Example 1
Last updated
Last updated
In a single-server queueing system the time of the arrivals and departures of the customers are given below. The simulation clock starts at 0 & ends at 10.5
Arrival times
0.4, 1.2, 2.2, 3.8, 4.0, 9.0, 9.8, 10.2
Departure times
2.4, 3.1, 4.4, 4.9, 8.6, 9.6, 10.0, 10.5
1
Chronological Ordering of Events
2
Show the graph of L(t) vs t. Where L(t) indicates system length at time t.
3
Show the graph of Q(t) vs t. Where Q(t) indicates queue length at time t.
4
Show the graph of B(t) vs t, Where B(t) indicates server busy at time t.
5
Average Interval (Inter Arrival) Time
6
Average Waiting Time of Those Who Wait
7
Average number of Customers in Queue or, Probability of Customers in Queue
8
Average Delay in Queue or, Average Waiting Time
9
Average Service Time
10
Average Time Spend in the System
11
Probability of Idle Server
12
Utilization of the Server
To draw the graphs we need to list all the events in ascending order of the event time in our system. Here Arrival Times and Departure Times of eight customers were given in the problem specification.
Arrival times
0.4, 1.2, 2.2, 3.8, 4.0, 9.0, 9.8, 10.2
Departure times
2.4, 3.1, 4.4, 4.9, 8.6, 9.6, 10.0, 10.5
Here we have two kinds of events, Arrival and Departure. In the following table, we listed all the events in the ascending order of event occurrence time:
SL
Time
Event Name
Length of the System
Length of the Queue
Server Busy Status
1
0.4
Arrival
1
0
1
2
1.2
Arrival
2
1
1
3
2.2
Arrival
3
2
1
4
2.4
Departure
2
1
1
5
3.1
Departure
1
0
1
6
3.8
Arrival
2
1
1
7
4.0
Arrival
3
2
1
8
4.4
Departure
2
1
1
9
4.9
Departure
1
0
1
10
8.6
Departure
0
0
0
11
9.0
Arrival
1
0
1
12
9.6
Departure
0
0
0
13
9.8
Arrival
1
0
1
14
10.0
Departure
0
0
0
15
10.2
Arrival
1
0
1
16
10.5
Departure
0
0
0
Show the graph of L(t) vs t. Where L(t) indicates system length at time t.
Here in the Graph, we denote the x-axis by t as time, and the y-axis by L(t) as the length of the system at time t. (Length of the system indicate the total number of customer in the system at a particular time)
When a new customer arrives in the system we increase the system length by one unit.
And if a customer leaves the system we decrease the system length by one unit.
N.B: If there is no customer at a particular time in the system, the system length will be zero at that time.
To draw the graph we need the event list in ascending order of event occurrence time, see the previous Table for Events List for reference.
For easier representation, we indicate arrival time and departure time in different colors and use an arrow sign towards the x-axis for arrival time and an arrow in opposite direction for departure time. N.B: For a hand-drawn graph one can use different shapes like an oval to enclose departure time and no shape for arrival time and use an arrow as described earlier. For other graphs, we will do the same.
Show the graph of Q(t) vs t. Where Q(t) indicates queue length at time t.
Here in the Graph, we denote the x-axis by t as time and the y-axis by Q(t) as the length of the queue at time t.
For Queue Length Graph when a new customer arrives in the system we need to check whether the server giving service to any customer at that time or the server is free.
If the server is free (which means no customer is being served by the server at that time) then the new customer will get the service directly from the server hence no change in the queue length.
If the server is busy (which means the server is serving a customer at that time) then the new customer will wait in queue hence we will increase the queue length by one unit.
If a customer leaves the system then queue length will decrease by one unit as one customer from the queue will get service from the server.
N.B: If there is only one customer in the system then the queue length will be zero at that time as the only customer will be receiving service from the server at that time. So there is no customer remaining in the waiting queue hence the queue length is zero at that time.
Show the graph of B(t) vs t, Where B(t) indicates server busy at time t.
Here in the Graph, we denote the x-axis by t as time, and the y-axis by B(t) indicates the server is busy at time t.
We can easily draw this graph if we already have the graph for system length. If the system length is one or more then the busy graph value is one unit. If the system length is zero then the busy graph is also zero.
If we don't have the system length graph then we can draw the graph by the following instruction:
For each new arrival, there is a customer present in the system at that time so the busy graph will be one unit. For each new arrival, we have to calculate and remember the currently present customer. For each new arrival, the total customer number will increase by one.
For each customer departure, we will decrease the total customer number by one unit. If there is no customer (which means the total customer number is zero) in the system then the graph will be zero at that time.
For easier representation, we can write the total number of customers present currently at the occurrence of each event. See the following graph. When the total number of customers in the system is one or more the server is busy. When no customer is in the system (which means the total number of customers is zero) the server is free.
We can use any one of the graphs to draw the busy graph.
Here the arrival times are given. We have to calculate the Inter Arrival Time. Inter Arrival Time = Arrival Time - Arrival Time of the previous customer For the first customer, no inter-arrival time is possible as Inter Arrival Time means a time difference between two respective arrival times. As there is no customer before the first customer hence no Inter Arrival Time is present for the first customer.
