There are 3 servers in the checkout area. The interarrival time of customers is 2 minutes. The processing time is 5 minutes. The
1 answer:
Answer:
As the one more server is added the Average number of customers in service reduces from ( 5.585 to 3.019 ) which is a reduction of 2.566
Explanation:
interval time of customers ( a ) = 2 minutes
processing time ( p ) = 5 minutes
coefficient of variation for arrival process ( Cva ) = 1
coefficient of variation for service process ( Cvp ) = 0.85
<u>Determine the impact on the average number of customers in service if one more server is added </u>
<em>For three (3) servers ( m ) </em> we will have
flow rate = 1 / a = 1/2 = 0.5
x = ( 2 (m+1) -1 ) ^0.5
where m = 3 ( number of servers )
hence X = 2.645
now lets determine
i) utilization = p / (m*a). where p = 5 , m = 3 , a = 2
hence utilization ( u ) = 0.833 , 1 - u = 1 - 0.833 = 0.167
ii) Time spent in queue
time spent in queue = 6.17 minutes
number of customers been served = mu = 3 * 0.833 = 2.500
number of customers waiting to be served = waiting time / arrival time
= 6.17 / 2 = 3.085
iii) Average customer in queue = number of customers been serviced + number of customers waiting to be served = 2.5 + 3.085 = 5.585
For 4 servers ( m = 4 )
x = ( 2 (m+1) -1 ) ^0.5
= ( 2 ( 5 ) - 1 ) ^ 0.5 = 3
utilization = p / (m*a). where p = 5 , m = 4 , a = 2
hence utilization = 0.625 ,
Time spent in queue = 1.039 minutes ( using equation 2 )
number of customers been serviced = m* u = 2.5
number of customers waiting to be served = waiting time / arrival time
= 1.039 / 2 = 0.520
Average customer in queue = number of customers been serviced + number of customers waiting to be served = 2.5 + 0.520 = 3.019
hence as the one more server is added the Average number of customers in service reduces from ( 5.585 to 3.019 ) which is a reduction of 2.566
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