CSIT 561 Computer Networks: An Internet Perspective Homework3
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1. Compute the routing table at node A using the link-state algorithm:
step
N’
D(B),P(B)
D(C),P(C)
D(D),P(D)
D(E),P(E)
D(F),P(F)
0
A
1, A
3, A
4, A
∞
∞
1
AB
2, B
4, A
11, B
5, B
2
ABC
3, C
11, B
5, B
3
ABCD
7, D
5, B
4
ABCDF
6, F
5
ABCDFE
2. Consider the following network topology. Consider the path information that reaches stub networks W, X, and Y. Based on the information available at W and X, what are their respective views of the network topology? Justify your answer. The topology view at Y is shown below.
3. Suppose nodes A and B are on the same 10 Mbps Ethernet segment, and the propagation delay between the two nodes is 225 bit times. Suppose node A begins transmitting a frame and, before it finishes, node B begins transmitting a frame. Can A finish transmitting before it detects that B has transmitted? Why or why not? If the answer is yes, then A incorrectly believes that its frame was successfully transmitted without a collision. Hint: Suppose at time t=0 bit times, A begins transmitting a frame. In the worst case, A transmits a minimum sized frame of 512+64 bit times. So A would finish transmitting the frame at t=512+64 bit times. Thus, the answer is no, if B’s signal reaches A before bit time t=512+64 bits. In the worst case, when does B’s signal reach A?
At time = 0, A transmits. At time = 576, A will finish transmitting, suppose B will transmit in the time=224, then time =224+225=449, 449 < 576, the first bit of B will reach to A. A will abort the transmitting.
So A can not finish transmitting before it detects that B has transmitted. It means that if A does not detect the host presence, then it will no other host begin transmitting when A is transmitting
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