Waiting Line Questions Render Essay

Custom Student Mr. Teacher ENG 1001-04 10 June 2016

Waiting Line Questions Render

1.A waiting line meeting the M/M/1 assumptions has an arrival rate of 4 per hour and a service rate of 12 per hour. What is the probability that the waiting line is empty? Po = 1 – λ/μ = 1 – 4/12 = 8/12 or 0.667. (The variety of queuing models, easy) {AACSB: Analytic Skills}

2.A waiting line meeting the M/M/1 assumptions has an arrival rate of 4 per hour and a service rate of 12 per hour. What is the average time a unit spends in the system and the average time a unit spends waiting? Ws = 1 / (μ – λ) = 1 / (12 – 4) = 1/8 or 0.125; Wq = λ / (μ*(μ-λ)) = 4 / (12*8) = 1/24 or 0.0417. (The variety of queuing models, easy) {AACSB: Analytic Skills}

3.A waiting line meeting the M/M/1 assumptions has an arrival rate of 10 per hour and a service rate of 12 per hour. What is the average time a unit spends in the system and the average time a unit spends waiting? Ws = 1 / (μ – λ) = 1 / (12 – 10) = 1/2 or 0.5; Wq = λ / (μ*(μ-λ)) = 10 / (12*2) = 10 / 24 or 0.4167. (The variety of queuing models, easy) {AACSB: Analytic Skills}

4.A waiting line meeting the M/M/1 assumptions has an arrival rate of 10 per hour and a service rate of 12 per hour. What is the probability that the waiting line is empty? Po = 1 – λ/μ = 1 – 10/12 = 2/12 or 0.1667. (The variety of queuing models, easy) {AACSB: Analytic Skills}

5.A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of λ = 7.5 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of μ = 10 vehicles per day with a repair time distribution that approximates an exponential distribution.

a. What is the utilization rate for this service system?
b. What is the average time before the facility can return a breakdown to service?
c. How much of that time is spent waiting for service?
d. How many vehicles are likely to be in the system at any one time?
(a) Utilization is ρ = 7.5 / 10 = .75 or 75 percent; (b) Ws = 1 / (10 – 7.5)
= 1 / 2.5 = 0.4 days; (c) Wq = 7.5 / 10*(10-7.5) = 0.3 days; (d) Ls = 7.5 / (10-7.5) = 7.5 / 2.5 = 3 units.
(The variety of queuing models, easy) {AACSB: Analytic Skills}

6.A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of λ = 7 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of μ = 11 vehicles per day with a repair time distribution that approximates an exponential distribution.

a. What is the utilization rate for this service system?
b. What is the average time before the facility can return a breakdown to service?
c. How much of that time is spent waiting for service?
d. How many vehicles are likely to be waiting for service at any one time? (a) Utilization is ρ = 7 / 11 = .636 or 64 percent; (b) Ws = 1 / (11-7) = 1/4 = 0.25 days; (c) Wq = 7 / 11*(11-7) = 7 / 44 = 0.16 days; (d) Lq = 7*7 / 11*(11-7) = 49 / 44 = 1.114 units

(The variety of queuing models, easy) {AACSB: Analytic Skills}

7.A crew of mechanics at the Highway Department garage repair vehicles which break down at an average of λ = 5 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of μ = 10 vehicles per day with a repair time distribution that approximates an exponential distribution.

a. What is the probability that the system is empty?
b. What is the probability that there is precisely one vehicle in the system?
c. What is the probability that there is more than one vehicle in the system?
d. What is the probability of 5 or more vehicles in the system?

(a)P0 = 1 – 5/10 = 0.50; (b) Pn>1 =(5/10)2 = 0.25; the probability of exactly one is .50 -.25 = .25; (c) 0.25 as previously calculated; (d) the probability of five or more is Pn>4 = (5/10)5 = 0.0313.

(The variety of queuing models, moderate) {AACSB: Analytic Skills}

8.A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of λ = 8 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of μ = 11 vehicles per day with a repair time distribution that approximates an exponential distribution. The crew cost is approximately $300 per day. The cost associated with lost productivity from the breakdown is estimated at $150 per vehicle per day (or any fraction thereof). What is the expected cost of this system? The number of vehicles out of service is Ls = 8 / (11-8) = 8/3 = 2.667. The cost of waiting is $150 x Ls = $150 x 2.667 = $400. Server cost is $300 per day for a total of $700.

(The variety of queuing models, moderate) {AACSB: Analytic Skills}

9.A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of λ = 8 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of μ = 10 vehicles per day with a repair time distribution that approximates an exponential distribution.

a. What is the probability that the system is empty?
b. What is the probability that there is precisely one vehicle in the system?
c. What is the probability that there is more than one vehicle in the system?
d. What is the probability of 5 or more vehicles in the system? (a) P0 = 1 – 8/10 = 0.20; (b) Pn>1 =(8/10)2 = 0.64; the probability of exactly one is .36 -.20 = .16; (c) 0.64 as previously calculated; (d) Pn>4 = (8/10)5 = 0.32768. (The variety of queuing models, moderate) {ACSB: Analytic Skills}

10.A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of λ = 8 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of μ = 11 vehicles per day with a repair time distribution that approximates an exponential distribution. The crew cost is approximately $300 per day. The cost associated with lost productivity from the breakdown is estimated at $150 per vehicle per day (or any fraction thereof). Which is cheaper, the existing system with one service crew, or a revised system with two service crews? Ls for the single server is 8 / (11-8) = 8/3 = 2.667. The single-server system server cost is $300 per day; wait cost is $150 x 2.667 = $400, for a total of $700. For the two-server system, Ls = 0.8381. The two-server system will double the server cost to $600, but reduce the wait cost to $150 x .8381 = $125.72, for a total of $725.72. The single-server system is cheaper.

