---
title: "Find required annual receipts A"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:required-annual-receipts-a-1000000"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/required-annual-receipts-a-1000000
site: BCT Engineering
---

# Find required annual receipts A

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

- Asked in: 2065 Kartik, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/required-annual-receipts-a-1000000
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Find the required annual receipts 'A' for the following investment proposal:
Initial investment = Rs. 10,00,000
Salvage value = Rs. 1,00,000
O & M expenses/year = Rs. 50,000
| End of year | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Benefits | A+70,000 | A+80,000 | A+90,000 | A+100,000 | A+110,000 |

## Short answer

Required A ≈ Rs 209,316/year at MARR 10% by setting present worth to zero. Benefits are uniform A plus gradient (base Rs 70,000, G=Rs 10,000), O&M Rs 50,000/year, salvage Rs 1,00,000 at year 5.

## Answer

The MARR is not given; assumption: MARR = 10% per year. "Required A" means the smallest A that makes the project just earn the MARR, i.e. PW = 0.

```text
 yr 0: -10,00,000
 yr 1..5: benefit A + 70,000, A + 80,000, A + 90,000, A + 1,00,000, A + 1,10,000
          O&M -50,000 every year
 yr 5: salvage +1,00,000
```
The benefits are a uniform A plus a gradient series with base 70,000 and G = 10,000.

Factors at 10%, 5 years: (P/A) = 3.7908, (P/G) = 6.8618, (P/F) = 0.6209

#### Equation
PW = 0:

−10,00,000 + A (P/A) + 70,000 (P/A) + 10,000 (P/G) − 50,000 (P/A) + 1,00,000 (P/F) = 0

- A × 3.7908 = 10,00,000 − 70,000 × 3.7908 − 10,000 × 6.8618 + 50,000 × 3.7908 − 1,00,000 × 0.6209
- A × 3.7908 = 10,00,000 − 265,355 − 68,618 + 189,539 − 62,092
- A × 3.7908 = 793,474
- A = 793,474 / 3.7908

**Answer: A ≈ Rs 209,316 per year at MARR = 10%.** Each year's benefit is then A + 70,000, A + 80,000 and so on.

Sensitivity: the required A changes with the MARR as follows.

| MARR | 8% | 10% | 12% | 15% |
|---|---|---|---|---|
| Required A (Rs) | 194,946 | 209,316 | 223,923 | 246,256 |

## Check yourself

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

1. What does 'required annual receipts A' mean?
   A. Largest A the firm can afford
   B. A that makes payback 1 year
   C. Smallest A making PW = 0 at the MARR
   D. A that makes salvage zero
   Answer: C. Required A is the minimum receipt that gives the project exactly the MARR, i.e., PW = 0. (Key point: Meaning of required A)

2. How are the benefits A+70,000 ... A+110,000 split for analysis?
   A. Geometric series
   B. Gradient of 70,000
   C. Uniform A plus uniform 70,000 plus gradient G = 10,000
   D. Uniform A only
   Answer: C. Uniform A, a base uniform of 70,000, and an arithmetic gradient of 10,000 per year. (Key point: Series decomposition)

3. In the PW = 0 equation, how are O&M and salvage treated?
   A. O&M subtracted using (P/A), salvage added using (P/F)
   B. Both ignored
   C. Both added
   D. O&M added, salvage subtracted
   Answer: A. O&M is a yearly cost (-50,000 (P/A)) and salvage is a year-5 inflow (+100,000 (P/F)). (Key point: Treatment of salvage and O&M)

4. At MARR = 10%, what is the required A?
   A. About Rs 2,09,316
   B. About Rs 1,94,946
   C. About Rs 7,93,474
   D. About Rs 3,79,000
   Answer: A. A = 793,474 / 3.7908 = about 209,316. (Key point: Result)

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