In: Operations Management
Problem 14-27 (Algorithmic) A perishable dairy product is ordered daily at a particular supermarket. The product, which costs $1.2 per unit, sells for $1.64 per unit. If units are unsold at the end of the day, the supplier takes them back at a rebate of $1 per unit. Assume that daily demand is approximately normally distributed with µ = 145 and ? = 30. What is your recommended daily order quantity for the supermarket? If required, round your answer to two decimal places. Q* = What is the probability that the supermarket will sell all the units it orders? If required, round your answer to four decimal places. P(Stockout) = In problems such as these, why would the supplier offer a rebate as high as $$1? For example, why not offer a nominal rebate of, say, 25¢ per unit? What happens to the supermarket order quantity as the rebate is reduced? The higher rebate the quantity that the supermarket should order.
We use the concept of newspaper vendor problem, use the Co and Cu and critical ratio as shown below:
m | mean | 145 | |
s | Std Dev | 30 | |
C | Cost | 1.2 | |
P | Price | 1.64 | |
V | Salvage | 1 | (10+0.4) |
Formula used | |||
Cu | Cost of under order | 0.44 | (P-C) |
Co | Cost of over order | 0.2 | (C-V) |
CR | Critical ratio | 0.6875 | (Cu/(Cu+Co) |
So actual order | m+Z*s | ||
159.6633 | NORMINV(CR,m,s) | ||
If | Optimal Order is | 160 | ANs |
X | X value | 160.00 | |
P value | 0.6915 | NORM.DIST(X,m,s,TRUE) | |
Probability of sale | 0.6915 | Ans |
If we take non rounded values
prob is
0.6875 |
The supplier would offer a rebale as high as $1 , so that the supermarket orders more and the terms are convincing to order higher volumes and the dairy is more preferred over other similar partners
If that is now $ 0.25
m | mean | 145 | |
s | Std Dev | 30 | |
C | Cost | 1.2 | |
P | Price | 1.64 | |
V | Salvage | 0.25 | (10+0.4) |
Formula used | |||
Cu | Cost of under order | 0.44 | (P-C) |
Co | Cost of over order | 0.95 | (C-V) |
CR | Critical ratio | 0.3165 | (Cu/(Cu+Co) |
So actual order | m+Z*s | ||
130.6787 | NORMINV(CR,m,s) | ||
If | Optimal Order is | 131 |
So the order is reduced from 160 to 131, hence a potential loss of 29* (1.2-1) = $ 5.8 per day.
Hence it is preferred to have a higher salvage value offered. as higher the rebate higher is the order and higher is the potential profit.