Waiting Line Models

Abstract

Waiting line or queuing systems are pervasive. Many of us remember the long lineups in front of stores in the Soviet Union and Vietnam, and we have all experienced lineups in banks and supermarkets, but there are many more instances with waiting lines: think, for instance, about traffic lights, where drivers line up and wait, files that wait for processing in the inbox at a clerk’s workstation, or telephone calls that are put in a queue. Queuing system were first examined by Agner Krarup Erlang (1878–1929). Erlang was a Danish mathematician, who worked for the Copenhagen Telephone Company. One of the questions he faced during this time was to determine the number of telephone circuits that are required to provide an acceptable level of service.

Exercises

Problem 1 (optimization of the number of channels and the service rate)

Customers arrive at a retail outlet a rate of 12 per hour. The total time that customers spend in the store contributes to their dissatisfaction. A wasted customer hour has been estimated to cost $20. A clerk at the checkout counter typically earns $8 and can serve up to 10 customers per hour.

• (a) What is the cost-minimal number of checkout counters?

• (b) Suppose now that an alternative to the system under (a) is to employ two clerks who have been retrained. Their retraining enables them to serve up to 15 customers per hour and they will earn $10 per hour. Is it worth considering this option?

Solution

• (a)
  $$ \begin{gathered} c{ = 2, }\lambda {\text{ = 6 each, }}\rho { = 6/10 = 0}{.6, }{L_s}{ = 1}{\text{.5, so that }}TC{ = 20(2)(1}{.5) + 2(8) = 76} \hfill \\ c{ = 3, }\lambda {\text{ = 4 each, }}\rho { = 4/10 = 0}{.4, }{L_s}{\text{ = 2/3, so that }}TC{ = 20(3)(2/3) + 3(8) = 64} \hfill \\ c{ = 4, }\lambda {\text{ = 5 each, }}\rho { = 3/10 = 0}{.3, }{L_s}{ = }{\text{.4286, so that }}TC{ = 20(4)(}{.4286) + 4(8) } \hfill \\ \quad \quad { = 66}{.29} \hfill \\ \end{gathered} $$

This implies that the optimal solution is to have 3 clerks. At optimum, the system will cost $64 per hour.

• (b) With μ = 15, we obtain ρ = .4 and L _s  = $2/3$. Then cost = 20(2)($2/3$) + 2(10) = $46.67, which is cheaper than the 3-clerk option in (a).

Problem 2 (optimization of the number of channels and sensitivity analysis)

Joe plans to open his own gas station “Joe’s Place.” He has planned to open from 7 a.m. to 11 p.m. He estimates that 15 customers will arrive each hour during the day to fill up their tanks. Doing so takes typically 4 min plus 1 min for paying the bill. Joe now has to decide how many pumps to install. He has read in the industry magazine “Full of Gas” that each hour that a customer waits in line costs $15 in terms of loss of goodwill (i.e., patronizing a different gas station in the future, buying smokes and other emergency items elsewhere, etc.). Also, he has determined that installing a pump costs $100 per day.

• (a) Determine the optimal number of pumps Joe should install.

• (b) Joe has also heard that there may be a possible gasoline shortage―or at least the perception of one―in the near future. Joe read that in the past, this meant that customers do not really change their driving habits, but fill up their tanks twice as often. Would that change his plans?

Solution

• (a) Arrival rate per hour λ = 15, service time 1/μ = 4 + 1 = 5 min, or μ = 12 customers per hour. Thus, we need at least c = 2 pumps for a steady state to exist.

$$ \begin{gathered} c{ = 2: }\lambda \prime{ = 7}{\text{.5 each, }}\rho { = }{.625, }{L_q}{ = 1}{\text{.0417, so that }}TC{(}c{ = 2) = 2(100) + 16(2)15(1}{.0417) = 700, } \hfill \\ c{ = 3: }\lambda \prime{\text{ = 5 each, }}\rho { = }{.4167, }{L_q}{ = }{\text{.2976, so that }}TC{(}c{ = 3) = 3(100) + 16(3)15(}{.2976) = 514, } \hfill \\ c{ = 4: }\lambda \prime{ = 3}{\text{.75 each, }}\rho { = }{.3125, }{L_q}{ = }{\text{.1420, so that }}TC{(}c{ = 4) = 4(100) + 16(4)15(}{.1420) = 536, } \hfill \\ \end{gathered} $$

so that it is optimal to install c = 3 pumps.

• (b) Arrival rate per hour λ = 30, service time 1/μ = 2 + 1 = 3 min (as the fill-up time is now only 2 min, since the customers fill up when the tank is half full), or μ = 20 customers per hour. Again, at least c = 2 pumps are needed.

Under these circumstances, it would be best for Joe to have c = 4 pumps. This represents a 9.34 % cost increase over the case without the perception of a shortage.

Problem 3 (comparing queuing systems with fast and slow service)

Customers arrive at a retail outlet at a rate of 30 customers per hour. The total time that customers spend in the store contributes to their dissatisfaction. A wasted customer hour has been estimated to cost $10. Management now has two options: either employ one fully trained fast clerk who is able to serve up to 50 customers per hour, or two less trained slower clerks, who can handle up to 30 customers per hour each. Each of the two clerks would have his own waiting line (the supermarket system). Each of the slow clerks earns $6 per hour, while the fast clerk is fully aware of his availability, and asks for $16 per hour.

• (a) Should we hire the two slower clerks or the one fast clerk?

• (b) A new applicant for the job offers his services. The company tried him out and it turned out that he is able to handle no less than 75 customers per hour. Based on the result under (a), what is the maximal amount that we would pay him?

Solution

• (a) The arrival rate is λ = 30. The fast clerk offers μ = 50, so that ρ = 30/50 = 0.6 and L _s  = λ/(μ − λ) = 30/20 = 1.5. The hourly costs are then (cost for clerk) + (costs for customers) = 16 + 1.5(10) = $31.

  In case of the two clerks, there are two M/M/1 systems, each with an effective arrival rate of λ′ = 15. With a service rate of μ = 30 each, we obtain ρ = 15/30 = 0.5 each, so that L _s  = 15/(30–15) = 1 each. The hourly costs are then (costs for two clerks) + 2(costs for customers in each system) = 2(6) + 2(1)(10) = 32. As a result, we should hire the fast clerk, even though he charges more than the two slow clerks together and can handle fewer customers than the two slower clerks combined.

• (b) Given a service rate of μ = 75, we obtain ρ = 0.4 and L _s  = $2/3$. With an unknown wage w, this results in costs of w + $2/3$(10) = 6$2/3$ + w. This amount should not exceed the costs of the best-known solution (a single fast clerk with hourly costs of $31), so that the bound on the superfast clerk’s wage is 6$2/3$ + w ≤ 31 or w ≤ $24.33.