1. Alternative Classification
Two Types:
Mutually Exclusive
Only one can be selected
Are compared against each other
Independent
More than one can be selected
Are compared only against do-nothing

2. Alternatives Example Solution:
For the alternatives shown below, which should be selected
if they are (a) Mutually exclusive, and (b) Independent
Project ID
Present Worth
A $30,000
B $12,500
C $4,000
D $2,000
Solution:
(a) Select project A
(b) Select projects A, B, & D

3. PW Analysis of Alternatives
Convert all cash flows to PW using MARR
Costs are preceded by minus sign; receipts plus
For mutually exclusive, select numerically largest

4. PW Example Solution: PWX = -20,000 - 9000(P/A,12%,5) + 5000(P/F,12%,5)
Alternative X has a first cost of $20,000, an operating cost of
$9,000 per year, and a $5,000 salvage value after 5 years.
Alternative Y will cost $35,000 with an operating cost of
$4,000 per year and a salvage value of $7,000 after 5 years.
At an MARR of 12% per year, which should be selected?
Solution:
PWX = -20, (P/A,12%,5) (P/F,12%,5)
=-$49,606
PWY = -35, (P/A,12%,5) (P/F,12%,5)
= -$45,447
Select alternative Y

5. Different Life Alternatives
Must compare alts for equal service(i.e. alts must end at the same time)
Two ways to compare for equal service:
(1) Least common multiple(LCM) of lives
(2) Specified planning period
(The LCM procedure is used unless otherwise specified)

6. Different Life Example
Compare the machines shown below on the basis of
their (a) present worth, and (b) future worth. Use i =10%
Machine A
Machine B
First cost,$
Annual cost,$/yr
Salvage value,$
Life, yrs
20,000
30,000
9000
7000
4000
6000 3 6
Solution:
(a) PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) (P/F,10%,6)
= -$68,961
PWB = -$30,000 – 7000(P/A,10%,6) (P/F,10%,6)
= -$57,100
(b) FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3)
= -$122,168
FWB = -30,000(F/P,10%,6) –7000(F/A,10%,6) +6000
= -$101,157
(both methods will always result in the same selection; in this case, machine B)

7. Capitalized Cost Refers to the present worth of an infinite series A
Basic equation is : P = A i
Ex: Cap cost of $2,000 per yr forever at i=10% is $20,000
For finite life alternatives, convert all cash flow into
an A value over one life cycle and then divide by i

8. Capitalized Cost Example
Compare the machines shown below on the basis of
their capitalized cost. Use i =10% per year.
Machine A
Machine B
First cost,$
Annual cost,$/yr
Salvage value,$
Life, yrs
20,000
300,000
9000
7000
4000
----- 3 ∞
First convert machine A cash flow into AW and then divide by i:
AWA = -20,000(A/P,10%,3) – (A/F,10%,3)
= -$15,834
Cap CostA = -15,834/ 0.10 = -$158,340
Cap CostB = -300,000 – 7000/ 0.10 = -$370,000
(Select machine A)

9. Payback Period
Payback period refers to the time it takes(i.e. n) to recover
the initial investment cost (i.e. P) of an investment.
General equation is: 0 = -P  A(P/A,i,n)  F(P/F,i,n)
( frequently requires trial and error solution)
Business persons sometimes use simple payback (ignoring interest).
Such a procedure,while ‘simple’, obviously yields a lower n value
than the correct one.

10. Payback Period Example
Example: A racing team purchased a transporter for $175,000.They will be able
to sell the truck at any time within the next 5 years for $90,000, after which it will
sell for $70,000. If they expect to win an average of $25,000 more per year because
of the truck (i.e. being able to go to more races), how long will it take to recover
their investment at (a) i = 0%, and (b) i = 12% per year?
Solution:
(a) 0 = -175, ,000(n) + 90,000
n = 3.4 ,or 4 years
(b) 0 =- 175, ,000(P/A,12%,n) + 90,000(P/F,12%,n) for n≤5
0 =- 175, ,000(P/A,12%,n) + 70,000(P/F,12%,n) for n>5
By trial and error, n =12.6 , or 13 years

11. Bonds
Bonds are IOU’s wherein entities get money now(V) and repay it later(n),
with interest paid(I) in between.
Important bond information:
b = bond interest rate/yr
c = no. of interest pmts/yr
n = bond maturity date
I = bond interest amt/ period
V = bond face value
The PW of a bond is represented by the following diagram:
where I =
(V)(b) c
PW=? I V 1 2 3 4 n

12. Bond Example Solution: (a) <$6000 (b) $6570 (c) $7120 (d) >$7500
A $10,000 bond with interest at 6% per year, payable semiannually is due in
20 years. The PW of the bond at 10% per year, comp’d semiannually is nearest:
(a) <$ (b) $ (c) $ (d) >$7500
Solution:
I =
(10,000)(0.06)
(2)
= $300 every six months
P = 300(P/A,5%,40) + 10,000(P/F,5%,40)
= $6568
Answer is (b)