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PW and IRR after tax with SL depreciation

Short answer

Straight-line depreciation = Rs 5 million/year; after-tax cash flow = Rs 6.5 million/year. PW at 10% = Rs 0.604 million > 0; IRR = 11.4% > MARR, so the project is acceptable.

A machine costs Rs 20 million with no salvage value. Rs 8 million revenues per year can be gained. Given: useful life = 4 years. Tax rate = 50%, MARR = 10%. Use straight line depreciation method to evaluate (i) PW (ii) IRR.

Data (Rs million): P = 20; S = 0; revenue = 8/yr for N = 4 years; tax rate = 50%; MARR = 10%; straight-line depreciation.

After-tax cash flow

Depreciation = (20 − 0)/4 = 5 per year
Taxable income = 8 − 5 = 3
Tax = 0.50 × 3 = 1.5
ATCF = 8 − 1.5 = 6.5 per year (years 1 to 4)

Cash-flow diagram:

Year    0     1     2     3     4
CF    −20   +6.5  +6.5  +6.5  +6.5   (Rs million)

(i) Present worth at 10%

PW = −20 + 6.5(P/A,10%,4) = −20 + 6.5(3.1699) = −20 + 20.604 = Rs 0.604 million

PW > 0, so the project is acceptable.

(ii) IRR (after tax)

Solve −20 + 6.5(P/A,i,4) = 0, i.e. (P/A,i,4) = 20/6.5 = 3.0769.

i = 11%: (P/A,11%,4) = 3.1024 → PW = +0.166
i = 12%: (P/A,12%,4) = 3.0373 → PW = −0.256
IRR = 11% + 1% × 0.166/(0.166 + 0.256) = 11.4%

Answer: After-tax PW = Rs 604,125 (about Rs 0.60 million) and after-tax IRR = 11.4% > MARR = 10%, so the machine is economically acceptable, though the margin is small. (Before tax the project is much better: IRR = 21.9%.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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