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Business Operations Management Production and Operations Analysis, Seventh Edition

(a) To find the expected time in the clinic, the expected time in the queue for advice, the expected number of students in the clinic and the expected number of students waiting for advice. The expected time in the clinicis 0.5 , the expected time in the queue for advice is 0.41667 , the expected number of students in the clinic is 5 and the expected number of students waiting for advice is 4.1667 . In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. Assume there is only one advisor serving the clinic. Thus, we have, λ = 10 per hour μ = 60 5 = 12 per hour Now, let us calculate the utilization rate as: p = λ μ = 10 12 = 5 6 Now, the expected number of students in the clinic is calculated as: l = p 1 − p = 5 / 6 1 − ( 5 / 6 ) = 5 And the expected time in the clinic is calculated as: w = l λ = 5 10 = 0.5 Thus, the expected number of students waiting for adviceis calculated as: l q = l p = 5 × 5 6 = 4.1667 And the expected time in the queue for advice is calculated as: w q = l q λ = 4.1667 10 = 0.41667 (b) To find out what is the probability that there are more than ten students in the clinic at a random point in time and what is the probability that the time spent in the clinic exceeds 30 minutes. The probability that there are more than ten students in the clinic at a random point in time is 0.1615 and what is the probability that the time spent in the clinic exceeds 30 minutes is 0.3679 . In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. Assume there is only one advisor serving the clinic. Thus, we have, λ = 10 per hour μ = 60 5 = 12 per hour Now, let us calculate the utilization rate as: p = λ μ = 10 12 = 5 6 Thus, the probability that there are more than ten students in the clinic at a random point in time is calculated as: P ( L > 10 ) = p 10 = ( 5 6 ) 10 = 0.1615 And the probability that the time spent in the clinic exceeds 30 minutes is calculated as: P ( W > t ) = P ( W > 30 60 ) = P ( W > 0.5 ) = e − ( μ − λ ) t = e − ( 12 − 10 ) 0.5 = 0.3679 (c) To repeat your answer for part (a) and what are the advisors’ utilizations and explain would you recommend this second advisor is hired. The expected time in the clinic is 0.07143 , the expected time in the queue for advice is 0.02976 , the expected number of students in the clinic is 0.7143 and the expected number of students waiting for advice is 0.2976 . In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. Assume that a second advisor is hired. Thus, we have, λ = 10 per hour μ = 60 5 = 12 per hour Now, let us calculate the utilization rate as: p = λ s μ = 10 2 × 12 = 5 12 Now, the expected number of students in the clinic is calculated as: l = p 1 − p = 5 / 12 1 − ( 5 / 12 ) = 0.7143 And the expected time in the clinic is calculated as: w = l λ = 0.7143 10 = 0.07143 Thus, the expected number of students waiting for advice is calculated as: l q = l p = 0.7143 × 5 12 = 0.2976 And the expected time in the queue for advice is calculated as: w q = l q λ = 0.2976 10 = 0.02976 Since the expected time in the clinic and the expected time in queue for advice decreases by hiring the second advisor then we recommend that the advisor should be hired. (d) To explain what are some ways that the advice clinic could improve students’ experiences in the advice clinic without hiring more staff. In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. There are some ways that the advice clinic could improve students’ experiences in the advice clinic without hiring more staff as they can improve the way of treating the students that come for advices by including extra-curricular activities or they can give them some activities to preform which can help the advisor also and which can become fun for students also.

(a) To find the expected time in the clinic, the expected time in the queue for advice, the expected number of students in the clinic and the expected number of students waiting for advice. The expected time in the clinicis 0.5 , the expected time in the queue for advice is 0.41667 , the expected number of students in the clinic is 5 and the expected number of students waiting for advice is 4.1667 . In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. Assume there is only one advisor serving the clinic. Thus, we have, λ = 10 per hour μ = 60 5 = 12 per hour Now, let us calculate the utilization rate as: p = λ μ = 10 12 = 5 6 Now, the expected number of students in the clinic is calculated as: l = p 1 − p = 5 / 6 1 − ( 5 / 6 ) = 5 And the expected time in the clinic is calculated as: w = l λ = 5 10 = 0.5 Thus, the expected number of students waiting for adviceis calculated as: l q = l p = 5 × 5 6 = 4.1667 And the expected time in the queue for advice is calculated as: w q = l q λ = 4.1667 10 = 0.41667 (b) To find out what is the probability that there are more than ten students in the clinic at a random point in time and what is the probability that the time spent in the clinic exceeds 30 minutes. The probability that there are more than ten students in the clinic at a random point in time is 0.1615 and what is the probability that the time spent in the clinic exceeds 30 minutes is 0.3679 . In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. Assume there is only one advisor serving the clinic. Thus, we have, λ = 10 per hour μ = 60 5 = 12 per hour Now, let us calculate the utilization rate as: p = λ μ = 10 12 = 5 6 Thus, the probability that there are more than ten students in the clinic at a random point in time is calculated as: P ( L > 10 ) = p 10 = ( 5 6 ) 10 = 0.1615 And the probability that the time spent in the clinic exceeds 30 minutes is calculated as: P ( W > t ) = P ( W > 30 60 ) = P ( W > 0.5 ) = e − ( μ − λ ) t = e − ( 12 − 10 ) 0.5 = 0.3679 (c) To repeat your answer for part (a) and what are the advisors’ utilizations and explain would you recommend this second advisor is hired. The expected time in the clinic is 0.07143 , the expected time in the queue for advice is 0.02976 , the expected number of students in the clinic is 0.7143 and the expected number of students waiting for advice is 0.2976 . In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. Assume that a second advisor is hired. Thus, we have, λ = 10 per hour μ = 60 5 = 12 per hour Now, let us calculate the utilization rate as: p = λ s μ = 10 2 × 12 = 5 12 Now, the expected number of students in the clinic is calculated as: l = p 1 − p = 5 / 12 1 − ( 5 / 12 ) = 0.7143 And the expected time in the clinic is calculated as: w = l λ = 0.7143 10 = 0.07143 Thus, the expected number of students waiting for advice is calculated as: l q = l p = 0.7143 × 5 12 = 0.2976 And the expected time in the queue for advice is calculated as: w q = l q λ = 0.2976 10 = 0.02976 Since the expected time in the clinic and the expected time in queue for advice decreases by hiring the second advisor then we recommend that the advisor should be hired. (d) To explain what are some ways that the advice clinic could improve students’ experiences in the advice clinic without hiring more staff. In the question, it is given that students arrive one at a time to an advice clinic at a rate of ten per hour. Students take on average five minutes of advice but there is wide variation in the time they need. There are some ways that the advice clinic could improve students’ experiences in the advice clinic without hiring more staff as they can improve the way of treating the students that come for advices by including extra-curricular activities or they can give them some activities to preform which can help the advisor also and which can become fun for students also.

Solution Summary: The author calculates the expected time in the clinic, the queue for advice, and the anticipated number of students in a clinic.

Production and Operations Analysis, Seventh Edition

Production and Operations Analysis, Seventh Edition

7th Edition

ISBN: 9781478623069

Author: Steven Nahmias, Tava Lennon Olsen

Publisher: Waveland Press, Inc.

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Production and Operations Analysis, Seventh Edition

Ch. 7.1 - Prob. 1P Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - Prob. 4P Ch. 7.1 - Prob. 5P Ch. 7.1 - Prob. 6P Ch. 7.1 - Prob. 7P Ch. 7.2 - Prob. 8P Ch. 7.2 - Prob. 9P Ch. 7.2 - Prob. 10P

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