Question

2.) Genie consumes steel screws at a steady rate of 600 per day. The screws cost 3 cents each. It costs Genie $12 to initiate an order, and the holding costs are based on an annual interest rate of 25%. Assume 260 days/year. a.) What is the optimal number of screw for the plant to purchase and the time between placement of orders? (5 points) b.) What is the yearly holding cost for this item? (5 points) c.) What is the year setup cost for the screws? (5 points) d.) Suppose we were talking about a bulky item such as large sheets of metals. What problem might there be with our analysis? (5 points)

Answer 1

(a): Optimal no. of screws = [2*annual demand*cost per order/Holding cost]^0.5

Now annual demand = 600 per day * 260 days = 156,000 screws

Thus optimal quantity = [2*156,000*12/25% of 0.03]^0.5

= 22,342.78 screws. This can be rounded off to 22,343 screws.

Time between placement of orders = Working days in a year/expected number of orders

= 260/(156,000/22,342.78)

= 37.24 days. This can be rounded to 37 days.

(b): Yearly holding costs = Q/2 * H

= 22,342.78/2 * 25% of 0.03

= $83.79

(c ): set up costs = D/Q * S

= 156,000/22,342.78 * 12

= $83.79

(d): When ordering bulky items like large sheet of metals the ordering will also have to consider storage requirements which the optimal order analysis fails to consider. In other words with regards to bulky items like large sheets of metals significant space costs are involved and this is not considered here.