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. **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 $90,000The task is to determine which process should be selected based on a future worth analysis at an interest rate of 12% per year.

We have, PROCESS T W First cost 750000 1,350,000 Operating costs 60000 25000 Salvage value 80,000 120000 Life (in years) 2 4 Updating cost (only for Process W) -- 90,000 at the end of 2nd year The rate of interest is 12%. As the life of both processes is unequal, use common multiple methods and convert the unequal life into equal life. The lowest common multiple of 2 and 4 = 4. Thus, Process T is to be repeated 2 times to make it equal to Process W.Future Worth of Process T FWT = -750,000 (F/P, 0.12, 4) - {750,000 – 80,000} (F/P, 0.12, 2) - 60,000 (F/A, 0.12, 4) + 80000 FWT = -750,000 (1.5735) - 670,000 (1.2544) - 60,000 (4.7793) + 80000 FWT = −1,180,125 −840,448 −286,758 +80000 FWT = -2,227,331 Future Worth of Process W FWW = -1,350,000 (F/P, 12%, 4) - 25,000 (F/A, 12%, 4) - 90,000 (F/P, 12%, 2) + 120,000 FWW = -1,350,000 (1.5735) - 25,000 (4.7793) - 90,000 (1.2544) + 120,000 FWW = −2,124,225−119,482.5 - 112896 + 120000 FWW = −2,236,603.5 or -2,236,604As per the above calculation, we select the Process T The difference between the FW of W and T = 9,273 (2,236,604 – 2,227,331)

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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

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

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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