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Simulation of multiple server queuing systems
Example 3: Bank system with two tellers
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Example 3 Bank system with two tellers:A bank has two tellers. Customers arrive at this teller at random from 1 to 4 minutes apart. Each possible value of interarrival time has the same probability of occurrence. The service times vary from 1 to 6 minutes with the same probabilities for each teller. The problem is to analyze the system by simulating the arrival and service of 20 customers.Solution:
A set of uniformly distributed random numbers is needed to generate the arrivals at the teller and another set is needed to generate service times for each teller. Random numbers have the following properties:1. The set of random numbers is uniformly distributed between 0 and 1.2. Successive random numbers are independent.
Distribution of time between Arrivals Service time Distribution
Time between Arrival
(Minutes)Probability Cumulative
probability Random-Digit Assignment
Service time (Minutes)
Probability Cumulative probability
Random-Digit Assignment
1 1
2 2
3 3
4 4
5 6
Simulation table : Bank system with two tellers
Customer
Time Since Last Arrival (Minutes)
Arrival Time
Teller 1 Teller 2
Time Customer Waits in Queue
Time customer spend in System
Idle Time of Server 1
Idle Time of Server 2Time
Service Begins
Service Time
Time service ends
Time Service Begins
Service Time
Time service ends
1 4 4 0 4 0 2 4 4 4 6 10 0 6 0 3 1 5 5 3 8 0 3 54 4 9 9 1 10 0 1 15 3 12 12 1 13 0 1 2 6 3 15 15 6 21 0 6 2 7 2 17 17 2 19 0 2 78 1 18 19 2 21 1 3 09 1 19 21 5 26 2 7 0
10 2 21 21 4 25 0 4 0
Tot 21 22 12 3 37 4 13
Av. 2.33 2.2 1.2 0.3 3.7
Pr. 0.2 0.15 0.5
• Average time between arrival– (sum of all interarrival times)/(number of arrivals-1)=21/9 =2.33
• Expected time between arrival E(T) =∑ ti*p(ti)= 1× 0.25+ 2× 0.25+ 3× 0.25+4× 0.25= 2.5• Average service time teller1:
– (total service time)/(total number of customer)=22/10=2.2• Average service time teller2
– (total service time)/(total number of customer)=12/10=1.2• Expected service time teller 1 & 2• Average waiting time
– (total waiting time in queue)/(total number of customers) =3/10=0.3• Probability of waiting
• (number of customers who wait)/(total number of customers) =2/10=0.2• Probability of idle server teller1
• (total idle server time)/(total run time of simulation) =4/26=0.15•The probability of the server being busy teller1 = 1- 0.15=0.85• Probability of idle server teller2
• (total idle server time)/(total run time of simulation) =13/25=0.52•The probability of the server being busy teller2 = 1- 0.52=0.48
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System statistics