Network Analysis – Report Sheet.

Name Mohamed Alsubaie

Course Computer and Communication Engineering

Date 15 March 2013

All information and results should be recorded in the boxes provided in this report sheet. When stating results always include the correct units.

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Section A (Initial study) - database application running on initial network

Recorded Results and Calculations

Database (DB) traffic

Using your own graphical results, measure the Client DB traffic sent in Bytes/sec - record this value in the box below

- (1)

Now measure and record the server DB traffic sent in Bytes/sec

- (2)

Network Load (Throughput)

Measure the Ethernet load in bits/sec

- (3)

Measure the point-to- point throughput in bits/sec

- (4)

Measure the point-to- point throughput in packets/sec

- (5)

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512 Bytes/sec

100,000 Bytes/sec

822400 Bits/sec

836960 Bits/sec

70 Packets/sec

Delays in network

Measure the DB query response time in seconds

- (6)

Measure the Server DB task processing time in seconds

- (7)

Measure the point to point queuing delay in seconds

- (8)

(4 marks for correct values)

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0.186 seconds

0.1 seconds

0.0011 seconds

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Discussion of Results

The results you have now recorded, together with knowledge of the network modeled, enable several important conclusions on the operation of a communication network to be reached. The questions that follow are intended to allow you to answer these.

Before considering these questions, the following points relating to the process you have modeled need to be considered.

The client/server communication across the network follows a layered approach (see my lecture notes and/or other sources on OSI 7-layer model).

More specifically the Database (DB) application modeled uses the TCP/IP protocols. (See additional notes at the end of the lab sheet on IP fragmentation and re-assembly).

The data link layer type used follows the Ethernet protocol and it is a shared access strategy running on a co-axial bus type physical layer.

The DB traffic data recorded (i.e. in Box 2) is the ‘raw’ data generated at the application layer.

The Ethernet load data recorded (i.e. in Box 3 ) is the load offered by the network layer to the datalink layer.

The bus throughput data recorded (i.e. in Box 4) is the load offered by the datalink layer to the physical layer (i.e. the average load on the point to point link).

Loads

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The diagram on the previous page provides a brief description of the data transfer process for the network you have modelled. First note the differences between the measured loads you have recorded at various points in the network in the Boxes numbered 2, 3 and 4.

As the application layer data is fragmented into blocks various overheads are added.

At the network layer, the standard overheads added to every packet are:

TCP overhead = 20 Bytes and IP overhead = 20 Bytes

When an IP packet is presented to the Data link (Ethernet) layer a fixed overhead of 26 Bytes per packet is added to produce a data frame

Delays

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Using the above information and the measured results you have recorded in Boxes 2 and 5 – calculate the load in bits/sec you expect after the Network layer overheads have been added (show working below). Then compare this with the measured result you have recorded in Box 3. (4 marks)

DB Traffic sent = 100, 000 Bytes/sNetwork Load Point-to-Point Throughput = 70 Packets/s

Load = (Total Overheads x Point-to-Point Throughput) + DB Traffic = [(20 + 20) x 70 x 8] + [100,000 x 8] = 22,400 + 800,000 = 822400 Bits/sec

The calculation result of the Load is the same as the measured results (simulation) in Box 3.

Now calculate the load in bits/sec you expect after the Data link layer overheads have been added (show working below). Then compare this with the measured result you have recorded in Box 4.

(2 marks)

Data Link Layer Overheads = 26 x 70 x 8 = 14560 Bits/sec

Total Load = 822, 400 + 14560 Bits/s = 836, 960 Bits/sec

The calculation result of the Total Load is the same as the measured results (simulation) in Box 4.

The Client database (DB) query response time is the total time elapsed between the instant the client sends a request to the server to the time it receives the entire database file.

Given that the database file size was set at 100,000 Bytes and given that the Ethernet based network you modeled had a data-rate of 10 Mbits/sec, calculate the minimum time that the file transfer should take. Show this calculation in the box below. (2 marks)

Compare this minimum calculated value with the actual response time you recorded in Box 6. Provide possible reasons for this difference in the box below. (Hint – consider any other delays you measured) (3 marks)

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This is only the minimum time that the file transfer should take.

The total time will required to measure other delays:

Latency = Processing Delay+ Queuing Delay + Transmission Delay + Propagation Delay

See the question below.

The minimum calculated value is much smaller than the response time.

0.08 second < 0.186 second.

As mention on previous question the response time (Latency) is the total of:

Latency = Processing Delay+ Queuing Delay + Transmission Delay + Propagation Delay

Processing time and Queuing time can be taken from box 7 & 8. Transmission time from

the calculated value. Where there is no vale for Propagation time.

Response time = 0.1 + 0.001 + 0.08 + ? = 0.181 seconds

Still this is not exactly equal the actual response time. This is due to the propagation

delay (which should have a value around 0.005 second).

General Questions

Explain the fundamental problem with measuring load in packets/sec. (2 marks)

State three reasons why it is usual in data communication networks to fragment large data messages into data packets of limited size. (3 marks)

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The fundamental problem with measuring load in packets/sec is that packets can vary in size.

Packet switching features delivery of variable-bit-rate data streams (sequences of packets) over a shared network. When traversing network adapters, switches, routers and other network nodes, packets are buffered and queued, resulting in variable delay and throughput depending on the traffic load in the network.

Prevent users from monopolising the use of data links by transmitting data continually.

 The packets move between nodes very quickly 

Helps overcome transmission problems. If a glitch occurs, only one packet may be affected. Then it is only necessary to retransmit that one packet rather than the entire file.

