Solution to Assignment 1:

 

9.1����� c = speed of light = 3 � 108 m/sec

Propagation delay�� = 10 m/(0.8 � 3 � 108 m/sec) = 4.17 � 10-8 sec

 

���������� For an efficiency of 99%, the transmission time per slot should be 99 times the propagation delay.

 

��������������������������������������������������� Transmission time�������� = 4.13 � 10-6 sec

��������������������������������������������������� Number of bits/slot������� = 4.13 � 10-6 sec � 500 � 106 bps� =� 2065 bits

 

 

9.2� Each telephone makes 0.5 calls/hour at 6 minutes each.� Thus a telephone occupies a circuit for 3 minutes per hour.� Twenty telephones can share a circuit (although this 100% utilization implies long queuing delays).� Since 10% of the calls are long distance, it takes 200 telephones to occupy a long distance (4 kHz) channel full time.� The interoffice trunk has 106/(4 � 103) = 250 channels.� With 200 telephones per channel, an end office can support 200 � 250 = 50,000 telephones. Source:� [TANE96]

 

9.3����� SS7 could be implemented using circuit switching. This would have the merit of providing minimum delay between control points for the exchange of control information. However, the circuits would have to be set up, which takes time. Alternatively, the circuits could be permanently maintained, which consumes resources.

 

 

Solution to Assignment 2:

 

10.10�� We show the results for starting from node 2.

 

 

M

L(1)

Path

L(3)

Path

L(4)

Path

L(5)

Path

L(6)

Path

1

{2}

3

2-1

3

2-3

2

2-4

�

�

�

�

2

{2, 4}

3

2-1

3

2-3

2

2-4

3

2-4-5

�

�

3

{2, 4, 1}

3

2-1

3

2-3

2

2-4

3

2-4-5

�

�

4

{2, 4, 1, 3}

3

2-1

3

2-3

2

2-4

3

2-4-5

8

2-3-6

5

{2, 4, 1, 3, 5}

3

2-1

3

2-3

2

2-4

3

2-4-5

5

2-4-5-6

6

{2, 4, 1, 3, 5, 6}

3

2-1

3

2-3

2

2-4

3

2-4-5

5

2-4-5-6

 

 

 

10.11�� We show the results for starting from node 2.

 

h

Lh(1)

Path

Lh(3)

Path

Lh(4)

Path

Lh(5)

Path

Lh(6)

Path

0

�

�

�

�

�

�

�

�

�

�

1

3

2-1

3

2-3

2

2-4

�

�

�

�

2

3

2-1

3

2-3

2

2-4

3

2-4-5

8

2-3-6

3

3

2-1

3

2-3

2

2-4

3

2-4-5

5

2-4-5-6

4

3

2-1

3

2-3

2

2-4

3

2-4-5

5

2-4-5-6

 

 

10.12�� a.�� We provide a table for node 1 of network a; the figure is easily generated.

 

 

M

L(2)

Path

L(3)

Path

L(4)

Path

L(5)

Path

L(6)

Path

1

{1}

1

1-2

�

�

4

1-4

�

�

�

�

2

{1,2}

1

1-2

4

1-2-3

4

1-4

2

1-2-5

�

�

3

{1,2,5}

1

1-2

3

1-2-5-3

3

1-2-5-4

2

1-2-5

6

1-2-5-6

4

{1,2,5,3}

1

1-2

3

1-2-5-3

3

1-2-5-4

2

1-2-5

5

1-2-5-3-6

5

{1,2,5,3,4}

1

1-2

3

1-2-5-3

3

1-2-5-4

2

1-2-5

5

1-2-5-3-6

6

{1,2,5,3,4,6}

1

1-2

3

1-2-5-3

3

1-2-5-4

2

1-2-5

5

1-2-5-3-6

 

������� b.��� The table for network b is similar in construction but much larger. Here are the results for node A:

 

A to B:� A-B

A to E:� A-E

A to H:� A-E-G-H

A to C:� A-B-C

A to F:� A-B-C-F

A to J:� A-B-C-J

A to D:� A-E-G-H-D

A to G:� A-E-G

A to K:� A--E-G-H-D-K

 

 

 

10.14      If there is a unique least-cost path, the two algorithms will yield the same result because they are both guaranteed to find the least-cost path. If there are two or more equal least-cost paths, the two algorithms may find different least-cost paths, depending on the order in which alternatives are explored.

 

 

10.17�� No.� Although it is true that the first packet to reach node 6 has experienced the minimum delay, this delay was experienced under a condition of network flooding, and cannot be considered valid for other network conditions.

 

 

 

10.20�� a.

