Question

Q. Daily demand for a product is normally distributed with a mean M= 12 units and standard deviation Q= 4 units. Lead time is 9 days. The order quantity for this item is set at a 12-day supply.

a. If a minimum probability of satisfying demand per cycle of 0.80 is desired, what is the reorder point?

b. If a fill rate of 9% is desired, what is the reorder point?

Answer 1

(a)

Cycle service level = 0.80; So, Z = normsinv(0.80) = 0.8416

Mean demand per day, d = 12
Stdev of daily demand, s = 4
Lead time, L = 9 days.

So, safety stock, SS = Z * s * √L = 0.8416 * 4 * √9 = 10 units

Reorder point = d*L + SS = 12*9 + 10 = 118 units.

(b)

The fill rate cannot be as low as 9%. There must be some typing mistake. Please check. Anyway, the computation is shown.

Order size Q = 12 day supply i.e. 12*12 = 144 units

Fill rate = 9% = 1 - (Expected Shortage per Cycle / Q)

or, 0.09 = 1 - ESC / 144

or, ESC = 144*0.91 = 131.04

We know that ESC = s * √L * E(Z) where E(Z) is the unit loss function

So, E(Z) = ESC / s*√L = 131.04 / (4*√9) = 10.92. For such a high E(Z), the corresponding Z vaue will be the negative. So, no safety stock (= Z * s * √L) wil be required in practice [note that negative safety stock does not have any significance].

So, the reorder point = d*L + Z * s * √L = d*L + 0 = 12*9+0 = 108 units