Question

Monsanto sells genetically modified seed to farmers. It needs to decide how much seed to put into a warehouse to serve demand for the next growing season. It will make one quantity decision. It costs Montanso $7 to make each kilogram (kg) of seed. It sells each kg for $46. If it has more seed than demanded by the local farmers, the remaining seed is sent overseas. Unfortunately, it only earns $3 per kg from the overseas market (but this is better than destroying the seed because it cannot be stored until next year). If demand exceeds its quantity, then the sales are lost—the farmers go to another supplier. As a forecast for demand, it will use a normal distribution with a mean of 325,000 and a standard deviation of 100,000. Use Table 13.4

A. How many kilograms should Monsanto place in the warehouse before the growing season? use table 13.2 and round up rule.

B. If Monsanto put 375000 kgs in the warehouse, what is the expected revenue (include both domestic revenue and overseas revenue)? Use table 13.4 and round-up rule.

C. How many kilograms should Monsanto place in the warehouse to minimize inventory while ensuring that stockout probability is no greater than 5%? Use table 13.4 and round-up rule.

D. What is maximum profit for this seed?

Answer 1

Solution:-

(a):-

Cu = Selling price - Cost = 46 - 7 = $39

Co = Cost - Overseas sale price = 7 - 3 = $4

F(z) = Cu/(Cu+Co) = 39/44 = 0.8863

In table, F(z) = 0.8863 round up to 0.9032,

z = 1.3

Optimal stock = mean + z *
= 325000 + 1.3*100,000 =455,000 kilograms

(b):-

z-statistic = (375000 - 325000) / 100000 = 0.5

For z = 0.5, L(z) = 0.6978

Use the L(z) value corresponding to Z = 0.5 which is 0.6978 (From the table 13.4)

Expected leftover inventory =
*L(z) = 100000*0.6978 = 69,780

Expected sales = Stock - Expected leftover = 375,000 - 69,780 = 305,220

Expected revenue = Expected sales * Local sales price + Expected leftover inv* Overseas sales price

Expected revenue = (305,220*46) + (69,780*3) =14,249,460

(c):-

Here the stockout prob is less than 5%

so, F(z) = 0.9452

The next higher value corresponding to F(z) = 0.9452 is 1.6 (From the table).

Quantity to be kept in warehouse = mean + z*

= 375000 + 1.6 *100000

= 535,000

(d):-

Maximum profit = Mean demand * Underage cost = 375000 * 39 = 14,625,000