Question

Consider the two mutually exclusive projects in the table below. Salvage values represent the net proceeds​ (after tax) from disposal of the assets if they are sold at the end of each year. Both projects B1 and B2 will be available​ (or can be​ repeated) with the same costs and salvage values for an indefinite period

В1 B2 Cash Flow Cash Flow Salvage Value -$22,000 Salvage Value n -$19,000 -2,500 -2,500 -2,500 7,000 4,500 2,000 1 -2,700 -2,.

Equal Payment Series Sinking Single Payment Compound Amount Capital Recovery Factor Compound Amount Present Present Worth Fun

  1. Assuming an infinite planning​ horizon, which project is a better choice at MARR = 11%? Use 15 years as the common analysis period.

    1. The present worth for project B1 is $______thousand.

    2. The present worth for project B2 is $______thousand.

  1. Which is better project choice?

  1. With a ten-year planning horizon, which project is a better choice at MARR = 11%?

    1. The present worth for project B1 is $______thousand.

    2. The present worth for project B2 is $______thousand.

4. Which is better project choice?

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Answer #1

Here the present worth analysis period is 15 years. The present worth of B1 can be calculated as follows

PWB1 = -22,000–2,700(P/A, 11%,15)–22,000(P/F, 11%,5)+ 5,000(P/F, 11%,5)–22,000(P/F, 11%, 10)+5,000(P/F, 11%, 10+ -5,000(P/F,

PWB1 = -22,000–2, 700(P/A, 11%,15)– 17,000(P/F,11%,5)– 17,000(P/F, 11%, 10) + 5,000(P/F, 11%, 15)

PWB1 = -22,000–2, 700(P/A, 11%,15)–17,000(A/F, 11%,5)(P/A, 11%, 10)+ 5,000(P/F, 11%, 15)

PWB1 = -22,000 - 2,700 x 7.19086-17,000 x 0.1606 x 5.8892 + 5,000 x 0.2090

S PWB1 = -22.000 – 19, 415.32 – 16,078.69 + 1.045

PWB1 $ 56, 449

Now calculate for machine B2

PWB2 = -19,000—2,500(P/A, 11%, 15)–17,000(A/F, 11%, 3)(P/A, 11%,12)+ 2,000(P/F, 11%, 15)

= PWB2 = -19,000 - 2,500 x 7.19086-17,000 x 0.2992 6.49235 + 2,000 x 0.209

PWR2 = -19.000 – 17,977.17 – 33, 024.16+ 418

PWB2 = $ 69583

Select B1 since it's present worth cost is less in comparison to B2.

When MARR =11% and useful life is 10 years

PWB1 = -22,000–2, 700(P/A, 11%, 10)-22,000(P/F, 11%,5)+ 5,000(P/F, 11%,5) + 5,000(P/F, 11%, 10)

PWB1 = –22,000–2,700(P/A, 11%, 10)–17,000(P/F, 11%,5)+ 5,000(P/F, 11%, 10)​​​

= PWB1 = -22,000 - 2,700 x 5.8892 – 17,000 x 0.5935+ 5,000 x 0.3522

PWB1 ~ $ 46,229

Now calculate the PW of B2

= PWB2 = -19,000–2,500(P/A, 11%, 10)–17,000(A/F, 11%,3)(P/A, 11%,9)+ 4,500(P/F, 11%, 10)

= PWB2 = -19,000 - 2,500 x 5.8892-17,000 x 0.2992 x 5.537 + 4,500 x 0.3522

PW12 ~ $ 60, 301

Select B1.

Please contact if having any query will be obliged to you for your generous support. Your help mean a lot to me, please help. Thank you.

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