Question

Q1. [40 marks] Answer the following multiple-choice questions (Please note that to score full marks in a question, you have to select the correct option and provide a complete correct solution. If you select the correct option while providing an incomplete solution, you will be awarded 50% of the mark. If you select the correct option without providing a solution or if you select an incorrect option, you will get a zero.): a. [10 marks] A supermarket chain buys loaves of bread from its supplier at $0.5 per loaf and sells them at $0.75 per loaf. The chain is considering two options to bake its own bread. Machine A $8,000 Capital investment Useful life (Years) Annual fixed costs Machine B $12,000 7 $3,500 $2,000 Neither machine has a market value at the end of seven years, and MARR is 10% per year. If the demand for bread at this supermarket is 20,000 loaves per year, what strategy should be adopted for acquiring bread? i. Continue buying from the supplier iii. Install Machine B ii. Install Machine A iv. Install both Machine A and Machine B

Answer 1

Purchasing from supplier:

Cost per loaf of bread = $0.50

Price of a loaf of bread = $0.75

Quantity of bread sold = 20,000 loaves per year

Annual cost = $0.50 * 20,000 = $10,000

Annual revenue = $0.75 * 20,000 = $15,000

Annual profit = Annual revenue - Annual cost = $15,000 - $10,000 = $5,000

Installing Machine A:

Capital investment = $8,000

Annual fixed cost = $2,000

Useful life = 7 years

MARR = 10% per year

From the compound interest factor table, we obtain

(A/P, 10%, 7) = 0.2054

Annualized cost of machine A = $8,000*(A/P, 10%, 7) + $2,000 = $8,000*0.2054 + $2,000 = $3,643.2

Annual revenue = Price * Quantity = $0.75 * 20,000 = $15,000

Annual profit = $15,000 - $3,643.2 = $11,356.8

Installing Machine B:

Capital investment = $12,000

Annual fixed cost = $3,500

Useful life = 7 years

Annualized cost of machine A = $12,000*(A/P, 10%, 7) + $3,500 = $12,000*0.2054 + $3,500 = $5,964.8

Annual revenue = Price * Quantity = $0.75 * 20,000 = $15,000

Annual profit = $15,000 - $5,964.8 = $9,035.2

From the above analysis, it can be observed that the annual profit is maximized when machine A is installed.

Ans: ii. Install Machine A