P1 
Let A: be automatic broom: PA = 35, ACA = 0, NA = 1
Let V be the vacuum cleaner: PV = 90, ACV = 10, NV = 4
Use study period of 4 years.
Present worth method:
Note that A requires investments at the beginning of the year.
NPWA = 35 + 35(P/A, 0.15, 3) = 35 + 35(2.283) = 114.91
NPWV = 90 + 10(P/A, 0.15, 4) = 118.55
Annual Cost method:
NACA = 35(A/P, 0.15, 1) = 35¥1.15 = 40.25.
NACV = 90(A/P, 0.15, 4)+ 10 = 90(0.3503) = 31.52 + 10 = 41.52
Both methods select the automatic broom.

P2 
Comparing the NPW of costs:
NPW1 = 1100 + 200(P/A, 0.15, 10)  100(P/F, 0.15, 10) = 2079
NPW2 = 1300 +152.3(P/A, 0.15, 10)  100(P/F, 0.15, 10) = 2039
The second machine has the smallest net present worth of cost and should
be chosen.
Comparing the NAW of costs or the NAC
NAC1 = 1100(A/P, 0.15, 10) + 200 100(A/F, 0.15, 10) = 414.25
NAC2 = 1300(A/P, 0.15, 10) +152.3  100(A/F, 0.15, 10) = 406.40
The second machine has the smallest net annual cost and should be chosen.

P3 
Compute the NAC of the first machine.
NAC = (P  S)(A/P, i, 10) + Si + A
= 1000(0.1992) + 100(0.15) + 500
= 199.2 + 15 + 500 = 714.2
Second machine
NAC = 1200(0.1992) + 100(0.15) + A = 714.2
A = 714.2  1200(0.1992)  100(0.15) = 714.2  239.04  15 = 460.16
An operating cost of $460 would make the two machines have approximately
the same NAC of 714.

P4 
Present Worth Cost Method
Type A
P = 500, A = 20, n = inf.
PW = 500 + 20(P/A, 0.06, inf.)
= 500 + 20/0.06 = 500 + 333.33
= 833.33
Type B
P = 700, A = 30 every two years, n = inf.
PW = 700 + 30(A/F,i,2)(P/A,i,inf.)
= 700 + 30*0.4854/0.06
= 700 + 242.72 = 942.72
Select type A with the lowest present worth of cost.
Annual Cost Method
NAC(A) = 500(A/P, 0.06, inf) + 20 = $50/year
NAC(B) = 700(A/P, 0.06, inf) + 30(A/F, 0.06, 2) = $56.56 per year
Select type A with lowest annual worth of cost.
Rate of Return Method
NAW(BA) = NAW(B)  NAW(A)
= (700i  30(A/F, i, 2))  (500i  20) = 0
By trial and error we find that ROR for B over A is a little more than
2%. The extra investment is not justified.

P5 
PW Method (study period 10 yrs)
A: PW = 20 + 1.5(P/A, 0.12, 10)  0.5(P/F, 0.12, 10) = 28.314
B: For 5 years
PW = 15 + 1(P/A, 0.12, 5)  0.5(P/F, 0.12, 5) = 18.321
For 10 years
PW = 18.321 + 18.321(P/F, 0.12, 5) = 28.716
Annual Cost Method
A: NAC = 20(A/P, 0.12, 10) + 1.5  0.5(A/F, 0.12, 10) = 5.011
B: NAC = 15(A/P, 0.12, 5) + 1  0.5(A/F, 0.12, 5) = 5.082
Rate of Return Method
NAW(A  B) = NAW(A)  NAW(B) =
= 20(A/P, i, 10  1.5 + 0.5(A/F, i 10)  (15(A/P, i, 5)  1 + 0.5(A/F,
i, 5)
By trial and error the ROR comes out to be slightly less than 14%. Accept
the extra investment in the higher priced car.
All analyses indicate that the higher priced car should be chosen.

P6 
Extra ceiling insulation
NPW = 500 + 100(P/A, 0.1, 7) = 500 + 486.84 = 13.16
NAW = 500(A/P, 0.1, 7) + 100 = 102.7 + 100 = 2.70
Extra insulation plus cooling fans
NPW = 1250 + 250(P/A, 0.1, 7) + 300(P/F, 0.1, 7) = 121
NAW = 1250(A/P, 0.1, 7) + 250 + 300(A/F, 0.1, 7) = 24.86.
The insulation alone does not have a 10% rate of return. The insulation
plus the fans does provide the appropriate return, so invest in that alternative.
The net annual worth of the savings is $25 per year.
ROR return method
Rank alternatives: None (N), Insulation(I), Insulation plus fans (IF)
Compare I  N
NPW = 500 + 100(P/A, i, 7) = 0
ROR is between 9% and 10%. Reject I
Compare IF  N
NPW = 1250 + 250(P/A, i, 7) + 300(P/F, i, 7) = 0
ROR is between 12% and 13%, so accept extra investment.
Choose to install insulation plus fans.

P7 
Using a NAC comparison
NAC(A) = 50(A/P, 0.12, 11)  10(A/F, 0.12, 11) + 5 = 50*0.1684  10*0.0484
+ 5
= 8.42  0.484 + 5 = 12.94/yr
NAC(B) = 40(A/P, 0.12, 10) + 2 = 40*0.1770 + 2 = 7.08 + 2 = 9.08/yr
Clearly B is better.
With the ROR method we would discover that the ROR(A  B) < 0%, so
surely we would reject the extra investment.

P8 
Present Worth Method
Use a study period of 18 years.
Machine A: PW Costs = 9000 + 5000(P/A, 0.1, 18) + 9000(P/F, 0.1, 6) +
9000(P/F, 0.1, 12)
PW Costs = 9000 + 5000(8.201) + 9000(0.5645) + 9000(0.3186) = 57,952.9
Machine B: PW Costs = 16000 + 4000(P/A, 0.1, 18) + 12000(P/F, 0.1, 9)
 4000(P/F, 0.1, 18)
PW Costs = 16000 + 4000(8.201) + 12000(0.4241)  4000(0.1799) = 53,173.6
SELECT B.
Rate of Return Method
Machine A: NAW(A) = 9000(A/P, i, 6)  5000
Machine B: NAW(B) = 16000(A/P, i, 9)  4000 + 4000(F/A, i, 9)
To find ROR of the extra investment set NAW(B  A) = NAW(B)  NAW(A)
= 0
By trial and error the interest rate or ROR is betwen 18 and 20%. Since
the MARR is 10%, the extra investment is surely justified. Select B.

P9 
Net Annual Cost Method
Manual: NAW = 90*700= 6300
Machine Tool: 10000*(A/P,15%,3) + 80*700 = $60,379
Automated: 70000*(A/P,15%,5) + 60*700 = 62,882
Choose the Machine Tool with the smallest net annual cost.
Rate of Return Method
Ranking in order of investment: Manual, Machine Tool, Automated
Find ROR of: (Machine Tool)  (Manual)
NAW(Machine  Manual) = 10000(A/P, i, 3)  56000  (63000)
or (A/P, i, 3) = 0.7
ROR = 49%, therefore accept machine tool over manual.
Find ROR of: (Machine Tool)  (Automated)
NAW(Automated  Machine) = 70000(A/P, i, 5)  42000  [10000(A/P, i,
3)  56000]
70000(A/P, i, 5) + 10000(A/P, i, 3) + 14000 = 0
Trial and error
i

NAW(i)

0

3333

0.05

1503

0.08

348

0.09

46

0.1

445

Rate of return is between 9 and 9%. Since MARR is 15%, reject the extra
investment. Choose the machine tool.
