Engineering economy effect of time and Interest.ppt

Engineering economics

1. How Time and Interest
Affects Money

2. Single Payment Factor (P/F and F/P)
To determine the amount of money F accumulated after n years (or periods)
from a single present worth P, with interest compounded one time per year
(or periods).

If an amount P is invested at time t=0, the amount F
1 accumulated 1 year
hence at an interest rate of i percent per year will be
F
1
= P + Pi
= P( 1 + i )
where the interest rate is expressed in decimal form.
For n number of years
F = P ( 1 + i )
n
Reordering the formulae gives the value of P








n
i
FP
)1(
1

3. F/P and P/F Factors
Standard notation for this factor

(F/P,6%,20)

(P/F,6%,20)
The first statement presents,

Find F, when P is given
6% is the interest rate

20 is the number of periods

4. Use of Tables for Interest rates i
To simplify routine engineering economy
calculations, tables of factor values have
been prepared for interest rates from 0.25 to
50% and time period from 1 to large n
values.
For a given factor, interest rate, and time,
the correct factor value is found at the
intersection of the factor name and n.

6. Example 2.1
John has accumulated $12,000 that he will invest now. He wants to
calculate the equivalent value after 24 years, when he plans to use all
the resulting money as the down payment on an island vacation home.
Assume a rate of return of 8% per year for each of the 24 years.
Find the amount he can pay down by using
 Standard notation

Factor formulae

Computer

7. Solution:
F = P ( F/P, i, n )
F = P ( 1 + i )
n
FV(8%,24,,12000)

8. Example 2.2
PTCL has completed a study indicating that $50,000, this year, resulted
from improved integrated circuit fabrication technology based on rapidly
changing designs
(a)If PTCL considers these types of savings worth 20% per year, find the
equivalent value of this result after 5 years.
(b)If the $50,000 maintenance savings occurs now, find its equivalent
value 3 years earlier with interest at 20% per year.
Solution
(a)P=$50,000 F=? i=20% per yearn=5 years
F = P ( F/P ,i, n )
(b) P=? F=$50,000 i=20% per year, n=3 years
P = F ( P/F ,i, n )

9. Example 2.3

10. Uniform series

11. Topics of Today
Uniform Series Compound Amount Factor
Sinking Fund Factor
Capital Recovery Factor
Uniform Series Present Worth Factor

12. Uniform Series Compound
Amount Factor (USCAF)
An amount A is deposited at the end of
each period for all periods in the
analysis and we seek to find the
equivalent amount at the end of the
transaction.

13. Sinking Fund Factor (SFF)
A certain amount of money is needed
at some point in time in the future, and
we want to find out the amount of
money that is required to be deposited
at the end of each period starting from
the present and ending at the point in
time in future.

14. Capital Recovery Factor (CRF)
An expenditure has occurred at
present.
  We want to find out the
amount of money that needs to be
deposited at the end of each period for
a given number of periods in the future
so that at the end of the transaction the
series of payments made is equivalent
to the expenditure made at the
beginning.

15. Uniform Series Present Worth
Factor (USPWF)
Given a series of equal amounts of
money received at the end of each
period for a given number of periods,
what is the investment that needs to
be made now so that the future series
of money can be received?
 

16. Summary of Compound
Interest Factors
Name Abbreviatio
n
NotationFormula
Single-payment present worth
factor
SPPWF
Single-payment compound-
amount factor
SPCAF
Uniform-series compound-
amount factor
USCAF
Sinking fund factor SFF
Uniform-series present-worth
factor
USPWF
Capital recovery factor CRF
 / , ,P F i n 1
n
i


 / , ,F P i n 1
n
i
 / , ,F A i n1 1
n
i
i
 
 / , ,A F i n
1 1
n
i
i 
 / , ,P A i n
1 1
n
i
i

 
 / , ,A P i n
1 1
n
i
i

 

18. SINGLE-PAYMENT COMPOUND
AMOUNT FACTOR (SPCAF)
1 2 3 4 n-1n-20
P= given
F= ??
(SPCAF)
n
1
n
F P i 
 / / ,, 1
n
F P F Pin i  

19. 1 2 3 4 n-1n-20
P= ??
F= given
(SPPWF)
n
SINGLE-PAYMENT PRESENT
WORTH FACTOR (SPPWF)
/ 1
n
P F i 
 

1
/ / , ,
1
n
P F P F i n
i
 


20. 1 2 3 4 n-1n-20
F= ??
(USCAF)
n
A =given
UNIFORM-SERIES COMPOUND-
AMOUNT FACTOR (USCAF)
 / , ,F A i n
1 1
*( )
n
i
F A
i
 


21. 11 

n
i
i
FA
 

,,
1 1
n
A i
A
i n
F
F i
 
 
1 2 3 4 n-1n-20
A = ??
(SFF)
n
F = given
SINKING FUND FACTOR (SFF)

22. Uniform-series present-worth factor (USPWF)
 

-
1-1P
P
= , i, n =
A
A
n
i
i


23. Capital-recovery factor (CRF)
 


1
, ,
1 -1
n
n
i iA
Ai n
P
P i

  


24. Uniform Series Present Worth Factor and
Capital Recovery Factor (P/A and A/P)










n
n
ii
i
AP
)1(
11









1)1(
)1(
n
n
i
ii
PA
The term in bracket is called the
uniform-series present worth present worth factor (USPWF)
The term in bracket is called the
Capital recovery factor (CRF)

25. Uniform Series Present Worth Factor and
Capital Recovery Factor (P/A and A/P)

26. Example 2.4
How much money should you be willing to pay now for a guaranteed
$600 per year for 9 years starting next year, at a rate of return of 16%
per year.
Solution:
P=600(P/A,16%,9)
For solution by computers use
PV(16%,9,600)