Ekt332 Computer Network

EKT332 COMPUTER NETWORK

TEST 2 - SCHEME

DATE : 24 MAY 2012 TIME : 12.00 PM – 01.00 PM

Instructions:

1. This is a close book test. No books and notes are allowed to be referred to during the test.

2. There are 4 (four) questions. Answer ALL questions. Please write your answer in the column provided.

3. Duration : 1.0 Hour.

4. Number of pages (including this page) : 6

TOTAL MARKS / 40

1 1. a) Distinguish which MAC protocol (ALOHA or CSMA-CD) has a higher efficiency in a LAN. What about in a WAN? Explain your answer. [6 marks]

The maximum efficiency achieved by the Slotted ALOHA is 0.368. The efficiency of CSMA-CD is given by 1/(1 + 6.4a), and is sensitive to a = tprop R/L, the ratio between delay-bandwidth product and frame length. In a LAN environment, the end-to-end distance is around 100m and the transmission rates are typically 10Mbps, 100Mbps and 1Gbps. An Ethernet frame has a maximum length of 1500 bytes = 12,000 bits. Shorter frame sizes predominate, e.g. 64 byte frames, then a increases by a factor of about 20. According to the above formula the efficiency of CSMA-CD at 1 Gbps then drops to about 0.7. The situation however is worse in that the minimum frame size at 1 Gbps needs to be extended to 512 bytes. In a WAN environment d is larger. Assuming 100 Km, a is larger by a factor of 103 resulting in an efficiency of 0.36, 0.05, and 0.005 respectively for 10 Mbps, 100 Mbps, and 1 Gbps transmission rates. In the case of 10 Mbps transmission rate the efficiency of CSMA-CD is close to the efficiency of ALOHA but in the other two cases it is much less than ALOHA.

b) Suppose that the ALOHA protocol is used to share a 56 kbps satellite channel. Suppose that frames are 1000 bits long. Calculate the maximum throughput of the system in frames/second. [4 marks]

Maximum throughput for ALOHA = Ge-2G = 0.184 Maximum throughput in frames/sec = Frames x Ge-2G = (56000 bits/sec) x (1 frame/1000 bits) x 0.184 = 10.304 The maximum throughput is approximately 10 frames/sec.

c) Suppose three terminals are attached to a hub in a star topology. The distance from each terminal to the hub is 50 meters, the transmission lines speed is 10 Mbps, all frames are of length 12,500 bytes, and the signal propagates on the line at a speed of 2.5 x 108 meter/second. Calculate the maximum network throughput achievable when the hub is implementing CSMA-CD. [6 marks]

Pmax = Max. Throughput = 1/(1+(2e+1)a)

where a = Tprop / X

8 -7 Tprop = distance / propagation speed = 50 / 2.5 x 10 = 2 x 10 s X = Frame length / bit rate = (12500 x 8 bits)/(10 x 106) = 0.01 s => a = 2 x 10-7/0.01 = 2 x 10-5

2 Therefore, Max. Throughput = 1/(1+(6.44 x 2 x 10-5) = 1/1.0000036 = 0.99987 2. Six stations (S1-S6) are connected to an extended LAN trough transparent bridges (B1 and B2), as shown in the following Figure. Initially, the forwarding tables are empty. Suppose the following stations transmit frames:

S2 transmits to S1 S5 transmits to S4 S3 transmits to S5 S1 transmits to S2 S6 transmits to S5

Fill in the forwarding tables with appropriate entries after the frames have been completely transmitted. [8 marks]

S1 S2 S3 S4 S5 S6

LAN1 LAN2 LAN3

B1 B2 Port 1 Port 2 Port 1 Port 2

Station Port Station Port S2 1 S2 1 S5 2 S3 2 S3 1 S1 1 S6 2

3 3. Assuming that a minimum-hop routing is used, produce the routing tables for each node for the datagram network shown in Figure 1. [6 marks]

Figure 1

Node 1: Destination Next Node 2 2 3 3 4 4 5 2 6 3

4 Node 4: Destination Next Node 1 1 2 2 3 3 5 5 6 3

4. Consider the network in Figure 2, use the Bellman-Ford algorithm to find the set of shortest paths from all nodes to destination node 2. [10 marks]

2 1 3 1 5 2 6 3 4

Figure 2

5 Iteration Node 1 Node 3 Node 4 Node 5 Node 6 Initial (-1,∞) (-1,∞) (-1,∞) (-1,∞) (-1,∞) 1 (2,3) (1,∞) (2,1) (2,4) (3,∞) (4,∞) (5,∞) (6,∞) 2 (2,3) (4,3) (2,1) (2,4) (5,6) 3 (2,3) (4,3) (2,1) (2,4) (3,4)

6 -oooOooo-