解析
主要是第6部分 。
test*: mode .txt, para .txt interarrival_.txt and service_.txt 輸出為兩個檔案。
mode *.txt ： random or trace
個人感覺随機的模式更麻煩，是以先嘗試trace
para *.txt
第1個：server數
第2個：是否是優先隊列的門檻值
第3個：随機模式裡面的time_end
interarrivial_*.txt
每一行都表示一個任務。
到達時間是：[1, 3, 5, 6, 7, 8, 9]
則檔案中是：[1, 2, 2, 1, 1, 1, 1]
如果是random:
λ, a2ℓ and a2u
p1, …, p5
service *.txt
每一行都表示一個任務。是以service .txt和interarrivial_.txt這兩個檔案的行數應該一樣。
第一個任務有一個子任務，執行時間是1.800。
numpy.loadtxt()
numpy.isnan()
For random mode, the file service *.txt contains one line, corresponding to the values of µ
and α.
cdf中的兩個參數，第一個是u，第2個a
這裡的cdf，如何生成随機數，需要看課件。
Generate a number u which is uniformly distributed in (0,1)
• Compute the number F-1(u)
輸出
輸出是分别計算每個隊列的響應時間和離開的時間。
在output檔案夾下面，每個測試對應一個響應時間，就是所有的任務的響應時間，而不是子任務的響應時間。
ref
中的檔案是參考檔案
對于departure time， 每一個子任務都有到達時間，和離開的時間。
按照離開時間排序。
random模式
測試
先運作代碼，再測試。有match比對的字樣，說明是正确的。有do not match，說明做錯了。
前4 個是trace mode,後3個是random模式。
老師的trace比較好做，而random模式測試的誤差範圍又很大，是以還是很容易的，關鍵是要早做。
windows:
python main.py 1
python cf_output_with_ref.py 1
python main.py 2
python cf_output_with_ref.py 2
Test 2: Mean response time matches the reference
Test 2: Departure times match the reference
python main.py 3
python cf_output_with_ref.py 3
python main.py 1
python cf_output_with_ref.py 1
linux :
./run_test.sh 1
python3 cf_output_with_ref.py 2
關于測試随機模式
在cse機器上運作
原文
- To use discrete event simulation to simulate a computer system
- To use simulation to solve a design problem
-
To use statistically sound methods to analyse simulation outputs
2
Support provided and computing resources
If you have problems doing this assignment, you can post your question on the course forum. We
strongly encourage you to do this as asking questions and trying to answer them is a
great way to learn. Do not be afraid that your question may appear to be silly, the
other students may very well have the same question!
If you need computing resources to run your simulation program, you can do it on the VLAB
remote computing facility provided by the School. Information on VLAB is available here: https:
//taggi.cse.unsw.edu.au/Vlab/
3
Server farm configuration for this project
Let us start with a short background. A server farm [1] consists of hundreds (or even thousands)
of servers. These server farms are used to process information in data centres [2]. They can also
form part of public cloud infrastructure where users can purchase computing time from cloud
service providers such as Amazon, Google and Microsoft [2]. When you purchase computing time,
1
Server 1
Server n
Arriving
jobs
Completed
sub-jobs
Dispatcher
Completed
sub-jobs
•
•
•
High priority queue ↓
Low priority queue ↑
Joiner
Completed
jobs
Memory
Figure 1: The server farm for this project.
Time
t1
t2
t3
Number of
servers
requested
Job 3
Job 1
Job 2
Sub-job (1,1)
Sub-job (1,2)
Sub-job (2,1)
Sub-job (2,2)
Sub-job (2,3)
Sub-job (3,1)
Status of
the 4
servers
and
queues
High pri. queue
Low pri. queue
(1,1)
(1,2)
(1,1)
(1,2)
(1,1)
(1,2)
(2,1)
(2,2)
(2,1)
(2,2)
(2,3)
(2,3)
(1,1)
(3,1)
(2,1)
(2,2)
(2,3)
t4
Sub-job (1,2)
completes at
t4
Figure 2: An illustration of the work load and the operation of the server farm. Each job consists
of one or more sub-jobs where each sub-job is to be processed by a server.
2
you may have to specify the type of servers that you want, how many of them and for how long.
The configuration of the server farm that you will use in this project is shown in Fig. 1. The
farm consists of a dispatcher, n servers (where n > 1) and a joiner. The dispatcher has two queues:
a high priority queue and a low priority queue. You can assume that both queues have infinite
queueing slots. There are no queues at the server nor joiner.
We will use the word job to refer to a request arriving at the server form. Each job will ask to
use the service of one or more servers. As an illustration, in Fig. 2, Job 1 asks for 2 servers, Job
2 asks for 3 servers and Job 3 asks for 1 server. If a job asks for k servers for service, then we say
that the job consists of k sub-jobs where each sub-job is to be processed by a server. For example,
in Fig. 2, Job 1 asks for 2 servers, so it consists of 2 sub-jobs which are referred to as using the
tuples (1,1) and (1,2). You can interpret the first coordinate of the tuple as the index to a job
and the second coordinate of the tuple as the index to a sub-job. You can see other examples in
Fig. 2.
We will now use Fig. 2 to illustrate the operation of the server farm assuming there are 4 (=
n) servers. We assume that the server farm is empty at time 0, i.e. all servers are idle and the
queues are empty. Job 1 arrives at time t1 and asks for 2 servers. Since there are sufficient number
of idle servers available, all the sub-jobs will head to the servers to receive their service. Fig. 2
shows the status of the servers and queues just after time t1. We assume that it takes negligible
time for the dispatcher to do its processing and to communicate with the server. This means we
can assume that Sub-jobs (1,1) and (1,2) arrive at the servers at time t1.
In Fig. 2, Job 2 arrives at time t2 and asks for 3 servers. However, only two servers are idle.
