Three mutually exclusive design alternatives are being considered. The estimated

Question

Three mutually exclusive design alternatives are being considered. The estimated cash flows for each alternative are given below. The MARR is 20% per year. At the end of the useful life, the investment will be sold. A decision-maker can select one of these alternatives or decide to select none of them Make a recommendation using the PW method. Investment cost Annual expenses Annual revenues Market value Useful life IRR $28,000 $15,000 $23,000 $5,500 10 years $54,000 $13,000 $28,000 $7,500 10 years $42,500 $23,000 $32,000 $9,000 10 years 26.3% 25.2% 17.8%

Explanation / Answer

Consider the given problem, here there are 3 possible investments alternatives and all the cash flows of these 3 investments are given in the question.

So, the PW of “A” is given by.

= (-28,000) + 5,500/(1+0.2)^10 + (23,000 – 15,000)*[{(1+0.2)^10-1}/{0.2(1+0.2)^10}]

= (-28,000) + 5,500/(1.2)^10 + 8,000*[{(1.2)^10-1}/{0.2(1.2)^10}]

= (-28,000) + 888.2807 + 8,000*(5.1917/ 1.2383)

= (-28,000) + 888.2807 + 33,540.8221 = 6,429.1028 = 6,429.

So, the PW of “B” is given by.

= (-54,000) + 7,500/(1.2)^10 + (28,000 – 13,000)*[{(1.2)^10-1}/{0.2(1.2)^10}]

= (-54,000) + 1,211.2918 + 15,000*[{(1.2)^10-1}/{0.2(1.2)^10}]

= (-54,000) + 1,211.2918 + 15,000*(5.1917/ 1.2383)

= (-54,000) + 1,211.2918 + 62,889.0414 = 10,100.3332 = 10,100 > 6,429.

So, the PW of “C” is given by.

= (-42,000) + 9,000/(1.2)^10 + (32,000 – 23,000)*[{(1.2)^10-1}/{0.2(1.2)^10}]

= (-42,000) + 1,453.55 + 9,000*[{(1.2)^10-1}/{0.2(1.2)^10}]

= (-42,000) + 1,453.55 + 9,000*(5.1917/ 1.2383)

= (-42,000) + 1,453.55 + 37,733.4248 = (-2,813.02) = (-2,813) < 10,100.

Now, if we compare all these alternatives we will get that the “B” project is more profitable, since it having max PW compared to the other alternatives. So, among all these alternative "B" most preferable alternatives.

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