Statistics

Question

A grocery store has four check-out counters. The average service rate for each check-out counter is 22 customers per hour. The average arrival rate is 82 customers per hour. Assuming it is a multiple-server waiting line model; determine the average number of customers waiting for a check-out counter and the average time a customer must wait for a check-out counter. What is the probability that there will be more than 4 customers in the system?

Answer 1

Anna,

 

There is a good explanation of the model I'm using below at

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCsQFjAA&url=http%3A%2F%2Fwww.csus.edu%2Findiv%2Fb%2Fblakeh%2Fmgmt%2Fdocuments%2FOPM101SupplC.pdf&ei=nf1qU9DkI8a2yAGszIGoCA&usg=AFQjCNFwTyVbOAy9I0d6FBBX_Q-YH78FPA&sig2=iSSeFjx-Sm1sHtVL8VAxjw&bvm=bv.66330100,d.aWc

 

The model I'm using starts around page C10, but the discussion before hand explains the theory.

 

First some definitions for your Single-line multi-server model (i.e. there is one entry point into the queue, but multiple servers who take customers from that line)

 

s = # of servers in the system = 4

λ = mean arrival rate of customers = 82 customers/hour

µ = mean service rate per server = 22 customers/hour

p = average utilization of the system = λ/sµ = 82/(4x22) = 0.932

 

First test that sµ > λ because if more customers arrive than are serviced, the line will eventually become infinitely long and the formulas will not work.  Here 4x22 = 88>82, so we're ok

 

We need to calculate Lq, Wq and P that more than 4 customers will be waiting in line.  To do this we need to calculate P(0) and use the formulas below.

 

P(0)= Probability that no customers are in the system =

[[from n=0 to s-1 the Σ [((λ/µ)^n/n!)]] + ((λ/µ)^s/s!)(1/(1-p)]^-1

 

Lq = average # of cust[mers waiting in line = [P(0)((λ/µ)^s)p]/[s!(1-p)^2]

 

Wq= the average time spent waiting in line = Lq/λ

 

Pn = Probability that n customers are in the system at a given time

 

for n≤s, Pn = P(0) • (λ/µ)^n/n!

 

for n>s, Pn = P(0) • (λ/µ)^n/(s!s^(n-s))

 

P(0) = [from n=0 to s-1 the [Σ [((λ/µ)^n/n!)] + ((λ/µ)^s/s!)(1/(1-p)]^-1
Since s=4, we need to this for the terms for n=0,1,2 & 3

n = 0: [((82/22)^0/0!) = (1/1) = 1 +

n = 1: [((82/22)^1/1!) = (82/22)/1 = 3.727+

n = 2: [((82/22)^2/2!) = (82/22)^2/2 = 6.946+

n = 3: [((82/22)^3/3!) = (82/22)^3/6 = 8.630+

= 20.303 + ((82/22)^4/4!)(1/(1-0.932)] = + (8.042)(14.706) = 20.303 + 118.265 = 138.568 = 1/P(0)

P(0)= 0.00722

 

Lq = [(0.00722)((82/22)^4)(.932)]/[4!(1-.932)^2] = 11.703 customers

 

Wq = 11.703 / 82 = 0.143 hours = 8.563 minutes

 

P(n) = Probability that n customers are in line.  I'm going to calculate the first 24 (about 2x the mean number in line), but they go on quite a while.

 

P(0) = 0.00722 [Already calculated]

P(1) = P(0) • (λ/µ)^n/n! = (0.00722)(84/22)^1/1! = 0.0276

P(2) = P(0) • (λ/µ)^n/n! = (0.00722)(84/22)^2/2! = 0.0526

P(3) = P(0) • (λ/µ)^n/n! = (0.00722)(84/22)^3/3! = 0.0670

P(4) = P(0) • (λ/µ)^n/n! = (0.00722)(84/22)^4/4! = 0.0639

p(5) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^5/(4!(4)^(5-4)) = 0.0610

p(6) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^6/(4!(4)^(6-4)) = 0.0583

P(7) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^7/(4!(4)^(7-4)) = 0.0556
P(8) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^8/(4!(4)^(8-4)) = 0.0531
P(9) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^9/(4!(4)^(9-4)) = 0.0507
P(10) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^10/(4!(4)^(10-4)) = 0.0484
P(11) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^11/(4!(4)^(11-4)) = 0.0462
P(12) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^12/(4!(4)^(12-4)) = 0.0441
P(13) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^13/(4!(4)^(13-4)) = 0.0421
P(14) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^14/(4!(4)^(14-4)) = 0.0402
P(15) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^15/(4!(4)^(15-4)) = 0.0383
P(16) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^16/(4!(4)^(16-4)) = 0.0366
P(17) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^17/(4!(4)^(17-4)) = 0.0349
P(18) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^18/(4!(4)^(18-4)) = 0.0333
P(19) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^19/(4!(4)^(19-4)) = 0.0318
P(20) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^20/(4!(4)^(20-4)) = 0.0304
P(21) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^21/(4!(4)^(21-4)) = 0.0290
P(22) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^22/(4!(4)^(22-4)) = 0.0277
P(23) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^23/(4!(4)^(23-4)) = 0.0264
P(24) = P(0) • (λ/µ)^n/(s!s^(n-s)) = (0.00722)(84/22)^24/(4!(4)^(24-4)) = 0.0252

So we could continue the table, eventually, the probabilities will get very very small.

 

We want "What is the probability that there will be more than 4 customers in the system?"

Note that because the probabilities will never end (i.e. there is a P(100) in line), we need to use the inverse probability, i.e. the probability that there will be more than 4 customers in the system is 1 - the probability that there will be 4 or fewer customers in the system. 

 

P(0) = 0.0072
P(1) = 0.0276
P(2) = 0.0526
P(3) = 0.0670
P(4) = 0.0639

P(x≤4) = P(0) + P(1) + P(2) + P(3) + P(4) = 0.2183

P(x>4) = 1 - P(x≤4) = 1 - 0.2183 = 0.7817

 

I hope this helps.  John