EC2
Polikar
Lecture 11
Engineering Economics – Part II
Annuities & Sinking Funds
Dr. R. Polikar
Recall
1. Draw proper cash-flow diagram (yearly, quarterly, etc.
compounding)
2. Do not add values unless they are located at the same
point on the time-line
3. Make sure your final answer is rounded to nearest $0.01
(no significant digits) - keep all significant digits the calculator
holds in any intermediate steps in the calculation
4. Make sure answer is reasonable
5. The compound interest relations are F = C P
r

C = 1 + 

n
nN
Annuities
Annuities involve a series of equal payments at regular intervals
- examples include installment payments, mortgage payments,
retirement benefits, sinking funds, etc
annuities
Annuities
0
A
A
A
A
1
2
3
4
year
F = A + (1+r)A + (1+r)2A + (1+r)3A
F = A [1 + (1+r) + (1+r)2 + (1+r)3]
F = A (1 + u + u2 + u3)
u=(1+r)
(1)
uF=A(u + u2 + u3 + u4) multiply (1) by u (2)
uF - F = A (u4-1)
subtract (1) from (2)
(1+r)4-1
u4-1
F= A u-1 = A
r
F
r = interest rate
Sinking Funds
F
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
0
1
year
2
3
Future worth:
F = A + (1+i)A + (1+i)2A + (1+i)3A + . . . + (1+i)nN-1 A
r: interest rate
n: compounding period
where
i=
r
n
Sinking Fund
F
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
Future worth = ?
i=
0
1
year
2
F = A + (1+i)A + (1+i)2A + (1+i)3A + . . . + (1+i)nN-1 A
3
r
n
multiply each side of the
equation by (1+i)
Subtract the first equation
F(1+i) = (1+i)A + (1+i)2A + (1+i)3A + . . . + (1+i)nN-1 A + (1+i)nN A from the second (sum of a
geometric series)
F(1+i) - F = i F = (1+i)nN A - A = A [(1+i)nN - 1] = A [(1+ r )nN - 1] = A (C - 1)
n
r
A(C − 1)
F = A(C − 1) ⇒ F =
n
rn
Sinking Funds
F
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
0
1
year
2
3
A (C - 1)
F=
r
n
or F = S A
where
S=
(C - 1)
sinking fund
factor
r
n

