Pramo is a large bio firm that sells genetically modified seed to farmers. They need to decide how much seed to put into a warehouse to serve demand for the next growing season. They will make one quantity decision. It costs Pramo $8 to make each kilogram (kg) of seed. They sell each kg for $45. If they have more seed than demanded by the local farmers, the remaining seed is sent overseas. Unfortunately, they only earn $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 their quantity, then the sales are lost - the farmers go to another supplier. As a forecast for demand they will use a normal distribution with a mean of 300,000 and a standard deviation of 106,000. How many kilograms should they place in this warehouse before the 2020 growing season? Choose the closest number.

Formula to calculate optimal order quantity is as below:

Optimal order quantity = mean + (z x standard deviation)

Optimal order quantity = 300,000 + (1.18 x 106,000)

Optimal order quantity = $425,080

Prior to optimal order quantity, calculate z value as below:

Critical ratio = Profit per sold unit / (Profit per sold unit + Loss per unsold)

Profit per sold unit = $45 - $8

Profit per sold unit = $37.

Profit per unsold unit = $8 - $3

Profit per unsold unit = $5.

Now;

Critical ratio = Profit per sold unit / (Profit per sold unit + Loss per unsold)

Critical ratio = 37 / 37+5

Critical ratio = 0.8809

Thus, the equivalent probability for 0.8809 is 1.18 (from Normal distribution table) z = 1.18

Now to get optimal order quantity as below:

Optimal order quantity = mean + (z x standard deviation)

Optimal order quantity = 300,000 + (1.18 x 106,000) = 425,080 kg

Therefore, Company Pramo needs to place 425,080 kg of seeds in warehouse before 2020 growing season.