Solved Problems1. Single-Server ModelThe new-accounts officer at the Citizens Northern Savings Bank enrolls all new customers in checking accounts. During the 3-week period in August encompassing the beginning of the new school year at State University, the bank opens a lot of new accounts for students. The bank estimates that the arrival rate during this period will be Poisson distributed with an average of 4 customers per hour. The service time is exponentially distributed with an average of 12 minutes per customer to set up a new account. The bank wants to determine the operating characteristics for this system to determine if the current person is sufficient to handle the increased traffic. Solution:Determine operating characteristics for the single-server system:

The average waiting time of 48 minutes and the average time in the system are excessive, and the bank needs to add an extra employee during the busy period. 2. Multiple-Server ModelThe Citizens Northern Bank wants to compute the operating characteristics if an extra employee was added to assist with new-accounts enrollments. Solution:Determine the operating characteristics for the multiple-server system:

The waiting time with the multiple-server model is 2.3 minutes, which is a significant improvement over the previous system; thus the bank should add the second new-accounts officer.

A commercial bank has three tellers counter for its customers. The services at these tellers are exponentially distributed with mean of 5 minutes per customer. The arrival of customers is Poisson distributed with mean arrival rate of 36 per hour. Analyse the system. Solution             Given;   λ = 36 per hour                         μ = 60/5 = 12 per hour                         λ/μ = = 36/12 = 3                         C = 4 tellers. (i)          Po = 1/[∑C–1

n=0 1/n! (λ / μ )n] + [1/C! (λ/μ)C μC / μ/Cλ]                                    = [1+(λ/μ) + 1/2(λ/μ)2 + 1/6(λ/μ)2] + [1/4!(λ/μ)4(μC / μC–λ)]                                      = [1+3 + 1/2 (3)2 + 1/6 (3)2] + 1/24(3)4 (12×4 / 2 × 4 – 26)                                     = 1 / 1 + 3 + 4.5 + 4.5 + 4.5 + 13.5 = 0.0377 (ii)        Average number of customers in the queue,                         Lq = λμ (λ/μ)C / (C–1)!(μC–λ)2 Po

                             = (36)(12)(36/12)4 / (4–1)![(12)4–36]2 × 0.0377                             = 1.53 (iii)       Average number of customers in the system,                         Ls = Lq + λ/μ = 1.530 + 3 = 4.53 (iv)       Average time, which a customer waits in queue,                         Wq = λμ(λ/μ)C / (C–1)!(μC–λ)2 Po

                              = 12(3)4 / (4–1)!(12×4–36)2 =  0.0377                             = 0.0424 hour = 2.54 minutes. (v)        Average time a customer spends in system,                         Ws = Wq + 1/μ                                     = 0.0424 + 1/12 = 0.1257 hour = 7.54 minutes. (vi)       Number of hours the tellers are busy during the 6-day week,                         Utilization factor, pc = λ/μC = 36 / 12 × 24 = 0.75 

Hence, if the bank works for 6 days on 6 hours daily basis, the teller is busy for 75% of time,             i.e., 0.75 × 6 × 6 = 27 hours per week.             (vii)      Expected number of tells idle at any point of time. For this, let us find the probability of no customer (P0), probability of 1 customer (P1), probability of two customers (P2) and probability of three customers (P3):             P0 = 0.0377   (already found earlier) as        Pn = 1/n! (λ/μ)n Po

             P1 = 1/1! (36/12) 0.0377 = 0.1131             P2 = 1/1! (36/12)2 0.0377 = 0.1696             P3 = 1/1! (36/12)3 0.0377 = 0.1696. Now,             When there is no customer, all the four tellers are idle. When there is one customer, one teller is occupied while three are idle. Similarly, for two customers, two tellers are idle and for three customers, one teller is idle.             Thus, expected number of idle tellers                         = P0 (4) + P1 (3) + P2 (2) + P3 (1)                         = 0.0377 × 4 + 0.1131 × 3 + 0.1696 × 2 + 0.1696                         = 0.9989.             Thus, on the average 0.9989 or one teller will remain idle at any point of time.

SOLVED EXAMPLES FOR THE OPERATING CHARACTERISTICS OF A QUEUE Example 1  Arrival of machinists at a tool crib is considered to be distributed as Poisson distribution with an average rate of 7 per hour. The service time at the tool crib is exponentially distributed with mean of 4 minutes. (a)       What is the probability that a machinist arriving at the tool crib will have to wait? (b)       What is the average number of machinists at the tool crib? (c)        The company made a policy decision that it will install a second crib if a machinist has to wait at least five minutes before being served. What should be additional flow of machinist to the tool crib to justify a second tool crib?

  Solution              Give    λ = 7 per hour = 7 / 60 = 0.117 machinist per minute                         μ = 1/4 per min = 0.25 machinist per minute                         r = λ/μ = 0.117/0.25 =  0.467              (a)       Probability of no machinists in the queue, Po = 1 – λ/μ = 0.533                          Hence; probability of at least one machinist in queue = 1 – Po

                                                                                                      = 1 – 0.533 = 0.467  The probability that a machinist has to wait would be the case when there is at least one machinist already present in the queue, which is 0.467.              (b)       Ls = λ / μ–λ = 0.117 / 0.25 – 0.117 = 0.875 amchinists,              (c)        Let the new arrival rate is λ’ when the average waiting time is 5 minutes. Since                          Wq = λ' / μ(μ–λ)                          5 = λ'/(0.25(0.25–λ')              Or,       0.3125 – 1.25λ' = λ'              Or,       λ’ = 0.3125 / 2.25 = 0.1389 per minute                                      = 0.1389 × 60 = 8.33 machinist per hour.  Example 2 

 At a telephone booth, arrivals are assumed to follow Poisson distribution with average time of 10 minutes between two calls. The average length of a telephone call is 4 minutes and it is assumed to be exponentially distributed. Find: (a)       Average number of calls (customers) in the system. (b)       Average number of calls waiting to served. (c)        Average time a call spends in the system. (d)       Average waiting time of a call before being served. (e)       Fraction of time during which booth is empty. (f)        Probability of at least one customer in the booth. (g)       Probability of more than three calls in the system.  Solution              Given;    λ = 1/10 = 0.1                          μ = 1/4 = 0.25                          Traffic intensity, p = λ/μ = 4/10 = 0.4              (a)       Ls = λ/μ–λ = 0.1/0.25–0.1 = 0.667              (b)       Lw = λ2 / μ(μ–λ)2 = (0.1)2/0.25(0.25–0.1) = 0.267              (c)        Ws = 1/μ–λ = 1 / 0.25 –0.1 = 6.67              (d)       Wq = p / μ–λ = 0.4/ 0.25–0.1 = 2.67              (e)       Po = 1 – p = 1 – 0.4 = 0.6              (f)        Pn (n > 0) = 1 – Po = 1 – 0.6 = 0.4             (g)       Pn (n > 3) = 1 – (Po + P1 + P2) = 1 – (Po + Por + Pop2) 

                                    = 1 – Po (1 + p + p2) = 1 – 0.6(1 + 0.4 + 0.16)                                     = 0.064.   Example 3  A repair shop is manned by a single worker. Customers arrive at the rate of 30 per hour. Time required providing service is exponentially distributed with mean of 100 seconds.   Find the mean waiting time of a customer, needing repair facilitity in the queue.  Solution              Given;   λ = 30 per hour                          μ = 60×60/100 = 36 Customer per hour              Mean waiting time of a customer in the queue,                          Wq = λ / μ(μ–λ) = 30 / 36(36–30) = 0.139 hour                                     = 8.33 minutes.