Two processes can be used for producing a poly- mer that reduces friction loss in engines. Process T will have a first cost of $750,000, an operating cost of $60,000 per year, and a salvage value of $80,000 after its 2-year life. Process W will have a first cost of $1,350,000, an operating cost of $25,000 per year, and a $120,000 salvage value after its 4-year life. Process W will also require updating at the

ENGR.ECONOMIC ANALYSIS
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Chapter1: Making Economics Decisions
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Problem 1QTC
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From the solution, where did the 670,000 come from?
The image presents a financial analysis comparing the future worth (FW) of two processes, T and W, using the formula for future worth calculations.

### Calculations for Process T (FW_T):

- **Initial Formula:**
  \[
  FW_T = -750,000(F/P, 12\%, 4) - 60,000(F/A, 12\%, 4) - 670,000(F/P, 12\%, 2) + 80,000
  \]

- **Substituting Factors:**
  \[
  = -750,000(1.5735) - 60,000(4.7793) - 670,000(1.2544) + 80,000
  \]

- **Calculation Result:**
  \[
  = -\$2,227,331
  \]

### Calculations for Process W (FW_W):

- **Initial Formula:**
  \[
  FW_W = -1,350,000(F/P, 12\%, 4) - 25,000(F/A, 12\%, 4) - 90,000(F/P, 12\%, 2) + 120,000
  \]

- **Substituting Factors:**
  \[
  = -1,350,000(1.5735) - 25,000(4.7793) - 90,000(1.2544) + 120,000
  \]

- **Calculation Result:**
  \[
  = -\$2,236,604
  \]

### Conclusion:

The analysis concludes by selecting process T, citing a small margin difference of only $9,273 in future worth (FW) favoring process T over process W. The point of this analysis is to determine the more favorable financial outcome by comparing the future worth of cash flows associated with each process.
Transcribed Image Text:The image presents a financial analysis comparing the future worth (FW) of two processes, T and W, using the formula for future worth calculations. ### Calculations for Process T (FW_T): - **Initial Formula:** \[ FW_T = -750,000(F/P, 12\%, 4) - 60,000(F/A, 12\%, 4) - 670,000(F/P, 12\%, 2) + 80,000 \] - **Substituting Factors:** \[ = -750,000(1.5735) - 60,000(4.7793) - 670,000(1.2544) + 80,000 \] - **Calculation Result:** \[ = -\$2,227,331 \] ### Calculations for Process W (FW_W): - **Initial Formula:** \[ FW_W = -1,350,000(F/P, 12\%, 4) - 25,000(F/A, 12\%, 4) - 90,000(F/P, 12\%, 2) + 120,000 \] - **Substituting Factors:** \[ = -1,350,000(1.5735) - 25,000(4.7793) - 90,000(1.2544) + 120,000 \] - **Calculation Result:** \[ = -\$2,236,604 \] ### Conclusion: The analysis concludes by selecting process T, citing a small margin difference of only $9,273 in future worth (FW) favoring process T over process W. The point of this analysis is to determine the more favorable financial outcome by comparing the future worth of cash flows associated with each process.
**Problem 5.27**

Two processes can be used for producing a polymer that reduces friction loss in engines. 

- **Process T**:
  - First cost: $750,000
  - Operating cost: $60,000 per year
  - Salvage value: $80,000 after its 2-year life

- **Process W**:
  - First cost: $1,350,000
  - Operating cost: $25,000 per year
  - Salvage value: $120,000 after its 4-year life
  - Requires updating at the end of year 2 at a cost of $90,000

The task is to determine which process should be selected based on a future worth analysis at an interest rate of 12% per year.
Transcribed Image Text:**Problem 5.27** Two processes can be used for producing a polymer that reduces friction loss in engines. - **Process T**: - First cost: $750,000 - Operating cost: $60,000 per year - Salvage value: $80,000 after its 2-year life - **Process W**: - First cost: $1,350,000 - Operating cost: $25,000 per year - Salvage value: $120,000 after its 4-year life - Requires updating at the end of year 2 at a cost of $90,000 The task is to determine which process should be selected based on a future worth analysis at an interest rate of 12% per year.
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