Dan McClure owns a thriving independent bookstore in artsy New Hope, Pennsylvani
ID: 376310 • Letter: D
Question
Dan McClure owns a thriving independent bookstore in artsy New Hope, Pennsylvania. He must decide how many copies to order of a new book, Power and Self Destruction, an exposé on a famous politician's lurid affairs. Interest in the book will be intense at first and then fizzle quickly as attention turns to other celebrities. The book's retail price is $20, and the wholesale price is $13. The publisher will buy-back the retailer's leftover copies at a full refund, but McClure Books incurs $5 in shipping and handling costs for each book returned to the publisher. Dan believes his demand forecast can be represented by a Normal distribution with mean 225 and standard deviation 60 If a part of the question specifies whether to use Table 13.4, or to use Excel, then credit for a correct answer will depend on using the specified method Dan will consider this book to be a blockbuster for him if it sells more than 380 and Self Destruction will be a blockbuster? (Round your answer to 4 decimal places.) Dan will consider a book a "dog" if it sells less than 50% of his mean forecast a. units. Using Table 13.4 and the round-up function, what is the probability Power 0.0049 0.0301 Using Excel, calculate the probability that this expose will be a dog (Round your answer to 4 decimal places.) Use Table 13.4 to determine the probability that demand for this book will be within ·20% of the mean forecast (Round your answer to 4 decimal places.) Using Table 13.4 and the round-up rule, calculate the quantity that maximizes 403 Dan's expected profit.Explanation / Answer
a). For 380 books, z-statistic = (380-225)/60 = 2.5833
From table 13.4, corresponding value of F(z) = 0.9951 (interpolation is used to find the exact value)
Probability of selling more than 380 books = 1 - 0.9951 = 0.0049
b) 50% of mean forecast = 225*50% = 112.5
z-statistic = (112.5-225)/60 = -1.8750
Corresponding value of F(z) = NORMSDIST(-1.8750) = 0.0304
Probability that this expose will be a dog = 0.0304
c)
Upper limit of 20% of forecast = 225*(1+20%) = 270
z-statistic = (270-225)/60 = 0.75 , From Table 13.4, corresponding value of F(z) = 0.7734
Lower limit of 20% of forecast = 225*(1-20%) = 180
z-statistic = (180-225)/60 = -0.75 , From Table 13.4, corresponding value of F(z) = 0.2266
Probability that the demand for the book will be within this range (180 to 270) = 0.7734 - 0.2266 = 0.5467
d) Underage cost, Cu = retail price - wholesale price = 20-13 = 7
Overage cost, Co = cost in shipping and handling for book returned to publisher = 5
Critical ratio F(z) = Cu/(Cu+Co) = 7/(7+5) = 0.5833
From Table 13.4, corresponding z-statistic = 0.2103
Optimal order quantity =µ + z* = 225 + 0.2103*60 = 238
e) For 95% in-stock probability, F(z) = 0.95
Corresponding to F(z) = 0.95, z-statistic = 1.645
Order quantity, Q = µ + z* = 225 + 1.645*60 = 323.75
Corresponding to F(z)=0.95, From Table 13.4, I(z) = 1.6657
Expected inventory, V = *I(z) = 60*1.6657 = 100
Expected sales, S = Q - V = 323.75 - 100 = 223.75
Expected lost sales, L = µ - S = 225 - 223.75 = 1.25
Stockout rate = L/µ = 1.25/225 = 0.0056
f) For Q=325, z-statistic = (Q-µ)/ = (325-225)/60 = 1.6667
From table 13.4, value of I(z) = 1.6865
Expected leftover inventory, V = *I(z) = 60*1.6865 = 101.19
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