Industrial Cost Analysis: MNET 414

Solution for Home work # 3

Chapter 5

5-2.

P = 30 + 20* (P/A,15%,2) + 30*(P/F,15%,3)
= 30 + 20* (1.626) + 30*(0.6575)
= $ 82.25

5-4.

    First let us find the F equivalent of the 50 uniform: F = 50*(F/A,12%,6)
Then Q is the P equivalent of this F in 4 periods behind,
So Q = F(P/F,12%,4) = 50*(F/A,12%,6)*(P/F,12%,4) = 50*(8.115)*(0.6355) = $257.85

5-14. How much would the owner of a building be justified in paying for a sprinkler system that will save $750 a year in insurance premiums if the system has to be replaced every twenty years and has a salvage value equal to 10% of its initial cost? Assume money is worth 7%.

If the present worth of the system is X then its salvage value would be 0.1X.  Then the present worth

X = 750(P/A, 7%,20) + 0.1X(P/F,7%,20)
OR,  X-0.1X(P/F,7%,20) = 750(P/A, 7%,20)
OR,  X- .02584X = 750(10.594)
OR, X = 750(10.594)/(1-.02584) = $8,156

5-69. An engineer has received two bids for an elevator to be installed in a new building.  The bids, plus his evaluation of of the elevators, are as follows:

The engineer will make a present worth analysis using a 10% interest rate.  Prepare the analysis and determine which bid should be accepted.

Cash flow diagram for Westinghome elevator for 30 years is as follows:

NPC (Net present cost) for Westinghome

= 45000+42000(P/F,10%,10)+42000(P/F,10%,20)-3000(P/F,10%,30)+2700(P/A,10%,30)
= 45000+42000(.3855)+42000(.1486)-3000(.0573)+2700(9.427) = $92,713.20

Cash flow diagram for Itis elevator for 30 years is as follows:

NPC (Net Present Cost)for iTIS

=54000+49500(P/F,10%,15)-4500(P/F,10%,30)+2850(P/A,10%,30)
= 54000+49500(.2394)-4500(.0573)+2850(9.427) = $92,459.40

As Itis has lesser Net Present Cost (NPC), it is the better choice.

5-16. 

Immediate $0.40 hourly wage increase means = 0.4*8*250 = $800.00 increase.
For $0.25 hourly wage increase means additional annual increase = 0.25*8*250 = $500.00 increase

So, assuming $500 increase will continue in next 10 years, the labor cost increases will be

Net Present worth of the above costs at 8% is:

P = 800+1300*(P/A,8%,10) + 500*(P/G,8%,10)
= 800+1300*(6.710) + 500*(25.977) = $22,511.50

5-27.

Monthly interest = 18%/12 = 1.5%

P = 250(P/A,1.5%,36) = 6915.25.

She can afford max 6915.25

5-72

Here no replacement will be made for B.   Consequently, the cash flow table for the alternatives B, C and D will be as following:

Net Present worth (NPW) of each of the alternatives are:

NPW(B) = -50+12*(P/A,10%,5) = -50+12*(3.791) = -4.508

NPW(C) = -30+4.5*(P/A,10%,10) =-30+4.5*(6.145) = -2.3475

NPW(D) = -40+6*(P/A,10%,10) =-40+6*(6.145) = -3.13

As the NPW's are negative, do nothing is the best alternative, as it has a NPW = 0.

5-67.

We know that the capitalized cost (P) = A/i, for a perpetual uniform series A.

Here the annual maintenance is occurring every year, hence, we can calculate its capitalized cost directly by the above formula.   Thus the capitalized costs, taking into account only the first cost and the annual maintenance cost (neglecting the periodic resurfacing cost), will be:

P1(A) = 500,000+350,00/0.12 = 791,666.67

P1(B) = 700,000+2,5000/0.12 = 908,333.33

The periodic resurfacing will occur every 10 years for A and every 15 years for B perpetually. To find its equivalent capitalized cost, we can find the effective interest rates for 10 years and 15 years respectively.

i(effective for 10 years) = (1+.12)^10 - 1 = 2.10585
i(effective for 15 years) = (1+.12)^15 - 1 = 4.473566

Thus the capitalized cost components for the periodic resurfacing are:

P2(A) = 350,000/2.10585 = 166203.81
P2(B) = 450,000/4.473566 = 100590.90

The total capitalized costs are P1+P2:

P(A) = 791,666.67 + 166203.81 = 957,870.48
P(B) = 908,333.33  + 100590.90 = 1,008,924.23

Thus based on capitalized cost Alternative A is more economical.

Note: Example 5-7 on Page 156 shows alternative methods of solving for capitalized cost when the recurring perpetual cost is occurring after n years.