Lecture 4 Rates of Return. Summary of comparison methods we know so far: Present worth Future worth Annual worth Still to come: Rate of return Cost/benefit.

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Lecture 4 Rates of Return

Summary of comparison methods we know so far: Present worth Future worth Annual worth Still to come: Rate of return Cost/benefit Payback period

The MARR This is your Minimum Acceptable Rate of Return If you’re in business at all, it’s because you think you can make more money from your investment than the bank can. So, your MARR must be at least as great as the bank rate.

The MARR If all the initial investment in the business is yours, your MARR is whatever you think it should be. If you’ve borrowed some of the money from the bank, your MARR must be enough to pay the interest on the loan.

If your business is owned by its shareholders… your rate of return must be high enough to keep them happy.

The Internal Rate of Return One way of comparing projects is to calculate their rates of return. If the project’s rate of return is less than your MARR, don’t do it. If you have several projects, the one with the highest rate of return might be the best…

The Internal Rate of Return If your project requires a single present investment, P, and yields a single future payout, F, in N years time, then its rate of return is the solution to: P = F(P/F, i, N)

The Internal Rate of Return This can be re-stated as: PW = P - F(P/F, i, N) = 0 More generally, the internal rate of return, or IRR, is the interest rate which makes the total present worth of the project cashflows equal to zero.

Example: We build a bicycle factory at an initial cost of \$800 000. We hire workers at \$100 000 a year, payable at year’s end. In the first five years of the project, our year-end sales are: End of YearSales (\$ 000) 180 2200 3500 41000 51200 What is the rate of return for the project?

A good way of solving these problems is to build a spreadsheet. We work out the value of: PW = - 800 - 20( P/F,i,1) + 100( P/F,i,2) + 400( P/F,i,3) + 900( P/F,i,4) + 1100( P/F,i,5) For different values of i and plot a graph.

i -800-20(P/F, i,1)100(P/F,i,2)400(P/F, i, 3)900(P/F, i,4)1100(P/F, i,5)PW 0.00-800-2010040090011001680 0.05-800-19913467408621220 0.10-800-1883301615683863 0.15-800-1776263515547583 0.20-800-1769231434442360 0.25-800-1664205369360182 0.30-800-155918231529637 0.35-800-1555163271245-81 0.40-800-1451146234205-179 0.45-800-1448131204172-260

This gives us a rate of return of about 32%, which is pretty good. So this is a good way of comparing investments, apart from a couple of possible traps…

Trap 1 Suppose we have a million dollars, and we can either build bicycles or unicycles. The rate of return per unicycle is 50%, and the rate of return per bicycle is 32%. Solution: build unicycles.

Trap 1 Suppose we have a million dollars. We can spend \$600 000 on a gold mine, and get back \$900 000 in a year. Or we can spend the whole million on a silver mine and get back \$1 400 000 in a year. There are no other investment opportunities available except putting money in the bank at 5% interest.

Trap 1 Gold Mine: After a year we have \$900 000 + \$420 000 = \$1 320 000 Rate of return on investment = 50% Silver mine: After a year we have \$1 400 000. Rate of return on investment = 40%

We can avoid this trap by considering the incremental rate of return: Consider three places we can put the \$400,000 left over from buying the gold mine: Old sock under mattress : End up with \$400,000, rate of return = 0% Bank : End up with \$420,000, rate of return = 20,000/400,000 = 5% Cancel gold mine investment, use the \$400,000 to upgrade to silver mine: End up with additional \$100,000*, rate of return = * \$1,400,000 - \$(900,000+400,000) 100 000/400 000 = 25% This is more than the MARR (5%), so we should do it.

General rule for evading Trap 1: Arrange the possible projects from cheapest to most costly. Choose the first one with a rate of return > MARR. Then consider the incremental return from upgrading to the next project. If this is greater than the MARR, choose that next project. Go on comparing the best candidate to the next most costly until you run out of projects.

Trap 2 Example: I have a small research company. A government agency offers me a contract: I get \$20 000 start-up funds now. If all goes well, I put in \$180 000 of my own money in three year’s time to develop a prototype. If the prototype works, I get \$200 000 in five years time. What is my rate of return if all goes well? Solve: PW (\$000) = 20 – 180( P/F,i, 3) + 200( P/F,i,5) = 0

We calculate a graphical solution using a spreadsheet. But it has two solutions, IRR = 16% and IRR = 80%. Which is right?

The second solution, IRR =80%, assumes that I can invest my start-up payment at 80% interest. But there is no reason to think that this is possible. To escape this trap, I have to assume that surplus funds can only be invested at a realistic interest rate, known as the auxiliary rate of return or explicit rate of return. (Often taken as equal to the MARR.) There are two ways of doing this, an exact way and an approximate way. The rate of return for the project that I calculate is called the external rate of return (exact or approximate).

External Rate of Return (Exact) Bring cash inflows forward at the auxiliary rate of return: So after three years the initial \$20 000 becomes 20 000 (F/P, 0.1, 3) = 26, 620 So the amount of my own cash I have to put in is 180 000 – 26 620 = 153 380 So my rate of return on this investment is i e, where 153 380(F/P, i e, 2) = 200 000 So i e = 14.2%

External Rate of Return (Approx) Bring cash inflows to project’s end at the auxiliary rate of return, bring cash outflows to project’s end at the (unknown) approximate external rate of return, equate the future worth to zero and solve for i ea 20(F/P, 0.1, 5) – 180(F/P, i ea, 2) + 200 = 0 (F/P, i ea, 2) = 232/180 i ea = 13.5%

The exact ERR is difficult to calculate if the cashflows are complicated. Both the exact and the approximate ERR’s are single-valued – we never get multiple solutions. Rate of return methods give the same ordering of projects as PW, FW, and AW methods, but make it easy to compare the profitability of projects of different size.

Sunk Costs Problem 1: Quarmby Electronics makes mobile phones. 100 000 batteries will be needed in the coming year. They can be produced in-house by two workers. The salaries of these two workers will be \$50 000 each, and the cost of raw materials is \$1.00 per battery. Alternatively, the batteries can be purchased from an external supplier at a unit cost of \$1.90 per battery. Which option should they choose?

Sunk Costs Problem 2: Quarmby Electronics makes mobile phones. They have just invested \$500 000 in a battery-making machine. This machine requires two workers to operate, and can produce 100 000 batteries per year. The salaries of these two workers will be \$50 000 each. Raw materials for the batteries cost \$1.00 per battery. The machine has no salvage value. Alternatively, the batteries can be purchased from an external supplier at a unit cost of \$1.90 per battery. Which option should they choose?

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