Question

Question 5: (5 Points) A local company uses an average of 650 items a day. The daily usage has a standard deviation of 23. The manager is willing to accept no more than a 1% chance of a stock out. The lead-time is 14 days. (Assuming the distribution of usage is normal). What is the re-order point? Question 6: (5 Points) A company has a daily demand of 850 shafts. The average lead-time is 12 days with a standard deviation of 2.5. The company desires a 99 percent service level. Calculate the reorder point. Question 7: (5 Points) A company orders an item with an average daily demand of 225 and a standard deviation of 8. It has an average lead-time of 22 days with a standard deviation of 4. They want a 99 percent service level. What is the re-order point?

Please kindly explain all the formulas and also how to find z value from service level in steps

Answer 1

5

Reoden paint = lead time demand + Standard denialion af demand lead time = lead lime x daily demand + Źx Standard devialion af demand X (lead time) ² = 14 X 650 + morminu Co.gg] x 23 (14) Esenice teren = 9300.201124

6

d = - T Si a 2= ROP = 850 shifts i 12 days = 2. ² gg%om 2.33 from 2 tonble) dx L + 2x qxd h 850 812+(-2.33x2.5 X850) 3 1 10200 + 4951252 15151:25 15251 urus

7

- ROP = d.LT + 2 = 2 ² LT + 2 12 d - agenage demand = 225 LT = lead time = 22 2 - Value pau 99% service level = 2.326 od & Stendard deviation of demand = 8 = Slandand deviation of lead hime=4 ROP = 2256 22 + 2.326 x 33 x 22 447 x122592 = ~ 7045.21 tous