# Upd Case

1 Pages

20 Downloads

Words: 225

Date added: 17-09-26

**Type:** Expository essay

**Category:** Education

Case: UPD Manufacturing Given, Demand, d = 6 Ordering Interval, OI = 89 Ordering cost, S = $32 Holding Cost/Carrying Cost, H = $. 08 As there is no demand variability, the formula for quantity is: Q = d (LT + OI) – A (as there is no safety stock) ------- A - ROP (Reorder point) We know, A = d * LT, so the fixed order interval order quantity equation Q becomes Q = (d * LT) + (d * OI) – (d * LT) * Q = d * OI = (6) (89) = 534 units Therefore, ordering at six-week intervals requires an order quantity of 534 units. Now, the optimal order quantity is determined by using EOQ equation. Q = sqrt(2dS/H) = sqrt[(2*89*32)/. 08] = 266. 833 (or) 267 The weekly total cost based on optimal order quantity EOQ is given below: TC of EOQ = (d/q)*s + (q/2)*H = (89/267)*32 + (267/2)*. 08 = 10. 6666+ 10. 68 = 21. 3466 The weekly total cost based on six-week fixed order interval (FOI) order quantity is given below: TC of FOI = (d/q)*s + (q/2)*H = (89/534)*32 + (534/2)*. 08 = 5. 3333+ 21. 36 = 26. 6933 Weekly savings of using EOQ rather than 6-week FOI is 26. 933 – 21. 3466 = $5. 3467 The annual savings = (52 weeks) * ($5. 3467 /week) = $278. 02 The total annual savings as a result of switching from six-week FOI to EOQ are relatively small and switching to the optimal order quantity may not be warranted. However, even though the absolute value of the savings is relatively small, the percentage of savings is approximately 25. 05% (5. 3467 / 21. 3466). Therefore if FOI approach is used with other parts or components as well, the total potential loss may be significant.

Read full document← View the full, formatted essay now!