Chapter 12:   Exercises 5, 8, & 9


Ex 5:
In Figure 12.9, assign the following costs to the links:

            A - B      5
            A - F      1
            B - C      2
            C - D      3
            C - F      1
            D - E      2
            E - F      2

(a) Determine the routing tables for each node, using the distance vector scheme.

Node A Node B Node C Node D Node E Node F
D L C D L C D L C D L C D L C D L C
A   0 A 2 4 A 2 2 A 2 5 A 1 3 A 1 1
B 2 4 B   0 B 1 2 B 1 5 B 1 5 B 2 3
C 2 2 C 2 2 C   0 C 1 3 C 1 3 C 2 1
D 2 5 D 2 5 D 3 3 D   0 D 2 2 D 3 4
E 2 3 E 2 5 E 2 3 E 2 2 E   0 E 3 2
F 2 1 F 2 3 F 2 1 F 2 4 F 1 2 F   0


(b) Determine the routing tables for node D, using link state routing.

Destination              Connection              Cost              Status
D              0              0             


Destination              Connection              Cost              Status
D              0              0             
C              1              3              T
E              2              2              T


Destination              Connection              Cost              Status
D              0              0             
C              1              3              T
E              2              2             
F              1              4              T


Destination              Connection              Cost              Status
D              0              0             
C              1              3             
E              2              2             
F              1              4              T
B              1              5              T


Destination              Connection              Cost              Status
D              0              0             
C              1              3             
E              2              2             
F              1              4             
B              1              5              T
A              1              5              T


Destination              Connection              Cost              Status
D              0              0             
C              1              3             
E              2              2             
F              1              4             
B              1              5             
A              1              5             



Ex 8:
Consider the journey of a packet of length 2048 octets.   The packet passes through networks with maximum packet sizes as shown.   Show what happens to the packet using intranet fragmentation and internet fragmentation.   Discuss the performance of each scheme.   (Assume that the packet header is 40 octets.)

            Network       Maximum Packet Length
            A                   2048 octets
            B                   512 octets
            C                   1024 octets
            D                   256 octets
            E                   1024 octets

Fragmenting a data unit requires replicating the header, providing header information regarding the fragments, and breaking up the data part of the data unit so that the resulting fragments meet the length requirements of the network.

Intranet fragmentation:
The data unit is reassembled as it leaves the network that required the fragmentation. This may result in wasted effort if the reconstructed data unit must again be fragmented at a later step in its path. On the other hand, because the complete data unit is available on entry to each network, the data unit can be divided into optimal size subunits for the next part of its journey.

Network A (2048 octets): The data unit remains intact.
Network B (512 octets): The data unit is divided into 5 data units, each with a 40 octet header. The first 4 are 512 octets, and the last is 200 octets. The data unit is reassembled into a 2048 octet as it leaves the network.
Network C (1024 octets): The data unit is divided into 3 data units, each with a 40 octet header. The first 2 are 1024 octets, and the last is 120 octets. The data unit is reassembled into a 2048 octet as it leaves the network.
Network D (256 octets): The data unit is divided into 10 data units, each with a 40 octet header. The first 9 are 256 octets, and the last is 144 octets. The data unit is reassembled into a 2048 octet as it leaves the network.
Network E (1024 octets): The data unit is divided into 3 data units, each with a 40 octet header. The first 2 are 1024 octets, and the last is 120 octets. The data unit is reassembled into a 2048 octet since this is the final destinaion.

Internet fragmentation:
The data unit remains fragmented until it reaches its destination. This saves the effort of reassembling data units as each intervening network is left, but means that smaller than optimal data units may be used.

Network A (2048 octets): The data unit remains intact.
Network B (512 octets): The data unit is divided into 5 data units, each with a 40 octet header. The first 4 are 512 octets, and the last is 200 octets. The data unit is not reassembled as it leaves the network.
Network C (1024 octets): The 5 data units from Network B pass through with no change. The data unit is not reassembled as it leaves the network.
Network D (256 octets): Each of the 5 data units are divided into 3 data units, each with a 40 octet header. The first 2 of each group of three are 256 octets, and the last 120 octets. This will result in a total of 15 data units. The data unit is not reassembled as it leaves the network.
Network E (1024 octets): The 15 data units from Network D pass through, and then the data unit is reassembled into a 2048 octet since this is the final destination.

Conclusion:
Intranet Fragmentation:
Total of 21 (5 + 3 + 10 + 3) fragmentations performed.
Total of 4 reassemblies performed.
Total of 22 (1 + 5 + 3 + 10 + 3) optimal sized data units passed.

Internet Fragmentation:
Total of 20 (5 + 15) fragmentations performed.
Total of 1 reassembly performed.
Total of 41 (1 + 5 + 5 + 15 + 15) data units passed.

Assuming that it takes the same amount of time to fragment the packets as it takes to reassemble them, the Internet has 4 fewer frags/reassembles than the Intranet. However, the Intranet passes 19 fewer data units and therefore would probably be more efficient assuming that the time it takes to pass a data unit is greater than or equal to the amount of time it takes to fragment or reassemble a data unit.


Ex 9:
In the scenario of Exercise 8, consider two cases:

(a) The packet is a data unit submitted to the Network layer by a connection-oriented transport service.
(b) The packet is a data unit submitted by a connectionless transport service.
Further, consider that network D is very unreliable.   What is the difference in the two cases in regard to network performance with the two fragmentation schemes?

The connection-oriented transport service has more overhead than connectionless since it must first establish a connection, and protect against old duplicates by internally generating ID numbers for request acknowledgements. It also transfers data, does status and error reporting, and releases the connection. Connection-oriented also guarantees that errors will be corrected.

The connectionless transport service will only transfer the data and will not attempt to correct errors. All information needed to deliver the data must accompany each data unit. The TTL (time to live) field is the length of time that the data unit may continue to exist in the internet.

Connection-oriented:
Intranet Fragmentation:
Has fewer connections than Internet and since data units are reassembled as they leave each network, errors can be detected earlier which saves time in error correcting (i.e., errors can be corrected before sending ot network E).

Internet Fragmentation
Has more connections than Intranet and data units are reassembled only after they reach their final destination. Since errors can only be detected after reassembly at the final destination, this is not as efficient as Connection-oriented Intranet Fragmentation.

Connectionless:
Intranet Fragmentation:
Is more efficient than Connectionless Internet Fragmentation for reasons as stated in the conclusion of EX 8 above. Whether it is more efficient than Connection-oriented would depend on the TTL duration and the reliability of the networks. Since network D is very unreliable, connectionless network performance would be worse than connection-oriented, even though it has less overhead and takes less time to actually transfer data units, since the data units would not likely arrive error free to network E.

Internet Fragmentation
Is the least efficient as shown above in Connectionless Intranet Fragmentation.