Question

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 62 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
1	48
2	13
3	100

Determine the order quantities for cycles 1, 2, and 3: (Round your answers to the nearest whole number)

Cycle	Units
1
2
3

Answer 1

Order interval(T) = 11 days

Lead time (L) = 3 days

Average daily demand(d-bar) = 62 units

Standard deviation of daily demand( 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) = 48 units

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

= [62(11+3)] + [2.05 X 6 X √ (11+3)] - 48

= (62×14) + (2.05 × 6 × √14) - 48

= 868 + (2.05 X 6 X 3.74) - 48

= 868 + 46.00 - 48

= 914-48

= 866 units

So the order quantity for cycle 1 is 866 units

For cycle 2 the on hand inventory(I) = 13 units

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

= [62(11+3)] + [2.05 X 6 X √ (11+3)] - 13

= (62×14) + (2.05 × 6 × √14) - 13

= 868 + (2.05 X 6 X 3.74) - 13

= 868 + 46.00 - 13

= 914-13

= 901 units

So the order quantity for cycle 2 is 901 units

For cycle 3 the on hand inventory(I) = 100 units

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

= [62(11+3)] + [2.05 X 6 X √ (11+3)] - 100

= (62×14) + (2.05 × 6 × √14) - 100

= 868 + (2.05 X 6 X 3.74) - 100

= 868 + 46.00 - 100

= 914-100

= 814 units

So the order quantity for cycle 3 is 814 units