Question

Polymer Molding, Inc. is considering two processes for manufacturing storm drains. Plan A involves conventional injection molding that will require making a steel mold at a cost of $2 million. The cost for inspecting, maintaining, and cleaning the molds is expected to be $60,000 per year. Since the cost of materials for this plan is expected to be the same as for the other plan, this cost is not included in the comparison. The salvage value for plan A is expected to be 10% of the first cost. Plan B involves using an innovative process known as virtual engineered composites wherein a floating mold uses an operating system that constantly adjusts the water pressure around the mold and the chemicals entering the process. The first cost to tool the floating mold is only $795,000, but because of the newness of the process, personnel and product-reject costs are expected to be higher than for a conventional process. The company expects the operating costs to be $85,000 for the first year and then decrease to $46,000 per year thereafter. There will be no salvage value with this plan. At an interest rate of 12% per year, which process should the company select on the basis of an annual worth analysis over a 3-year study period?

Answer 1

To determine which plan to select, we need to calculate the annual worth (AW) of each plan. The AW is the total amount of money that would need to be set aside each year to account for the costs of the plan.

For Plan A:
- First cost is $2 million.
- Annual operation and maintenance cost is $60,000.
- Salvage value after 3 years is 10% of first cost, which is $200,000.
- Using the formula for AW (AW = (P - S) * (A / P, i, n) + S * (A / F, i, n) + A), where P is the first cost, S is the salvage value, A is the annual operation and maintenance cost, i is the interest rate, and n is the number of years, we get AW = ($2,000,000 - $200,000) * (A / P, 12%, 3) + $200,000 * (A / F, 12%, 3) + $60,000.

For Plan B:
- First cost is $795,000.
- Annual operation and maintenance cost is $85,000 for the first year and $46,000 for the next two years.
- There is no salvage value.
- Using the formula for AW (AW = (P - S) * (A / P, i, n) + S * (A / F, i, n) + A), where P is the first cost, S is the salvage value, A is the annual operation and maintenance cost, i is the interest rate, and n is the number of years, we get AW = ($795,000 - $0) * (A / P, 12%, 3) + $0 * (A / F, 12%, 3) + $85,000 for the first year and $46,000 for the next two years.

The company should select the plan with the lower AW. To get the exact figures, you'll need to plug in the values into the formula and compute the AW for each plan.

Answer 2

We must compute the annual worth of each process and compare them in order to determine which one the organization should choose based on an examination of annual worth during a three-year research period.

PLAN A

• First cost (C_A) = $2,000,000
• Annual operating cost (A_A) = $60,000
• Salvage value (S_A) = 10% of the first cost = 0.10 * $2,000,000 = $200,000

FORMULA PLAN A

• AW_A = A_A + (A_A - S_A) * (P/A, i, n)

SOLUTION

• i = 12% per year
• n = 3 years
• P/A, 12%, 3 = 1 - (1 + 0.12)^-3 / 0.12 = 1 - (1.12)^-3 / 0.12 ≈ 2.4013

PLAN A

AW_A = $60,000 + ($60,000 - $200,000) * 2.4013 ≈ -$240,780.20

PLAN B

First cost (C_B) = $795,000

Annual operating cost for year 1 (A_B1) = $85,000

Annual operating cost for year 2 and beyond (A_B2) = $46,000

FORMULA PLAN B

• AW_B = A_B1 + (A_B2 - C_B) * (P/A, i, n)

Calculate the present worth factor (P/A, i, n) using the same values as before:

P/A, 12%, 3 ≈ 2.4013

PLAN B

AW_B = $85,000 + ($46,000 - $795,000) * 2.4013 ≈ $159,983.20

COMPARE

• AW_A ≈ -$240,780.20

• AW_B ≈ $159,983.20

Plan B is the better choice as it has a positive annual worth of approximately $159,983.20 over the 3-year study period, while Plan A has a negative annual worth of approximately -$240,780.20.