Permata Bank: One Teller, Two Tellers or a Faster Teller
Permata Bank has one teller at a suburban branch. Customers arrive as a Poisson process at an average of 18 per hour. Service times are exponential with a mean of 3 minutes. The branch is considering two changes: (i) opening a second teller, each with the same service rate, or (ii) replacing the present teller with a faster one whose mean service time would be 2.5 minutes, at an extra cost of RM20 per hour. A teller costs RM60 per hour. Management values customer waiting time at RM25 per customer per hour waited before service. Customers who arrive and find a long queue do not balk in the current data. The branch operates below saturation. You may treat the system as M/M/1 or M/M/2 as appropriate and ignore switching costs.
1. For the current single teller, calculate utilisation, average number in the queue, average time in the queue and average time in the system.
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Arrival rate lambda = 18 per hour. Service rate mu = 60/3 = 20 per hour.
Utilisation rho = lambda/mu = 18/20 = 0.90. The teller is busy 90% of the time, so the system is stable but highly utilised.
For M/M/1: Lq = rho^2/(1 - rho) = 0.81/0.10 = 8.1 customers. Wq = Lq/lambda = 8.1/18 = 0.45 hours = 27 minutes. W = Wq + 1/mu = 0.45 + 0.05 = 0.50 hours = 30 minutes. L = lambda W = 18 x 0.50 = 9 customers.
The long queue is a direct consequence of rho being close to 1, not of an unusually high arrival rate.
2. Evaluate the two proposals using total hourly cost, where total cost = teller cost + waiting cost. Which is cheaper at the stated waiting cost?
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Two tellers: M/M/2 with lambda = 18, mu = 20, r = lambda/mu = 0.9, rho = r/2 = 0.45. P0 = [1 + r + r^2/(2(1-rho))]^-1 = [1 + 0.9 + 0.81/(2 x 0.55)]^-1 = [1.9 + 0.73636]^-1 = 0.37931. Lq = P0 r^2 rho / [2(1-rho)^2] = 0.37931 x 0.81 x 0.45 / (2 x 0.3025) = 0.2285. Wq = 0.2285/18 = 0.01270 hours = 0.762 minutes. Total cost = 2 x 60 + 25 x 0.2285 = 120 + 5.71 = RM125.71 per hour.
Faster single teller: mu = 60/2.5 = 24 per hour, rho = 18/24 = 0.75. Lq = 0.75^2/(1 - 0.75) = 0.5625/0.25 = 2.25. Total cost = (60 + 20) + 25 x 2.25 = 80 + 56.25 = RM136.25 per hour.
Current single teller: 60 + 25 x 8.1 = RM262.50 per hour.
At the stated waiting cost, two tellers is cheapest at RM125.71/hour. The faster single teller is second at RM136.25/hour. Both are far cheaper than the current system, mainly because the current queue is very long.
3. The RM25 waiting cost is only an estimate. Find the waiting cost per customer-hour at which the faster single teller becomes cheaper than two tellers, and comment on the sensitivity of the choice.
Read the worked answer
Let c be the waiting cost per customer-hour. Waiting cost per hour equals Lq x c.
Two tellers: 120 + 0.2285c. Faster single: 80 + 2.25c.
Set equal: 120 + 0.2285c = 80 + 2.25c 40 = 2.0215c c = 19.79.
If c is below about RM19.79 per customer-hour, the faster single teller is cheaper. If c is above RM19.79, two tellers are cheaper. At the stated RM25, two tellers win, but the margin is only about RM10.54 per hour (125.71 versus 136.25). Since c is an accounting estimate and the M/M assumptions may not hold exactly, the recommendation is moderately sensitive. If waiting is valued lower, or if actual service times are less variable than exponential, the single faster teller may be preferred. If service quality and reputation matter beyond the accounting cost, the much shorter queue with two tellers may also be worth a premium.