---
title: "Repeatability, best of A and B"
subject: "Engineering Economics"
chapter: 4
key: "semester_6/engineering_economics:repeatability-a-b-400000-700000"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/repeatability-a-b-400000-700000
site: BCT Engineering
---

# Repeatability, best of A and B

Engineering Economics · Chapter 4: Comparative Analysis of Alternatives

- Asked in: 2069 Poush
- 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/repeatability-a-b-400000-700000
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4

## Question

Recommend the best project from the following information by using repeatability assumption when MARR is 12%.
| Project | A | B |
|---|---|---|
| Initial investment (Rs.) | 4,00,000 | 7,00,000 |
| Annual revenue (Rs.) | 1,75,000 | 2,50,000 |
| Annual cost (Rs.) | 25,000 | 35,000 |
| Salvage value (Rs.) | 40,000 | 70,000 |
| Useful life (year) | 6 | 8 |

## Short answer

Repeatability at 12% MARR: AW(A) = Rs. 57,639/yr, AW(B) = Rs. 79,779/yr. Both positive. Project B has higher annual worth. Select Project B.

## Answer

Repeatability assumption: each project can be repeated indefinitely with the same cash flows, so the comparison is made on annual worth (AW) of one life cycle. MARR = 12%.

AW = -I(A/P, 12%, N) + (Revenue - Cost) + S(A/F, 12%, N)

#### Project A (N = 6)
(A/P, 12%, 6) = 0.2432, (A/F, 12%, 6) = 0.1232
AW(A) = -400,000(0.2432) + (175,000 - 25,000) + 40,000(0.1232)
      = -97,290 + 150,000 + 4,929 = **Rs. 57,639**

#### Project B (N = 8)
(A/P, 12%, 8) = 0.2013, (A/F, 12%, 8) = 0.0813
AW(B) = -700,000(0.2013) + (250,000 - 35,000) + 70,000(0.0813)
      = -140,912 + 215,000 + 5,691 = **Rs. 79,779**

| Project | Life | AW (Rs./yr) |
|---|---|---|
| A | 6 | 57,639 |
| B | 8 | 79,779 |

**Answer: Select Project B**, it has the larger annual worth (Rs. 79,779 against Rs. 57,639), and both are above zero so both meet the MARR.

## Check yourself

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

1. Why use AW here?
   A. MARR is zero
   B. No salvage
   C. Lives are equal
   D. Lives differ (6 and 8) and repeatability applies
   Answer: D. AW over one cycle each works for repeated projects. (Key point: Method)

2. Which AWs?
   A. Both negative
   B. A 79,779; B 57,639
   C. A 29,413; B 37,765
   D. A 57,639; B 79,779
   Answer: D. Computed with 12% factors. (Key point: AW values)

3. Selection?
   A. Project B, higher AW
   B. Project A, lower investment
   C. Neither
   D. Project A
   Answer: A. B has the larger AW and both exceed zero. (Key point: Decision)

## Related in Engineering Economics

- [Repeatability assumption, select best of three projects](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Arepeatability-abc-14) — Engineering Economics, chapter 4, asked 2×
- [AW sensitivity to investment and net annual cash flow](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-aw-500000-75000) — Engineering Economics, chapter 6
- [Sensitivity to investment, revenue and life](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-investment-revenue-life-500000) — Engineering Economics, chapter 6
- [Best project for 4-year and 8-year service periods](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aselect-best-project-fw-4yr-irr) — Engineering Economics, chapter 4
- [Sensitivity on investment, revenue and life; interpret](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-initial-100000-useful-10) — Engineering Economics, chapter 6
- [Repeatability comparison of projects A and B](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Arepeatability-a-b-200000-300000) — Engineering Economics, chapter 4

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