Question 8 of 9: Queueing Theory — Tool-Crib Attendant Staffing

Question 8: Queueing Theory — Tool-Crib Attendant Staffing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. M/M/1 (or M/M/2) queue; mean inter-arrival time 4 min $\Rightarrow\lambda=15$/hr; mean service time 3 min $\Rightarrow\mu=20$/hr per attendant; attendant wage $10/hr; mechanic wage (idle/waiting cost) $15/hr.

Find. Whether the expected hourly cost (attendant wages + mechanics' waiting-time cost) is lower with 1 or 2 attendants.

Approach. Compute the expected number of mechanics in the system, $L$, for the M/M/1 case (1 attendant) and the M/M/2 case (2 attendants), then form total hourly cost $=$ (number of attendants)$\times\$10$ + $L\times\$15$ for each, and compare.

1. Traffic intensity, 1 attendant. $\rho=\lambda/\mu=15/20=0.75<1$ (stable). For M/M/1: $L=\dfrac{\rho}{1-\rho}=\dfrac{0.75}{0.25}=3$ mechanics in the system on average.
2. Cost, 1 attendant. $$\text{Cost}_1 = \$10 + \$15\times L = 10+15(3) = \boxed{\$55/\text{hr}}.$$
3. M/M/2 performance measures. With $a=\lambda/\mu=0.75$, $s=2$, $\rho=a/s=0.375$: $$P_0=\left[\sum_{n=0}^{1}\frac{a^n}{n!}+\frac{a^2}{2!(1-\rho)}\right]^{-1}=\left[1.75+0.45\right]^{-1}=0.4545,$$ $$L_q=\frac{P_0\,a^s\,\rho}{s!(1-\rho)^2}=\frac{0.4545(0.5625)(0.375)}{2(0.625)^2}=0.1227,\qquad L=L_q+a=0.1227+0.75=0.8727.$$
4. Cost, 2 attendants. $$\text{Cost}_2 = 2(\$10)+\$15\times L = 20+15(0.8727)=\boxed{\$33.09/\text{hr}}.$$
5. Compare. $\$33.09/\text{hr} < \$55.00/\text{hr}$: adding a second attendant saves about $\$21.91$/hr — the reduction in mechanics' expected waiting cost (from $L=3$ down to $L\approx0.87$) far outweighs the extra $\$10$/hr wage.

Conclusion: Yes — hiring a second tool-crib attendant is advisable; it lowers total expected hourly cost by roughly $22/hr.