Customer No
Arrival Times
Inter Arrival Time
1
0.4
-
2
1.2
0.8
3
2.2
1.0
4
3.8
1.6
5
4.0
0.2
6
9.0
5.0
7
9.8
0.8
8
10.2
0.4
Total
Customer
Total Inter
Arrival Time
8
9.8
Therefore, Total Interval (Inter Arrival) Time = 9.8 or, Total Time between Arrival = 9.8
Average Interval (Inter Arrival) Time = (Total Interval Time)/(Total Customer - 1) = 9.8/(8-1) = 9.8/7 = 1.4
Average Time between Arrival = (Total Time between Arrival)/(Total Customer - 1) = 9.8/(8-1) = 9.8/7 = 1.4
To calculate the Delay/Waiting Time we need to know the Service Time Begin for each customer. Here, Service Time Begin = MAX(Arrival Time of Customer, Departure Time of the Previous Customer) And, Delay/Waiting Time = ABS(Service Time Begin - Arrival Time)
Customer No
Arrival Times
Departure Time
Service Time Begin
Delay/ Waiting Time
1
0.4
2.4
0.4
0
2
1.2
3.1
2.4
1.2
3
2.2
4.4
3.1
0.9
4
3.8
4.9
4.4
0.6
5
4.0
8.6
4.9
0.9
6
9.0
9.6
9.0
0
7
9.8
10.0
9.8
0
8
10.2
10.5
10.2
0
Total
Customer
Total Delay in Queue
8
3.6
Therefore, Total Delay in Queue = 3.6 or, Total Waiting Time in Queue = 3.6 and, Total Customers Who Wait in the Queue = 4
Using data from Table for Calculating Delay/Waiting Time:
Average Waiting Time of Those Who Wait = (Total Waiting Time) / (Total Customer who wait) = 3.6/4 = 0.9
Using data from Table for Calculating Delay/Waiting Time:
The average number of Customers in the Queue = (Total Customer in Queue)/(Total Customer in System) = 4/8 = 0.5
Probability of Customer in Queue = (Total Customer Wait in Queue)/(Total Customer in System) = 4/8 = 0.5
Using data from Table for Calculating Delay/Waiting Time:
Average Waiting Time = (Total Waiting Time)/(Total Customer) = 3.6/8 = 0.45
Average Delay in Queue = (Total Delay)/(Total Customer) = 3.6/8 = 0.45
To calculate Service Time we need to know the Service Time Begins of each customer. We calculate Service Time Begin here as in the earlier table. Here, Service Time = ABS(Departure Time - Service Time Begin)
Customer No
Arrival Times
Departure Time
Service Time Begin
Service Time
1
0.4
2.4
0.4
2.0
2
1.2
3.1
2.4
0.7
3
2.2
4.4
3.1
1.3
4
3.8
4.9
4.4
0.5
5
4.0
8.6
4.9
3.7
6
9.0
9.6
9.0
0.6
7
9.8
10.0
9.8
0.2
8
10.2
10.5
10.2
0.3
Total
Customer
Total Service Time
8
9.3
Therefore, Total Service Time = 9.3
Average Service Time = (Total Service Time)/(Total Customer) = 9.3/8 = 1.16
Previously we calculate, Average Waiting Time in Queue = 0.45 and, Average Service Time = 1.16 By using those values:
Average Time Spend in System = Average Waiting Time in Queue + Average Service Time = 0.45 + 1.16 = 1.61
We can calculate the Average Time spent in the System with an alternate approach shown below:
Previously we calculate, Total Service Time = 9.3 and, Total Delay in Queue = 3.6 Therefore, Total Spend Time in System by All Customer = Total Service Time + Total Delay in Queue = 9.3 + 3.6 = 12.9
Average Time Spend in System = (Total Spend Time in System by All Customer)/(Total Customer) = 12.9/8 = 1.61
Here, Server Idle Time = ABS (Service Time Begin - Departure Time of the Previous Customer) For First Customer, Server Idle Time = ABS (Service Time Begin - Server Start Time)
Customer No
Arrival Times
Departure Time
Service Time Begin
Server Idle Time
1
0.4
2.4
0.4
0.4
2
1.2
3.1
2.4
0.0
3
2.2
4.4
3.1
0.0
4
3.8
4.9
4.4
0.0
5
4.0
8.6
4.9
0.0
6
9.0
9.6
9.0
0.4
7
9.8
10.0
9.8
0.2
8
10.2
10.5
10.2
0.2
Total
Customer
Total Idle Time
8
1.2
Using data from Table for Server Idle Time Calculation
Here, Total Runtime = Server End Time - Server Start Time = 10.5-0 = 10.5 N.B: Server End Time & Server Start Time was given in the problem specification.
Probability of Idle Server = (Total Idle Time)/(Total Runtime) = 1.2/10.5 = 0.11
Using data from Table for Server Idle Time Calculation
Previously we calculated Server Running Time = 10.5
Utilization of the Server = (Server Run Time - Idle Time)/(Server Run Time) = (10.5-1.2)/10.5 = 9.3/10.5 = 0.89
We can alternately calculate the utilization of the server using the following approach:
Utilization of the Server = 1-Probability of Idle Server = 1-0.11 = 0.89
