---
title: "Modified BCR with co-termination, AW"
subject: "Engineering Economics"
chapter: 4
key: "semester_6/engineering_economics:modified-bcr-cotermination-aw"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/modified-bcr-cotermination-aw
site: BCT Engineering
---

# Modified BCR with co-termination, AW

Engineering Economics · Chapter 4: Comparative Analysis of Alternatives

- Asked in: 2072 Asoj, 6 marks
- Video: [Benefit cost ratio | Conventional & Modified method | PW.AW.FW | 2018 PU](https://www.youtube.com/watch?v=p8TJ_rz1GqQ) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/modified-bcr-cotermination-aw
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4

## Question

Co-terminating both projects at 5 years and select the best project by modified BCR (using AW formulation). Take MARR = 15%. Salvage value of each project = 10% of first cost.
| PROJECT | First Cost | Annual Benefits | Annual O & M Costs | Useful Life |
|---|---|---|---|---|
| A | Rs. 4,00,000 | Rs. 175,000 | Rs. 25,000 | 6 years |
| B | Rs. 7,00,000 | Rs. 250,000 | Rs. 35,000 | 8 years |

## Short answer

Co-terminated at 5 years, salvage 10% first cost, MARR 15%. Modified B/C: A=1.323, B=1.083. Incremental B/C (B over A) = 0.764 < 1. Select Project A.

## Answer

Co-termination: both projects are used for the same study period of 5 years (the shorter of the two lives is 6, and the study period is stated as 5). The projects are truncated at year 5 and the asset is assumed sold then for its salvage value, taken as **10% of first cost** (assumption: this is the market value at year 5 for both).

Modified B/C (AW formulation) = (AW of benefits - AW of O&M) / CR, where
CR = I(A/P, 15%, 5) - S(A/F, 15%, 5) is the capital recovery cost.
(A/P, 15%, 5) = 0.2983, (A/F, 15%, 5) = 0.1483

#### Project A (I = 400,000, S = 40,000)
CR = 400,000(0.2983) - 40,000(0.1483) = 119,326 - 5,933 = 113,394
Modified B/C = (175,000 - 25,000) / 113,394 = **1.323**

#### Project B (I = 700,000, S = 70,000)
CR = 700,000(0.2983) - 70,000(0.1483) = 208,821 - 10,382 = 198,439
Modified B/C = (250,000 - 35,000) / 198,439 = **1.083**

#### Incremental analysis (B over A)
Extra net benefit = 215,000 - 150,000 = 65,000 per year
Extra CR = 198,439 - 113,394 = 85,045
Incremental B/C = 65,000 / 85,045 = 0.764 < 1

Both are above 1, so each is acceptable alone, but the extra investment in B does not pay for itself.

| Project | CR (Rs./yr) | Modified B/C | AW (net) |
|---|---|---|---|
| A | 113,394 | 1.323 | 36,606 |
| B | 198,439 | 1.083 | 16,561 |

**Answer: Select Project A** (higher modified B/C and incremental B/C of B over A is only 0.76).

## Check yourself

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

1. What does co-terminating at 5 years mean?
   A. Costs ignored
   B. Both repeated
   C. Both are truncated at year 5 and sold at salvage
   D. Both run to 6 years
   Answer: C. A common study period is used with salvage assumed at year 5. (Key point: Co-termination)

2. What is CR?
   A. I(P/A) - S
   B. I/N
   C. I(A/P) - S(A/F)
   D. I(F/P) + S
   Answer: C. CR = I(A/P,15%,5) - S(A/F,15%,5). (Key point: Capital recovery)

3. Modified B/C values for A and B?
   A. A 1.323, B 1.083
   B. Both 1.0
   C. A 0.76, B 1.32
   D. A 1.083, B 1.323
   Answer: A. A = 150,000/113,394 = 1.323; B = 215,000/198,439 = 1.083. (Key point: Ratios)

4. Why is A selected?
   A. A has higher life
   B. B has higher cost
   C. B is rejected for B/C < 1
   D. Higher modified B/C and B's incremental B/C is 0.76 < 1
   Answer: D. The extra investment in B does not pay for itself. (Key point: Selection)

## Related in Engineering Economics

- [IRR to select best of A, B, C (15 years)](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-select-abc-15yr) — Engineering Economics, chapter 4
- [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
- [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
- [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×
- [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
- [Best of three mutually exclusive projects, 10-year study](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Amutually-exclusive-abc-study-10-years) — Engineering Economics, chapter 4

## Related in other subjects

- [Feasibility using two methods, MARR 15%](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Afeasibility-two-methods-marr-15) — Project Engineering, chapter 6
- [RoR best installed capacity (NRs, 45 years)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-nrs-10-percent) — Hydropower Engineering, chapter 3