Question

23. Ross White’s machine shop uses 2,500 brackets dur- ing the course of a year, and this usage is relatively constant throughout the year. These brackets are purchased from a supplier 100 miles away for $15 each, and the lead time is 2 days. The holding cost per bracket per year is $1.50 (or 10% of the unit cost) and the ordering cost per order is $18.75. There are 250 working days per year.

    1. (a) What is the EOQ?

    2. (b) Given the EOQ, what is the average inventory?

       What is the annual inventory holding cost?

    3. (c) In minimizing cost, how many orders would be made each year? What would be the annual or-

       dering cost?

    4. (d) Given the EOQ, what is the total annual inven-

       tory cost (including purchase cost)?

    5. (e) What is the time between orders?

    6. (f) What is the ROP?

Ross White (see problem above) wants to reconsider his decision of buying the brackets and is considering making the brackets in-house. He has determined that setup costs would be $25 in machinist time and lost production time, and 50 brackets could be produced in a day once the machine has been set up. Ross estimates that the cost (including labor time and materials) of producing one bracket would be $14.80. The holding cost would be 10% of this cost.

(a) What is the daily demand rate?

(b) What is the optimal production quantity?
(c) How long will it take to produce the optimal quantity? How much inventory is sold during this time?
(d) If Ross uses the optimal production quantity, what would be the maximum inventory level? What would be the average inventory level? What is the annual holding cost?

(e) How many production runs would there be each year? What would be the annual setup cost?

(f) Given the optimal production run size, what is the total annual inventory cost?

(g) If the lead time is one-half day, what is the ROP?

Answer 1

Annual demand (D) = 2500 units

Number of days per year = 250 days

Average daily demand (d) = D/number of days per year = 2500/250 = 10 units

Holding cost(H) = $1.50

Ordering cost (S) = $18.75

Purchase cost per unit(p) = $15

Lead time(L) = 2 days

a) EOQ = √(2DS/H)

= √[(2 × 2500 × 18.75) / 1.50]

= √(93750/1.50)

= √62500

= 250 units

b) Average inventory = EOQ/2 = 250/2 = 125 units

Annual inventory holding cost = (EOQ/2) H = (250/2) 1.50 = 125 × 1.50 = $187.5

C) Number of orders per year = (D/EOQ) = 2500/250 = 10

Annual ordering cost = (D/EOQ) S = (2500/250) 18.75 = 10 × 18.75 = $187.5

d) Purchase cost = p × D = $15 × 2500 = $37500

Total annual inventory cost = Annual ordering cost+Annual holding cost+purchase cost

= $187.5 + $187.5 + $37500

= $37875

e) Time between orders =( EOQ/D) number of days per year

= (250/2500) 250

= 0.1 × 250

= 25 days

f) Reorder point = d × L = 10 × 2 = 20 units