Question

A Ana An advertising campaign will cost $ 200 000 for planning and $ 40 000 in each of the next six years. It is expected to increase revenues permanently by $ 40 000 per year. Additional revenues will be gained in the pattern of an BUY ME arithmetic gradient with $ 20 000 in the first year, declining by $ 5000 per year to zero in the fifth year. What is the IRR of this investment? If the company's MARR is 20 percent, is this a good investment?

engineering economics question, please show all steps

Answer 1

Initial cost = $ 200,000

Operation cost = $ 40,000

Annual increase in revenue = $ 40,000

Additional revenue = $ 20,000

The present worth of the cost is

PWc = 200,000 + 40,000(P/A,i%,6)

The present worth of income is

PW = 40,000/i + 20,000(P/A,i%,5) - 5,000(P/G,i%,5)

The present worth of cost and income must be equal

200,000 + 40,000(P/A,i%,6) = 40,000/i + 20,000(P/A,i%,5) - 5,000(P/G,i%,5)

Simplifying the above equation we get

200 + 40(P/A,i%,6) = 40/i + 20(P/A,i%,5) -5(P/G,i%,5)

40i + 8i(P/A,i%,6) -4i(P/A,i%,5) +i(P/G,i%,5) = 8

Let us assume i = 12%

LHS = 7.7842

Now assume i = 13%

LHS = 8.3319

Now through linear interpolation we get

IRR = 12 + (13 -12 ) ×(8 - 7.7842)/(8.3319 - 7.7842)

IRR = 12 + 0.2158/0.5477

IRR = 12 + 0.39

IRR = 12.39%

Since, the IRR is less than MARR thus, the project is not beneficial.

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