000 10001000 1000

+

PG

1 2 3

0

10

900

100 200

Geometric SeriesGeometric Series

Suppose we start a new computer consultant business. We assume that we will start our business with a modest income but that the business will grow at a rate of 10% per year. If we assume an initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows.

1 2 3

0

4

10001100

12101331


P = A1(1 + i)-1 + A2(1 + i)-2 +......+An(1 + i)-n

= A(1 + i)-1 + A(1 + j)(1 + i)-2 + A(1 + j)2(1 + i)-3

+..... + A(1 + j)n - 1(1 + i)-n

= A ( ) ( )1 11

1

j it t

t

n

P

1 2 3

0

n

A

A(1+j)A(1+j)2

A(1+j)n-1


Special Case: For the special case where i = j, we have

P = A

= A

( ) ( )1 11

1

i it t

t

n

( )1 1

1

it

n


Special Case: For the special case where i = j, we have

P = A

= A

P =

( ) ( )1 11

1

i it t

t

n

( )1 1

1

it

n

nA

i( )1

Example; GeometricExample; Geometric

Computer business has initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows. If i = 10%,

1 2 3

0

4

10001100

12101331

ExampleExample

Example: Individual deposits of $1000 in end of year 1 and increases by 10% each year for 30 years. If the account earns 10% per year, how much will he have at end of 30 years?

SummarySummary

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

P = Fn(1 + i)-n = F(P/F,i,n)

Fn

1 2 3 40 n

100

. . . .

SummarySummary

= A(F/A,i,n)

= Fn(A/F,i,n)

0 1 2 3 4

. . . .

n

A

Fn

F Ai

in

n

( )1 1

A Fi

in n

( )1 1

SummarySummary

= A(P/A,i,n)

= P(A/P,i,n)

0

1 2 3 4

. . . .

n

A

P

P Ai

i i

n

n

( )

( )

1 1

1

A Pi i

i

n

n

( )

( )

1

1 1

SummarySummary

PG = G = G(P/G,i,n)

A= G = G (A/G, i, n)

1 2 3

0

n

(n-1)G

G 2G

1

i 1 1

n

i i n

[( )

1 1 12

( )( )ni i

i

n

SummarySummary

Case i jP = A = A(P/A,i,j,n)

F = A = A(F/A,i,j,n)

P

1 2 3

0

n

A

A(1+j)A(1+j)2

A(1+j)n-1

( ) ( )1 1

i j

i j

n n

1 1 1

( ) ( )j i

i j

n n

Break TimeBreak Time

Geometric DerivationGeometric Derivation

0 1 2 3 4

. . . .

n

A

Fn

A i k

k

n

( )10

1

F

Recall the geometric series

xx

xk

k

n n

0

1 1

1

Letting x = (1+i)k

F x kk0

n 1

1

Geometric DerivationGeometric Derivation

0 1 2 3 4

. . . .

n

A

Fn

1

1

x

x

n

F x kk0

n 1

1 1

1 1

1( )

( )

i

i

n 1( )i

i

n

A

F Ai

iA F A i n

n

( )

( , , )1

TM 661 Engineering Economics for Managers Unit 1 Time Value of Money.

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