Question 6 of 6: Break-Even Analysis and the Present Value of an Equipment Investment

Question 6: Break-Even Analysis and the Present Value of an Equipment Investment (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A product whose cost structure is stated over a defined output range, at two alternative selling prices; and a machine whose first cost, life, salvage value, annual operating cost and interest rate are all stated.

Find. (a) the break-even output at each of the two selling prices and the effect of the price increase, and (b) the present value of the expenditures charged to the new machine.

Approach. Part (a) equates revenue with total cost, so the break-even output is the fixed cost divided by the contribution per unit; part (b) discounts each cash flow to time zero with the standard series and single-payment factors and sums them.

1. Part (a) — write the break-even condition. At break-even, revenue equals total cost, so $$pQ = FC + vQ \quad\Longrightarrow\quad Q_{BE} = \frac{FC}{p-v}$$ The denominator is the contribution per unit, the money each unit leaves behind after paying its own variable cost.
2. Evaluate at the original price. With a contribution of $20-10 = 10$ dollars per unit, $$Q_{BE,1} = \frac{25{,}000}{20-10} = 2{,}500\ \text{units}$$ $$\boxed{Q_{BE,1} = 2{,}500\ \text{units, at a revenue of 50,000 dollars}}$$ This sits exactly at the top of the 1,500 to 2,500 unit range over which the cost data are stated, so the plant only just breaks even at the highest output for which the answer is valid.
3. Evaluate at the increased price. Raising the price to 25 dollars raises the contribution to 15 dollars per unit, half as much again, so $$Q_{BE,2} = \frac{25{,}000}{25-10} = 1{,}666.7\ \text{units}$$ $$\boxed{Q_{BE,2} = 1{,}667\ \text{units (rounded up), at a revenue of 41,667 dollars}}$$ A fractional unit cannot be sold, so the break-even quantity is always rounded up: at 1,666 units the operation is still a few dollars short.
4. State the effect of the price increase. The break-even output falls by $2{,}500-1{,}667 = 833$ units, that is by 33.3 per cent, and the break-even revenue falls from 50,000 to 41,667 dollars. The reason the quantity falls proportionally further than the price rises is that the price increase of 25 per cent lands entirely on the contribution, which is only half the price, so contribution rises by 50 per cent and the break-even quantity falls by the reciprocal factor. Equivalently, at the 2,500-unit ceiling the higher price now yields a profit of $2{,}500(25) - \left[25{,}000+10(2{,}500)\right] = 12{,}500$ dollars instead of nothing at all.
5. Part (b) — identify the cash flows and their factors. Three flows are charged to the machine: the first cost of 24,000 dollars at time zero, an operating cost of 3,000 dollars at the end of each of eight years, and a salvage receipt of 4,000 dollars at the end of year eight. At 10 per cent, $$(P/A,10\%,8) = \frac{1-(1.10)^{-8}}{0.10} = 5.3349 \qquad (P/F,10\%,8) = (1.10)^{-8} = 0.46651$$
6. Discount and sum. The operating series is worth $3{,}000(5.3349) = 16{,}004.78$ dollars today and the salvage receipt is worth $4{,}000(0.46651) = 1{,}866.03$ dollars today, so $$PV = 24{,}000 + 16{,}004.78 - 1{,}866.03$$ $$\boxed{PV = 38{,}138.75\ \text{dollars}}$$ The salvage value is a receipt and therefore carries the opposite sign to the two disbursements.
7. Express the answer as an equivalent annual cost. Spreading that present value uniformly over the eight-year life gives $$EUAC = PV\,(A/P,10\%,8) = \frac{38{,}138.75}{5.3349} = 7{,}148.88\ \text{dollars per year}$$ which is the figure to compare against a rival machine of a different life, since present values may only be compared over equal study periods.