Fragmentation allows higher level protocols to be unconcerned with the characteristics of the transmission channel, and to send data in conveniently sized pieces.

Fragmentation allows the source host to deal with routes having different MTUs without having to know what path packets are taking.

Fragmentation allows protocols to optimize performance for high bandwidth connections

Section B (further work) – Extended Network results/analysis/amendments

(N.B To support your discussion you must include and refer to the appropriately labeled graphs included in a separate Appendix),

1. State and explain the underlying reasons for any significant performance problems you have found when increasing the number of stations on the network from 1 to 10.

(10 Marks)

To compare the network performance for both scenarios of the following network in

figure-1, it is necessarily to consider the point-to-point utilization / throughout and

response time.

Utilization is effectively a measure of how busy a link is measured over a set of time

period (channel efficiency). Where, throughout is the average measure of the transferred

data over a communication channel. The simulation result, for both point-to-point

utilization and throughout, shows that the scenario with 10 works station (expansion) is

approximately 10 times the initial scenario with 1 workstation. It can be clearly seen on

figure-2, that the initial scenario has only 8% utilization wile it about 80% on the

expansion scenario. According to figure-3, initial scenario was having a throughput of

836,960 bits/sec. But when the number of workstations was increased to 10 the

throughput became approximately 8,000,000 bits/sec. The results of the expansion

scenario for both point-to-point utilization and throughout are evidently shows that now a

lot more data is being transferred through the network and the network is being used to

almost maximum capacity.

According to the pervious results of the expansion scenario, the time response will be

poor since the data rate is 10 times larger. In theory, a higher throughput and utilization

will cause a higher response time, which is mean that the network will be slow. Figure-4

shows the networks DB Query Response Time, which is the time taken by the network to

respond to a particular query, for both scenarios. The response time for the initial

scenario was 0.186 second. However, the response time for expansion scenario has

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unstable results. It keeps on increasing for every extra bit of data being transfer resulting

an unstable network and not suitable for communication.

Another reason behind the poor performance is the network topology. The sheared network that been used in both scenario has a “Bus Topology”. In such a LAN topology, all the nodes are connected directly to a single cable. Since the bus topology consists of only one wire, it is rather inexpensive to implement when compared to other topologies. Even though it has its certain advantages, the low cost of implementing the technology is offset by the high cost of managing the network. Additionally, since only one cable is utilized, it can be the single point of failure. If the network cable breaks, the entire network will be down. The main disadvantage is that the efficiency of Bus network reduces, as the number of devices connected to it increases, which is not suitable for networks with heavy traffic. Generally, it only used in small scale network due to problems with collision at high loading levels.

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Figure-1: Expansion scenario network. Figure-2: Utilization for expansion scenario.

Figure-3: Throughput for expansion scenario. Figure-4: Response time for expansion scenario.

2. Describe and explain the reasons for network changes you made to address the performance problems (include new network topology diagrams). Justify these improvements by comparing and discussing the results you obtained from the different scenarios.

(10 Marks)

The network type used on both initial and expansion scenario has “Bus Topology”. As

discussed before, Increasing the number of work stations from 1 to 10 is leading to have

unstable network and not suitable for communication, because bus topology is not

suitable for networks with heavy traffic. 

In order to improve the network performance, it is necessarily to change the network

topology. In this case a “Star Topology” can solve the problem. All devices and

computers on this network are connected to a central “Hub” or “Switch". A major

disadvantage of this network topology is the high dependence on the central hub. If the

central hub fails, all computers connected to that hub would be disconnected. However, it

can effectively stop collision occurring and allowed increasing the network size and

capacity. So increasing the number of workstation to 10 will not be a problem using a star

topology.

To implement this on the software, a new scenario has been created with changing the

10BaseT LAN to 10BaseT-Switched LAN (Star Topology) see figure-5. The graphs

below show the result for this network. It can be clearly seen that the performance has

improved and the network became stable. The point to point utilization and throughout

increased slightly to about 84% and 8,400,000 bits/sec respectively, comparing to the

pervious results for expansion scenario which is 80% and 8,000,000. However, the time

response has a steady result which is approximately 0.93 seconds.

Reducing the point to point utilization can be done through changing the data like from

10BaseT with 100BaseT. The results are shown in the same graphs. It is obvious that

point to point utilization reduce to almost the same as the initial design and the response

time decreased to about 0.23 seconds. Where the throughout is still having the same

results as when using a link of 10BaseT.

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An extra improvement can be done through using a LAN with higher data transfer rate.

The previous network has 10BaseT-Switched LAN, which is 10Mbits/sec. by changing

this to 100BaseT-Switched LAN (see figure-9), the response time for this scenario is

approximately 0. 186 seconds, which is even better than the initial scenario.

Other improvements:

Increase the switching speed and upgrading the nodes (computers): this will make a little

improvement to the network performance.

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Figure-6: Utilization for 10BaseT-Switched LAN scenario.

Figure-4: 10BaseT-Switched LAN scenario network.

Figure-7: Throughput for 10BaseT-Switched LAN scenario.

Figure-8: Response time for 10BaseT-Switched LAN scenario.

Figure-9: 100BaseT-Switched LAN scenario network.

Figure-10: Utilization for 100BaseT-Switched LAN scenario.

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Figure-11: Throughput for 100BaseT-Switched LAN scenario.

Figure-12: Response time for 100BaseT-Switched LAN scenario.