MODULE 5 – CHAPTER 11 MODULE 5 – ASSIGNMENT CHAPTER 11 – QUESTIONS 1, 2, 3, 6, 14, 15, 18, 19, 24 1. a) What is the mean or expected number of customers that will arrive in a five-minute period? The mean or expected number of customers to arrive in a 5-minute period is: 0.40 probability of 1 customer per minute x 5 minutes = 5 x 0.40 = 2 2 customers are expected. b) Assume that the Poisson probability distribution can be used to describe the arrival process. Use the arrival rate in part (a) and computer the probabilities that exactly 0, 1, 2, and 3 customers will arrive during a 5-minute period: ๐(0) = ๐(1) = ๐(2) = ๐(3) = (2.0)0 โ −2.0 0! (2.0)1 โ −2.0 1! (2.0)2 โ −2.0 2! (2.0)3 โ −2.0 3! = 0.13533 Number of Arrivals Probability 0 0.13533 1 0.27067 2 0.27067 3 0.18045 = 0.27067 = 0.27067 = 0.18045 c. Delays are expected if more than three customers arrive during any 5-minute period. What is the probability that delays will occur? P(x > 3, 2) = 1 - P(0; 2) + P(1; 2) + P(2; 2) + P(3; 2) P(x > 3, 2) = 1 – ( (2.0)0 โ −2.0 0! )+( (2.0)1 โ −1.0 1! )+( (2.0)2 โ −2.0 2! )+( P(x > 3, 2) = 1 - (0.13533 + 0.27067 + 0.27067 + 0.18045) P(x > 3, 2) = 0.14288 0.14288 is the probability a delay will occur (2.0)3 โ −2.0 3! ) MODULE 5 – CHAPTER 11 2. Assume that the service times for the drive-up teller follow an exponential probability distribution with a service rate of 36 customers per hour, or 0.6 customers per minute. Use the exponential probability distribution to answer the following questions: a) What is the probability that the service time is one minute or less? P(service time ≤ 1.0 min.) = 1 − โ −0.60(1.0) = ๐. ๐๐๐๐๐ ๐. ๐๐๐๐๐ is the probability the service time is 1 minute or less b) What is the probability that the service time is two minutes or less? P(service time ≤ 2.0 min.) = 1 − โ −0.60(2.0) = ๐. ๐๐๐๐๐ ๐. ๐๐๐๐๐ is the probability the service time is 2 minutes or less c) What is the probability that the service time is more than two minutes? 1- ๐. ๐๐๐๐๐ = ๐. ๐๐๐๐๐ ๐. ๐๐๐๐๐ is the probability the service time is more than 2 minutes 3. Determine the following operating characteristics for the system: a. The probability that no customers are in the system ๐ 0.40 ๐0 = 1 − ๐ข = ๐0 = 1 − 0.60 = 0.33333 The probability that no units are in the system is 0.33333 b. The average number of customers waiting ๐2 0.402 ๐ฟ๐ = ๐ข(๐ข−๐) = ๐ฟ๐ = 0.60(0.60−0.40) = 1.33333 The average number of customers waiting is 1.33333 c. The average number of customers in the system ๐ 0.40 ๐ฟ = ๐ฟ๐ + ๐ข = ๐ฟ = 1.33333 + 0.60 = ๐ The average number of customers in the system is 2 d. The average time a customer spends waiting ๐ฟ๐ 1.33333 ๐๐ = ๐ = ๐๐ = 0.40 = ๐. ๐๐๐๐ The average time a customer spends waiting is ๐. ๐๐๐๐ ๐ฆ๐ข๐ง๐ฎ๐ญ๐๐ฌ e. The average time a customer spends in the system 1 1 ๐ = ๐๐ + = ๐ = 3.33333 + = ๐. ๐๐๐๐๐ ๐ข 0.60 The average time a customer spends in the system is 5 minutes f. The probability that arriving customers will have to wait for service ๐ 0.40 ๐๐ค = ๐ข = ๐๐ค = 0.60 = ๐. ๐๐๐๐๐ The probability that arriving customers will have to wait for service is ๐. ๐๐๐๐๐ MODULE 5 – CHAPTER 11 6. During the weeknight evenings, customers arrive at Movies Tonight with an arrival rate of 1.25 customers per minute. The checkout clerk has a service rate of 2 customers per minute. a. What is the probability that no customers are in the system? ๐ 1.25 ๐0 = 1 − ๐ข = ๐0 = 1 − 2.00 = 0.375 The probability that no customers are in the system is 0.375 b. What is the average number of customers waiting for service? ๐2 1.252 ๐ฟ๐ = ๐ข(๐ข−๐) = ๐ฟ๐ = 2.00(2.00−1.25) = 1.04167 The average number of customers waiting is 1.04167 c. What is the average time a customer waits for service to begin? ๐ฟ๐ 1.04167 ๐๐ = ๐ = ๐๐ = 1.25 = ๐. ๐๐๐๐๐ The average time a customer spends waiting is ๐. ๐๐๐๐๐ ๐ฆ๐ข๐ง๐ฎ๐ญ๐๐ฌ or 50 seconds d. What is the probability that an arriving customer will have to wait for service? ๐ 1.25 ๐๐ค = ๐ข = ๐๐ค = 2.00 = ๐. ๐๐๐ The probability that arriving customers will have to wait for service is ๐. ๐๐๐ e. Do the operating characteristics indicate that the one-clerk checkout system provides an acceptable level of service? A customer having a 62.5% chance of having to wait approximately 50 seconds is an acceptable level of service in my opinion. 