---
title: "Sensitivity in 10% steps; interpret"
subject: "Engineering Economics"
chapter: 6
key: "semester_6/engineering_economics:sensitivity-30-step-10-20000"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/sensitivity-30-step-10-20000
site: BCT Engineering
---

# Sensitivity in 10% steps; interpret

Engineering Economics · Chapter 6: Risk Analysis

- Asked in: 2065 Kartik, 8 marks
- Video: [Sensitivity Analysis using PW method ||Sensitivity Graph |Board Exam Solution in NEPALI Economics](https://www.youtube.com/watch?v=K1f9_a5tE7E) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/sensitivity-30-step-10-20000
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-6

## Question

Perform sensitivity analysis over ±30% (varying in increment of 10%) in (i) initial investment (ii) annual net revenue (iii) useful life. Draw sensitivity diagram and interpret the result.
Initial investment = Rs. 20,000
Useful life = 10 years
Revenues/Year = Rs. 6,000
Expenses/Year = Rs. 2,000

## Answer

**Assumptions:** MARR is not given; take 10%. No salvage value. Net annual revenue = 6,000 − 2,000 = 4,000. PW = −I + R(P/A,10%,N).

#### Base case
PW₀ = −20,000 + 4,000(P/A,10%,10) = −20,000 + 4,000(6.1446) = **Rs 4,578**

#### PW with changes in steps of 10%
| Change | PW: Initial investment | PW: Net annual revenue | PW: Useful life |
|---|---|---|---|
| -30% | 10,578 | -2,795 | -526 |
| -20% | 8,578 | -337 | 1,340 |
| -10% | 6,578 | 2,120 | 3,036 |
| +0% | 4,578 | 4,578 | 4,578 |
| +10% | 2,578 | 7,036 | 5,980 |
| +20% | 578 | 9,494 | 7,255 |
| +30% | -1,422 | 11,952 | 8,413 |

#### Sensitivity diagram
```text
    12.0 |                                             N
         |   I
    10.1 |
         |                                      N
     8.3 |          I                                  U
         |                               N      U
     6.4 |                 I             U
         |
     4.6 |                        *
         |
     2.7 |                 U             I
         |                 N
     0.9 |          U                           I
    -0.0 |----------N--------------------------------------
    -1.0 |   U
         |                                             I
    -2.8 |   N
         +-------------------------------------------------
            -30    -20    -10     0     10     20     30   
          % change   (y: PW in Rs '000; I = Initial investment, N = Net annual revenue, U = Useful life)
```
#### Interpretation
- **Net annual revenue:** PW changes from -2,795 to 11,952 over the range (swing 14,747). PW becomes zero at about -18.6% change (inside the range tested).
- **Initial investment:** PW changes from -1,422 to 10,578 over the range (swing 12,000). PW becomes zero at about +22.9% change (inside the range tested).
- **Useful life:** PW changes from -526 to 8,413 over the range (swing 8,940). PW becomes zero at about -27.3% change (inside the range tested).

- Base PW = Rs 4,578, so the project is acceptable.
- A 10% fall in revenue costs about Rs 2,458 of PW; a 10% rise in investment costs exactly Rs 2,000; a 10% shorter life costs about Rs 1,542.
- PW becomes zero if revenue falls 19%, investment rises 23% or life falls to about 7.3 years (27% shorter).

**Answer:** The project is acceptable at the base values and PW is **most sensitive to the net annual revenue**, then the initial investment, then the useful life. It turns unacceptable if revenue falls by more than about 19%, investment rises by more than about 23%, or life falls by more than about 27%.

## Check yourself

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

1. How is the PW set up when MARR is not given?
   A. PW = -I + R(P/A,10%,N) with MARR taken as 10%, net revenue 4,000, no salvage
   B. PW = -I + 6,000(P/A,10%,N)
   C. PW = -I + 4,000(A/P,10%,N)
   D. PW = -I + 4,000(P/F,10%,N)
   Answer: A. Net annual revenue is 6,000 - 2,000 = 4,000 and it is discounted as an annuity. (Key point: Assumptions)

2. What is the base PW?
   A. Rs 24,578
   B. Rs 4,578
   C. Rs -4,578
   D. Rs 2,000
   Answer: B. -20,000 + 4,000(6.1446) = 4,578, positive so acceptable. (Key point: Base case)

3. At what change does PW become zero for net annual revenue?
   A. About +22.9%
   B. About -27.3%
   C. About -18.6%
   D. About +18.6%
   Answer: C. Revenue is the most sensitive variable and PW hits zero at about -19%; investment at +23% and life at -27%. (Key point: Break-even changes)

4. What is the order of sensitivity of PW?
   A. Initial investment, then revenue, then life
   B. Useful life, then investment, then revenue
   C. Initial investment, then life, then revenue
   D. Net annual revenue, then initial investment, then useful life
   Answer: D. Swings are 14,747 (revenue), 12,000 (investment) and 8,940 (life). (Key point: Ranking)

## Related in Engineering Economics

- [Sensitivity on investment, revenue and life; interpret](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-initial-100000-useful-10) — Engineering Economics, chapter 6
- [Sensitivity to investment, revenue and life](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-investment-revenue-life-500000) — Engineering Economics, chapter 6
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- [Sensitivity of PW to investment and MARR](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-investment-marr-10-50) — Engineering Economics, chapter 6, asked 2×
