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22-Mec-B4 Integrated Manufacturing Systems · December 2019

Question 6 of 7: Break-Even Analysis and the Profit Effect of a Cost Trade

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2019 — 16-Mec-B4 Integrated Manufacturing Systems. Three hours, OPEN BOOK, any non-communicating calculator permitted. Seven questions are printed and any five constitute a complete paper; all questions are of equal value, so each is worth 20 marks of the 100 available. Note 1 of the paper invites the candidate to state any assumption made where a question is open to interpretation, and this solution uses that licence wherever the source withholds a datum. Every one of the seven questions is worked below, because the set is a study resource rather than a three-hour sitting.

Reference texts (22-Mec-B4 Integrated Manufacturing Systems).

Question 6: Break-Even Analysis and the Profit Effect of a Cost Trade (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.

PartQuantityValue
(a)Fixed cost, $F$15,000 CAD per year
(a)Selling price, $p$20.00 CAD per unit
(a)Present variable manufacturing cost, $v$12.50 CAD per unit
(a)Reduced variable cost, $v'$11.80 CAD per unit
(b)Sales revenue, $S$1,000,000 CAD per year
(b)Profit150,000 CAD per year
(b)Fixed cost250,000 CAD per year
(b)Effect of the new systemvariable cost down 20 per cent, fixed cost up 10 per cent
(b)Sales growth consideredup 5 per cent

Find. (a) the break-even volume before and after the cost reduction, and the price that would restore the original break-even volume; (b) the change in annual profit at constant sales, and the profit if sales also rise 5 per cent.

Approach. Both parts are the same linear cost model, $\pi = (p-v)Q - F$, used in two directions. In (a) the volume is the unknown and then the price is the unknown; in (b) the variable cost is not given directly and must first be recovered by difference from the income statement.

  1. Part (a) — state the break-even condition. Break-even is the volume at which contribution exactly covers fixed cost, $$Q_{BE} = \frac{F}{p - v}$$ where $p - v$ is the contribution margin per unit. Nothing else in the model matters at that point.
  2. Compute the present break-even volume. The present margin is $20.00 - 12.50 = 7.50$ CAD per unit, so $$Q_{BE} = \frac{15{,}000}{7.50}$$ $$\boxed{\ Q_{BE} = 2{,}000\ \text{units per year}\ }$$
  3. Recompute it at the lower manufacturing cost. The margin rises to $20.00 - 11.80 = 8.20$ CAD per unit, so $$Q_{BE}' = \frac{15{,}000}{8.20} = 1{,}829.3$$ $$\boxed{\ Q_{BE}' = 1{,}830\ \text{units per year}\ }$$ a fall of 8.5 per cent. Put the other way, at the old volume of 2,000 units the 70-cent saving is now worth $8.20(2{,}000) - 15{,}000 = 1{,}400$ CAD of profit where before there was none.
  4. Invert the relation to find the price that holds the old break-even point. Requiring $Q_{BE} = 2{,}000$ with the new variable cost gives $$p' = v' + \frac{F}{Q_{BE}} = 11.80 + \frac{15{,}000}{2{,}000} = 11.80 + 7.50$$ $$\boxed{\ p' = 19.30\ \text{CAD per unit}\ }$$ The whole 70-cent cost saving is passed to the customer, which is exactly what holding the break-even volume constant means: the contribution margin has to stay at 7.50 CAD per unit. Rebuilding the income statement at 2,000 units confirms it, $19.30(2{,}000) - [15{,}000 + 11.80(2{,}000)] = 38{,}600 - 38{,}600 = 0$.
  5. Part (b) — recover the variable cost from the income statement. The question gives revenue, profit and fixed cost but not variable cost, so take it by difference, $$V = S - F - \pi = 1{,}000{,}000 - 250{,}000 - 150{,}000 = 600{,}000\ \text{CAD}$$ which is a variable-cost ratio of 0.60 and a contribution ratio of 0.40. This step is the one the question is really testing.
  6. Apply the two changes the new system brings. The variable cost falls by 20 per cent and the fixed cost rises by 10 per cent, $$V' = 0.80(600{,}000) = 480{,}000, \qquad F' = 1.10(250{,}000) = 275{,}000\ \text{CAD}$$ so at unchanged sales $$\pi' = S - V' - F' = 1{,}000{,}000 - 480{,}000 - 275{,}000$$ $$\boxed{\ \pi' = 245{,}000\ \text{CAD, an increase of } 95{,}000\ \text{CAD}\ }$$ that is 63.3 per cent above the present profit. The change can be read directly as the variable saving less the fixed increase, $120{,}000 - 25{,}000 = 95{,}000$ CAD, which is the check worth writing down.
  7. Add the 5 per cent sales increase. Variable cost moves with volume while fixed cost does not, so $$S'' = 1.05(1{,}000{,}000) = 1{,}050{,}000, \qquad V'' = 1.05(480{,}000) = 504{,}000$$ $$\pi'' = 1{,}050{,}000 - 504{,}000 - 275{,}000$$ $$\boxed{\ \pi'' = 271{,}000\ \text{CAD}\ }$$ The extra 26,000 CAD is simply the contribution on the extra sales, $50{,}000 \times (1 - 0.48) = 26{,}000$ CAD, and total profit is 80.7 per cent above the original 150,000 CAD.
  8. Note what the trade has done to risk. Break-even sales in dollars are $F/(1 - V/S)$, which moves from $250{,}000/0.40 = 625{,}000$ CAD to $275{,}000/0.52 = 528{,}800$ CAD. So this particular trade improves profit and lowers the break-even point, because the variable saving outweighs the fixed increase. That is not automatic — substituting fixed cost for variable cost usually raises operating leverage and therefore raises break-even — and it is worth stating explicitly in the recommendation.
QuantityValue
(a) Present contribution margin7.50 CAD per unit
(a) Present break-even volume2,000 units
(a) New contribution margin8.20 CAD per unit
(a) New break-even volume1,830 units (1,829.3)
(a) Price that holds the old break-even point19.30 CAD per unit
(b) Variable cost recovered600,000 CAD (60 per cent of sales)
(b) New variable and fixed cost480,000 and 275,000 CAD
(b) Profit at constant sales245,000 CAD (up 95,000, or 63.3 per cent)
(b) Profit with sales up 5 per cent271,000 CAD
(b) Break-even sales, before and after625,000 and 528,800 CAD