---
title: "Both types of BCR (FW) for Machine A"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:bcr-fw-machine-a-10yr"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/bcr-fw-machine-a-10yr
site: BCT Engineering
---

# Both types of BCR (FW) for Machine A

Engineering Economics · Chapter 3: Basic Methodologies of Engineering Economic Analysis

- Asked in: 2071 Bhadra, 4 marks
- Video: [Scenario analysis |Future formulation |Engineering Economics |Board Exam Solution in NEPALI](https://www.youtube.com/watch?v=csZ0F12UsF8) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/bcr-fw-machine-a-10yr
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Machine A: initial investment Rs. 6000, annual benefits Rs. 3000, O & M cost Rs. 1000, salvage value Rs. 1500, MARR 10%. Evaluate both types of BCR (FW formulation). Take useful life = 10 years.

## Short answer

Use future worth factors: FW(Benefits)=Rs 47,812, FW(O&M)=Rs 15,937, FW(Investment)=Rs 15,562, salvage=Rs 1,500. Conventional B/C=1.59, modified B/C=2.27 (both >1, acceptable).

## Answer

Data: I = 6,000; annual benefit 3,000; annual O&M 1,000; S = 1,500 (at year 10); n = 10; MARR = 10%.

```text
 yr 0: -6,000      yr 1..10: +3,000 benefit, -1,000 O&M      yr 10: +1,500
```
Factors: (F/A, 10%, 10) = 15.9374, (F/P, 10%, 10) = 2.5937

- FW(B) = 3,000 × 15.9374 = Rs 47,812
- FW(O&M) = 1,000 × 15.9374 = Rs 15,937
- FW(I) = 6,000 × 2.5937 = Rs 15,562
- S = Rs 1,500 (already at year 10)

**Conventional B/C** = FW(B) / [FW(I) + FW(O&M) − S] = 47,812 / (15,562 + 15,937 − 1,500) = 47,812 / 30,000 = **1.594**

**Modified B/C** = [FW(B) − FW(O&M)] / [FW(I) − S] = (47,812 − 15,937) / (15,562 − 1,500) = 31,875 / 14,062 = **2.267**

**Answer: conventional B/C = 1.59, modified B/C = 2.27. Both exceed 1, so Machine A is acceptable.**

## Check yourself

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

1. How is FW of the 3,000 annual benefit found?
   A. 3,000 x (P/A, 10%, 10)
   B. 3,000 x (F/A, 10%, 10) = 47,812
   C. 3,000 x 10
   D. 3,000 x (F/P, 10%, 10)
   Answer: B. Equal annual amounts use the F/A factor. (Key point: FW of benefits)

2. Why is salvage used as 1,500 without a factor?
   A. It occurs at year 10, the FW point
   B. It is negligible
   C. It occurs at year 0
   D. It is multiplied by MARR later
   Answer: A. The FW date is year 10, which is when salvage arrives. (Key point: Salvage in FW)

3. What is the conventional B/C?
   A. 2.27
   B. 1.00
   C. 1.59
   D. 0.63
   Answer: C. 47,812 / (15,562 + 15,937 - 1,500) = 47,812/30,000. (Key point: Conventional B/C)

4. What is the modified B/C?
   A. 1.59
   B. 3.40
   C. 0.44
   D. 2.27, computed as (47,812 - 15,937)/(15,562 - 1,500)
   Answer: D. O&M is subtracted from benefits; both ratios exceed 1, so accept. (Key point: Modified B/C)

## Close variants set in other exams

- [Both types of payback period for Machine A](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Apayback-both-machine-a-5yr) — Engineering Economics, chapter 3

## Related in Engineering Economics

- [Both types of BCR with FW, growing income](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-fw-growing-income) — Engineering Economics, chapter 3
- [Both BCR with FW, gradient revenue](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-fw-gradient-revenue-250000) — Engineering Economics, chapter 3
- [Both types of BCR, gradient income and cost](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-both-growing-12years) — Engineering Economics, chapter 3
- [IRR by AW, both BCR by FW, and ERR](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-aw-bcr-fw-err-250000) — Engineering Economics, chapter 3
- [Sensitivity on initial cost and life; interpret](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-initial-cost-life-25) — Engineering Economics, chapter 6
- [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

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