Question

Monsanto sells genetically modified seed to farmers. It needs to decide how much seed to put...

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 $8 to make each kilogram (kg) of seed. It sells each kg for $45. 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 300,000 and a standard deviation of 100,000.

A. How many kilograms should it place in the warehouse before the growing season?

B. If it puts 400,000 kg in the warehouse, what is their expected revenue (include both domestic revenue and overseas revenue)?

C. How many kilograms should it place in the warehouse if it wants to minimize its inventory while ensuring that the stockout probability is no greater than 10 percent?

D. What is the maximum profit for this seed?

Homework Answers

Answer #1

A) Underage cost, Cu = selling price - cost = 45 - 8 = $ 37

Overage cost, Co = cost - overseas selling price = 8 - 3 = $ 5

Critical fractile = Cu/(Cu+Co) = 37 / (37+5) = 0.8810

z = NORMSINV(0.8810) = 1.18

Optimal quantity to place in the warehouse, Q = mean demand + Std dev of demand * z

= 300000 + 1.18 * 100000

= 418,000 kg

B) For Q = 400000, value of z = (400000-300000)/100000 = 1

For, z = , value of I(z) = 1.08332

Expected unsold inventory, V = std dev of demand * I(z) = 100000*1.08332 = 108332

Expected sales, S = Q - V = 400000 - 108332 = 291668

Expected domestic revenue = 45*291668 = $ 13,125,060

Expected overseas revenue = 3*108332 = $ 324,996

Total expected revenue = 13125060+324996

= $ 13,450,056

c) For stockout probability of 10%, z = NORMSINV(1-10%) = 1.2816

Optimal stock = 300000+100000*1.2816 = 428,160 kg

d) Maximum profit = Average demand * Cu

= 300000*37

= $ 11,100,000

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