1. The fast-food restaurant McDonalds operates on national basis. The store manager of Mcdonalds on Jesse street has to decide how many ‘Cheese Dog’s to make every day. He learns that daily demand for ‘Cheese Dog’s follows a normal distribution of mean 100 and standard deviation 16. (a) The manager estimates that he will be able to make profit if he can sell more than 136 ‘Cheese Dog’s every day. The manager expects to get incentive only if he makes profit. What is the probability that the manager will receive incentive? (b) The manager is also aware that if the sale drops below 60, then he receives a pay-cut. What is the probability that he will receive a pay cut? (c) What is the probability that the ‘Cheese Dog’ sales is within 20% of the average value? (d) The store manager must plan to get breads ready to cook ‘Cheese Dog’ everyday morning. How many breads should he prepare for any given days so that he can keep the stock-out probability within 10%? (e) If the manager has materials available to make only 128 ‘Cheese Dog’s for a given day, how many will the expected lost sales be, assuming there is no way that additional raw materials cannot be refilled during the day? (f) If the manager wants to keep the expected lost sales to be 15 or lower, then How many breads (equivalent to how many ‘Cheese Dog’s) must he store at the beginning of the day?

Mean demand, m = 100
Std deviation of demand, s = 16
(a) z value = (136-100)/16 = 2.25
Corresponding probability = 1 – NORMSDIST(2.25) = 0.0122 or 1.22 %
Probability that the manager will receive incentive = 1.22 %
(b) z value = (60-100)/16 = -2.5
Corresponding probability = NORMSDIST(-2.5) = 0.0062 or 0.62 %
Probability that the manager will receive incentive = 0.62 %
(c) 20% more than mean demand = 100+20 = 120
z value = (120-100)/16 = 1.25, Corresponding probability that sales is less than 120 = NORMSDIST(1.25) = 0.8944
20% less than meand demand = 100-20 = 80
z value = (80-100)/16 = -1.25, Corresponding probability that sales is less than 80 = NORMSDIST(-1.25) = 0.1056
Probability that the sales is within 20% of the average (mean demand) value = 0.8944 – 0.1056 = 0.7888 = 78.88 %
(d) For stock-out probability within 10%, target stock-in probability = 90%
Corresponding z value = NORMSINV(0.90) = 1.28
Quantiy of breads to prepare = m+zs = 100+1.28*16 = 120.5 ~ 121 breads
(e) z = (128-100)/16 = 1.75
L(z) = 0.0162 (taken from standard loss table corresponding to z = 1.75)
Expected lost sales = s*L(z) = 16*0.0162 = 0.26
(f) Target L(z) = 15/16 = 0.9375
Corresponding value of z = -0.8219
Quantity of breads to store = m+zs = 100 + (-0.8219)*16 = 86.85 ~ 87
He must store a minimum of 87 breads
 
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