Telecommunications software and multimedia laboratory

Telecommunications architectures

2001: Home assignment 4 - Model answer

Notice! The role of this model answer is instructional only. The answers might not cover everything in detail that the questions asks. The answers give you the content of the answer and idea what it should be. The things that are important in the grading of the homework have been emphasized. If there is a mistake in the model answer in your opinion, please let the course staff know!

1. Terms & Acronyms (4 p)

Write out the acronyms and explain the meaning of each term briefly.
  1. TSAP
    TSAP is Transport Service Access Point. It is the place that the application layer can access the service offered by Transport layer.

  2. MTU
    MTU is the Maximun Transfer Unit. It is the maximun size of the segment limited by the network. If segment is larger than MTU, it should be fragmented into smaller ones.

  3. Convergence Sublayer
    Convergence Sublayer is the upper part of ATM adaptation layer to provide the interface to the application.

  4. AAL
    AAL is ATM Adaptation Layer. It provides services to application programs and to shield them from the mechanics of chopping data up into cells at the source and reassembling them at the destination.

  5. Option Negotiation
    Option negotiation is a process to decide what is the value of parameters agreed on when a connection is requested.

  6. Credit Message
    Credit Message is a flow control mechanism to avoid providing and managing buffers within the transport entity. The receiver sends credit message to the sender's transport entity. The sender will not send the message unless credit message has arrived.

  7. Bandwidth-delay Product
    Bandwidth-delay Product is the result of multiplying the bandwidth and the round-trip delay time. It shows the capacity of the pipe from the sender to receiver and back.

  8. Well-known Port
    Any TCP or UDP port between 0 and 1023. These ports are controlled by system (or root) and them are reserved for standard services. (At the begin, only ports 0-255 were well known ports. Both answers, 0-255 and 0-1024, are accepted.)
Each question is 0.5 point

2. Flow-based routing (8 p)


Each sub-question is 2 points. If you use 24000bps per weight as the assumed condition. You will be decreased 1 point for the whole question if the following answers are right according to your assumption.

The traffic matrix of this network looks like this (traffic in packets/sec):
- A B C D E F
A - 8 6 3 4 1
B 8 - 6 2 2 1
C 6 6 - 9 8 5
D 3 2 9 - 8 5
E 4 2 8 8 - 7
F 1 1 5 5 7 -
Assume that all link capacities and traffic is symmetric and the mean packet size is 800 bits. The capacities are counted by 24000bps/weight.

  1. Determine routes for all traffic classes (entries in the routing matrix) for this network using Dijkstra's algorithm. If there are several possible shortest routes, take the one with the minimum number of hops.

    The traffic and routing matrix is as following:
    A BCDEF
    A -- 8
    AB     		6
    AC     		3
    ACD     		4
    AE     		1
    ACDF
    B 8
    BA     		-- 6
    BC     		2
    BCD     		2
    BAE     		1
    BCDF
    C 6
    CA     		6
    CB     		-- 9
    CD     		8
    CDE     		5
    CDF
    D 3
    DCA     		2
    DCB     		9
    DC     		-- 8
    DE     		5
    DF
    E 4
    EA     		2
    EAB     		8
    EDC     		8
    ED     		-- 7
    EDF
    F 1
    FDCA     		1
    FDCB     		5
    FDC     		5
    FD     		7
    FDE     		--

  2. Calculate the mean delay for each entry in the matrix.
    i Line Actual trafficCapacityTWeight
    1 AB 10 15 200 0.094
    2 AC 6+3+1=10 15 200 0.094
    3 AE 4+2=6 10 250 0.056
    4 BC 6+2+1=9 10 1000 0.085
    5 CD 3+1+2+1+9+8+5=29 30 1000 0.274
    6 DE 8+8+7=23 30 142.86 0.216
    7 DF 1+1+5+5+7=19 30 90.91 0.179
    Total -- 106 -- -- 1
    The delay for each entry is adding all link delays belonging to a certain route.

  3. Calculate the mean delay for the entire network. The entire subnet mean delay time is the weighted sum of each line.
    Mean delay time=457.67ms

  4. Using Dijkstra's algorithm, determine routes for all traffic classes using calculated delays as weights. If there are several possible shortest routes, take the one with the minimum number of hops.

    A BCDEF
    A -- AB AC AED AE AEDF
    B BA -- BAC BAED BAE BAEDF
    C CA CAB -- CAED CAE CAEDF
    D DEA DEAB DEAC -- DE DF
    E EA EAB EAC ED -- EDF
    F FDEA FDEAB FDEAC FD FDE --

3. Bits and pieces (3 p)

Each question is 1 point.

  1. What is the difference between flow control and congestion control?

    1. The flow control solves the problem that the sender is fasting than the receiver.
      The congestion control solves the problem that the network is overloaded.
    2. Congestion control is a global issue related to all hosts, all routers, and all other factors tend to diminish the capacity of subnet.
      Flow control is related only to point-to-point traffic between a given sender and a given receiver.

  2. Why is UDP needed? Isn't plain IP enough?

    UDP is a way for applications to send the raw IP datagrams without establishing a connection. It is used for unreliable transmission, in which the prompt delivery is more important than accurate delivery. In some situations, the subnet doesn't support the connection-oriented service, so we can only use UDP as the transport protocol.