So, Sub-jobs (2,1) and (2,2) will head to the two idle servers, while sub-job (2,3) will join the
queue. This illustrates that the sub-jobs are ordered and they are processed in an order according
to their indices.
We will now explain the operation of the queues. The server farm uses a threshold h to decide
whether a sub-job will join the high or low priority queue. If a sub-job comes from a job that asks
for h or less servers, then that sub-job will join the high priority queue; otherwise, it will join the
low priority queue. For the example in Fig. 2, we assume h = 2. Since Sub-job (2,3) comes from
Job 2 and Job 2 asks for 3 servers, so the number of servers that Job 2 asks for is higher than the
threshold h = 2 and this means Sub-job (2,3) will head to the low priority queue. Fig. 2 shows
the status of the servers and queues just after time t2, where you can see that Sub-jobs (2,1) and
(2,2) are in the servers but Sub-job (2,3) is in the low priority queue.
Before moving on, we need to further explain how the queues operate at the dispatcher. The
queueing discipline in the dispatcher is non-preemptive and we will explain that in Week 7. You
can also read about it on p. 500 of [1]. Since we hope that you can get your project started early,
we will explain here the rules for the queues at the dispatcher:
-
If a sub-job arrives when all the servers are busy, you will need to determine whether the
sub-job is of a high or low priority. After that, a high priority sub-job will go to the back of
the high priority queue, and similarly a low priority sub-job will go to the back of the low
priority queue.
-
When a server finishes the processing of a sub-job, it will ask the dispatcher for the next
sub-job. One of the following three possibilities will happen:
(a) If the high priority queue is non-empty, the dispatcher will send the first sub-job in
the high priority queue to the available server. (Note that if the high priority queue is
non-empty, you do not need to check the status of the low priority queue.)
3
(b) If the high priority queue is empty and the low priority queue is non-empty, the dis-
patcher will send the first sub-job in the low priority queue to the available server.
© If both queues are empty, the server will become idle.
We now continue on the example in Fig. 2 where Job 3 arrives at time t3 and asks for a server.
Since all the servers are busy at the time that Job 3 arrives and Job 3 has a high priority, the
only sub-job of Job 3 will join the high priority queue. Fig. 2 shows the status of the server and
queues just after the arrival of Job 3.
Next, we assume that Sub-job (1,2) is completed at time t4, see Fig. 2. A completed sub-job
will depart the server and head to the joiner. The available server will contact the dispatcher for
a possible next sub-job. Since the high priority queue is non-empty, the sub-job at the head of
the queue — which is Sub-job (3,1) — will head to the available server. Fig. 2 shows the status
of the server and queues just after Sub-job (3,1) has gone to the server.
We have now explained how the dispatcher and the servers operate, we will now explain the
operation of the joiner. The joiner has memory to store the results of the sub-jobs. Once all the
results of all the sub-jobs of a job have arrived at the joiner, the jointer will combine all the results
and send them to the user who submitted the job. We say that the departure time of a job from
the server farm is the time at which the joiner sends all the results of a job to an user. We assume
that it takes negligible time for a server to send results to the joiner, it takes negligible time for
a joiner to combine all the results from the sub-jobs, and there is sufficient memory at the joiner
to store the results from the sub-jobs. Let us consider the example in Fig. 2. Since the server
completes the processing of Sub-job (1,2) at time t4, the result of Sub-job (1,2) will be at the
joiner at time t4. Let us assume that Sub-job (1,1) will be done at the server at time t5 (and we
assume t5 > t4), so this is also the time that the result of Sub-job (1,1) arrives at the joiner. This
means that, at time t5, all the sub-jobs of Job 1 have been processed. The joiner will combine
these results and send them to the user who submitted Job 1. Since it takes negligible time for
the joiner to combine the results, we therefore consider that the server farm completes Job 1 at
time t5. In general, consider a job that has k sub-jobs and the times at which these k sub-jobs
are completed at the servers are c1, c2, …, and ck. Let cmax be the maximum of c1, c2, …, and
ck. This means that all the sub-jobs of this job will have been completed at the time cmax. This
means that cmax is the time that the joiner sends the results of this job to the user and the time
cmax is the departure time of this job from the server farm. Consequently, the response time of a
job is its departure time minus its arrival time.
We want to make a few remarks concerning this server farm. First, there are no rejections or
losses in this server farm since we assume the dispatcher has infinite number of queueing slots and
the joiner has sufficient memory. Second, the response time of the server farms depends only on
the queues and the servers. This is because it takes negligible processing time at the dispatcher
and at the joiner.
We have now completed our description of the operation of the server farm. We will provide
a number of numerical examples to further explain their operation in Section 4. Note that in the
illustration in this section, we have assumed particular values for the number of servers n and the
threshold h for deciding whether a sub-job is considered high or low priority. However, note that
n and h are parameters, and we will vary them in the simulation.
You will see that for a given number of servers n, we can use the value of threshold h to
influence the mean response time of the server farm. So, a design problem that you will consider
in this project is to determine the value of the threshold h to minimise the mean response time
of the server farm. You can read in [1] how priority queueing can be used to reduce the mean
response time of computer systems.
4
4
Examples
We will now present two examples to illustrate the operation of the server farm that you will
simulate in this project. In all these examples, we assume that the system is initially empty.
4.1
Example 1: number of servers n = 4 and threshold h = 1
In this example, we assume the there are n = 4 servers in the server farm and the threshold h for
determining whether a sub-job is of low or high priority is 1.
In this example, each job may ask for the service of either 1 server or 2 servers. Table 1 shows,
for each job, its arrival time and the service times of its sub-jobs. If there is only one service time,
then it means the job is only asking for one server. If there are two service times, then the job is
asking for two servers.
Job index
Arrival time
Service times of the sub-jobs
1
1
1.8
2
3
8.0, 6.1
3
5
3.9
4
6
2.1, 3.1
5
7
1.9
6
8
5.0, 4.1
7
9
3.8
Table 1: Data for Example 1.
