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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?
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1 Answer

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