Question

The La Puerta Bank of Stockholm City has one outside drive-up teller. It takes the teller an average of 4 minutes to serve a bank customer. Customers arrive at the drive-up window at a rate of 12 per hour. The bank operations officer is currently analyzing the possibility of adding a second drive-up window, at an annual cost of Php 1,000,000. It is assumed that arriving cars would be equally divided between both windows. The operations officer estimates that each minute's reduction in customer waiting time would increase the bank's revenue by Php 100,000 annually. Should the second drive-up window be installed? Show computations that will prove your claim.

Answer 1

Here, for Single Window

λ = 12; µ = 60/4 = 15

Utilization(p) = λ/µ = 12/15 = 0.8; Average Time Spent in System(W) = 1/ (µ - λ) = 1/ (15-12) = 1/3 = 0.33hours

Average time spend waiting in line = pW = 0.8 x 0.33 = 0.24 hours = 14.4 Mins

For 2 windows

Utilization(p) = λ/sµ = 12/(2 x 15) = 0.4

Probability that no customers are waiting(P₀) = [∑(0.8)⁰/0 + (0.8)¹/1 + (0.8)²/3(1/(1-0.4)]^-1 = [0.8 + 0.21(1 / 0.6)]^-1

= (0.8 + 0.35)^-1 = 0.87

Average number of Customers in Line(L_Q) = (P₀(0.8)² x p) / s!(1 - p)² = (0.87 x 0.6 x 0.4) / 2(1 - 0.4)² = 0.21 / 0.72 = 0.29

Average time spent Waiting in Line(W_Q) = L_Q / λ = 0.29 / 12 = 0.024 hours = 1.44

Hence, the waiting minutes are reduced by 14.4 - 1.4 = 13 mins, So increase in revenue = 13 x 100,000

= Php 13,00,000

Conclusively, if a new window is installed at Php 1,000,000, the bank will still gain a revenue of Php 300,000.

So, the second drive-up window should be installed.