Use Table 134 Montanso sells genetically modified seed to farmers. It needs to d

Question

Use Table 134 Montanso 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 $44. If Montanso has more seed than demanded by the local farmers, the remaining seed is sent overseas. Unfortunately, only $2 per kg is earned 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, Montanso will use a normal distribution with a mean of 275000 and a standard deviation of 150000. If a part of the question specifies whether to use Table 13.4, or to use Excel, then credit for a correct answer will depend on using the specified method. How many kilograms should Montanso place in the warehouse before a. the growing season? Use Table 13.4 and round-up rule. If Montanso put 450000 kgs in the warehouse, what is its expected Table 13.4 and round-up rule How many kilograms should Montanso place in the warehouse to greater than 10%? Use Table 13.4 and round-up rule. b. revenue (include both domestic revenue and overseas revenue)? Use c. minimize inventory while ensuring that the stockout probability is no d. What is maximum profit for this seed?

Explanation / Answer

mean = 275000
std. dev. = 150000

(a)
As profit margins are very high, we assume that Monsanto will try to fulfill 99% of its demand.

z-value = 2.4 (from the table corresponding value for the F(z) = 0.9918)

Using central limit theorem, we have

xbar = mean + z*sigma
xbar = 275000 + 2.4*150000 = 635000

b)
If we consider 99% of 450000kgs is sold in domestic market and remaining 1% in overseas market.

Expected revenue = 0.99*450000*44 + 0.01*450000*10 = $19647000

c)
For 10% of value, respective z-value = 1.3 (value for F(z) = 0.9032 in the table)

Using central limit theorem, we have

xbar = mean + z*sigma
xbar = 275000 + 1.3*150000 = 470000

470000kgs of seeds should be stored in inventory

d)
Maximum profit would happen if all of the seeds produced are sold in domestic market and the production is according to highest possible demand

z-value = 3.8

xbar = mean + z*sigma
xbar = 275000 + 3.8*150000 = 845000

Profit earned = 845000 * (44 - 8) = $30420000

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