Question

A piece of equipment used in a manufacturing plant that costs 220,000 to purchase has the economic data shown in the table below including the market value if the equipment was sold at the end of the period of use. Apply an interest rate of 9.0% to answer the questions that follow the data.

Period	1	2	3	4	5
Operating Cost	45,000	56,000	70,000	85,000	98,000
Market Value	160,000	125,000	105,000	45,000	15,000

a. What is the period cost of using the equipment just one period?

b. What is the lowest period equivalent cost for keeping the equipment in operation for the economic service life?

c. What is the economic service life?

Answer 1

The attached excel screenshot explains the answers in detail. For calculating the period cost and economic service life, the best way is to use excel. It is very important to understand the discounted factor which in this problem is calculated using the interest rate of 9%

Discounted factor formula is 1/(1+r)^n

Now let's answer the questions:

PeriodCapital Cost Market Value ationg Cost (OC) | Discounted Factor (DF) | OC"DF 5000.00 51376.15 8917.60 155293.75 65635.60 220929.34 69425.67 290355.01 | Sum of (OC*DF) | Period Cost! Cuma lative Discount | weightage Average Coast 45000.00 105000.00 191376.15 270293.75 395929.34 495355.01 160000 45000 0.9174 0.8417 0.7722 0.7084 105000.00 99808.61 97964.06 112120.17 116836.73 96376.15 1.9174 2.7591 3.5313 4.2397 220000 105000 70000 85000 98000 5000

a. Period cost of using equipment for 1 period is $105,000 (as can be seen in the above chart)

b. Lowest period equivalent cost for keeping the equipment in operation for the economic service life= Lowest value in column "Weightage Average Cost"= $97,964.06

c. To answer this, we find the lowest point in the graph, and the corresponding period is the economic service life:

Period WAC 105000 99809 97964 112120 116837 120 115 110 105 100 95 90 112 2 105 100 98 3 4 Period

Hence, the economic service life = 3