Math Problem Statement

Akash Uni-Safe in Chennai, India, makes Terminator fire extinguishers. The company needs replacement equipment to form the neck at the top of each extinguisher during production.

Machine D E First cost, $ −84,000 −94,000 AOC, $ per year −18,000 −21,000 Salvage value, $ 8,000 10,000 Life, years 4 6

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.

Select between two metal-constricting machines. Use the corporate MARR of 15% per year with future worth analysis using tabulated factors.

The future worth of machine D is $− , and the future worth of machine E is $− .

Solution

To solve this problem, we need to calculate the future worth (FW) of each machine, considering all relevant costs and the corporate Minimum Attractive Rate of Return (MARR) of 15% per year.

We'll use the formula for the future worth of cash flows:

FW=First Cost×(F/P,15%,n)+AOC×(F/A,15%,n)+Salvage Value×(F/F,15%,n)\text{FW} = \text{First Cost} \times (F/P, 15\%, n) + \text{AOC} \times (F/A, 15\%, n) + \text{Salvage Value} \times (F/F, 15\%, n)

Where:

  • F/PF/P is the future worth factor for a single payment at the present
  • F/AF/A is the future worth factor for a uniform series of payments (AOC in this case)
  • F/FF/F is the future worth factor for a future payment (Salvage value in this case)

The factors can be calculated from standard future worth factor tables or formulas. Let’s calculate for each machine.

Machine D (4 years life)

Given:

  • First cost: −84,000
  • AOC: −18,000
  • Salvage value: 8,000
  • Life: 4 years
  1. First cost: FW of First Cost=84,000×(F/P,15%,4)\text{FW of First Cost} = -84,000 \times (F/P, 15\%, 4)
  2. AOC (Annual Operating Cost): FW of AOC=18,000×(F/A,15%,4)\text{FW of AOC} = -18,000 \times (F/A, 15\%, 4)
  3. Salvage value: FW of Salvage Value=8,000×(F/F,15%,4)\text{FW of Salvage Value} = 8,000 \times (F/F, 15\%, 4)

Machine E (6 years life)

Given:

  • First cost: −94,000
  • AOC: −21,000
  • Salvage value: 10,000
  • Life: 6 years
  1. First cost: FW of First Cost=94,000×(F/P,15%,6)\text{FW of First Cost} = -94,000 \times (F/P, 15\%, 6)
  2. AOC (Annual Operating Cost): FW of AOC=21,000×(F/A,15%,6)\text{FW of AOC} = -21,000 \times (F/A, 15\%, 6)
  3. Salvage value: FW of Salvage Value=10,000×(F/F,15%,6)\text{FW of Salvage Value} = 10,000 \times (F/F, 15\%, 6)

Let me now calculate these values based on the future worth factors.The future worth (FW) values for the two machines are:

  • Future worth of Machine D: $−228,797.27
  • Future worth of Machine E: $−391,256.22

Based on the future worth analysis, Machine D has a higher (less negative) future worth, making it the better choice financially compared to Machine E.

Would you like to explore further details or have any questions?

Here are 5 related questions you could explore:

  1. How does changing the MARR affect the future worth analysis?
  2. What is the impact of increasing the salvage value on future worth?
  3. How would a shorter or longer machine life affect future worth?
  4. How would inflation impact these future worth calculations?
  5. What are the advantages of using future worth analysis over present worth analysis?

Tip: Always compare machines over the same time frame; use equivalent annual cost (EAC) for machines with different lives.

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Math Problem Analysis

Mathematical Concepts

Engineering Economics
Future Worth Analysis
Time Value of Money

Formulas

FW = First Cost × (F/P, MARR, n) + AOC × (F/A, MARR, n) + Salvage Value × (F/F, MARR, n)

Theorems

Time Value of Money Theorem

Suitable Grade Level

Undergraduate Engineering