Question

Eat & Gas convenience store operates a two-pump gas station. The lane leading to the pumps can house at most 3 cars, excluding those being served. Arriving cars go elsewhere if the lane is full. The distribution of arriving cars is Poisson with mean 20 per hour. The time to fill up and pay for the purchase is exponential with mean 6 minutes. What is the percentage of cars that will seek business elsewhere? What is the percentage of time one pump is in use? What is the percent utilization of the two pumps? What is the probability that an arriving car will not start service immediately but will find an empty space in the lane? What is the capacity of the lane that will ensure that, on the average, no more than 10% of the arriving cars are turned away? What is the capacity of the lane that will ensure that the probability that both pumps are idle is .05 or less?

Answer 1

This is an M/M/s server with -

N = waiting capacity = 3
? = arrival rate = 20 per hour
? = service rate of one server = 1 in 6 min = 10 per hour
So, ? = 2
c = number of servers (pumps) = 2
So, ?/c = 1

(a)

Basically, we have to find Probability that the waiting queue is more than 3.

or, Probability that the system has more than 3+2 = 5 cars.

Use the following formula for P0

P0 = (1+2 + (2²*(5-2+1)/2!))^-1 = 1/11 = 0.0909

P1 = (?¹ / 1!)P0 = (2¹ / 1!)P0 = 2P0 = 0.18
P2 = (?² / (2!*2^2-2))P0 = 2P0 = 0.18
P3 = (?³ / (2!*2^3-2))P0 = 2P0 = 0.18
---
P5 = 2P0 = 0.18

So. % of cars that will seek business elsewhere = 1 - (0.09 + 5*0.18) = 0.01

(b)

% of time only one pump is used = P1 = 18%

(c)

?eff = (1 – PN) ? = (1 – P5) 20 = 16.36. Average number of busy servers, c1 = ?eff/? = 1.63636. Percent utilization of pumps =c1/ c = 81.8%.

(d)

Probability that an arriving car will not start service immediately but will find an empty space in the lane = P2 + P3 + P4 = 54%