In: Operations Management
A retail outlet sells a perishable product for $12 per unit. The cost of the product is $6 per unit. The product not sold has the salvage value of $1.
a) Compute the overage cost and underage cost per unit.
b) If the demand follows the normal distribution with the average of 400, and the standard deviation of 100. What is the optimal order quantity to maximize the expected profit?
c) What is the optimal ordering quantity if the retailer uses the following discrete distribution for demand?
quantity 200 250 300 350 400 450 500 550 600
probability 0.05 0.05 0.15 0.15 0.20 0.15 0.15 0.05 0.05
d) What is the profit or loss for the quantity obtained from c)?
a.
Overage cost = incremental cost per unit for items not sold = cost
- salvage price = $6 - $1 = $5
Underage cost = incremental cost per unit for not meeting demand of
items = price of item - cost of item = $12 - $6 =
$6
b. Normal distribution of Demand
Average of demand ()=
400
Standard deviation ()
= 100
Optimal order quantity (EOQ) =
+ (Z*)()
where Z* = Normsinv((Underage cost)/(Underage cost+overage
cost))
Note, normsinv is an excel function to calculate Z*, alternatively
you can use statistical tables to find value of Z*
So, here Z* = NORMSINV(5/11) = -0.1142
So, EOQ = 400 + (-0.1142)(100) = 388.581 i.e. 389 (as items
have to be ordered in integer value)
c. Discrete distribution of Demand
Demand | Probability | Cumulative Probability of demand less than this value (F(x)) | Cumulative Probability of demand less than this value(1-F(x)) |
200 | 0.05 | 0.05 | 0.95 |
250 | 0.05 | 0.1 | 0.9 |
300 | 0.15 | 0.25 | 0.75 |
350 | 0.15 | 0.4 | 0.6 |
400 | 0.2 | 0.6 | 0.4 |
450 | 0.15 | 0.75 | 0.25 |
500 | 0.15 | 0.9 | 0.1 |
550 | 0.05 | 0.95 | 0.05 |
600 | 0.05 | 1 | 0 |
Probability we are able to sell Q +1 unit is given by critical limit i.e. (Overage cost/(Overage cost + Underage cost)) | |||
Pr(Sell Q +1st unit) = 5/11 i.e. 0.45 | |||
We compare this value with the Cumulative probability of demand being less than this value | |||
We see that .45 comes between .6 and .4 I.e. the optimal order quantity will be 400 |
d. Profit/ Loss for above quantity
Here we need to take cases for all the demands given and then calculate the expected profit for Q = 400
Demand | Probability |
Overage (400 - D) |
Revenue Price*D + Overage*Salvage value |
Expected revenue Probability * Revenue |
200 | 0.05 | 200 | $ 2,600 | $ 130 |
250 | 0.05 | 150 | $ 3,150 | $ 158 |
300 | 0.15 | 100 | $ 3,700 | $ 555 |
350 | 0.15 | 50 | $ 4,250 | $ 638 |
400 | 0.2 | 0 | $ 4,800 | $ 960 |
450 | 0.15 | 0 | $ 5,400 | $ 810 |
500 | 0.15 | 0 | $ 6,000 | $ 900 |
550 | 0.05 | 0 | $ 6,600 | $ 330 |
600 | 0.05 | 0 | $ 7,200 | $ 360 |
Total expected revenue | $ 4,840 | |||
Total expected profit | Expected revenue - Cost | |||
= $4840 - $6*400 | ||||
Total expected profit | $ 2,440 |