---
title: "PW and IRR after tax with SL depreciation"
subject: "Engineering Economics"
chapter: 7
key: "semester_6/engineering_economics:pw-irr-sl-tax-20-million"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/pw-irr-sl-tax-20-million
site: BCT Engineering
---

# PW and IRR after tax with SL depreciation

Engineering Economics · Chapter 7: Depreciation and Corporate Income Taxes

- Asked in: 2068 Jestha, 10 marks
- Video: [Present Worth of After Tax Cash Flows of an Investment - Engineering Economics Lightboard](https://www.youtube.com/watch?v=2HDrJDcRkgQ) — Engineering Economics Guy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/pw-irr-sl-tax-20-million
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-7

## Question

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.

## 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.

## Answer

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
```text
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:
```text
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.
```text
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%.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With P = 20, S = 0, N = 4 and revenue 8 per year, what is the yearly taxable income under straight-line depreciation?
   A. 8
   B. 5
   C. 3
   D. 6.5
   Answer: C. Depreciation = 20/4 = 5; taxable income = revenue - depreciation = 8 - 5 = 3. (Key point: Taxable income)

2. What is the after-tax cash flow (ATCF) per year at a 50% tax rate?
   A. 6.5 million
   B. 4 million
   C. 5 million
   D. 8 million
   Answer: A. Tax = 0.5 x 3 = 1.5, so ATCF = 8 - 1.5 = 6.5 million. (Key point: After-tax cash flow)

3. What is the after-tax PW at MARR 10%, and what does it show?
   A. -0.604 million; reject
   B. +0.604 million; accept
   C. +6.5 million; accept
   D. +0.604 million; reject
   Answer: B. PW = -20 + 6.5(3.1699) = +0.604 million, which is positive, so the machine is acceptable. (Key point: Present worth)

4. Which equation is solved to find the after-tax IRR?
   A. (P/A,i,4) = 20/8
   B. (P/A,i,4) = 20/5
   C. (P/A,i,4) = 20/1.5
   D. (P/A,i,4) = 20/6.5
   Answer: D. Setting -20 + 6.5(P/A,i,4) = 0 gives (P/A,i,4) = 3.0769, which interpolates to IRR of about 11.4%, above MARR. (Key point: IRR condition)

## Related in Engineering Economics

- [PW of machine with tax at 50%](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Apw-machine-save-4-million-tax) — Engineering Economics, chapter 7
- [After-tax cash flow, sinking fund depreciation](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aaftertax-cash-flow-sinking-fund-25000) — Engineering Economics, chapter 7
- [MACRS schedule and after-tax IRR by AW](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Amacrs-after-tax-irr-aw-45000) — Engineering Economics, chapter 7
- [ATCF and IRR with given depreciation schedule](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aatcf-irr-pw-macrs-style-1000000) — Engineering Economics, chapter 7
- [Depreciation and book value by four methods](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adepreciation-sl-db-soyd-sf-7000) — Engineering Economics, chapter 7
- [After-tax cash flow, DDB with switchover](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aafter-tax-cash-flow-ddb-switchover) — Engineering Economics, chapter 7

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