Goofy owns and manages hot dog stand

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Reference no: EM131797189

Goofy owns and manages a hot dog stand near Walt Disney World. Although Goofy can serve 30 customers per hour on average (μ), he only gets 20 customers per hour (λ). Because Goofy could wait on 50% more customers than actually visit his stand, it doesn’t make sense to him that he should have any waiting lines.

Goofy hires you to examine the situation and determine some characteristics of his queue. After looking into the problem, you find this to be an M/M/1 system.

Find L, W, Lq, wq, and the probabilities of 0, 1, 2, and 3 customers waiting in line.

A. L= (1/30)/(1/20)-(1/30)=2

Lq=(1/30)^2/(1/20) [(1/20)-(1/30)]=4/3 or 1.33

W=1/{(1/20)-(1/30)]= 6hrs

Wq=(1/30) /(1/20) [(1/20)-(1/30)]= 4hrs

Probability of 0 customers is 1-2/3= 1/3 or .33

Probability of 1 customer is 2/3 * (1-2/3)= 2/9 or .22

Probability of 2 customers is 4/9*(1-2/3)= 4/27 or .15

Probability of 3 customers is 8/27*(1-2/3)= 8/81 or .10

Gawrsh! After looking over your data from (a) you and Goofy are surprised by the length of the lines. Goofy estimates the value of the customer’s waiting time (in the queue) at 10 cents per minute. During the 12 hours that he is open he gets (12 x 20) = 240 customers. The average customer is in a queue 4 minutes, so the total customer waiting time is (240 x 4) = 960 minutes. The value of 960 minutes is ($.10)(960 minutes) = $96. You tell Goofy that not only is 10 cents per minute quite conservative, but he could probably save most of that $96 of customer ill will if he hired another salesclerk. After much haggling, Good agrees to provide you with all the chili dogs you can eat during a week in exchange for your analysis of the results of having two clerks wait on the customers.

Assuming that Goofy hires one additional clerk whose service rate is equal to his own, find L, W, Lq, wq, the total customer waiting time and the total cost of waiting (again assuming $.10 per minute).

Lq=

Wq=

Wait time=

Wait cost=

Reference no: EM131797189

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