In Table 1, Job 1 has one sub-job, which can be represented as the tuple (1,1), and this sub-job
requires a service time of 1.8. Job 2 has two sub-jobs and its two sub-jobs are represented by the
tuples (2,1) and (2,2). The service times required by Sub-jobs (2,1) and (2,2) are, respectively, 8.0
and 6.1.
The events in the server farm are the arrival of a job to the dispatcher and the departure of a
completed sub-job from a server. We will illustrate how the simulation of the server farm works
using “on-paper simulation”. The quantities that you need to keep track of are:
•
Next arrival time is the time of the next job arrival
•
For each server, we keep track of the following information:
– Server status, which can be either busy or idle.
– Next departure time is the time at which the sub-job that is being processed will
depart from the server. If the server is idle, the next departure time is ∞. Note that
there is a next departure time for each server.
– For a sub-job in the server, the time that this sub-job arrives at the server farm
– The tuple to identify the sub-job in the server
For example, if the server status is “Busy, 7.5, 3.4, (1,2)”, this means the server is busy
serving the sub-job identified by the tuple (1,2), and this sub-job is scheduled to depart at
time 7.5 and has arrived at the server farm at time 3.4. If the server is idle, we will write
the status as “Idle, ∞” where the departure time is ∞.
•
The contents of the high and low priority queues. Each sub-job in the queue is identified by
3 fields: the tuple identifying the sub-job, the sub-job’s arrival time to the server farm, the
sub-job’s service time. For example, we write a sub-job in a queue as
5
[ (4,2), 6, 3.1 ]
which means Sub-job (4,2) arrives at the server farm at time 6 and requires a service time
of 3.1.
The “on-paper simulation” is shown in Table 2. The notes in the last column explain what
updates you need to do for each event.
6
Master
clock
Event
type
Next
arrival
time
Server
1
Server
2
Server
3
Server
4
High
priority
queue
Low
priority
queue
Notes
–
1
Idle,
∞
Idle,
∞
Idle,
∞
Idle,
∞
–
–
We assume the servers are idle and queues are
empty at the start of the simulation. The next
departure times for all servers are ∞. The “–”
indicates that the queues are empty.
1
Arrival
3
Busy,
2.8,
1,
(1,1)
Idle,
∞
Idle,
∞
Idle,
∞
–
–
The event is the arrival of Job 1 with one sub-
job (1,1). Since all the servers are idle before this
arrival, the sub-job can be sent to any one of the
idle servers. We have chosen to send this sub-job
to Server 1. Sub-job (1,1) requires a service time
of 1.8 and it starts to receive service at time 1,
so its departure time is 2.8. Lastly, we need to
update the next arrival time, which is 3.
2.8
Departure
Server 1
3
Idle,
∞
Idle,
∞
Idle,
∞
Idle,
∞
–
–
The event is a departure from Server 1.
Sub-
job (1,1) has now been completed with an arrival
time of 1 and a departure time of 2.8. Since both
queues are empty, Server 1 becomes idle.
The
server status has been updated accordingly.
3
Arrival
5
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Idle,
∞
Idle,
∞
–
–
The event is the arrival of Job 2 which consists
of two sub-jobs (2,1) and (2,2). Since there are
4 idle servers before the arrival of these two sub-
jobs, they can be sent to any two idle servers. We
have chosen to send the two sub-jobs to Servers
1 and 2.
Sub-job (2,1) requires a service time
of 8.0 and it starts to receive service at time 3,
so its departure time is 11.0. Similarly, Sub-job
(2,2) will depart at time is 9.1. Lastly, we need
to update the next arrival time, which is 5.
7
5
Arrival
6
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Busy,
8.9,
5,
(3,1)
Idle,
∞
–
–
The event is the arrival of Job 3 which consists of
one sub-job (3,1). Since there are 2 idle servers
before the arrival of this sub-job, it can be sent to
any of the idle servers. We have chosen to send
this sub-job to Server 3. Sub-job (3,1) requires a
service time of 3.9 and it starts to receive service
at time 5, so its departure time is 8.9. Lastly, we
need to update the next arrival time, which is 6.
6
Arrival
7
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Busy,
8.9,
5,
(3,1)
Busy,
8.1,
6,
(4,1)
–
[ (4,2), 6,
3.1 ]
The event is the arrival of Job 4 which consists
of two sub-jobs (4,1) and (4,2).
Since there is
only one idle server before the arrival of these
two sub-jobs, the first sub-job (i.e., Sub-job (4,1))
will go to the idle server (i.e., Server 4).
The
status of Server 4 has been updated accordingly.
The second sub-job (i.e., Sub-job (4,2)) will join
the queue. Since Job 4 has two sub-jobs and the
threshold h = 1, therefore Sub-job (4,2) will enter
the low priority queue. We enclose the details of
each sub-job in the queue within a pair of square
brackets. In this case, the sub-job details are [
(4,2), 6, 3.1 ] which refers to Sub-job (4,1) with
an arrival time of 6 and a service time of 3.1.
Lastly, we need to update the next arrival time,
which is 7.
7
Arrival
8
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Busy,
8.9,
5,
(3,1)
Busy,
8.1,
6,
(4,1)
[ (5,1), 7,
1.9 ]
[ (4,2), 6,
3.1 ]
The event is the arrival of Job 5 which consists
of one sub-job (5,1). Since all servers are busy,
Sub-job (5,1) will be sent to the queue.
Since
Job 5 has one sub-job and the threshold h = 1,
therefore Sub-job (5,1) will enter the high priority
queue.
The sub-job details [ (5,1), 7, 1.9 ] are
added to the high priority queue. Lastly, we need
to update the next arrival time, which is 8.