 

 

From Node

 

 

1

2

3

4

5

6

 

1

�

1

5

5

2

3

 

2

2

�

5

5

2

3

To

3

2

5

�

5

3

3

Node

4

2

5

5

�

4

3

 

5

2

5

5

5

�

3

 

6

2

5

6

5

3

�

���������� b.

 

 

From Node

 

 

A

B

C

D

E

F

G

H

J

K

 

A

�

B

B

H

A

C

E

G

C

D

 

B

B

�

B

H

A

C

C

G

C

F

To

C

B

C

�

H

G

C

C

G

C

F

Node

D

E

C

G

�

G

K

H

D

D

D

 

E

E

A

G

H

�

K

E

G

D

D

 

F

B

C

F

K

G

�

C

D

D

F

 

G

E

C

G

H

G

K

�

G

D

D

 

H

E

C

G

H

G

K

H

�

D

D

 

J

B

C

J

J

G

K

H

D

�

D

 

K

E

C

F

K

G

K

H

D

D

�

 

 

 

10.22�� Yes. A large noise burst could create an undetected error in the packet. With an N-bit CRC, the probability of an undetected error is on the order of 2-N. If such an error occurs and alters a destination address field or virtual circuit identifier field, the packet would be misdelivered.

 

10.23�� The layer 2 flow control mechanism regulates the total flow of data between DTE and DCE.� Either can prevent the other from overwhelming it.� The layer 3 flow control mechanism regulates the flow over a single virtual circuit.� Thus, resources in either the DTE or DCE that are dedicated to a particular virtual circuit can be protected from overflow.

 

10.24      Yes. Errors are caught at the link level, but this only catches transmission errors. If a packet-switching node fails or corrupts a packet, the packet will not be delivered correctly. A higher-layer end-to-end protocol, such as TCP, must provide end-to-end reliability, if desired.

 

11.4� a.�� The reception of a valid BOM is required to enter reassembly mode, so any subsequent COM and EOM cells are rejected. Thus, the loss of a BOM results in the complete loss of the PDU.

������� b.�� Incorrect SN progression between SAR-PDUs reveals the loss of a COM, except for the cases covered by (c) and (d) below. The result is that at least the first 40 octets of the AAL-SDU is correctly received, namely the BOM that originally began the reassembly, and any subsequent COMs up to the point where cell loss occurred. Data from the BOM through the last SAR-PDU received with a correct SN can be passed up to the AAL user as a partial CPCS-PDU.

������� c.�� The SN wraps around so that there is no incorrect SN progression, but the loss of data is detected by the CPCS-PDU being undersized. In this case, only the BOM may be legitimately retrieved.

������� d.�� The same answer as for (c).

 

11.5� a.�� If the BOM of the second block arrives before the EOM of the first block, then the partially reassembled first block must be released. The entire partially reassembled CPCS-PDU received to that point is considered valid and passed to the AAL user along with an error indication. This mechanism works when just the EOM is lost or when a cell burst knocks out some COMs followed by the EOM.

������� b.�� This might be detected by the SN mechanism. If not, the loss will be detected when the Btag and Etag fields fail to match. The Length indication may fail to pick up this error if the cell bust loses as many cells as are added by concatenation the two CPCS-PDU fragments. In this case only the first BOM may be legitimately retrieved.

 

11.6� a.�� Single bit errors are not picked up until the CPCS-PDU CRC is calculated, and they result in the discarding of the entire CPCS-PDU.

������� b.�� Loss of a cell with SDU=0 is detected by an incorrect CRC. If the CRC fails to catch the error, the Length field mismatch ensures that the CPCS-PDU is discarded.

������� c.�� Loss of a cell with SDU=1 is detectable in three ways. First, the SAR-PDUs of the following CPCS-PDU may be appended to the first, resulting in a CRC error or Length mismatch being flagged when the second CPCS-PDU trailer arrives. Second, the AAL may enforce a length limit which, if exceeded while appending the second CPCS-PDU, can flag an error and cause the assembled data to be discarded. Third, a timer may be attached to the CPCS-PDU reassembly; if it expires before the CPCS-PDU is completely received, the assembled data is discarded.

 

���������� See [ARMI93] for a further discussion of the issues raised in problems 11.4 through 11.6.

12.3���� Throughput is as follows:

 

 

0 � l � 1

1 � l � 10

10 � l

A-A' traffic

0.8

(1.8 � 0.8)/(0.8 + l)

1.44/10.8 = 0.13

B-B' traffic

l

1.8l/(0.8 + l)

18/10.8 = 1.67

Total throughput

0.8 + l

1.8

1.8

 

���������� The plot:

���������� For 1 � l � 10, the fraction of throughput that is A-A' traffic is 0.8/(0.8 + l).

 

12.4���� The terms are roughly similar, but the mathematical definitions differ, so one would expect that the practical application of the two sets of concepts would produce different results.