---
title: "Both types of BCR with FW, growing income"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:bcr-fw-growing-income"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/bcr-fw-growing-income
site: BCT Engineering
---

# Both types of BCR with FW, growing income

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

- Asked in: 2076 Shrawan, 8 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-growing-income
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Calculate both types of BCR using FW formulation when:
- Initial investment is Rs 50000;
- Income is Rs 10000 at the end of first year and increasing by 10% per year;
- Annual expenditure is Rs 2000 at the end of first year and increasing by Rs 200 per year;
- Useful life is 6 years;
- Salvage value is Rs 20000;
- MARR = 15%.

## Short answer

Future worth of benefits and costs are computed at MARR=15%. The conventional B/C ratio is 0.927 and the modified B/C is 0.911, both below 1, so the project should be rejected.

## Answer

Definitions used (Sullivan, FW form):

- Conventional B/C = FW(benefits) / [FW(investment) + FW(O&M) − salvage value]
- Modified B/C = [FW(benefits) − FW(O&M)] / [FW(investment) − salvage value]

Data: I = Rs 50,000; benefit B₁ = 10,000 growing 10% per year; O&M M₁ = 2,000 + 200 per year; n = 6; S = 20,000 (at year 6, so FW = 20,000); i = 15%.

```text
 Benefit : 10,000 -> 11,000 -> 12,100 ... (x1.10 each year)
 O&M     :  2,000 ->  2,200 ->  2,400 ... (+200 each year)
 Year 6  : salvage +20,000
```
| Yr | Benefit | O&M | (F/P,15%,6-t) | FW benefit | FW O&M |
|---|---|---|---|---|---|
| 1 | 10,000.00 | 2,000 | 2.0114 | 20,113.57 | 4,022.71 |
| 2 | 11,000.00 | 2,200 | 1.7490 | 19,239.07 | 3,847.81 |
| 3 | 12,100.00 | 2,400 | 1.5209 | 18,402.59 | 3,650.10 |
| 4 | 13,310.00 | 2,600 | 1.3225 | 17,602.48 | 3,438.50 |
| 5 | 14,641.00 | 2,800 | 1.1500 | 16,837.15 | 3,220.00 |
| 6 | 16,105.10 | 3,000 | 1.0000 | 16,105.10 | 3,000.00 |
| | | | Total | 108,299.95 | 21,179.13 |

Check by formulas: FW(B) = 10,000 [(1.15)^6 − (1.10)^6] / (0.15 − 0.10) = 108,299.95.
FW(I) = 50,000 (1.15)^6 = 50,000 × 2.31306 = 115,653.04.

**Conventional B/C** = 108,300 / (115,653 + 21,179 − 20,000) = 108,300 / 116,832 = **0.927**

**Modified B/C** = (108,300 − 21,179) / (115,653 − 20,000) = 87,121 / 95,653 = **0.911**

**Decision: both ratios are less than 1, so the benefits do not cover the costs at MARR = 15%. The project should be rejected.**

## Check yourself

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

1. What is conventional B/C in FW form?
   A. [FW(B) - FW(O&M)] / [FW(I) - S]
   B. FW(benefits) / [FW(investment) + FW(O&M) - salvage]
   C. FW(I) / FW(B)
   D. FW(B) / FW(I)
   Answer: B. O&M cost is placed in the denominator in the conventional ratio. (Key point: Conventional B/C)

2. What is modified B/C in FW form?
   A. [FW(B) - FW(O&M)] / [FW(I) - S]
   B. FW(B) / [FW(I) + FW(O&M) - S]
   C. [FW(B) + FW(O&M)] / FW(I)
   D. FW(B) / FW(O&M)
   Answer: A. O&M is subtracted from benefits, leaving investment less salvage in the denominator. (Key point: Modified B/C)

3. Benefit grows 10% per year, MARR 15%. Which formula gives FW(B)?
   A. 10,000 x (F/A, 15%, 6)
   B. 10,000 x 6 x 1.10
   C. 10,000[(1.15)^6 - (1.10)^6]/(0.15 - 0.10)
   D. 10,000[(1.15)^6 - 1]/0.15
   Answer: C. For a geometric series FW = B1[(1+i)^n - (1+g)^n]/(i - g) = 108,300. (Key point: Geometric income)

4. Results are 0.927 and 0.911. What is the decision?
   A. Accept; both are near 1
   B. Accept on the modified ratio only
   C. Reject because salvage is too large
   D. Reject; both ratios are below 1
   Answer: D. B/C < 1 means benefits do not cover costs at MARR = 15%. (Key point: Salvage and decision)

## Related in Engineering Economics

- [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
- [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 (FW) for Machine A](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-fw-machine-a-10yr) — 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
- [PW sensitivity to investment, income, life by ±25%](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-pw-400000-60000-25) — 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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