8
8
Arrival
9
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Busy,
8.9,
5,
(3,1)
Busy,
8.1,
6,
(4,1)
[ (5,1), 7,
1.9 ]
[ (4,2), 6,
3.1 ]
[ (6,1), 8,
5.0 ]
[ (6,2), 8,
4.1 ]
The event is the arrival of Job 6 which consists of
two sub-jobs (6,1) and (6,2). Since all servers are
busy, both sub-jobs will be sent to the low priority
queue. The sub-job details [ (6,1), 8, 5.0 ] and [
(6,2), 8, 4.1 ] have been added to the low-priority
queue. Lastly, we need to update the next arrival
time, which is 9.
8.1
Departure
Server 4
9
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Busy,
8.9,
5,
(3,1)
Busy,
10.0,
7,
(5,1)
–
[ (4,2), 6,
3.1 ]
[ (6,1), 8,
5.0 ]
[ (6,2), 8,
4.1 ]
The event is a departure from Server 4.
Sub-
job (4,1) has now been completed with an arrival
time of 6 and a departure time of 8.1. Since the
high priority queue is non-empty, the first sub-job
in the queue will advance to the available server.
The sub-job heading to the queue has a service
time of 1.9 and the current time (= master clock)
is at 8.1, so the departure time of the sub-job will
be 8.1 + 1.9 = 10.0. The server status has been
updated accordingly.
8.9
Departure
Server 3
9
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Busy,
12.0,
6,
(4,2)
Busy,
10.0,
7,
(5,1)
–
[ (6,1), 8,
5.0 ]
[ (6,2), 8,
4.1 ]
The event is a departure from Server 3. Sub-job
(3,1) has now been completed with an arrival time
of 5 and a departure time of 8.9. Since the high
priority queue is empty and the low priority queue
is non-empty, the first sub-job in the low priority
queue will advance to the available server. The
server status and queue status have been updated
accordingly.
9
Arrival
∞
Busy,
11.0,
3,
(2,1)
Busy,
9.1,
3,
(2,2)
Busy,
12.0,
6,
(4,2)
Busy,
10.0,
7,
(5,1)
[ (7,1), 9,
3.8 ]
[ (6,1), 8,
5.0 ]
[ (6,2), 8,
4.1 ]
The event is the arrival of Job 7 which consists of
one sub-job (7,1). Since all servers are busy, Sub-
job (7,1) will be sent to the high-priority queue
and the queue status has been updated. Lastly,
we need to update the next arrival time to ∞
because there are no more arrivals.
9
9.1
Departure
Server 2
∞
Busy,
11.0,
3,
(2,1)
Busy,
12.9,
9,
(7,1)
Busy,
12.0,
6,
(4,2)
Busy,
10.0,
7,
(5,1)
–
[ (6,1), 8,
5.0 ]
[ (6,2), 8,
4.1 ]
The event is a departure from Server 2. Sub-job
(2,2) has now been completed with an arrival time
of 3 and a departure time of 9.1. Since the high
priority queue is non-empty, the first sub-job in
the high priority queue will advance to the avail-
able server. The server status and queue status
have been updated accordingly.
10
Departure
Server 4
∞
Busy,
11.0,
8
3,
(2,1)
Busy,
12.9,
9,
(7,1)
Busy,
12.0,
6,
(4,2)
Busy,
15,
8,
(6,1)
–
[ (6,2), 8,
4.1 ]
The event is a departure from Server 4. Sub-job
(5,1) has now been completed with an arrival time
of 7 and a departure time of 10.0. Since the high
priority queue is empty and the low priority queue
is non-empty. the first sub-job in the low priority
queue will advance to the available server. The
server status and queue status have been updated
accordingly.
11
Departure
Server 1
∞
Busy,
15.1,
8,
(6,2)
Busy,
12.9,
9,
(7,1)
Busy,
12.0,
6,
(4,2)
Busy,
15,
8,
(6,1)
–
–
The event is a departure from Server 1. Sub-job
(2,1) has now been completed with an arrival time
of 3 and a departure time of 11.0. Since the high
priority queue is empty and the low priority queue
is non-empty. the first sub-job in the low priority
queue will advance to the available server. The
server status and queue status have been updated
accordingly.
12
Departure
Server 3
∞
Busy,
15.1,
8,
(6,2)
Busy,
12.9,
9,
(7,1)
Idle,
∞
Busy,
15,
8,
(6,1)
–
–
The event is a departure from Server 3.
Sub-
job (4,2) has now been completed with an arrival
time of 6 and a departure time of 12.0.
Since
both queues are empty, Server 3 becomes idle.
The server status has been updated accordingly.
12.9
Departure
Server 2
∞
Busy,
15.1,
8,
(6,2)
Idle,
∞
Idle,
∞
Busy,
15,
8,
(6,1)
–
–
The event is a departure from Server 2.
Sub-
job (7,1) has now been completed with an arrival
time of 9 and a departure time of 12.9.
Since
both queues are empty, Server 2 becomes idle.
The server status has been updated accordingly.
10
15.0
Departure
Server 4
∞
Busy,
15.1,
8,
(6,2)
Idle,
∞
Idle,
∞
Idle,
∞
–
–
The event is a departure from Server 4.
Sub-
job (6,1) has now been completed with an arrival
time of 8 and a departure time of 15.0.
Since
both queues are empty, Server 4 becomes idle.
The server status has been updated accordingly.
15.1
Departure
Server 1
∞
Idle,
∞
Idle,
∞
Idle,
∞
Idle,
∞
–
–
The event is a departure from Server 1.
Sub-
job (6,2) has now been completed with an arrival
time of 8 and a departure time of 15.1.
Since
both queues are empty, Server 1 becomes idle.
The server status has been updated accordingly.
Table 2: Table illustrating the updates in the server farm.
11
The above description has not explained what happens if an arrival and a departure are at the
same time. We will leave it unspecified. If we ask you to simulate in trace driven mode, we will
ensure that such situation will not occur. If the inter-arrival time and service time are generated
randomly, the chance of this situation occurring is practically zero so you do not have to worry
about it.
