ENSC 201 Assignment 3: Model Answers

3.1

You have a small company that has grown by 15% every year for the last three years. You sell that company for $500000 in order to start a new company. You borrow $500000 at 10% interest, and sell $500000 worth of stock in the new company, to members of the public who were impressed by the performance of your last company. What is the weighted cost of capital to the new company, and what should the pre-tax MARR be?

The new company has $1 500 000: 500 000 at 10% interest, 500 000 from investors who expect 15% growth, and $500 000 from you, where you also expect 15% growth. So your weighted cost of capital (WCC) is

WCC = (1×0.15 + 0.5×0.1)/1.5 = 13.33%

Your MARR should be at least as great as the WCC.

3.2

You own a small company, developing and marketing a unique technology. Over the past few years it has grown in value to $100 000. You need a large influx of capital, $900 000, to take the company to the next stage. If you can get this capital, then over the next year, you expect the company to increase in value to $2 000 000. You consider two alternative methods: you can sell $400 000 worth of non-voting stock in the company, retaining control and buying $100 000 of stock yourself, using cash from your personal savings. Then you can borrow the remaining $400 000 from the bank at 12% interest. If you follow this plan and the company performs as you expect, how much will you have at the end of the year after paying off the debt?

You decide this is not enough reward for your work, and decide to increase your leverage: you will only sell $100 000 of stock, award yourself $100 000 of stock corresponding to the current assets of the company, and raise the remaining $800 000 from the bank, again at 12% interest. How much more money can you expect to make this way?

Unexpectedly, another company goes into business during the year, selling the same technology. By the end of the year, your company is worth only $900 000. After paying off the debt, how much money would you have under each of the two financial strategies you considered?

Under your first strategy, the initial equity in the company is $600,000: $100,000 from the small company that you own, $100,000 from your personal savings, and $400,000 from the investors. So you own a third of the company and your investors own two-thirds.

Your debt to the bank by the end of the year is $400000(F/P,0.12,1) = $448 000. If the company has grown as you expected, then after paying this off, the company’s assets are $2 000 000 – 448 000 =

$1 552 000. You own one-third of the company, so your personal wealth is now $517 333.

If you use the second strategy, the initial equity in the company is $200 000: $100 000 from the original company, which you own, and $100 000 from your investors. At the end of the year, you pay off a debt of $896 000, leaving the company with $1 104 000. You own one-half of this, so you have $553 000.

If, on the other hand, the company falls in value over the year, the first, low-leverage strategy will leave the company with $900 000 – 448 000 = $452 000. Your one-third share of this is worth $150 666.

If you’ve followed the high-leverage strategy, your company is worth $4 000 at the end of the year, and your half-share of this is worth $2 000. Also, your fellow-investors are not happy with you.

It is also possible to read the question as saying that under the first strategy, you put in just $100 000 (the value of the company) and the other investors put in $400 000. In this case you would own one-fifth of the equity, and have $310 000 if everything went well, and $90 400 if the company drops in value to $900 000.

3.3

A company expects a new machine to save $15 000 per year for eight years, after which it will have a salvage value of $2 000. 80% of the firm's capital is represented by common stock that sells for $30/share, pays annual dividends of $2.70, but has not increased in selling price over the last four years. The other 20% of its capital comes from long-term debt on which the annual interest rate averages 12%. How much can the company pay for the new machine if the investment is to earn twice the cost of capital?

First, the company must find the weighted cost of capital. The effective interest rate on the common stock is 9%, so the WCC = 0.8 × 0.09 + 0.2 × 0.12 = 9.6%.

So if the investment is to earn twice the cost of capital, it has to earn 19.2% interest. So the most it can pay for the machine is 15 000(P/A,0.192,8) + 2 000(P/F,0.192,8) = 58,950 + 491 = $59,441.

It is acceptable to round-off the 19.2% to 20%, since the text doesn’t have 19% interest tables. If we do this, the above calculation becomes:

15 000(P/A,0.2,8) + 2 000(P/F,0.2,8) = 57,555 + 465 = $58,020.

