---
title: "Repeatability, 4-year study, infinite life"
subject: "Engineering Economics"
chapter: 4
key: "semester_6/engineering_economics:select-best-repeatability-study-infinite"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/select-best-repeatability-study-infinite
site: BCT Engineering
---

# Repeatability, 4-year study, infinite life

Engineering Economics · Chapter 4: Comparative Analysis of Alternatives

- Asked in: 2065 Kartik, 16 marks
- Video: [Repeatability assumption |Economics |Board Exam Solution in NEPALI](https://www.youtube.com/watch?v=OfLkWltuILI) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/select-best-repeatability-study-infinite
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4

## Question

Select the best project.
|  | Project A | Project B |
|---|---|---|
| Initial investment (Rs.) | 3,50,000 | 5,00,000 |
| Annual revenues (Rs.) | 1,90,000 | 2,50,000 |
| Annual expenses (Rs.) | 64,500 | 1,38,000 |
| Useful life (years) | 4 | 8 |
| Salvage value at the end of useful life | 0 | 0 |
Use
a) Repeatability assumption
b) Study period is 4 years
c) Infinite project life

## Short answer

Repeatability: AW A = Rs10,268, AW B = Rs11,349; select B. For4-year study with B's salvage Rs2,50,000, AW B = -309; select A. Infinite life: capitalized worth A = Rs85,566, B = Rs94,571; select B.

## Answer

Net annual revenue = revenue − expenses: A = 1,90,000 − 64,500 = 1,25,500; B = 2,50,000 − 1,38,000 = 1,12,000. Salvage values are zero. **MARR is not given; assume 12%.**

#### a) Repeatability assumption (compare AW over own lives)
```text
AW(A) = −3,50,000(A/P,12%,4) + 1,25,500
      = −3,50,000(0.32923) + 1,25,500 = 10,268
AW(B) = −5,00,000(A/P,12%,8) + 1,12,000
      = −5,00,000(0.20130) + 1,12,000 = 11,349
```
B has the higher AW. **Select B.**

#### b) Study period = 4 years
Only the first 4 years of B are used, so B must be sold at year 4. Its value then is not given; assume it falls straight-line from 5,00,000 to 0 over 8 years, i.e. Rs 2,50,000 at year 4.
```text
AW(A, 4 yr) = 10,268
AW(B, 4 yr) = −5,00,000(A/P,12%,4) + 2,50,000(A/F,12%,4) + 1,12,000
            = −164,617 + 52,309 + 1,12,000 = -309
```
With no resale value at all, AW(B) would be even lower (−52,617). Either way A is clearly better. **Select A.**

#### c) Infinite project life
Capitalized worth CW = AW / i (projects repeated for ever).
```text
CW(A) = 10,268 / 0.12 = 85,566
CW(B) = 11,349 / 0.12 = 94,571
```
**Select B.**

| Basis | AW(A) | AW(B) | Choice |
|---|---|---|---|
| a) Repeatability | 10,268 | 11,349 | B |
| b) 4-year study | 10,268 | -309 | A |
| c) Infinite (CW) | 85,566 | 94,571 | B |

**Answer:** B under repeatability and for infinite life; A when only 4 years of service are needed.

## Check yourself

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

1. What net annual revenue is used for A and B?
   A. A = 1,25,500; B = 1,12,000
   B. A = 1,90,000; B = 2,50,000
   C. A = 64,500; B = 1,38,000
   D. A = 1,12,000; B = 1,25,500
   Answer: A. Net annual revenue = revenue - expenses: 1,90,000 - 64,500 and 2,50,000 - 1,38,000. (Key point: Net annual revenue)

2. Under repeatability, how are projects compared and which wins?
   A. PW over own lives; A wins
   B. Payback; A wins
   C. AW over own lives; B wins (11,349 vs 10,268)
   D. AW over own lives; A wins
   Answer: C. AW of one cycle is compared; B's AW is higher so B is chosen. (Key point: Repeatability result)

3. Why does A win when the study period is 4 years?
   A. B's life is shorter
   B. B must be sold at year 4 (its truncated use is costly) and its AW falls to about zero or below
   C. A has salvage value
   D. MARR changes to 0
   Answer: B. Using B for only 4 of its 8 years leaves much of its investment unrecovered, so AW(B) is far below A's 10,268. (Key point: 4-year study period)

4. How is the infinite-life comparison made?
   A. CW = AW x i; A wins
   B. CW = AW / i; B wins (94,571 vs 85,566)
   C. CW = AW / i; A wins
   D. Compare initial investments only
   Answer: B. Capitalized worth is AW divided by i, and B's larger AW gives the larger CW. (Key point: Infinite life)

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