Table 3 summarises the arrival and departure times of all the sub-jobs. The two sub-jobs of Job
2 complete at times 9.1 and 11.0. Hence, the departure time of Job 2 from the server farm is 11.0,
which is the later time of the two departure times. Since Job 2 arrives at time 3, the response time
of Job 2 is 11–3 = 8. The departure and response times of other jobs can be similarly computed.
The arrival and departure times of all the jobs for Example 1 are given in Table 4. The mean
response time of the 7 jobs in this example is 33.7
7
= 4.8143.
Sub-jobs
Arrival time
Departure time
(1,1)
1.0
2.8
(4,1)
6.0
8.1
(3,1)
5.0
8.9
(2,2)
3.0
9.1
(5,1)
7.0
10.0
(2,1)
3.0
11.0
(4,2)
6.0
12.0
(7,1)
9.0
12.9
(6,1)
8.0
15.0
(6,2)
8.0
15.1
Table 3: The arrival and departure times of the sub-jobs for Example 1.
Job
Arrival time
Departure time
1
1.0
2.8
2
3.0
11.0
3
5.0
8.9
4
6.0
12.0
5
7.0
10.0
6
8.0
15.1
7
9.0
12.9
Table 4: The arrival and departure times of the jobs for Example 1.
Note that in this example, we have chosen to obtain the departure times of the sub-jobs and
perform post-processing to obtain the departure times of the jobs. Alternatively, the computation
of the departure times of the jobs can be incorporated into the simulation exemplified in Table 2.
You can use either of the methods.
12
4.2
Example 2: number of servers n = 4, threshold h = 2
This example is identical to Example 1 except that the threshold h = 2. In other words, for this
example, the server farm uses n = 4 and h = 2, and the arrivals to the server farm are given in
Table 1. Note that the setting h = 2 essentially means all sub-jobs go into the high priority queue
and the low priority queue is not used at all.
Table 5 summarises the arrival and departure times of all the sub-jobs while Table 6 summarises
the arrival and departure times of all the jobs. The mean response time of the 7 jobs in this example
is 35.4
7
= 5.0571, which is higher than that of Example 1. This demonstrates that priority queueing
can be used to influence the mean response time of the server.
Sub-jobs
Arrival time
Departure time
(1,1)
1.0
2.8
(4,1)
6.0
8.1
(3,1)
5.0
8.9
(2,2)
3.0
9.1
(5,1)
7.0
10.8
(2,1)
3.0
11.0
(4,2)
6.0
11.2
(6,1)
8.0
14.1
(7,1)
9.0
14.8
(6,2)
8.0
14.9
Table 5: The arrival and departure times of the sub-jobs for Example 2.
Job
Arrival time
Departure time
1
1.0
2.8
2
3.0
11.0
3
5.0
8.9
4
6.0
11.2
5
7.0
10.8
6
8.0
14.9
7
9.0
14.8
Table 6: The arrival and departure times of the jobs for Example 2.
13
5
Project description
This project consists of two main parts. The first part is to develop a simulation program for
the server farm in Fig. 1. The system has already been described in Section 3 and illustrated in
Section 4. In the second part, you will use the simulation program that you have developed to
solve a design problem.
5.1
Simulation program
You must write your simulation program in one (or a combination) of the following languages:
Python (either version 2 or 3), C, C++, or Java. All these languages are available on the CSE
system.
We will test your program on the CSE system so your submitted program must be able to
run on a CSE computer. Note that it is possible that due to version and/or operating system
differences, code that runs on your own computer may not work on the CSE system. It is your
responsibility to ensure that your code works on the CSE system.
Note that our description uses the following variable names:
-
A variable mode of string type. This variable is to control whether your program will run
simulation using randomly generated arrival times and service times; or in trace driven mode.
The value that the parameter mode can take is either random or trace.
-
A variable time_end which stops the simulation if the master clock exceeds this value. This
variable is only relevant when mode is random. This variable is a positive floating point
number.
Note that your simulation program must be a general program which allows different param-
eter values to be used. When we test your program, we will vary the parameter values. You can
assume that we will only use valid inputs for testing.
For the simulation, you can always assume that the system is empty initially.
5.1.1
The random mode
When your simulation is working in the random mode, it will generate the inter-arrival times
and the workload of a job in the following manner.
-
We use {a1, a2, . . . , ak, . . . , …} to denote the inter-arrival times of the jobs arriving at the
dispatcher. These inter-arrival times have the following properties:
(a) Each ak is the product of two random numbers a1k and a2k, i.e ak = a1ka2k ∀k = 1, 2, …
(b) The sequence a1k is exponentially distributed with a mean arrival rate λ requests/s.
© The sequence a2k is uniformly distributed in the interval [a2l, a2u].
Note: The easiest way to generate the inter-arrival times is to multiply an exponentially
distributed random number with the given rate and a uniformly distributed random number
in the given range. It would be more difficult to use the inverse transform method in this
case, though it is doable.
-
The workload of a job is characterised by the number of sub-jobs and the service times of the
sub-jobs. The first step to generate the workload of a job is to generate a random positive
integer which is the number of sub-jobs for that job. You will be given a sequence of J
non-negative real numbers p1, p2, …, pk, … , pJ with the property �J
k=1 pk = 1. Given
these numbers, we want the probability that a job consists of k sub-jobs to be equal to pk,
14
for k = 1, …, J.
For example, if you are given the sequence 0.5, 0.2, 0.3. This means for the jobs in this
workload, you have
(a) Prob[a job consists of exactly 1 sub-job] = 0.5
(b) Prob[a job consists of exactly 2 sub-jobs] = 0.2
© Prob[a job consists of exactly 3 sub-jobs] = 0.3
Note that you may interpret J is the maximum number of servers that a job can request.
