TM 661 Engineering Economics for Managers

1
TM 661 Engineering Economics for Managers

• Unit 2
• Time Value of Money

2
Time Value of Money

• The value of money changes over time. 1000
  now has more value that 1000 two years from now.
  
• This is because of the interest rate.

3
Time Value-of-Money

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

F
0 1 t
100
4
Time Value-of-Money

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

F1
F1 100 interest 100 100 (.05)
105
0 1 t
100
5
Time Value-of-Money

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

F2
F2 105 interest 105 105 (.05)
110.25
0 1 2 t
100
105
6
Time Value-of-Money

• In General

F1
F1 P interest P iP P(1i)
0 1 2 t
P
7
Time Value-of-Money

• In General

F2
F1 P interest P iP P(1i)
F2 F1 interest F1 iF1 F1(1i)
0 1 2 t
P
8
Time Value-of-Money

• In General

F2
F1 P(1i) F2 F1 interest F1
iF1 F1(1i) P(1i)(1i) P(1i)2
0 1 2 t
P
9
Time Value-of-Money
Fn

• In General

Fn P(1i)n
0 1 2 3 n
P
10
Example

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

Fn P(1i)n
11
Example

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

Recall, Fn P(1i)n
12
Example

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

Recall, Fn P(1i)n Then, P
Fn(1i)-n
13
Example

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

Then, P Fn(1i)-n 20,000(1.1)-15
20,000(0.2394) 2,394
14
Future Worth Given Annuity

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

15
Annuities Given Future Worth
F A(1i)n-1 A(1i)n-2 A(1i)n-3...
A(1i) A
16
Annuities Given Future Worth
17
Annuities Given Future Worth
18
Annuities Given Future Worth
19
Annuities Given Future Worth
iF A(1i)n - 1

-
n
i
F
)
1
1
(


F
A
A
i
n
(
,
,
)
i
A
20
Annuities Given Future Worth
21
Annuities Given Future Worth
22
Annuities Given Present Worth
, for A1 A2 A3 ... A
23
Annuities Given Future Worth
F
1
2
3
4
n
0
. . . .
A1
A2
A3
An
A4
24
Annuities Given 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.

25
Annuities Given Present Worth

• Recall
• Fn A
• and
• P Fn(1 i) -n

26
Annuities Given Present Worth

• Recall
• and
• P Fn(1 i) -n

27
Annuities Given Present Worth

• Inverting and solving for A gives

28
Car Example

• We have an initial loan of 15,000 at a rate of
  i1 per month for n36 months. The monthly loan
  payment is then

A 15,000.01(1.01)36/(1.0136 - 1)
498.22
29
Car Example Alternative

• We have an initial loan of 15,000 at a rate of
  i1 per month for n36 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

30
Car Example Tables
31
Example Home Mortgage

• 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.

32
Example Home Mortgage

• 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.
• Using the Formula
• A 75,000
• 603.47

33
Home Mortgage

• Using the Table
• A P(A/P, i, n)
• 75,000(A/P, 9, 30)
• 75,000(.0973)
• 7,297.5 / year
• 608.12 / month

34
Gradient 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.

35
Gradient Equivalent Flows

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

36
Gradient Derivation

• PW PA PG
• PA A (P/A, i, n)
• PG 0(1i)-1 G(1i)-2 2G(1i)-3 ...

37
Gradient Derivation

• (1i)PG G (1i)-1 2(1i)-2 3(1i)-3
  4(1i)-4
• PG G (1i)-2 2(1i)-3
  3(1i)-4 4(1i)-5

38
Gradient Derivation

• (1i)PG G (1i)-1 2(1i)-2 3(1i)-3
  4(1i)-4
• PG G (1i)-2 2(1i)-3
  3(1i)-4 4(1i)-5
• iPG G (1i)-1 (1i)-2 (1i)-3
  (1i)-4 - 4(1i)-5

39
Gradient Derivation

• (1i)PG G (1i)-1 2(1i)-2 3(1i)-3
  4(1i)-4
• PG G (1i)-2 2(1i)-3
  3(1i)-4 4(1i)-5
• iPG G (1i)-1 (1i)-2 (1i)-3
  (1i)-4
• A Miracle Occurs
• PG G( 1 - (1ni)(1i)-n )/ i2

40
Example

• 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

41
Depreciation Example

• 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

42
Example (cont.)

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

43
Gradient 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

44
Geometric 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.

45
Geometric Series

• 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

46
Geometric Series

• Special Case For the special case where i j,
  we have
• P A
• A

47
Geometric Series

• Special Case For the special case where i j,
  we have
• P A
• A
• P

48
Geometric Series
??

• Case i j For the case when the interest rate i
  and the growth rate j are not equal, we have
• P A

Miracle 2 Occurs
49
Example 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,
• P 1000(P/A,i,j,n)
• 1000(P/A,10,10,4)
• 1000(3.6363)
• 3,636

50
Example

• 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?

51
Example (Cont).

• Solution
• F A(F/A, i, j, n)
• 1000((F/A, 10, 10, 30)
• 1000(475.8934)
• 475,893

52
Summary
1
2
3
4
0
n
. . . .
100

• Fn P(1 i)n P(F/P,i,n)
• P Fn(1 i)-n F(P/F,i,n)

53
Summary

• A(F/A,i,n)
• Fn(A/F,i,n)

54
Summary
P
1
2
3
4
n
0
. . . .
A

• A(P/A,i,n)
• P(A/P,i,n)

55
Summary
(n-1)G
2G
G

0
1
2
3
n

• PG G G(P/G,i,n)
• A G G (A/G, i, n)

56
Summary

• Case i j
• P A A(P/A,i,j,n)
• F A A(F/A,i,j,n)

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