Question

The beer distributor also sells cases of Brew With The Unnecessarily Pretentious Long Name beer (“BWTUPLN” for short), made by a small regional brewery. The demand for BWTUPLN is normally distributed with a mean of 12 cases and a standard deviation of 4 cases per day. When placing a replenishment order (directly with the brewer), it takes an average of 4 days for the order to arrive, with a standard deviation of 2 days (assume the delivery time is normally distributed). The distributor uses an (s,Q) policy to manage its inventory of the cases of beer.

a) If the distributor desires a 97% service level, what should the reorder point be?

b) If the standard deviation of the lead-time was 1 day (instead of 2), what would the reorder point be?

c) If the lead-time was exactly 4 days, what would the reorder point be?

Answer 1

Q1:

Parameters		Quantity
Cycle service Level	CSL	0.97
Z-value	z =+NORM.S.INV(0.97)	1.88
Lead time	L (days)	4.00	Days
Standard Deviation of lead time	σL	2	days
Average demand during day	d	12	0
Standard deviation of demand	σd	4	gallons per week
Continuous review model
Safety Stock	SS = z√(Lσd2 + dσL2)	1.88*√(4*42 + 12*22) = 19.89
ROP	R = d*L + SS	12*4 + 19.89 = 67.89

Q2:

Parameters		Quantity
Cycle service Level	CSL	0.97
Z-value	z =+NORM.S.INV(0.97)	1.88
Lead time	L (days)	4.00	Days
Standard Deviation of lead time	σL	1	days
Average demand during day	D	12	0
Standard deviation of demand	σd	4	gallons per week
Continuous review model
Safety Stock	SS = z√(Lσd2 + dσL2)	1.88*√(4*42 + 12*12) = 16.389
ROP	R = d*L + SS	12*4 + 10.58 = 64.389

Q3:

Parameters		Quantity
Cycle service Level	CSL	0.97
Z-value	z =+NORM.S.INV(0.97)	1.88
Lead time	L (days)	4.00	Days
Standard Deviation of lead time	σL	0	days
Average demand during day	D	12	0
Standard deviation of demand	σd	4	gallons per week
Continuous review model
Safety Stock	SS = z√(Lσd2 + dσL2)	1.88*√(4*42 + 12*02) = 14.4
ROP	R = d*L + SS	12*4 + 14.4 = 62.4