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UPSC MainsCivil Engineering (Optional)InfrastructurePractice question

Equipment Selection Using Present Worth Method

There are two alternatives for purchasing a concrete mixer. Both the alternatives have same useful life. The cash flow details of alternatives are as follows: Alternative-1: Initial purchase cost = Rs. 300,000, annual operating and maintenance cost = Rs. 20,000, expected salvage value = Rs. 125,000, useful life = 5 years. Alternative-2: Initial purchase cost = Rs. 200,000, annual operating and maintenance cost = Rs. 35,000, expected salvage value = Rs. 70,000, useful life = 5 years. Using the present worth method, find out which alternative should be selected, if the rate of interest is 10% per year.

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How to approach

Identify the given cash flow parameters for both alternatives and the interest rate. Calculate the relevant discount factors (P/A and P/F) at 10% for a 5-year life. Compute the Net Present Worth of costs for each alternative and select the alternative with the lower present equivalent cost.

Model answer

400 words

Introduction

The Present Worth (PW) method evaluates competing engineering investment alternatives by discounting all future cash flows—including operating expenditures and salvage value—to the present time (time t = 0) at a specified Minimum Attractive Rate of Return (MARR). For mutually exclusive cost-only alternatives having equal service lives, the alternative that minimizes the present worth of total net costs is deemed the economically superior option.

Given Parameters and Discount Factors

Interest rate (i) = 10% per year; Useful life (n) = 5 years.

  • Uniform Series Present Worth Factor (P/A, 10%, 5):
    (P/A, 10%, 5) = [(1 + 0.10)5 - 1] / [0.10(1 + 0.10)5] = (1.61051 - 1) / (0.161051) = 3.7908
  • Single Payment Present Worth Factor (P/F, 10%, 5):
    (P/F, 10%, 5) = 1 / (1 + 0.10)5 = 1 / 1.61051 = 0.6209

1. Evaluation of Alternative-1

For Alternative-1:

  • Initial purchase cost (P1) = Rs. 3,00,000
  • Annual operating & maintenance cost (A1) = Rs. 20,000
  • Salvage value (S1) = Rs. 1,25,000

The Present Worth of Cost (PW1) is given by:

PW1 = Initial Cost + Annual O&M × (P/A, 10%, 5) - Salvage Value × (P/F, 10%, 5)

PW1 = 3,00,000 + 20,000(3.7908) - 1,25,000(0.6209)

PW1 = 3,00,000 + 75,816 - 77,612.50 = Rs. 2,98,203.50 (or approximately Rs. 2,98,203)

2. Evaluation of Alternative-2

For Alternative-2:

  • Initial purchase cost (P2) = Rs. 2,00,000
  • Annual operating & maintenance cost (A2) = Rs. 35,000
  • Salvage value (S2) = Rs. 70,000

The Present Worth of Cost (PW2) is given by:

PW2 = Initial Cost + Annual O&M × (P/A, 10%, 5) - Salvage Value × (P/F, 10%, 5)

PW2 = 2,00,000 + 35,000(3.7908) - 70,000(0.6209)

PW2 = 2,00,000 + 1,32,678 - 43,463 = Rs. 2,89,215

Comparative Summary

  • Present Worth of Cost for Alternative-1: Rs. 2,98,203
  • Present Worth of Cost for Alternative-2: Rs. 2,89,215

Since both alternatives provide equivalent service over equal service lives, the decision criterion is to minimize the total present value of net outflows (PW of costs).

Conclusion

Comparing the present worth of net costs reveals that PW2 is less than PW1 (Rs. 2,89,215 < Rs. 2,98,203). Alternative-2 results in a lower net equivalent cost over the 5-year study period, generating a net savings of Rs. 8,988 in present worth terms. Therefore, Alternative-2 should be selected.

Key facts to remember

definition
Present Worth Method

An engineering economic evaluation technique in which all cash inflows and outflows across the life cycle of a project are discounted to their equivalent value at time t = 0 using a specified discount rate.

definition
Equal-Service Comparison Rule

When comparing mutually exclusive alternatives via the present worth method, the comparison must be conducted over identical service lifespans or a common study period (e.g., least common multiple of lives).

Frequently asked questions

Why is salvage value subtracted in the present worth cost formula?

Salvage value represents a terminal cash inflow (recovery of capital) at the end of the asset's useful life, which reduces the total net cost incurred over the life cycle.

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