3.4

A newspaper company in Pretoria, South Africa, has annual labour costs of R 12 000 000. The proprietor, John Dube, is considering replacing the company's old equipment with an advanced computer-based system. He can get R 900 000 in scrap metal costs for the old equipment, and it will cost R 6 600 000 to purchase the new system. Both these cash flows would occur in the present moment. Once the new system is in place, he can reduce his labour costs to R 10 000 000 every year. However, he expects it to take a year before the new system is working smoothly, and during that year he will have to go on paying his current labour costs and also pay R 2 000 000 for skilled engineers to manage the transition. Once the system is set up, he will have to pay an additional R 2 000 000 every five years to upgrade it, but as long as he keeps upgrading, the system should last indefinitely. Dube's MARR is 10%.

On the basis of a calculation of equivalent annual cost, decide whether it is worth replacing the system.

The difference in annual costs between the old system and the new system is R 2 000 000 a year. So Dube should express all the one-time costs incurred by the replacement as their annual equivalents and see if they come to less than R 2 000 000.

We are not told what study period to use. It should certainly be longer than five years, but perhaps it would not be reasonable to think that a computer-based system will still be useable in fifty years. We decide to do a first analysis assuming a thirty-year life; it might then be prudent to consider the results for a twenty-year and a forty-year life and see if the answer changes.

Note that we usually treat wages as being paid in a lump sum at the end of the year – this is not realistic, but it is a common convention, and as long as we apply the same convention to all alternatives, it should not lead to major errors. If you treat them as being paid in the middle of the year (the `Mid-Period Convention’), this is more accurate and will give a slightly different answer (though the same conclusion) as the analysis below.

We don’t need to find out the EUAC of both alternatives, keeping the old equipment and buying the new equipment; we can just find the EUAC of the upgrade. Then, if this is less than zero, we upgrade, otherwise we don’t.

The wage saving from the upgrade is R 2 000 000 a year, but this doesn’t start until the end of year 2 (since we consider wages to be paid at year’s end, and we don’t start saving money on wages till the end of year 1.) So we need to turn this into an equivalent EUAC starting at the end of year 1. There are several ways to do this: one is to multiply it by a factor (P/F,i,1) to move it a year closer to the present. Another is to say that it is an EUAC starting at the end of year 1, but that we have an outward cashflow of –R 2 000 000 that cancels it out in the first year. In this example we’ll do the latter.

We express all figures in R 000's:

EUAC = (6 600 – 900)(A/P,0.1, 30) (this is the initial cost, spread out)

+ 2 000(P/F, 0.1, 1) (A/P, 0.1, 30) (this is the cost of the skilled engineers, brought back to the present, then spread out over 30 years)

+ 2 000(P/F, 0.1, 1) (A/P, 0.1, 30) (this is the cost of retaining the old labour force for the first year , brought back to the present, then spread outover 30 years)

+ 2 000((P/F,0.1,6) + (P/F,0.1,11) + (P/F,0.1,16) + (P/F,0.1,21)+ (P/F,0.1,26))(A/P,0.1,30)

(this is the cost of the maintenance every five years)

= (5 700)(0.1061) + (2 000+2 000)(0.909)(0.1061)

+ 2 000(0.5645 + 0.3505 + 0.2176 + 0.1351+ 0.8391)(0.1061)

= 605 + 386 + 287 = 1 278

This cost is much less than the annual saving of R 2 000 000, so John Dube will certainly save money by replacing the system. The margin is sufficiently large that we would not expect the conclusion to change with a ten-year change to the study period.

