question archive The Snowy Peaks ski resort closes for the season at the end of February
The Snowy Peaks ski resort closes for the season at the end of February
Subject:StatisticsPrice:3.87 Bought7
The Snowy Peaks ski resort closes for the season at the end of February. In addition to supplying ski equipment, food, and many other things to its customers, it has to heat their hotel rooms. Since the number of customers in February is unknown, the amount of heating oil needed is unknown. They have to decide how much oil they should order. The Snowy Peaks marketing department feels that the amount of oil needed will be determined by a triangular distribution with a = 1000 gallons, c = 2000 gallons, and b = 4000 gallons.
The resort's oil supplier will charge them $3.00 per gallon in February. They estimate that each gallon that is consumed creates $6.00 in revenue. Since February is the last month of the ski season, any oil that is left at the end of the month must be discarded in an environmentally responsible manner. Consequently, any oil that remains on March 1 costs $2.00 to remove. In other words, it has a negative salvage value: -$2.00 per gallon.
They would like to maximize their net profit. How much should they order for February?
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Answer:
2063.5
Step-by-step explanation
For triangular demand
Cu = cost of underage = revenue - cost = 6 - 3 = 3
Co = cost of overage = cost - salvage value = 3 - (-2) = 5
So, the critical factor = Cu / (Co+Cu) = 3 / (5+3) = 0.375
Assume the mean is 2000 gallons and standard deviation is 200 gallons.
The demand, x is following the following cumulative distribution function:
For a <= x <= c, putting the values of a, b, and c, we get
?D(x)=(x−1000)2/(4000−1000)∗(2000−1000)=(x−1000)2/3000000?
For optimal order quantity, critical ratio should be equal to D(x).
So, ?(x−1000)^2/3000000=0.375?
or, ?(x−1000)^2=1125000?
or, x = 1060.66 + 1000 = 2060.66.
But this is not in the range [a, c]. So, this is infeasible.
Let us check the other part of D(x) i.e. for c <= x <= b
?D(x)=1−(4000−x)^2/(4000−1000)×(4000−2000)=0.375?
?=(4000−x)^2/(4000−1000)×(4000−2000)=1−0.375=0.625?
?=(4000−x)^2=3750000?
?x=4000−1936.5=2063.5.?
This is a feasible solution. So, x = 2063.5 is the optimal order quantity.