14. The customer calls follow a Poisson probability distribution, with an arrival rate of five calls per hour. Qn average, it takes 7.5 minutes for a consultant to answer a customer s questions. a. What is the service rate in terms of customers an hour? 60 ๐๐๐๐ข๐กโ ๐ 7.5 ๐๐๐๐ข๐กโ ๐ ๐โ ๐ ๐๐๐๐ =8 The service rate is 8 customers per hour b. What is the probability that no customers are in the system and the consultant is idle? ๐ 5 ๐0 = 1 − ๐ข = ๐0 = 1 − 8 = 0.375 The probability that no customers are in the system is 0.375 MODULE 5 – CHAPTER 11 c) What is the average number of customers waiting for a consultant? ๐2 52 ๐ฟ๐ = ๐ข(๐ข−๐) = ๐ฟ๐ = 8(8−5) = 1.04167 The average number of customers waiting is 1.04167 d) What is the average time a customer waits for a consultant? ๐ฟ๐ 1.04167 ๐๐ = ๐ = ๐๐ = = ๐. ๐๐๐๐๐๐ 5 The average time a customer spends waiting is ๐. ๐๐๐๐๐๐ ๐ก๐จ๐ฎ๐ซ๐ฌ or 12 minutes and 30 seconds e) What is the probability that a customer will have to wait for a consultant? ๐ 5 ๐๐ค = ๐ข = ๐๐ค = 8 = ๐. ๐๐๐ The probability that arriving customers will have to wait for service is ๐. ๐๐๐ f) If Ocala customer service guidelines state that no more than 35% of customers should have to wait for technical support and the waiting time should be 2 minutes or less, Ocala is not meeting its customer service guidelines as 62.5% of customer have to wait and the average wait time is over 12 minutes. Ocala needs to either hire more consultants or make the system more efficient. They could consider adding self-service prompts to receive prerecorded instructions for common technical problems with the option to escalate to a consultant if required. 15. What effect would adding a second consultant have on customer service? Would two consultants allow Ocala to meet its customer service guidelines of 35% or less customers wait and the wait time is 2 minutes or less? ๐ ๐0 = 0.52395 ๐๐ ๐โ ๐ ๐ก๐๐๐โ 11.4: ๐ข = 5 8 = ๐. ๐๐๐ Average table values for 0.60 and 0.65 = (0.5385+0.5094)/2 = 0.52395 ๐ท๐ = ๐. ๐๐๐๐๐ The probability that an arriving unit has to wait for service: 1 ๐ ๐ ๐๐ข 1 5 2 2(8) ๐๐ค = ๐! (๐ข) (๐๐ข−๐) ๐0 = 2! (8) (2(8)−5) 0.52395 = ๐. ๐๐๐๐๐๐ The probability that a customer would have to wait with 2 servers would decrease to 0.148849, which is within the Ocala customer service guidelines of 35% or less. The average number of units in the waiting line: (๐โ๐ข)๐ ๐๐ข (5โ8)2 5(8) ๐ฟ๐ = (๐−1!)(๐๐ข−๐)2 ๐0 = (2−1!)(2(8)−5)2 0.52395 = 0.06766 MODULE 5 – CHAPTER 11 The average time a customer spends waiting: ๐๐ = ๐ฟ๐ ๐ = 0.013532 = or 0.81191 minutes or approximately 49 seconds. A 49 second wait time is within the Ocala customer service guidelines of 2 minutes or less. 18. The airport has three screening stations available. The service rate for processing passengers at each screening station is 3 passengers per minute. The arrival rate is 5.4 passengers per minute. Assume that processing times at each screening station follow an exponential distribution and that arrivals follow a Poisson distribution. a. Suppose two of the three screening stations are open on Monday morning. Compute the operating characteristics for the screening facility. ๐ = 5.4 ๐๐๐ ๐ โ ๐๐โ ๐๐ ๐โ ๐ ๐๐๐๐ข๐กโ ๐ข = 3 ๐๐๐ ๐ โ ๐๐โ ๐๐ ๐โ ๐ ๐๐๐๐ข๐กโ ๐โ ๐ ๐๐๐๐ ๐ ๐ = 2 1. ๐โโ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ ๐๐ ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โ โ ๐๐๐๐. ๐0 = 0.0526 (5.4/3 = 1.80 ๐/ ๐ข ๐๐๐ก๐๐ − ๐โ ๐ ๐ก๐๐๐โ 11.4 ๐๐ ๐กโโ ๐กโ ๐ฅ๐ก: = ๐. ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ ๐๐ ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โ โ ๐๐๐๐. 2. the average number of customers in line: (๐โ๐ข)๐ ๐๐ข (5.4โ3)2 5.4(3) ๐ฟ๐ = (๐−1!)(๐๐ข−๐)2 ๐0 = (2−1!)(2(3)−5.4)2 0.0526 = ๐. ๐๐๐๐ There is an average of 7.6691 customers in line. 3. The average number of customers in the system: ๐ ๐ฟ = ๐ฟ๐ + ๐ข = 7.6691 + 1.8 = 9.4691 average number of customers in the system There is an 9.4691 average number of customers in the system 4. The average time a customer spends in the waiting line: ๐๐ = ๐ฟ๐ ๐ = 7.6691 5.4 = 1.4202 The average time a customer spends in the waiting line is 1.4202 minutes 5. The average time a customer spends in the system is: ๐ = ๐๐ + 1 ๐ข = 1.4202 + 0.33333 = 1.75353 The average time a customer spends in the system is: 1.75353 minutes MODULE 5 – CHAPTER 11 b. Because of space considerations, the facility manager’s goal is to limit the average number of passengers waiting in line to 10 or fewer. Will the two-screening-station system be able to meet the manager’s goal? Yes, the average number of passengers in line with 2 screening stations is 7.6991. c. What is the average time required for a passenger to pass through security screening? 1.75353 minutes or 1 minute, 45 seconds. 19. When security level is raised to high, the service rate for processing passengers is reduced to 2 per minute at each screening station. The arrival rate is 5.4 passengers per minute. ๐ = 5.4 ๐๐๐ ๐ โ ๐๐โ ๐๐ ๐โ ๐ ๐๐๐๐ข๐กโ ๐ข = 2 ๐๐๐ ๐ โ ๐๐โ ๐๐ ๐โ ๐ ๐๐๐๐ข๐กโ ๐โ ๐ ๐๐๐๐ ๐ ๐ = 3 1. ๐โโ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ ๐๐ ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โ โ ๐๐๐๐. ๐0 = 0.0526 (5.4/3 = 2.70 ๐/ ๐ข ๐๐๐ก๐๐ − ๐โ ๐ ๐ก๐๐๐โ 11.4 ๐๐ ๐กโโ ๐กโ ๐ฅ๐ก: = ๐. ๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ ๐๐ ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โ โ ๐๐๐๐. 2. the average number of customers in line: (๐โ๐ข)๐ ๐๐ข (5.4โ2)3 5.4(2) ๐ฟ๐ = (๐−1!)(๐๐ข−๐)2 ๐0 = (3−1!)(3(2)−5.4)2 0.02525 = ๐. ๐๐๐๐๐ There is an average of 7.45494 customers in line. 3. The average number of customers in the system: ๐ ๐ฟ = ๐ฟ๐ + ๐ข = 7.45494 + 2.70 = 10.15494 average number of customers in the system There is an 10.15494 average number of customers in the system 4. The average time a customer spends in the waiting line: ๐๐ = ๐ฟ๐ ๐ = ๐.๐๐๐๐๐ 5.4 = 1.38054 The average time a customer spends in the waiting line is 1.38054 minutes 5. The average time a customer spends in the system is: ๐ = ๐๐ + 1 ๐ข = 1.38054 + 0.50 = 1.75353 The average time a customer spends in the system is: 1.88054 minutes MODULE 5 – CHAPTER 11 a. The facility manager’s goal is to limit the average number of passengers line to 10 or fewer. How many screening stations must be open in order to manager's goal? 3 Screen stations must be opened with an average of ๐. ๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐. b. What is the average time required for a passenger to pass through security screening? 1.88054 minutes or 1 minute and 53 seconds. 24. The average time between the arrival of a customer at the food-service counter and his or her departure with a filled order is 10 minutes. During the game, customers arrive at the rate of four per minute. The food-service operation requires an average of 2 minutes per customer order. ๐ = 4 ๐๐ข๐ ๐ก๐๐โ ๐๐ ๐โ ๐ ๐๐๐๐ข๐กโ ๐ข = 0.50 ๐๐ข๐ ๐ก๐๐โ ๐๐ ๐โ ๐ ๐๐๐๐ข๐กโ a. What is the service rate per server in terms of customers per minute? ๐ข = 1 ๐๐๐๐ข๐กโ /2 ๐๐๐๐ข๐กโ ๐ ๐โ ๐ ๐๐ข๐ ๐ก๐๐โ ๐ = 0.50 ๐๐ข๐ ๐ก๐๐โ ๐๐ ๐โ ๐ ๐๐๐๐ข๐กโ b. What is the average waiting time in the line prior to placing an order? W = 10 ๐๐๐๐ข๐กโ ๐ = ๐๐ + 1 0.50 = Wq = 10 minutes – 2 minutes = 8 the total time waiting in line is 8 minutes. c. On average, how many customers are in the food-service system? Wq=Lq / λ 8= Lq / 4 Lq = 32 L= Lq + λ/ μ L= 32 + 4/0.5 L=40 (therefore 40 is the average number of units in the system)