Question

A breakeven analysis for net present worth is performed with a MARR of 15% and a useful life of 20 years with the following data:
Initial Cost: $750,000
Annual Cost (O&M): $60,000/ year with an annual increase of $5,000 each year
Annual Revenue: $80,000/ year with an annual increase of $1,000 each year
Salvage Value: $475,000
Determine the following:
a. If the project is viable (i.e. if the profit is larger than the cost using net present worth).
b. The gradient annual revenue increase to make the project viable.
c. The salvage value to make the project viable.

Answer 1

A.

R = 15%

Time = 20 years

Uniform annuity of O&M cost = $60000

Arithmetic gradient of O&M cost = $5000

Arithmetic gradient of annual revenue = $1000

Uniform annuity of Annual revenue = $80000

Net present worth = -750000 + (80000-60000)*(P/A, 15%, 20) - 5000*(P/G, 15%, 20) + 1000*(P/G, 15%, 20) + 475000*(P/F, 15%, 20)

Net present worth = -750000 + (80000-60000)*6.2593 - 5000*33.582 + 1000*33.582 + 475000*.0611

Net present worth = -$730120

Since net present worth is negative, then project is not viable, and cannot be accepted in a given scenario.

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B.

To make project to be viable, let new annual gredient of revenue = P

Then,

0 = -750000 + (80000-60000)*6.2593 - 5000*33.582 + P*33.582 + 475000*.0611

P = (750000 - (80000-60000)*6.2593 + 5000*33.582  - 475000*.0611)/33.582

P = $22741.39

So, increase in arithmetic gradient of revenue = $22741.39

Or

Net increase in arithmetic gradient of revenue = 22741.39 - 1000

Net increase in arithmetic gradient of revenue = $21741.39

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C.

Let salvage value = S

0 = -750000 + (80000-60000)*6.2593 - 5000*33.582 + 1000*33.582 + S*.0611

S = (750000 - (80000-60000)*6.2593 + 5000*33.582 - 1000*33.582)/.0611

S = $12424582.7 or $12424583

So, the salvage value should be $12424582.7 or $12424583