Question

Case 1: “Grocery_Store”

Planning for a major redesign, Beth collected data at her store on several consecutive Saturday mornings. She noticed that customers arrived at the checkout at a rate of approximately 80 per hour. Fully 25 percent of the customers had 10 items or less. Those people took about 2 minutes to serve on average, while customers with more than 10 items took about 4 minutes to process. Beth expects service time to improve when new scanners are installed in the revised design. Help Beth with her design for the system. She would like to achieve a service level of approximately 80%-90% and an average waiting time in any line approximately between 3-7 minutes.

Hint: Note that the service level equals the system utilization coefficient ρ = λ/(Sμ). Therefore, if you assume ρ = 0.8 (or 0.9), you can determine the initial/desired S (the number of channels in an M/M/S system) after you plug in your relevant λ and μ into the formula for ρ.

Answer 1

Average arrival rate, λ = 80 per hour
Weighted average service time, t_e = 25% * 2 + 75% * 4 = 3.5 minutes

So, the average service rate, μ = 60/3.5 = 17.14 per hour

u = λ/(mμ) = 0.80

or, m = 80/(17.14*0.80)

or, m = 5.83 or 6

If we take u = λ/(Sμ) = 0.9 then m = 80/(17.14*0.90) = 5.18 i.e. 5

So, based on the service level requirement, the number of servers should be 5 to 6.

Assuming Poisson arrival, t_a = 60/80 = 0.75 minutes and c_a = 1

t_e = 3.5 minutes

The variance of service time = =0.25*(2-3.5)^2 + 0.75*(4-3.5)^2 = 0.75 ans stdev, s_e = sqrt(0.75) = 0.866

So, c_e = s_e / t_e = 0.866 / 3.5 = 0.25

Use the following formula to compute the average waiting time (CT_q):

For m = 5, u = 80/(5*17.14) = 0.933, CT_q = ((1^2 + 0.25^2)/2)*0.933^(SQRT(2*(5+1))-1)*3.5 / (5*(1-0.933) = 4.7 minutes

For m = 6, u = 80/(6*17.14) = 0.778, CT_q = ((1^2 + 0.25^2)/2)*0.778^(SQRT(2*(6+1))-1)*3.5 / (6*(1-0.778) = 0.7 minutes

So,

The summar table is as follows:

# servers	Service level	Waiting time
5	93.3%	4.7 minutes
6	77.8%	0.7 minutes

So, it seels that it is best to keep 5 servers.