Equivalence

Next: Lecture 3

Assignment for Week 4

Read Chapter 7 and Chapter 10 of Fraser, Bernhardt, Jewkes and Tajima; do Problems 7.16, 7.17, 7.22, 7.37, 10.1, 10.5, 10.10, and 10.23 (If you have the first edition of the text, Fraser, Bernhardt and Jewkes, you should read Chapters 7 and 11 and do Problems 7.10, 7.11, 7.16, 7.28, 11.1, 11.5, 11.9, and 11.10.)

Model Answers

Question 7.22:

First, let's make a table showing the cost of computers having the same computational power at different 18-month periods, where Period Zero is the present:

Period	Cost
-2	$320,000
-1	$160,000
0	$80,000
1	$40,000
2	$20,000

So the current computer would cost $80,000 if we wanted to buy one, and the old one we've already got has a salvage value of $40,000.

We consider four possible plans:

Not replace now, keep 3 more years
Replace now, keep 3 more years
Replace in 18 months
Replace now and in 18 months

Plan 1

In 3 years time, the salvage value of the current computer will be $5,000 So the capital cost of keeping it will be the $40,000 salvage we forego by not selling it for salvage right now, less the current value of the salvage cost:

PW(CC)=40,000-5,000(P/F,0.12,3)=$36,441

The maintenance costs will be:

Year 1: accumulated depreciation = 300,000, so maintenance = $30,000

Year 2: accumulated depreciation = 310,000, so maintenance = $31,000

Year 3: accumulated depreciation = 315,000, so maintenance = $31,500

So total present worth is:

PW=36,441+30,000(P/F,0.12,1)+31,000(P/F,0.12,2)+31,500(P/F,0.12,3)=$110,361

Plan 2

A new computer now would cost $80,000, and in three years it would have a salvage value of $10,000. In calculating the capital cost of this, we must not include a credit for the salvage value of the current computer; by charging $40,000 for foregoing the salvage value in Plan 1, we're implicitly assuming that the common background to all four plans is that we do sell the old computer for salvage. (Professors Fraser, Bernhardt, Jewkes and Tajima get this wrong in their book of model answers.)

PW(CC)=92,000-10,000(P/F,0.12,3)=$84,882

The maintenance costs will be:

Year 1: accumulated depreciation = 40,000, so maintenance = $4,000

Year 2: accumulated depreciation = 60,000, so maintenance = $6,000

Year 3: accumulated depreciation = 70,000, so maintenance = $7,000

So total present worth is:

PW=84,882+4,000(P/F,0.12,1)+6,000(P/F,0.12,2)+7,000(P/F,0.12,3)=$98,600

Plan 3

To carry out calculations for a computer bought in 18 months, it is convenient to work out the 18-month interest rate and depreciation rate:

MARR₁₈=(1+MARR₁₂)^1.5=1.12L^1.5-1 = 18.53%

and

d₁₈=1-(1-d₁₂)^1.5=0.646

So the salvage value of the current computer in 18 months will be $40,000(1-0.646)=$14,142. The cost of a new computer in 18 months will be $40,000, and the salvage value of that new computer 18 months later will be $40,000(1-0.646)=$14,142.

So the capital cost of the current computer is the $40,000 foregone by not selling it for salvage immediately, less the present value of the salvage income we get for it in 18 months:

PW(CC1)=40,000-14,142(P/F,0.1853,1)=28,069

Then we add on the capital cost of buying a new computer in 18 months:

PW(CC2)=(46,0000-14,142(P/F,0.1853,1))(P/F,0.1853,1)=28,743

So PW(CC)=PW(CC1)+PW(CC2)=$56,812

And the maintenance costs will be:

Period 1: accumulated depreciation = 305,840, so maintenance = $45,871

Period 2: accumulated depreciation = 25,840, so maintenance = $3,876

So total present worth of Plan 3 is $98,271.

Plan 4: Replace current computer now and in 18 months

Cost of first new computer today = $80,000.

Salvage value of first new computer in 18 months = $28,284

Cost of second new computer after 18 months = $40,000

Salvage value of second new computer after 18 more months= $14,142

PW of first new computer, including installation costs:

PW(CC1)=(96,000-40,000)-28,284(P/F,0.1853,1)=32,138

PW of second new computer, including installation costs:

PW(CC1)=(46,000-14,142(P/F,0.1853,1))(P/F,0.1853,1)=28,743

So PW(CC)= $60,881

And the maintenance costs will be:

Period 1: accumulated depreciation = 51,680, so maintenance = $7,752

Period 2: accumulated depreciation = 25,840, so maintenance = $3,876

So total present worth of Plan 3 is $70,180.

Thus the best choice overall is to replace the defender at once, and again in 18 months.

Question 10.5

Section a

The current dumping of BOD5 is

20,000(1.0)+(9000)(1.5)=33,500 kg/day

If both plants limited their dumping to 0.81 kg/steer, dumping would be (29,000)(0.81)=23,490 kg/day.

The cost of this reduction is as follows:

Plant A: [(0.05)(0.1)+(0.08)(0.09)](20,000)=$244/day

Plant B: [(0.15)(0.5)+(0.35)(0.1)](9,000)=$1354.5/day

For a total cost of $1598.5/day.

Section b

(This is probably the trickiest bit.) If there's a tax of $0.16/kg dumped, each plant will implement only those measures which cost less than $0.16/kg to put into place. So Plant A will reduce dumping by 0.3 kg per steer, and Plant B will reduce dumping by 0.5 kg per steer.

The total reduction in BOD5 will therefore be:

Plant A: 0.3(20000) = 6,000 kg/day

Plant B: 0.5(9000) = 4,500 kg/day

The cost will be:

Plant A: (0.05 + 0.08 + 0.12)(0.1)(20000)=$500/day

Plant B: (0.5)(0.5)(9000) = $675/day

The total is $1,175/day, which is a lower total than with regulation, and yet there's a greater reduction in the total amount dumped.

Sections c and d

The reductions in the amounts dumped would be the same under the subsidy scheme as under the tax. Therefore, the costs would be the same. The behaviour of the two plants would be the same because it would cost A less than $0.16/kg to reduce dumping by 0.3 kg/steer. Greater reductions would cost more than $0.16/kg. It would cost B less than $0.16/kg to reduce dumping by 0.5 kg/steer. Greater reductions would cost more than $0.16/kg/steer. Avoiding paying tax that the company would otherwise pay is a benefit equivalent to gaining a subisdy.

Section e

The tax and subsidy schemes encourage Plant A, whihc has lower costs of reducing BOD5, to reduce dumping more than Plant B, which has higher costs.

John Jones
Fri, Jan 28, 2000