Economic Analysis of Manufacturing Alternatives: Method R vs. Method S

QUESTION

A company is considering two alternatives for manufacturing a certain part. Method R will have a first cost of $40,000, an annual operating cost of $25,000, and a $10,000 salvage value after its five-year life. Method S will have an initial cost of $100,000, an annual operating cost of $15,000 increasing by $600 each year thereafter, and a $13,000 salvage value after its 10-year life. At an interest rate of 12% per year, which alternative should be chosen? Use both present worth comparison and annual worth comparison (show your calculations).

ANSWER

Economic Analysis of Manufacturing Alternatives: Method R vs. Method S

Introduction

When businesses face the decision to invest in equipment or methods for manufacturing certain parts, they must consider various factors, including initial costs, annual operating expenses, salvage values, and the prevailing interest rate. In this scenario, we will evaluate two manufacturing alternatives: Method R and Method S, using both the present worth and annual worth comparison methods. The goal is to determine which alternative is the most financially viable option at an interest rate of 12% per year.

Method R

Method R has an initial cost of $40,000, an annual operating cost of $25,000, and a $10,000 salvage value after its five-year life. To calculate the present worth of Method R, we need to find the present value of both the initial cost and the annual operating costs over five years, considering the salvage value.

Present Worth of Method R

Initial Cost: $40,000

Annual Operating Cost: $25,000

Salvage Value: $10,000

Using the present worth formula:

PW(R) = Initial Cost – Annual Operating Cost * (P/A, 12%, 5) + Salvage Value * (P/F, 12%, 5)

PW(R) = $40,000 – $25,000 * (3.6048) + $10,000 * (0.5674) PW(R) = $40,000 – $90,120 + $5,674 PW(R) = -$44,446

Method S

Method S entails an initial cost of $100,000, an annual operating cost of $15,000 increasing by $600 each year, and a $13,000 salvage value after its 10-year life. To compute the present worth of Method S, we need to determine the present value of both the initial cost, the increasing annual operating costs, and the salvage value over ten years.

Present Worth of Method S:

Initial Cost: $100,000

Annual Operating Cost (Year 1): $15,000

Annual Operating Cost (Year 2): $15,600 (increasing by $600)

Salvage Value: $13,000

Using the present worth formula:

PW(S) = Initial Cost – Annual Operating Costs * (P/A, 12%, 10) + Salvage Value * (P/F, 12%, 10)

PW(S) = $100,000 – [$15,000 * (6.3522) + $15,600 * (5.6509)] + $13,000 * (0.322)

PW(S) = $100,000 – [$95,283 + $87,860] + $4,186 PW(S) = $100,000 – $183,143 + $4,186 PW(S) = -$78,957

Comparison

Present Worth Comparison

The present worth of Method R is -$44,446, while the present worth of Method S is -$78,957. Since the goal is to minimize costs, Method R is the preferable option as it has a less negative present worth compared to Method S.

Annual Worth Comparison

To calculate the annual worth of each method, we can use the following formula:

AW = PW * (A/P, i, n)

For Method R: AW(R) = -$44,446 * (0.1625) AW(R) = -$7,236

For Method S: AW(S) = -$78,957 * (0.1625) AW(S) = -$12,846

Conclusion

In conclusion, when considering an interest rate of 12% per year, Method R is the economically superior choice for manufacturing the certain part, both in terms of present worth and annual worth. It is essential for businesses to carefully analyze such decisions, as they can significantly impact their financial performance over time.

 

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