Question

Kiwidale Dairy is considering purchasing a new ice-cream maker. Two models, Smoothie and Creamy, are available and their information is given below. (perform all calculation using 5 significant figures, and give your final answer to 1 decimal place).

(a) What is Kiwidale's MARR that makes the two alternatives equivalent? Use a present worth comparison.

	Smoothie	Creamy
First Cost	19,000	38,000
Service Life	12 years	12 years
Annual profit	4,500	11,200
Annual operating cost	1,200	3,920
Salvage value	2,250	5,400

A MARR which makes the two alternatives equivalent in term of PW is?

Answer 1

Ans.
Cash flow in year 0 = -First Cost

Cash flow in year 1 to 11 = Annual Profit - Annual Operating cost

Cash flow in year 12 = Annual profit - Annual Operating Cost + Salvage value

Thus, present worth of smoothie at MARR = i%, we get,

PWs = -19000 + 3300/(1+i) + 3300/(1+i)^2 + 3300/(1+i)^3 + 3300/(1+i)^4 + 3300/(1+i)^5 + 3300/(1+i)^6 + 3300/(1+i)^7 + 3300/(1+i)^8 + 3300/(1+i)^9 + 3300/(1+i)^10 + 3300/(1+i)^11 + 5550/(1+i)^12

and present worth of creamy at MARR = i%, we get,

PWc = -38000 + 7280/(1+i) + 7280/(1+i)^2 + 7280/(1+i)^3 + 7280/(1+i)^4 + 7280/(1+i)^5 + 7280/(1+i)^6 + 7280/(1+i)^7 + 7280/(1+i)^8 + 7280/(1+i)^9 + 7280/(1+i)^10 + 7280/(1+i)^11 + 12680/(1+i)^12

If the two alternatives have equivalent PW, then,

PWs = PXc

=> -19000 + 3300/(1+i) + 3300/(1+i)^2 + 3300/(1+i)^3 + 3300/(1+i)^4 + 3300/(1+i)^5 + 3300/(1+i)^6 + 3300/(1+i)^7 + 3300/(1+i)^8 + 3300/(1+i)^9 + 3300/(1+i)^10 + 3300/(1+i)^11 + 5550/(1+i)^12 = -38000 + 7280/(1+i) + 7280/(1+i)^2 + 7280/(1+i)^3 + 7280/(1+i)^4 + 7280/(1+i)^5 + 7280/(1+i)^6 + 7280/(1+i)^7 + 7280/(1+i)^8 + 7280/(1+i)^9 + 7280/(1+i)^10 + 7280/(1+i)^11 + 12680/(1+i)^12

=> -19000 + 3980/(1+i) + 3980/(1+i)^2 + 3980/(1+i)^3 + 3980/(1+i)^4 + 3980/(1+i)^5 + 3980/(1+i)^6 + 3980/(1+i)^7 + 3980/(1+i)^8 + 3980/(1+i)^9 + 3980/(1+i)^10 + 3980/(1+i)^11 + 7130/(1+i)^12 = 0

=> i = 18.655%

Thus, at MARR of 18.655%, the two alternatives will be equivalent in terms of PW

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