---
title: "Feasibility using two methods, MARR 15%"
subject: "Project Engineering"
chapter: 6
key: "semester_7/project_engineering:feasibility-two-methods-marr-15"
source: https://www.bctengineer.com/q/semester_7/project_engineering/feasibility-two-methods-marr-15
site: BCT Engineering
---

# Feasibility using two methods, MARR 15%

Project Engineering · Chapter 6: Introduction to Project Financing

- Asked in: 2070 Chaitra (old course), 2+6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/project_engineering/feasibility-two-methods-marr-15
- Whole chapter: https://www.bctengineer.com/md/semester_7/project_engineering/chapter-6

## Question

Why is capital budgeting important? Determine the feasibility of the following project using any two methods. MARR is 15%.

| Initial investment | Annual income | Annual O & M | Useful life | Salvage value |
|---|---|---|---|---|
| 1,00,00,000 | 25,00,000 | 4,50,000 | 5 years | 40,00,000 |

## Short answer

Capital budgeting is important because large irreversible funds are committed, affecting growth, risk, and profit. Net annual cash flow is Rs 20,50,000. At MARR 15%, present worth is Rs -11,39,375 and IRR is 10.83%, which is below 15%; thus the project is not feasible.

## Answer

#### Importance of capital budgeting

1. Large funds are committed for a long period.
2. Decisions are irreversible; a wrong decision may cause heavy loss.
3. It decides the future growth and profit of the firm.
4. It affects the firm's risk and its cost of capital.
5. Funds are limited, so projects have to be ranked.
6. It needs a long-term forecast of demand, cost and technology.

#### Feasibility of the project

Net annual cash flow = 25,00,000 - 4,50,000 = Rs 20,50,000. The salvage value of Rs 40,00,000 is received at the end of year 5. MARR = 15%.

**Method 1: Present worth (net present value)**

$$
\begin{aligned}
(P/A, 15\%, 5) &= 3.3522, \quad (P/F, 15\%, 5) = 0.4972\\
PW &= -1{,}00{,}00{,}000 + 20{,}50{,}000 \times 3.3522 + 40{,}00{,}000 \times 0.4972\\
&= -11{,}39{,}375
\end{aligned}
$$

(With exact factors; rounded table factors give about -11,39,190.)

PW is negative, so the project is **not feasible** at MARR = 15%.

**Method 2: Internal rate of return**

Solve $-1{,}00{,}00{,}000 + 20{,}50{,}000\,(P/A,i,5) + 40{,}00{,}000\,(P/F,i,5) = 0$.

- At $i = 10\%$: PW = 2,54,798 (positive)
- At $i = 15\%$: PW = -11,39,375 (negative)

So the IRR lies between 10% and 15%. By trial (interpolating between 10% and 11%), **IRR = 10.83%**, which is less than MARR = 15%, so the project is **not feasible**.

(Check by benefit-cost ratio: $B/C = \dfrac{PV\ \text{of benefits}}{PV\ \text{of costs}} = \dfrac{88{,}60{,}625}{1{,}00{,}00{,}000} = 0.886 < 1$; the annual worth = -3,39,893.)

**Answer: PW = Rs -11,39,375 and IRR = 10.83% < 15%; the project is not feasible.**

## Check yourself

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

1. Net annual cash flow is:
   A. 25,00,000
   B. 20,50,000
   C. 4,50,000
   D. 29,50,000
   Answer: B. Annual income 25,00,000 minus O&M 4,50,000 = 20,50,000. (Key point: Net annual cash flow)

2. The present worth at MARR 15% is negative, so the project is:
   A. Feasible
   B. Break-even
   C. Not feasible
   D. Feasible only with more O&M
   Answer: C. PW = -11,39,375 < 0, so not feasible. (Key point: Present worth)

3. IRR = 10.83% against MARR 15% means:
   A. Accept
   B. Accept since IRR > 10%
   C. Cannot decide
   D. Reject since IRR < MARR
   Answer: D. A project is feasible only if IRR exceeds MARR. (Key point: IRR comparison)

4. Salvage value Rs 40,00,000 is:
   A. Received at the end of year 5 and discounted by (P/F,15%,5)
   B. Received at year 0
   C. Subtracted from cost
   D. Ignored
   Answer: A. It is a year-5 inflow discounted with (P/F,i,5). (Key point: Salvage treatment)

## Related in Project Engineering

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## Related in other subjects

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