r
=
+


C 1
 n
nN
Sinking Funds
Sinking funds are often used to accumulate sufficient money to replace worn-out
or obsolete equipment.
Example: How much money would be accumulated by a sinking fund to
purchase new equipment if $90 is deposited at the end of each month for 3
years at a rate of return of 10 percent (values are compounded monthly)?
F
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
A = 90: n = 12
N = 3: r = 0.10
0
1
C = (1 +
0.1
12
year
)12(3) = 1.348 1818
F = S A = $3 760.36
2
S= C-1
0.1/12
3
= 41.781 821
Example
Installment Loan Example: You want to purchase a new car and you can afford
to spend $200 per month for 3 years compounded monthly. You have a $3 000
down payment. The interest rate is 4%. What price car can you afford (today)?
P1
P2 = $3 000
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
A = 200: n = 12
N = 3: r = 0.04
0
1
year
C = (1 +0.04 12 )12(3) = 1.127272
2
S=
3
C −1
= 38.18157
0.04
12
S
A = $6 774.15
C
purchase price = 6 774.15 + 3 000 = $9 774.15
P1 =
Recall:
F=P*C
F=S*A
Analysis of Alternatives
Analysis of Alternatives: Engineers must make comparison of alternative projects.
Definitions:
First cost - this is the initial cost of the purchase and includes items such as
freight, sales tax, and installation
Life - this refers to the number of years of service the user expects from the item
or property
Salvage - this is the net sum to be realized from the disposal of an item or property
after service. It normally includes recovered value less removal costs, freight out,
etc.
Alternatives must be compared for the same length of time, and replacements due
to short-life expectancies must be considered. The method of comparisons
include:
comparison of present worth called capitalized cost
comparison of uniform costs called average annual cost (or a corresponding
uniform period of time such as average monthly cost)
comparison of future worths
Example
Consider the purchase of two computer-aided-design (CAD) systems. Assume
the annual interest rate is 12% compounded yearly.
Initial cost (P1)
System 1
System 2
$100 000
$65 000
Maintenance and operating cost (A2)$4 000 / year
$8 000 / year
Salvage after 5 years (P3)
$5 000
$18 000
F
P1
0
A2
A2
1
2
year
n = 1: r = 0.12: N1 = 5: N2 = 5: N3 = 0
A2
A2
3
4
A2
P3
5
Example
Sinking fund
Initial cost (P1)
System 1
System 2
$100 000
$65 000
Maintenance and operating cost (A2)$4 000 / year
$8 000 / year
Salvage after 5 years (P3)
$5 000
P1
0
$18 000
A2
A2
1
2
year
A2
A2
3
4
F
A2
P3
5
n = 1: r = 0.12: N1 = 5: N2 = 5: N3 = 0
F = F1 + F2 + F3 = P1 C1 + A2 S2 + P3 C3
C1 = (1 +
0.12)5
= C2 = 1.7623417 : C3 = 1:
S2 =
C2 -1
0.12
= 6.3528475
Example
option 1
F = F1 + F2 + F3 = P1 C1 + A2 S2 + P3 C3
= 100 000(1.7623417) + 4 000(6.3528475) + 18 000
F = $219,645.56
option 2 (lower cost option)
F = F1 + F2 + F3 = P1 C1 + A2 S2 + P3 C3
= 65 000(1.7623417) + 8 000(6.3528475) + 5 000
F = $170,374.99
Example
option 1
F = F1 + F2 + F3 = P1 C1 + A2 S2 + P3 C3
= 100 000(1.7623417) + 4 000(6.3528475) + 18 000
F=
P=
A=
$219,645.56
F
C = $124 632.79
F = $34,574.35 yr -1
S
option 2 (lower cost option)
F = F1 + F2 + F3 = P1 C1 + A2 S2 + P3 C3
= 65 000(1.7623417) + 8 000(6.3528475) + 5 000
F=
P=
A=
$ 170,374.99
F
C = $96,675.34
F
-1
S = $26,818.68 yr
Annuity Summary
Given A, n, N, r ; solve for F, the sinking fund
Given A, n, N, r ; solve for P, the retirement plan
Given P, n, N, r ; solve for A, the mortgage, installment payments
r

C = 1 + 
 n
S=
nN
C −1
;
r
n
; F = CP
F = SA
Solutions of the sinking fund equation for N
C −1
 r
; C = 1+ 
F = S A ; F = CP ; S =
r
 n
n
nN
given r,n,P,a
solve for N
Eliminate F, S, C from these equations:
CP C − 1
r

S A = CP; S =
; C =  1+ 
=
r

A
n
n
−1
C=
=
 rP 

 −1
 nA 
r

 1+ 

n
nN
nN
rP  r 
1−
=  1+ 
n A  n
Solution for N if r, n, P, A are known

rP 
ln  1 −

nA

N= −
r

n ln  1 + 

n
Summary
− nN
Summary
Solutions of the sinking fund equation for r
C −1
 r
; C = 1+ 
F = S A ; F = CP ; S =
r
 n
n
nN
Solution for r if N, n, P, A are known,
no direct solution can be found.
Eliminate F, S, C from these equations:
CP C − 1
r

S A = CP; S =
; C =  1+ 
=
r

A
n
n
−1
C=
=
 rP 

 −1
 nA 
r

 1+ 

n
nN
rP 
r
1−
− 1+ 
n A  n
nN
rP  r 
1−
=  1+ 
n A  n
− nN
=0
− nN
Engineering Economics
F=CP
compound interest formula: used for single payments
F=SA
sinking fund formula: used for uniform payments
Help
r
C ≡ (1 + n )nN
S≡
(C - 1)
r
n
compound interest factor
sinking fund factor
F = future worth: P = present worth: A = uniform amount
r = rate of return: n = compounding time per year: N = years for the project
All calculations in engineering economics (at least for ENG102) can be
done by combinations of these four relations! MAKE SURE YOU HAVE
THEM ON YOUR REFERENCE SHEET and know how to use them!!!