(The variety of queuing models, difficult) {AACSB: Analytic Skills}

11.A dental clinic at which only one dentist works is open only two days a week. During those two days, the traffic is uniformly busy with patients arriving at the rate of three per hour. The doctor serves patients at the rate of one every 15 minutes. a. What is the probability that the clinic is empty (except for the dentist)? b. What percentage of the time is the dentist busy?

c. What is the average number of patients in the waiting room? d. What is the average time a patient spends in the office (wait plus service)? e. What is the average time a patient waits for service?

(a) Po = 1 – 3/4 = 0.25; (b) The dentist is busy when the clinic is not empty, or 1 – .25 = 0.75 or 75 percent of the time; (c) Lq = 3*3 / 4*(4-3) = 2.25; (d) Ws = 1 / (4-3) = 1 hour; (e) Wq = 3 / 4*(4-3) = 0.75 hours. (The variety of queuing models, easy) {AACSB: Analytic Skills}

12.A dental clinic at which only one dentist works is open only two days a week. During those two days, the traffic arrivals follow a Poisson distribution with patients arriving at the rate of three per hour. The doctor serves patients at the rate of one every 15 minutes. a. What is the probability that the clinic is empty (except for the dentist)? b. What is the probability that there are one or more patients in the system? c. What is the probability that there are four patients in the system? d. What is the probability that there are four or more patients in the system?

(a) Po = 1 – 3/4 = 0.25; (b) The probability that there are one or more patients is Pn>0 = 3/4 or .75; (c) The probability of exactly four patients is Pn>3 − Pn>4=.3164 – .2373 = .0791; (d) .3164 as previously calculated. (The variety of queuing models, moderate) {AACSB: Analytic Skills}

13.At the order fulfillment center of a major mail-order firm, customer orders, already packaged for shipment, arrive at the sorting machines to be sorted for loading onto the appropriate truck for the parcel’s address. The arrival rate at the sorting machines is at the rate of 100 per hour following a Poisson distribution. The machine sorts at the constant rate of 150 per hour. a. What is the utilization rate of the system?

b. What is the average number of packages waiting to be sorted? c. What is the average number of packages in the sorting system? d. How long must the average package wait until it gets sorted? e. What would Lq and Wq be if the service rate were exponential, not constant? (a) The utilization rate is ρ = 100/150 = 0.67 or 67 percent; (b) Lq = 100*100 / (2*150*50) = .67; (c) Ls = .67 + 100/150 = 1.33; (d) Wq = 100 / (2*150*50) = 0.0067 hours, or 0.4 minutes. (e) Both values would be exactly doubled from the constant service rate results: Lq = 1.33 and Wq = .0133. (The variety of queuing models, moderate) {AACSB: Analytic Skills}

14.At the order fulfillment center of a major mail-order firm, customer orders, already packaged for shipment, arrive at the sorting machines to be sorted for loading onto the appropriate truck for the parcel’s address. The arrival rate at the sorting machines is at the rate of 140 per hour following a Poisson distribution. The machine sorts at the constant rate of 150 per hour.

a. What is the utilization rate of the system?b. What is the average number of packages waiting to be sorted? c. What is the average number of packages in the sorting system? d. How long must the average package wait until it gets sorted? (a) The utilization rate is ρ = 140/150 = 0.9333 or 93.33 percent; (b) Lq = 6.53; (c) Ls = 7.47; (d) Wq = 0.0467 hours, or less than 3 minutes. Parts (b)-(d) are supported by the excerpt from ExcelOM results below.

Results
Average server utilization(r)0.933333
Average number of customers in the queue(Lq)6.533333
Average number of customers in the system(L)7.466667
Average waiting time in the queue(Wq)0.046667
Average time in the system(W)0.053333
Probability (% of time) system is empty (P0)0.066667
(The variety of queuing models, moderate) {AACSB: Analytic Skills}

15.A waiting-line system that meets the assumptions of M/M/1 has λ = 1, μ = 4. Calculate Po. Build a table showing the probability of more than 0, 1, 2, 3, 4, 5, 6,and 7 units in the system. Round to six decimal places in your work Number in systemProbability of n or more

0.25Po = .75
1.0625
2.015625
3.003906
4.000977
5.000244
6.000061
7.000015
(The variety of queuing models, moderate) {AACSB: Analytic Skills}

16.Genco, Inc., a small manufacturer of diesel-generator sets has four shearing machines. Because of the age of these machines, they need minor repairs after 30 hours of use. Analysis of previous breakdowns indicates that breakdowns follow a Poisson distribution. The facility employs one repairman specifically to repair these machines. Average repair time is two hours following an exponential distribution.
a.What is the service factor for this system? b.What is the average number of these machines in service?
c.What is the impact of adding a second repairman?
(a) X = 2/(2+30) = .0625; (b) 4 – .2987 = 3.7123 machines; (c) 4 – .2514 = 3.7486, there is a slight improvement in availability of these machines. The table below summarizes the software results from ExcelOM.

One server
Two servers
Average server utilization (r)0.246753Average server utilization (r)0.124954 Average number of customers in the queue (Lq)0.051957Average number of customers in the queue (Lq)0.001464 Average number of customers in the system (L)0.29871Average number of customers in the system (L)0.251373 Average waiting time in the queue (Wq)0.421129Average waiting time in the queue (Wq)0.011717 Average time in the system (W)2.421129Average time in the system (W)2.011717 Probability (% of time) system is empty (P0)0.753247Probability (% of time) system is empty (P0)0.772099 Effective arrival rate0.123376Effective arrival rate0.124954 (The variety of queuing models, difficult) {AACSB: Analytic Skills}

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