3.5

The Board of Governors at the University of Gollawalloo in Northern Australia is considering several possible projects for the coming year. In increasing order of first cost, these are:

  1. Hire a former Olympic swim coach for the University swim team. This will cost $120 000 a year, but is expected to increase enrolments by 100 students per year for the next 10 years – that is, the University expects total enrolment to go up by 100 students in the first year, 200 in the second year, and so on. Each additional student increases the University’s profits by $500.
  2. Convert the University’s Philosophy Department to sheep pens. This requires an initial investment of $400 000, but will save $100 000 a year on salaries. This saving will start to be realised after the first year, since the incumbent Philosophy professors must be given a year’s notice. The sheep-breeding business is expected to bring in $50 000 in profit every year. Enrolment will not be significantly affected.
  3. Convert the University’s Olympic-size swimming pool to a sheep dip. This has an initial cost of $450 000, but use of the sheep-dip can be offered to local farmers, bringing in a net profit of $60000 per year. If the University is also running a sheep-breeding business, having a sheep dip on campus will increase the profitability of this business to $90 000 per year, in addition to the income from the local farmers.
  4. Lease the University football fields to the Flinders Mining Company, which will mine them for uranium. Flinders will make an immediate payment of $350 000 for this, but the University will have to agree to restore the fields to their original condition after ten years, at a cost of $200 000. If the football fields are converted to uranium mines, enrolment will drop by 10%, which will lose the University $50 000 a year for the next ten years. This drop is expected to occur even if the former Olympic swim coach is hired for the swim team.

Organize the possible projects into independent projects and a number of mutually exclusive sets. If the University’s MARR is 10%, evaluate each of these in terms of their present worth and recommend which, if any, should be carried out.

The feasible alternatives are:

  1. Hire swim coach
  2. Build sheep pens
  3. Build sheep dip
  4. Build sheep dip and sheep pens
  5. Lease fields

The combinations (1,2), (2,5), (3,5) and (4,5) are also feasible. Their present worths can be calculated by summing the present worths of their component projects.

We proceed to calculate the present worth of each alternative, taking a study period of ten years. All figures are given in $ 000’s:

1. Hire Swim Coach:

PW = - 120(P/A,0.1,10) + (50 +50(A/G,0.1,10))(P/A,0.1,10)

We assume that the increase in enrolment starts at the end of Year 1, and treat it as a base annuity of $50,000, with an arithmetic gradient of $50,000 starting at the end of Year 2.

= - 120(6.1446) + (50 +50(3.7255))( 6.1446)

= 714.5

So the present worth of hiring the swim coach is $714 500.

(Some people used an alternative method of calculating the present worth of the arithmetic gradient, taking the base annuity as zero and having the gradient start in Year 2, then including a factor of (F/P, 0.1, 1) to move the start of the gradient back to Year 1. This method gives a present worth of $521, 700. However, this answer is wrong: the problem is that when the start of the gradient is moved back to Year 1, the end moves back to Year 9, so we miss out on the Year 10 income.)

2. Build Sheep Pens:

PW = - 400 + 150(P/A,0.1,10) = -400 + 150(6.1446) = 521.7

So the present worth of building sheep pens is $521 700.

3. Build Sheep Dip:

PW = -450 + 60(P/A,0.1,10) ) = -450 + 60(6.1446) = -81.3

So the present worth of converting the pool to a sheep dip is -$81 300

4. Build Sheep Pens and Sheep Dip:

PW = - 850 + (100+60+90) (P/A,0.1,10) = -850 + 250(6.1446) = 686.1

So the present worth of building sheep pen and a sheep dip is $686 100.

5. Lease Fields:

PW = 350 - 50(P/A,0.1,10) – 200(P/F,0.1,10)

= 350 – 50(6.1446) – 200(0.3855) = -34.3

So leasing the fields to the Flinders Mining Company has a net present worth of -$34 300.

Since project 5 has a negative present worth, the only combination worth considering is (1, 2). Projects 1 and 2 have no effect on each other’s profitability, so the present worth of the combination is the sum of the individual present worths, which is $1 236 200. This is better than the second-best alternative, building sheep pens and a sheep dip, so the best choice – from the economic point of view - is to hire the swim coach and convert the University’s Philosophy Department to sheep pens.