-
If a job consists of k sub-jobs, then you will need to generate k random service times for
the k sub-jobs. These k service times are independent and they all come from the same
probability distribution. The cumulative distribution function (CDF) S(t) of the service
time t for a sub-job is:
S(t)
=
1 − exp(−(µ t)α)
(1)
where the parameters µ and α are positive real numbers.
As an example, if a job consists of 3 sub-jobs, then you will need to generate 3 random
numbers which come from a probability distribution whose CDF is given by S(t).
5.1.2
The trace mode
When your simulation is working in the trace mode, it will read the list of inter-arrival times
and the list of service times of the sub-jobs from two separate ASCII files. We will explain the
format of these files in Sections 6.1.3 and 6.1.4 .
An important requirement for the trace mode is that your program is required to simulate
until all jobs have departed. You can refer to Table 2 for an illustration.
Hint: Do not write two separate programs for the random and trace modes because they share
a lot in common. A few if–else statements at the right places are what you need to have both
modes in one program.
5.2
Determining the threshold h that minimises the mean response time
After writing your simulation program, your next step is to use your simulation program to de-
termine the threshold h that can minimise the mean response time.
For this design problem, you will assume the following parameter values:
•
Number of servers: n = 10
•
For inter-arrival times: λ = 1.8, a2ℓ = 0.7, a2u = 0.9
•
For the number of sub-jobs per job: the sequence p1, p2, p3, p4, p5 is 0.4, 0.25, 0.15, 0.11, 0.09.
•
For the service time per job: µ = 0.9, α = 0.9.
In solving this design problem, you need to ensure that you use statistically sound methods
to compare systems. You will need to consider parameters such as length of simulation, number
of replications, transient removals and so on. You will need to justify in your report on how you
determine the value of the threshold h.
15
6
Testing your simulation program
In order for us to test the correctness of your simulation program, we will run your program using
a number of test cases. The aim of this section is to describe the expected input/output file format
and how the testing will be performed.
Each test is specified by 4 configurations files. We will index the tests from 1. If 12 tests are
used, then the indices for the tests are 1, 2, …, 12. The names of the configuration files are:
•
For Test 1, the configuration files are mode_1.txt, para_1.txt, interarrival_1.txt and
service_1.txt. The files are similarly named for indices 2, 3, …, 9.
•
For Test 10, the configuration files are mode_10.txt, para_10.txt, interarrival_10.txt
and service_10.txt. The files are similarly named if the test index is a 2-digit number.
We will refer to these files using the generic names mode *.txt, para *.txt etc. We will describe
the format of the configuration files in Section 6.1
Each test should produce 2 output files whose format will be described in Section 6.2. We will
explain how testing will be conducted in Sections 6.3 and 6.5.
6.1
Configuration file format
Note that Test 1 is the same as Example 1 discussed in Section 4.1. We will use this test to
illustrate the file format.
6.1.1
mode *.txt
This file is to indicate whether the simulation should run in the random or trace mode. The file
contains one string, which can either be random or trace.
6.1.2
para *.txt
If the simulation mode is trace, then this file has two lines. The first line is the value of n (=
number of servers) and the second line has the value of h (= threshold for priority queueing). If
the test is Example 1 in Section 4.1, then the contents of this file are:
4
1
These values are in the sample file para_1.txt.
If the simulation mode is random, then the file has three lines. The meaning of the first two
lines are the same as above. The last line contains the value of time_end, which is the end time of
the simulation. The contents of the sample file para_5.txt are shown below where the last line
indicates that the simulation should run until 2000.00.
5
2
2000.00
You can assume that we will only give you valid values. You can expect n to be a positive
integer greater than 1. You can expect h to be a non-negative integer. For time_end, it is a
strictly positive integer or floating point number.
16
6.1.3
interarrival *.txt
The contents of the file interarrival *.txt depend on the mode of the test. If mode is trace,
then the file interarrival *.txt contains the interarrival times of the jobs with one interarrival
time occupying one line. You can assume that the list of interarrival times is always positive. For
Example 1 in Section 4.1, the arrival times are [1, 3, 5, 6, 7, 8, 9] which means the inter-arrival times
are [1, 2, 2, 1, 1, 1, 1]. The inter-arrival times will be specified by a file whose contents are:
1.000
2.000
2.000
1.000
1.000
1.000
1.000
If the mode is random, then the file interarrival *.txt contain 2 lines. The first line contains
three values corresponding to the parameters λ, a2ℓ and a2u. The second line contains the the
values for the sequence p1, …, pJ. As an example, the contents of interarrival 5.txt are:
1.400 0.600 0.800
0.400 0.300 0.200 0.050 0.050
For this example, the values of λ, a2ℓ and a2u are respectively 1.400, 0.600 and 0.800. The
values of p1, …, p5 are 0.400, 0.300, 0.200, 0.050, 0.050. This means that you can infer the value
of J by counting the number of values found in the second line of the file.
6.1.4
service *.txt
For trace mode, the file service *.txt contains the service times of the sub-jobs. As an illus-
tration, the service times of the sub-jobs for Example 1 in Section 4.1 will be specified by a file
whose contents are:
1.800 NaN
8.000
6.100
3.900 NaN
2.100
3.100
1.900 NaN
5.000
4.100
3.800 NaN
where you will find the service times of the sub-jobs of each job in a line of the file. Note that
the symbol NaN is a Python floating point number to denote not a number and is often used to
indicate an absence of numbers. In this example, if there are two numbers on the line, the job is
requesting two servers; if there is a number and an NaN, the job is requesting for one server.
In general, if the maximum number of servers that a job can request is J, then every line of
service *.txt will have a total of J numbers and NaN’s. The following is the first three lines of
service 4.txt where J = 5.
