1. An urn contains nine white balls and 11 black balls. A ball is drawn and repl
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1. An urn contains nine white balls and 11 black balls. A ball is drawn and replaced. If the ball is white your opponent pays you 25 cents. If it is black you pay him 25 cents. You have one dollar and your opponent has 50 cents. Play continues until one of you is broke. What is the probability that you lose all your money? What is the expected number of times that you play prior to the game ending? What is the answer to these two questions if both you and your opponent started with 75 cents? 2. The manager of the Hippon Drug Store is remodeling his facility and is considering three ways of organizing his branded-items business. The first way is to employ a fast clerk to wait on all customers. Assume such a clerk serves customers exponentially at an average rate of two minutes per customer. The second way is to employ two moderately fast clerks to wait on all customers. Assume each serves customers exponentially at an average rate of four minutes per customer. The third way is to use a self-service system in which each customer waits on himself. Self-service, at an average rate of six minutes, is slower than clerk service. The manager wants to calculate the average number of customers in the store, the average time each spends in the store, and the average time each spends waiting for service, under each way of organizing. Assume that customers arrive completely randomly, one at a time, at the rate of 15 per hour. Calculate the operating characteristics of interest to the manager.Explanation / Answer
1. Note that, as we have $1 and the opponent has $.50 and each game is $.25, this game is equivalent to starting with n=4 (1/.25) and going to either 6=(1+.5)/.25 or 0 As there are 9 white balls and 11 black balls, and we are playing with replacement, p = 9/20 and q = 11/20 We thus have difference equations sn = qsn-1 + psn+1, where s0 = 0 and s6 = 1 It is shown in http://www.math.harvard.edu/~ctm/papers/home/text/class/harvard/154/course/course.pdf page 70 that the solution to this problem is (11/9)^4 - 1 / (11/9)^6 - 1 This is 0.527755378511757 2. We are asked for average number of customers in the store, average time in store, and average waiting time. For this problem, my source for the equations is http://irh.inf.unideb.hu/user/jsztrik/education/09/english/3f.html We start with an M/M/1 system service time is 2 minutes, customers arrive at 15/hr equals every 4 minutes. Thus, mu = 1/2 and lambda = 1/4 rho = lambda/mu = 1/4/(1/2) = 1/2 Then, from the formulas, average number number of customers = rho/(1- rho) = 1/2/(1-1/2) = 1/2/(1/2) = 1 Ws, average service time, is 1/mu = 2minutes Average time in the store W is Ws/ 1 - rho = 2/(1 - 1/2) = 2/(1/2) = 4 minutes Average waiting time Wq = W - Ws = 4 minutes - 2 minutes = 2 minutes Next, we have an M/M/2 system where mu is now 1/4 rho remains 1/2 (It is now calculated as (1/4)/(1/4)/2 ) L = 2 rho/(1 - rho ^ 2) = 2 * 1/2/(1- (1/2)^2) = 1/(3/4) = 4/3 W = Ws/(1 - rho ^ 2) = 4/(1 - (1/2)^2) = 4/(3/4) = 16/3 minutes = 5 1/3 minutes Wq = W - Ws = 5 1/3 - 4 = 1 1/3 minutes Finally, we have M/M/infinity L = 1/4/(1/6) = 3/2 Total waiting time W = Ws = 6 minutes Waiting time Wq is now 0
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