question archive A furniture store chain with 25 stores is stocking a fashionable chair in each of their stores with a normal demand distribution with a mean of 250 and a standard deviation of 125
A furniture store chain with 25 stores is stocking a fashionable chair in each of their stores with a normal demand distribution with a mean of 250 and a standard deviation of 125
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A furniture store chain with 25 stores is stocking a fashionable chair in each of their stores with a normal demand distribution with a mean of 250 and a standard deviation of 125. The cost of the chair is $750, and it is sold for $1,500. Chairs that are not sold by the end of the season are sold for $250 per chair.
a. What are the values of cu and co?
b. What is the optimal service level for an individual store?
c. What is the Z value?
d. What is the optimal purchasing quantity for an individual store?
e. If the chain is purchasing and keeping the chairs centrally (for all 25 stores), what is the demand distribution at the center?
F. If the chain is purchasing and keeping the chairs centrally (for all 25 stores), what is the optimal order quantity?
g. What is the probability of having some shortage at the center?
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Answer:
Step-by-step explanation
Mean = 250
Standard deviation (s) = 125
Cost of chair = 750$
Selling price of chair = 1500$
Salvage value of chair = 250$
a)
Cost of under supply (Cu) = Selling price of chair - Cost of chair = 1500 - 750 = 750$
Cost of over supply (Co) = Cost of chair - Salvage value of chair = 750 - 250 = 500$
b)
Service level = Cu/(Cu+Co) = 750/(750+500) = 0.60
60%
c)
z value for 0.6 using z table or NORMSINV function in excel is:
NORMSINV(0.6) = 0.25
d)
Optimal purchasing quantity = Mean + z*s = 250 + 0.25*125 = 281.25 units
Round off = 281 units
(e)
Mean of the demand for the 25 stores = 25*250 = 6250
Standard deviation of the demand for 25 stores = 25*125 = 3125
Hence the demand distribution is with mean of 6250 and standard deviation of 3125
(F)
From the earlier calculations:
Optimal purchasing quantity = Mean + z*s = 6250 + 0.25*3125 = 7031.25
Round off = 7031 units
(g)
Probability of shortage at the center = 1-0.6 = 0.40 = 40 percentage.