1.904 NaN NaN NaN NaN
1.828 0.298 1.303 NaN NaN
2.537 0.016 NaN NaN NaN
You can conveniently load the contents of this file by using the function numpy.loadtxt() into
a numpy array. You may also find the function numpy.isnan() useful.
17
For random mode, the file service.txt contains one line, corresponding to the values of µ
and α.
You can assume that the data we provide for trace mode are consistent in the following way:
The number of inter-arrival times and the number of lines of service times are equal.
6.2
Output file format
In order to test your simulation program, we need two output files per test. One file containing
the mean response time. The other file contains the departure times of the sub-jobs from the
servers.
We want to start by clarifying what we mean by mean response time. When we talk about
mean response time, we are referring to the mean response time of the jobs of the server farm. We
are not referring to the mean response times of the sub-jobs. For trace mode, the mean response
time will be calculated using all the jobs provided in the interarrival .txt and service .txt.
For random mode, the mean response time should be calculated using all the jobs that have been
completed by the end time of simulation as specified by time_end. Note that you do not have to
consider transient removal for the mean response before you write the result to the output file.
However, you should consider transient removal when you do your design.
The mean response time should be written to a file whose filename has the form mrt_.txt.
For Example 1 in Section 4.1, the contents of this file are:
4.8143
The other file dep_.txt contains the departure times of of the sub-jobs from the servers. For
Example 1 in Section 4.1, the contents of dep_.txt are:
1.0000
2.8000
6.0000
8.1000
5.0000
8.9000
3.0000
9.1000
7.0000
10.0000
3.0000
11.0000
6.0000
12.0000
9.0000
12.9000
8.0000
15.0000
8.0000
15.1000
Note the following requirements for the file containing the departure times:
-
Each line contains the arrival time and departure time of a sub-job. The arrival time is
printed first, followed by blank spaces or tab, then the departure time from the server
processing the sub-job.
- The sub-jobs must be ordered according to ascending departure times.
-
If the simulation is in the trace mode, we expect the simulation to finish after all sub-jobs
have been processed. Therefore, the number of lines in dep_*.txt should be equal to the
total number of sub-jobs.
-
If the simulation is in the random mode, the file should contain all the sub-jobs that have
left the server by time_end.
All numbers in mrt_.txt and dep_.txt should be printed as floating point numbers to
exactly 4 decimal places.
18
6.3
The testing framework
When you submit your project, you must include a Linux bash shell script with the name
run_test.sh so that we can run your program on the CSE system. This shell script is required
because you are allowed to use a computer language of your choice.
Let us first recall that each test is specified by a four configuration files and should produce
two output files. For example, test number 1 is specified by the configuration files mode_1.txt,
interarrival_1.txt, service_1.txt and para_1.txt; and test number 1 is expected to produce
the output files mrt_1.txt and dep_1.txt.
We will use the following directory structure when we do testing.
the directory containing run test.sh
config/
output/
We will put all the configuration files for all the tests in the sub-directory config/. You should
write all the output files to the sub-directory output/.
To run test number 1, we use the shell command:
./run_test.sh 1
The expected behaviour is that your simulation program will read in the configuration files for
test number 1 from config/, carry out the simulation and create the output files in output/.
Similarly, to run test number 2, we use the shell command:
./run_test.sh 2
This means that the shell script run_test.sh has one input argument which is the test number
to be used.
Let us for the time being assume that you use Python (Version 3) to write your simulation
program and you call your simulation program main.py. If the file main.py is in the same directory
as run_test.sh, then run_test.sh can be the following one-line shell script:
python3 main.py $1
The shell script will pass the test number (which is in the input argument $1) to your simula-
tion program main.py. This also implies that your simulation program should accept one input
argument which is the test number.
Just in case you are not familiar with shell script, we have provided two sample files: run_test.sh
and main.py to illustrate the interaction between a shell script and a Python (Version 3) file. You
need to make sure run_test.sh is executable. If you run the command ./run_test.sh 2, it
will read the sample service_2.txt in the config/ directory and write a file with the name
dummy_2.txt to the directory output/. You can also try using input arguments 1, 3 or 4 for the
sample shell script. You can use these sample files to help you to develop your code.
If you use C, C++ or Java, then your run_test.sh should first compile the source code and
then run the executable. You should of course pass the test number to the executable as an input.
You can put your code in the same directory that contains run_test.sh or in a subdirectory
below it. For example, you may have a subdirectory src/ for your code like the following:
19
the directory containing run test.sh
config/
output/
src/
6.4
Sample files
You should download the file sample_project_files.zip from the project page on the course
website. The zip archive has the following directory structure:
Base directory containing cf output with ref.py, run test.sh and main.py
config/
output/
ref/
Details on the zip-archive are:
•
The sub-directory config/ contains configuration files that you can use for testing.
– The files mode_1.txt, mode_2.txt, …, mode_6.txt and mode_7.txt. Note that Tests
1–4 are for trace mode while Tests 5–7 are for random mode.
– The files para_.txt, interarrival_.txt and service_.txt for * from 1 to 7, as
the input to the simulation.
– Note that Tests 1 and 2 are the same as respectively, Example 1 and Example 2, in
Section 4.
•
The sub-directory output/ is empty. Your simulation program should place the output files
in this sub-dirrectory.
•
The sub-directory ref/ contains the expected simulation results.
– The files mrt_ref.txt and depref.txt for * from 1 to 7, as the reference files for
the output. For Tests 1–4, you should be able to reproduce the results in mrtref.txt
and dep*_ref.txt. However, since Tests 5–7 are in random mode, you will not be
able to reproduce the results in the output files. They have been provided so that you
can check the expected format of the files.
•
The Python file cf_output_with_ref.py which illustrates how we will compare your output
against the reference output. This file takes in one input argument, which is the test number.
For example, if you want to check your simulation outputs for test 2, you use:
python3 cf_output_with_ref.py 2
Note the following:
– The file cf_output_with_ref.py expects the directory structure shown earlier.
