Question

The Wod Chemical Company produces a chemical compound that is used as a lawn fertilizer. The compound can be produced at a (fixed) rate of 15,000 pounds per day. Annual demand for the compound is 0.6 million pounds per year. The fixed cost of setting up for a production run of the chemical is $1000, and the variable cost of production is $6.50 per pound. The company uses an annual interest rate of 25 percent to account for the cost of capital, and the costs of storage and handling of the chemical.

a. What is the optimal size of the production run for this particular compound?

b. What proportion of each production cycle consists of uptime (i.e. time that the compound is being produced) and what proportion consists of downtime (i.e. time that the compound is not being produced)?

c. What is the average annual cost of holding and setup attributed to this item?

Answer 1

a. Calculation of holding cost: (0.22+0.12)(3.5) = (0.34)(3.5) = $1.19/unit/year. The modified holding rate:

h’ = h(1-D/P) = (1.19)(1-600,000/(250*10,000)) = (1.19)(1-6/25) = $0.9044/unit/year. To substitute the same into the EOQ formula and using h’ for h,

Q* = SQRT(2AD/h’) = SQRT(2*1,500*600,000/0.9044) = 44,612.44 = 44,612 units

b. Cycle Time,T = Q/D = 44,612/600,000 = 0.07435 year. The uptime = T1 = Q/P = 44,612/2,500,000 = 0.01784 year, and the down time is T2 = T-T1 = 0.07435-0.01784 = 0.05651. Thus , the proportion of uptime is 0.01784/0.07435 = 24% and downtime is 76%

c. The average annual cost of holding and setup = Underroot (2*1500*600000*0.9044) = $40,347.5

d. Profit per unit is 3.90 – 3.50 = $0.40/pound. Total Revenue= (0.40)(600,000)-1500 (fixed cost) = $238500
Annual Profit = Total Revenue - setup and annual holding cost
= $238500 - $40347.5 = $198152.5