Provide your detailed answers below every part of each question.
The selling price of a particular food item is $20 per packet. The cost to the store is $14 per packet. The unsold packets must be salvaged at the end of the day for $3 per packet. The store manager believes that the shortage cost is $2 per packet. How many packets should the store buy each day to maximize profit if demand is:
a) Normal with mean 400 and standard deviation 60?
b) Uniform with lower limit to be 200 and upper limit to be 600?
c) Demand for the last 400 days was as shown below
Demand200250350500
Frequency 80 200 80 40
d) Calculate the average daily profit for part c)
a)
For a normal distribution with mean 400 and standard deviation 60, we can use the Newsvendor model to determine the optimal order quantity that maximizes profit.
Given data:
Selling price = $20/packet
Cost to the store = $14/packet
Salvage value = $3/packet
Shortage cost = $2/packet
Mean demand = 400
Standard deviation = 60
To calculate the optimal order quantity, we use the following formula:
Optimal order quantity = Mean demand + (Z * Standard deviation)
Z is the z-score corresponding to the desired service level. Let's assume a service level of 0.95, which corresponds to a z-score of approximately 1.645.
Optimal order quantity = 400 + (1.645 * 60) = 400 + 98.7 ≈ 499.7
Since we can't order fractional packets, the optimal order quantity is rounded up to the nearest whole number, which is 500 packets.
b) For a uniform distribution with a lower limit of 200 and an upper limit of 600, we can use the Newsvendor model to determine the optimal order quantity that maximizes profit.
Given data:
Selling price = $20/packet
Cost to the store = $14/packet
Salvage value = $3/packet
Shortage cost = $2/packet
Lower limit of demand = 200
Upper limit of demand = 600
To calculate the optimal order quantity, we use the following formula:
Optimal order quantity = (Selling price - Salvage value) / (Selling price - Cost to the store) * (Upper limit - Mean demand)
Optimal order quantity = (20 - 3) / (20 - 14) * (600 - 200) = 17 / 6 * 400 ≈ 1133.33
Since we can't order fractional packets, the optimal order quantity is rounded up to the nearest whole number, which is 1134 packets.
c) Given the demand and frequency data for the last 400 days, we can calculate the average demand and use that as the mean demand in the Newsvendor model.
Given data:
Demand: [200, 250, 350, 500]
Frequency: [80, 200, 80, 40]
To calculate the average demand, we use the following formula:
Average demand = (Demand1 * Frequency1 + Demand2 * Frequency2 + ... + Demandn * Frequencyn) / Total frequency
Average demand = (200 * 80 + 250 * 200 + 350 * 80 + 500 * 40) / (80 + 200 + 80 + 40) = (16000 + 50000 + 28000 + 20000) / 400 = 114000 / 400 = 285
Using the average demand of 285 as the mean demand, we can apply the Newsvendor model to determine the optimal order quantity.
d) To calculate the average daily profit for part c), we need to consider all possible scenarios based on the demand frequencies.
Given data:
Demand: [200, 250, 350, 500]
Frequency: [80, 200, 80, 40]
Selling price = $20/packet
Cost to the store = $14/packet
Salvage value = $3/packet
Shortage cost = $2/packet
For each demand scenario, we can calculate the profit using the following formula:
Profit = (Selling price - Cost to the store) * Quantity sold - Shortage cost * Quantity short - Salvage value * Quantity salvage
Let's calculate the average daily profit:
Profit1 (Demand: 200) = (20 - 14) * 200 - 2 * (285 - 200) - 3 * (200 - 285) ≈ $1200
Profit2 (Demand: 250) = (20 - 14) * 250 - 2 * (285 - 250) - 3 * (250 - 285) ≈ $1800
Profit3 (Demand: 350) = (20 - 14) * 350 - 2 * (285 - 350) - 3 * (350 - 285) ≈ $2900
Profit4 (Demand: 500) = (20 - 14) * 400 - 2 * (285 - 400) - 3 * (400 - 285) ≈ $4700
Average daily profit = (Profit1 * Frequency1 + Profit2 * Frequency2 + Profit3 * Frequency3 + Profit4 * Frequency4) / Total frequency
Average daily profit ≈ ($1200 * 80 + $1800 * 200 + $2900 * 80 + $4700 * 40) / (80 + 200 + 80 +40) ≈ $2466.67