– For trace mode, we will check your mean response time and the departure times. Note
that we are not looking for an exact match but rather whether your results are within
a valid tolerance. The tolerance for the trace mode is 10−3 which is fairly generous
for numbers with 4 decimal places.
– For random mode, we will only check the mean response time. You can see from the
sample file that we check whether the mean response time is within an interval. We
obtain this interval using the following method: (i) we first simulate the system many
times; (ii) we then use the simulation results to estimate the maximum and minimum
mean response times; (iii) we use the estimated maximum and minimum values to form
an interval; (iv) in order to provide some tolerance due to randomness, we enlarge this
interval further.
20
– Note that we use a very generous tolerance so if your mean response time does not pass
the test, then it is highly likely that your simulation program is not correct.
•
The files run_test.sh and main.py as mentioned in Section 6.3.
6.5
Carrying out your own testing on the CSE system
It is important for you to note the assumption on directory structure mentioned in Section 6.3.
You must ensure your shell script and program files are written with this assumption in mind.
Since we will be testing your work on the CSE system, we strongly advise you to carry out the
following on the CSE system before submission.
•
Create a new folder in your CSE account and cd to that folder. We will refer to this directory
as the base directory.
– Copy your shell script run_test.sh and program files to the base directory.
– Copy the config and ref directories, as well as their contents, to the base directory
– Create a empty directory output
•
Make sure your shell script is executable.
•
Run your shell script for each test one by one.
Make sure that each run produces the
appropriate output files for that test in the output directory.
•
Copy cf_output_with_ref.py to the base directory. Run it to compare your output against
the reference output.
These steps are the same as those that we will use for testing. It is important to know that
we will create an empty output/ directory before we run your code. This means your code does
NOT have to create the the output/ directory.
7
Project requirements
This is an individual project. You are expected to complete this project on your own.
7.1
Submission requirements
Your submission should include the following:
-
A written report
(a) Only soft copy is required.
(b) It must be in Acrobat pdf format.
© It must be called ”report.pdf”.
-
Program source code:
(a) For doing simulation
(b) The shell script run_test.sh, see Section 6.3.
-
Any supporting materials, e.g. logs created by your simulation, scripts that you have written
to process the data etc.
21
The assessment will be based on your submission and running your code on the CSE system.
It is important that you submit the right version of the code and make sure that it runs on the
CSE system.
It is important that you write a clear and to-the-point report. You need to aware that you
are writing the report to the marker (the intended audience of the report) not for yourself. Your
report will be assessed primarily based on the quality of the work that you have done. You do
not have to include any background materials in your report. You only have to talk about how
you do the work and we have provided a set of assessment criteria in Section 7.2 to help you to
write your report. In order for you to demonstrate these criteria, your report should refer to your
programs, scripts, additional materials so that we are aware of them.
7.2
Assessment criteria
We will assess the quality of your project based on the following criteria:
-
The correctness of your simulation code. For this, we will:
(a) Test your code using test cases
(b) Look for evidence in your report that you have verified the correctness of the inter-arrival
probability distribution, probability distribution of the number of sub-jobs, and service
time distribution. You can include appropriate supporting materials to demonstrate
this in your submission.
© Look for evidence in your report that you have verified the correctness of your simulation
code. You may derive test cases such as those in Section 4 to test your code. You can
include appropriate supporting materials to demonstrate this in your submission.
-
You will need to demonstrate that your results are reproducible. You should provide evidence
of this in your report.
-
For the part on determining a suitable value of the threshold h that minimises the mean
response time, we will look for the following in your report:
(a) Evidence of using statistically sound methods to analyse simulation results
(b) Explanation on how you choose your simulation and data processing parameters, e.g
lengths of your simulation, number of replications, end of transient etc.
7.3
How to submit
You should “zip” your report, shell script, programs and supporting materials into a file called
“project.zip”. The submission system will only accept this filename. Please ensure that you
run zip in the directory containing your run_test.sh. If you need to store directories
when zipping, you need to use the -r switch. The last point is especially important
if you put your program code in a sub-directory under the base directory; in this
case, the relative path between your run_test.sh and your program code need to be
preserved.
You should submit your work via the course website. Note that the maximum size of your
submission should be no more than 20MBytes.
You can submit multiple times before the deadline. A later submission overrides the earlier
submissions, so make sure you submit the correct file. Do not leave until the last moment to
submit, as there may be technical or communication error and you will not have time to rectify.
22
8
Plagiarism
You should be aware of the UNSW policy on plagiarism. Please refer to the course web page for
details.
8.1
Project Conditions
Furthermore, you should be aware of the following conditions for this project.
•
Joint work is not permitted on this project.
– This is an individual project.
– You are allowed to use the code provided by the teaching staff of COMP9334 but other
than that, the work you submit must be entirely your own work.
– Do not request help from anyone other than the teaching staff of COMP9334.
– Do not post your project code to the course forum.
– Project submissions are routinely examined both automatically and manually for work
written by others.
•
Sharing, publishing, or distributing your project work is not permitted.
– Do not provide or show your project work to any other person, other than the teaching
staff of COMP9334. For example, do not message your work to friends.
– Do not publish your project via the Internet. For example, do not place your project
in a public GitHub repository.
•
Sharing, publishing, or distributing your project work after the completion of COMP9334
is not permitted. For example, do not place your assignment in a public GitHub repository
after this offering of COMP9334 is over.
Violation of any of the above conditions may result in an academic integrity investigation,
with possible penalties up to and including a mark of 0 in COMP9334, and exclusion from future
studies at UNSW. For more information, read the UNSW Student Code, or contact the course
account.
References
[1] Mor Harchol-Balter. Performance Modeling and Design of Computer Systems. Cambridge
University Press (2013).
[2] Mor Harchol-Balter. Open problems in queueing theory inspired by datacenter computing.
Queueing Systems, 97:3-27, 2021.
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