Montanso is a large bio firm that sells genetically modified seed to farmers. Mo
ID: 431937 • Letter: M
Question
Montanso is a large bio firm that sells genetically modified seed to farmers. Montanso needs 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 Montanso $16 to make each kilogram (kg) of seed. They sell each kg for $90. If they have more seed than demanded by the local farmers, the remaining seed is sent overseas. Unfortunately, they only earn $6 per kg from the overseas market (but this is better than destroying the seed because it cannot be stored until next year). If local demand exceeds their quantity, then the sales are lost – the farmers go to another supplier. As a forecast for local demand they will use a normal distribution with a mean of 300,000 kgs and a standard deviation of 100,000 kgs.
1. Newsvendor Quantity: How many kilograms (kgs) of seed should they place in their warehouse before the next growing season to maximize their expected profit?
(i) A senior executive at the company asks the following, “Suppose we were to place 500,000 kilograms in this warehouse. What is the probability that our total revenue will be greater than $35,000,000?” This question is concerned with actual revenue values, not expected revenue or the average revenue. Don’t forget that total revenue comes from both local sales and overseas sales, which depend on the realized value of local demand and the supply of seeds placed in the warehouse. Furthermore, you are concerned with revenue and not profit, so you can ignore costs. [Hint: Write the total revenue as a function of the realized local demand value and the quantity placed in the warehouse.]
(ii) Suppose Montanso places 400,000 kilograms in their warehouse. Calculate the expected local sales of seed and the expected overseas sales of seed. Would these sales values increase, decrease, or remain unchanged if the standard deviation of demand decreases?
Explanation / Answer
Answer:
Revenue/kg=r=$90
Cost /kg=c=$16
Salvage value/kg=s=$6
Let Q be the optimum quantity to be produced to maximise expected profits
Using the newspaper vendor model,
P(D<=Q)= Service level =SL= (r-c)/(r-s)
=(90-16)/(90-6)=0.88
Therefore to maximize profit a service level of 88% needs to be maintained
As per forcasting mean demand =300000 kg
standard deviation =100000 kg
SL (as derived)=0.88
Service factor =NORMSINV(SL)= 1.174987
Considering lead time as 1 period(growing season). Therefore lead time to forecast time factor=1
Reorder point =Demand during lead time + standard dev.*service factor*lead time factor
=300000+ 100000*1.174987*1
=417498.7
Therefore the quantity of seeds that should be placed in warehouse to maximise profit is 417498.7 Kgs
(1) Total Revenue = Revenue from local sales +Revenue from overseas sales
Assuming that the local demand is less than the quantity stored in warehouse.
Let local sales be X kgs and as per given in question total 500000 kgs are placed in warehouse
Therefore Total revenue = 90*X +(500000-X)*6
=90X+3000000-6X=84X+3000000
Equating total revenues to 35000000$ we solve the value of X
84X+3000000=35000000
Therefore x=380952.38 Kgs
Calculating the corresponding Z value as (X- mu)/sigma = (380952.38 - 300000) / 100000
=0.80952 =0.81 approx
Using the standard Z table the area under the curve with Z=0.81 is 0.2910 ie 29.10%
Hence the probability that Revenue is more than 3500000$ is (100- 29.1)% =70.90%
[For revenue to be greater than 35000000$ the value of X has to be greater than 380952.38. So we need to find the probability of X greater than 380952.38]
(ii) We first calculate the service level when the seeds stored in warehouse is 400000 kgs
Reorder point = Mean demand+ Std. Dev *Service factor *lead time factor
400000=300000 + 100000*Service factor *1
Service factor =1
The service level calculated is approximately 0.84 or 84% [NORMSINV(0.84) =1 appprox]
With increase in standard deviation the service level would decrease ,other factors remaining same.P(D<=Q) =SL . Hence the probability that demand is less than the optimum storage /production will decrease and there will be more probability of loss of sales opportunity [P(D>=Q) = 1 - P(D<=Q)]. Therefore overall sales value or revenue would decrease.Remember the more the local sales , more is the revenue.
If demand is D and quantity stored in warehouse is 400000 then net sales contribution
=r*D - c*Q +s*(Q-D) if D<=Q
and r*Q -c*Q if D>=Q
ie 90*D - 16*400000 + 6*(400000-D) if D<=Q
and 90*400000 - 16*400000 if D>=Q
If we are concerned only about sales , meaning ignoring the costs, the net sales expected
= [90D +6*(400000-D) ]*0.84 + [90*400000]*0.16