    The plain IP is not enough. IP is a host-to-host protocol that delivers a packet from one physical device to another. But the receiver won't know which upper application it should put the packets in. UDP is port-to-port communication. It adds both source port and destination port, checksum error control and length information to the data from the upper layer. Souce port and destination port will identify the end points of within the souce and destination machine.

  3. How is two-army-problem handled in TCP?

    The solution for two-army-problem is to add the timer in both sender and recerver. After sender sends disconect request (DR) to the receiver, it starts a timer. Then after the receiver gets the DR, it also starts a timer, then sends back the DR. When the sender receives the DR, it releases connection and sends ACK back. When receiver gets the ACK, it releases the connection. Any error occurring during the procedure will lead to the timer out. In such case, the connection will also be released.

4. Shaping (6 p)

Each question is 2 point. Error in calculation will be decreased 0.5 point. Error in concept will be given 0 point.

  1. What is the percentage of lost data if
    1. a leaky bucket (queue length 15, output rate 10)
    2. a token bucket (token bucket size 15, token rate 10, queue length infinite)
    3. a token bucket (token bucket size 15, token rate 10, queue length 15)
    is used, the queuing discipline is FIFO and the incoming data is:
    5, 7, 3, 10, 16, 7, 20, 18, 25, 12, 7, 3, 1
    If the queue is full, all incoming packets are discarded.
  2. What are the mean transmitted rates?
  3. What are the mean queuing delays? If there are packets in the queue in the end of the trace, ignore them.

If you use Token generate----waiting in queuing line-----overflow in token queuing line----transmission as order, then the answer should be:

For Leak Bucket:
Time interval Arrival QueuingLost
0-1 5 0 0
1-2 7 0 0
2-3 3 0 0
3-4 10 0 0
4-5 16 6 0
5-6 7 3 0
6-7 20 13 0
7-8 18 15 6
8-9 25 15 15
9-10 12 15 2
10-11 7 12 0
11-12 3 5 0
12-13 1 0 0

For Token Bucket (Infinite):
Time interval Arrival Token leftqueuingLost
0-1 5 5 0 0
1-2 7 8 0 0
2-3 3 12 0 0
3-4 10 5 0 0
4-5 16 0 1 0
5-6 7 2 0 0
6-7 20 0 8 0
7-8 18 0 16 0
8-9 25 0 31 0
9-10 12 0 33 0
10-11 7 0 30 0
11-12 3 0 23 0
12-13 1 0 14 0

For Token bucket(Finite):
Time interval Arrival Token leftqueuingLost
0-1 5 5 0 0
1-2 7 8 0 0
2-3 3 12 0 0
3-4 10 5 0 0
4-5 16 0 1 0
5-6 7 2 0 0
6-7 20 0 8 0
7-8 18 0 15 1
8-9 25 0 15 15
9-10 12 0 15 2
10-11 7 0 12 0
11-12 3 0 5 0
12-13 1 4 0 0

Methods Percentage of lost Mean rateMean queuing delay
Leak Bucket 17.2% 8.54 1.31
Token Bucket(Finite) 13.4% 8.92 1.16
Token Bucket(Infinite) 0 9.23 1.82

If you use token generate-----token used in transmission----waiting in token queuing line---overflow in token queuing line as the order, the answer should be:

For Leak Bucket:
Time interval Arrival QueuingLost
0-1 5 0 0
1-2 7 0 0
2-3 3 0 0
3-4 10 0 0
4-5 16 6 0
5-6 7 3 0
6-7 20 13 0
7-8 18 15 6
8-9 25 15 15
9-10 12 15 2
10-11 7 12 0
11-12 3 5 0
12-13 1 0 0

For Token Bucket (Infinite):
Time interval Arrival Token leftqueuingLost
0-1 5 5 0 0
1-2 7 8 0 0
2-3 3 15 0 0
3-4 10 15 0 0
4-5 16 9 0 0
5-6 7 12 0 0
6-7 20 2 0 0
7-8 18 0 6 0
8-9 25 0 21 0
9-10 12 0 23 0
10-11 7 0 20 0
11-12 3 0 13 0
12-13 1 0 4 0

For Token bucket(Finite):
Time interval Arrival Token leftqueuingLost
0-1 5 5 0 0
1-2 7 8 0 0
2-3 3 15 0 0
3-4 10 15 0 0
4-5 16 9 0 0
5-6 7 12 0 0
6-7 20 2 0 0
7-8 18 0 6 0
8-9 25 0 15 6
9-10 12 0 15 2
10-11 7 0 12 0
11-12 3 0 5 0
12-13 1 0 0 0

Methods Percentage of lost Mean rateMean queuing delay
Leak Bucket 17.2% 8.54 1.31
Token Bucket(Finite) 6% 9.7 1.3
Token Bucket(Infinite) 0 10 1.74

Mean transmission rate = Total packet transmitted/total time
Mean delay = [Total of (Each packet * its waiting time)]/total packets have exited in the queuing line.

We accept all these two methods.

5. Feedback (1 p)

  1. How many hours did you spend doing this assignment?
  2. Was the assignment too hard or too easy? Why?
  3. What topics should this assignment cover next year?

Returning your answer

Return your answer as explained in the general instructions for returning your answer.

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The page has last been updated 2.10.2001.
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