Question

Problem 13-31 A drugstore uses fixed-order cycles for many of the items it stocks. The manager wants a service level of .98. The order interval is 11 days, and lead time is 3 days. Average demand for one item is 65 units per day. and the standard deviation of demand is 6 units per day. Given the on-hand inventory at the reorder time for each order cycle shown in the following table. Use Table Cycle On Hand 39 104 Determine the order quantities for cycles 1, 2, and 3: (Round your answers to the nearest whole number) Cycle Units 917 934 844

last two of my answers are incorrect. please help!

Answer 1

Order interval(T) = 11 days

Lead time (L) = 3 days

Average daily demand(d-bar) = 65 units

Standard deviation of daily demand($\sigma$d) = 6 units

With aservice level of 0.98 or 98% the value of Z = 2.05

For cycle 1 the on hand inventory(I) = 39 units

So,order quantity for cycle 1 = [d-bar(T+L)] + [Z X $\sigma$d X √(T+L)] - I

= [65(11+3)] + [2.05 X 6 X √ (11+3)] - 39

= (65×14) + (2.05 × 6 × √14) - 39

= 910 + (2.05 X 6 X 3.74) - 39

= 910 + 46.00 - 39

= 956-39

= 917 units

So the order quantity for cycle 1 is 917 units

For cycle 2 the on hand inventory(I) = 14 units

So,order quantity for cycle 2 = [d-bar(T+L)] + [Z X $\sigma$d X √(T+L)] - I

= [65(11+3)] + [2.05 X 6 X √ (11+3)] - 14

= (65×14) + (2.05 × 6 × √14) - 14

= 910 + (2.05 X 6 X 3.74) - 14

= 910 + 46.00 - 14

= 956-14

= 942 units

So the order quantity for cycle 2 is 942 units

For cycle 3 the on hand inventory(I) = 104 units

So,order quantity for cycle 3 = [d-bar(T+L)] + [Z X $\sigma$d X √(T+L)] - I

= [65(11+3)] + [2.05 X 6 X √ (11+3)] - 104

= (65×14) + (2.05 × 6 × √14) - 104

= 910 + (2.05 X 6 X 3.74) - 104

= 910 + 46.00 - 104

= 956-104

= 852 units

So the order quantity for cycle 3 is 852 units