Answer:
Lq = 1.680
time taken by customer 20.03 min
Step-by-step explanation:
Given data:
m = 6
a = 3 min
p = 15 minutes
[tex]u = \frac{p}{am} [/tex]
[tex]u = \frac{15}{3\times 6} = \frac{15}{18} = 0.833[/tex]
CVa = 1
[tex]CVp =\frac{5}{15} = 1/3[/tex]
Number of customer waiting in line =[tex]Lq = utilization ^{\sqrt{\frac{2 \times (number \ of servers + 1 ) \times (CVa^2 + CVs^2)}{2\times (1- utilization)}}[/tex]
[tex]Lq = 0.833^{\sqrt{\frac{2\times (6+1)\times (1^2+(1/3)^2}{2\times (1-0.833}}[/tex]
Lq = 1.679158
Lq = 1.680
FROM QUENNING FORMULA WE GET Tq = 5.039
so, time taken by customer = 5.039 + 15 = 20.03 min