ENGINEERING ECONOMICS

1
ENGINEERING ECONOMICS

• TIME VALUE FACTORS

2
TIME VALUE FACTORS

• SINGLE PAYMENT COMPOUND AMOUNT FACTOR
• P AMOUNT INVESTED AT TIME t 0
• Fn FUTURE WORTH AFTER n PERIODS
• i INTEREST RATE PER PERIOD

3
TIME VALUE FACTORS

• F1 P P i
• P (1i)
• F2 F1 F1 i
• P(1i) P(1i) i
• P(1i) 1 i
• P(1i)2

4
TIME VALUE FACTORS

• F3 F2 F2 i
• P(1i)2 P(1i)2 i
• P(1i)2 1 i
• P(1 i)3
• (1i)n SPCAF
• SINGLE PAYMENT COMPOUND AMOUNT FACTOR

5
TIME VALUE FACTORS

• SINGLE PAYMENT COMPOUND AMOUNT FACTOR
• F P SPCAF
• SINGLE PAYMENT PRESENT WORTH FACTOR
• P F ( 1\ SPCAF )
• 1/ SPCAF SPPWF

6
TIME VALUE FACTORS
P F SPPWF
F P SPCAF
7
TIME VALUE FACTORS

• UNIFORM SERIES PRESENT WORTH FACTOR
• P A SPPWF(i,1) A SPPWF(i,2) A SPPWF(i,3)
• .... A SPPWF(i,n)
• P A SPPWF(i,1) .........
  SPPWF(i,n) Eq I
• P(1/(1i)) A SPPWF(i,2) ......
  SPPWF(i, n1) Eq II
• subtracting
• P/(1i) - P A SPPWF(i,n1) - SPPWF(i,1)

8
TIME VALUE FACTORS

• P ( -i / (1i) ) A (1/ (1i) )
  1/(1i)n - 1
• algebraic manipulation results in
• P A (1i)n - 1 /
  i (1i)n
• P A USPWF
• UNIFORM SERIES PRESENT WORTH FACTOR

9
TIME VALUE FACTORS

• CAPITAL RECOVERY FACTOR
• REARRANGING THE USPWF
• A P i (1i)n
  / (1i)n - 1
• A P CRF

10
TIME VALUE FACTORS
P
A
A
A
A
A
P A USPWF
A P CRF
11
TIME VALUE FACTORS

• SINKING FUND FACTOR
• A P i (1i)n / (1i)n - 1
• BUT PF ( 1 /
  (1i) n )
• A F i /
  ((1i)n - 1)
• A F SFF

12
TIME VALUE FACTORS

• UNIFORM SERIES COMPOUND AMOUNT FACTOR
• F A 1/ SFF
• F A
  ( (1i)n - 1) / i
• F A USCAF

13
TIME VALUE FACTORS
F
A
A
A
A
A
A
F A USCAF
A F SSF
14
TIME VALUE FACTORS

• STANDARD NOTATION
• (X Y, i,n)
• X VARIABLE TO BE
  CALCULATED
• Y KNOWN VARIABLE
• i INTEREST RATE
  PER PERIOD
• n NUMBER OF PERIODS

15
TIME VALUE FACTORS

• SINGLE PAYMENT PRESENT WORTH (PF,
  i, n)
• SINGLE PAYMENT COMPOUND AMOUNT (FP, i,
  n)
• UNIFORM SERIES PRESENT WORTH
  (PA, i, n)
• CAPITAL RECOVERY
  (AP, i, n)
• SINKING FUND
  (AF, i, n)
• UNIFORM SERIES COMPOUND AMOUNT (FA, i,
  n)

16
TIME VALUE FACTORS

• STANDARD NOTATION MAY BE USED PSEUDO
  ALGEBRAICALLY
• (XY, i, n) (YZ, i, n) (XZ, i, n)

17
EXAMPLE 2.1

• Bill wishes to save enough money to by a new car.
  He will place a sum in a savings account today
  and again in three years in anticipation of
  spending 20,000 in five years. What should the
  amount be?

18
EXAMPLE 2.1

• 20 K S (FP, 12, 5) S (FP, 12, 2)
• S (1.7623) S
  (1.2544)
• S (3.0167)
• S 20 K / 3.0167 6630

19
EXAMPLE 2.2

• Bill wants to make deposits each year for five
  years to buy the 20,000 car. His first payment
  will be one year from today. How big must the
  deposits be if interest is 12 per year?

20
EXAMPLE 2.2

• 20 K A (FA, 12, 5)
• A
  (6.353)
• A 20
  K / 6.353 3148.12

A
A
A
A
A
1
2
3
4
5
20 K
21
TIME VALUE FACTORS

• UNIFORM GRADIENT PRESENT WORTH FACTOR
• G GRADIENT, THE AMOUNT A CASH FLOW CHANGES EACH
  PERIOD STARTING WITH THE END OF PERIOD TWO.

