---
title: "PW with differential inflation rates"
subject: "Engineering Economics"
chapter: 8
key: "semester_6/engineering_economics:pw-constant-dollars-differential-inflation"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/pw-constant-dollars-differential-inflation
site: BCT Engineering
---

# PW with differential inflation rates

Engineering Economics · Chapter 8: Inflation and Its Impact on Project Cashflows

- Asked in: 2070 Magh, 5 marks
- Video: [✅2080 to 2069 | Inflation full chapter| 5+ question solved| engineering economics](https://www.youtube.com/watch?v=564K0T_Ws-k) — Revise with Roshan 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/pw-constant-dollars-differential-inflation
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-8

## Question

Evaluate the PW of the following project.
Initial investment = Rs. 1,00,000 (in constant dollars)
Annual sales income = Rs. 40,000 (in constant dollars)
Annual labour cost = Rs. 3,000 (in constant dollars)
Annual material X = Rs. 2,000 (in constant dollars)
Annual material Y = Rs. 1,000 (in constant dollars)
Salvage value = 20% of initial investment (in constant dollars)
Inflation rate for sales income, labour cost, materials X, material Y and salvage value are 5%, 8%, 0%, 6% and 3% respectively for the project period. Take market interest rate = 20%, project life is 4 years.

## Short answer

Inflate each item by its own rate: sales 5%, labour 8%, materials 0% and 6%, salvage 3%. Discount net actual cash flows at 20%: PW ≈ Rs 9,300. The project is marginally acceptable.

## Answer

Each item inflates at its own rate (differential inflation), so each is converted to actual dollars as constant × (1 + f_item)^n and then all net actual cash flows are discounted at the market interest rate of 20%. The initial investment (year 0) needs no conversion.

Salvage value = 20% × 1,00,000 = 20,000 constant dollars, inflating at 3%: 20,000(1.03)⁴ = 22,510.

| Item | Constant \$ | Inflation f |
|---|---|---|
| Sales income | 40,000 | 5% |
| Labour | 3,000 | 8% |
| Material X | 2,000 | 0% |
| Material Y | 1,000 | 6% |
| Salvage | 20,000 | 3% |

| Year | Sales | Labour | Material X | Material Y | Salvage | Net actual CF | PW at 20% |
|---|---|---|---|---|---|---|---|
| 1 | 42,000 | 3,240 | 2,000 | 1,060 | 0 | 35,700 | 29,750 |
| 2 | 44,100 | 3,499 | 2,000 | 1,124 | 0 | 37,477 | 26,026 |
| 3 | 46,305 | 3,779 | 2,000 | 1,191 | 0 | 39,335 | 22,763 |
| 4 | 48,620 | 4,081 | 2,000 | 1,262 | 22,510 | 63,786 | 30,761 |

Net actual CF = sales − labour − X − Y + salvage.
```text
PW = −1,00,000 + 29,750 + 26,026 + 22,763 + 30,761
   = Rs 9,300
```
**Answer:** PW ≈ **Rs 9,300** (positive but small), so the project is just acceptable at the 20% market rate.

## Check yourself

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

1. How are the items converted to actual dollars?
   A. All items at the same average inflation rate
   B. Each item multiplied by (1 + its own inflation rate)^n
   C. Only the sales income is inflated
   D. Each item divided by (1 + f)^n
   Answer: B. Items inflate at different rates, so each uses its own f_item. (Key point: Differential inflation)

2. At what rate are the net actual cash flows discounted?
   A. The market interest rate of 20%
   B. The average inflation rate
   C. An inflation-free rate of 5%
   D. 0%
   Answer: A. Actual-dollar flows are discounted at the market rate, 20%. (Key point: Discount rate)

3. What is the actual-dollar salvage at year 4?
   A. 20,000
   B. 22,510
   C. 24,000
   D. 21,200
   Answer: B. 20% of 1,00,000 = 20,000 inflated at 3%: 20,000(1.03)^4 = 22,510. PW total is about Rs 9,300, slightly acceptable. (Key point: Salvage)

## Related in Engineering Economics

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