P
1
2
3
4
G
2G
3G
22
TIME VALUE FACTORS

• P G (PF, i, 2) 2G (PF, i, 3) ... (n -1)
  (PF, i, n)
• P G ?2ltjltn ( j-1) (PF,
  i, j) Eq 1
• multiplying by (1i)
  and noting that
• (PF, i, n)
  (1i) (PF, i, n-1)
• P(1i) G ?2ltjltn (j-1)
  (PF, i, j-1) Eq 2

23
TIME VALUE FACTORS

• SUBTRACTING EQ 1
  FROM EQ 2
• P(1i-1) G ?1ltkltn-1 ( 1 / (1i) k) - G
  (n-1) / (1i)n
• 
  MULTIPLYING BY (1/ i)
• P (G/i)
  ?1ltkltn (1 / (1i)k ) - ( G/i) (n / (1i)n )
• RECOGNIZING THAT
  S1/(1i) (PA,i,n) for A1
• P (G/i)
  (PA1, n, i) - (G/i) (n / (1i)n)
• P G 1/i
  (PA1, n, i) - n / (1i) n
• UNIFORM GRADIENT PRESENT WORTH FACTOR
• (PG, i, n)

24
TIME VALUE FACTORS

• UNIFORM GRADIENT ANNUAL WORTH FACTOR
• (AG, i, n) (PG, i, n) (AP, i, n)

25
TIME VALUE FACTORS

• UNIFORM GRADIENT FUTURE WORTH FACTOR
• (FG, i, n) (PG, i, n) (FP, i, n)

26
EXAMPLE 2.3

• 100 is deposited in a saving account one year
  from today, 200 two years from now, 300 three
  years from now, ... , 1000 ten years from now.
  What is the value of the account in ten years if
  interest is 7.
• F 100(FA, 7, 10) 100 (PG, 7, 10)
  (FP,7,10)
• 100 (13.816 (27.716) (1.9672))
  6833.89

27
TIME VALUE FACTORS

• LINEAR INTERPOLATION

TABLE
a / b c / d
i
factor
tabulated
value 1
c d (a / b)
a
c
desired
unlisted
d
b
unlisted value 1 c if value 2 gt value 1
tabulated
value 2
unlisted value 1 - c if value 2 lt value 1
28
EXAMPLE 2.4

• Find (PF, 4, 48)
• (PF, 4, 45) .1712
• (PF, 4, 50) .1407
• a 3, b 5, d .0205
• c .0205 ( 3 / 5) .0123
• (PF, 4, 48) .1712 - .0123 .1589

29
EXAMPLE 2.5

• How much should you be willing to pay for a bond
  which will yield 500 per year for 10 years
  starting a year from today if interest is 10 per
  year?
• P 500 (PA, 10, 10)
• 500 (6.1446) 3022.30

30
EXAMPLE 2.6

• If 1000 is deposited today and 5000 in three
  years, what must the interest be such that the
  value of the account after 8 years is 10,000.

10 K
1000
5000
1
2
3
4
5
6
7
8
31
EXAMPLE 2.6

• 10 K 1000 (FP, i, 8) 5000 (FP, i, 5)
• i (FP, i, 8) (FP, i, 5)
  F
• 9 1.9926 1.5386
  9.68 K
• 10 2.1436 1.6105
  10.19 K

c (a/b) d ( .32 / .51) 1 .627
i 9 .627 9.63
32
EXAMPLE 2.7

• If 1000 is deposited today and 5000 three years
  from now with interest a 8 per year, how long
  must the money be invested until 10 K can be
  withdrawn?

5000
10000
1000
1
2
3
n
33
EXAMPLE 2.7

• 10 K 1000 (FP, 8, n) 5000 (FP, 8, n -3)
• n (FP, 8, n) (FP, 8, n -3)
  F
• 9 1.999 1.5869
  9.93
• 10 2.1589 1.7138
  10.73

c (.07 / .80) 1 .0875
n 9 .0875 9.09
34
PROJECT BALANCE NET WORTH METHOD

• Another method of evaluating the value of a cash
  flow diagram is called the Project Balance Net
  Worth Method. Starting at time zero, PBNWs are
  calculated by PBk1 PBk(1i) Fk1. Where i
  is the interest rate and Fk is the cash flow
  during period k. BPn F, the future value of the
  cash flows.

P
F
1
2
3
n
35
PROJECT BALANCE NET WORTH METHOD

• End of Project
• Period Net Worth
• 0 P
  
• 1 P(1i)
• 2 P(1i)
  (1i) P(1i)2
• n
  P(1i)n -1 (1i) P(1i)n
• 
  F P (1i)n
• Single Payment
  Future Worth Factor

36
PROJECT BALANCE NET WORTH METHOD

• Consider the future worth of a uniform series.

A
A
A
A
A
1
2
3
4
n
F
37
PROJECTED NET WORTH METHOD

• End of Projected
• Period Net Worth
• 0 0
• 1 A
• 2 A(1i) A A
  1 (1i)
• 3 A(1i) A(1i) A
• A 1 (1i) (1i)2
• n A 1 (1i) (1i)2
  (1i)n -1

38
PROJECT BALANCE NET WORTH METHOD