Chapter 6, Problem P10P

a.

Explanation of Solution

Given that:

There are two nodes named “A” and “B”. The node “A” contains more data to transmit to node “B”. The retransmission probability of node “A” is greater than node “B”.

To find:

• Formula for node “A’s” average throughput.
• Total efficiency

Solution:

Formula for node “A’s” average throughput:

The node “A’s” average throughput is determined by multiplying retransmission probability of A and (1 – retransmission probability of B)...

b.

Explanation of Solution

Given that:

There are two nodes named “A” and “B”. The node “A” contains more data to transmit to node “B”. The retransmission probability of node “A” is greater than node “B”.

Solution:

Throughput of A: $\text{pA(1-pB) = 2pB(1-pB)}$

Multiply 2pB inside the brackets =>${\text{pA(1-pB) = 2pB1-2(pB)}}^2$ (1)

Throughput of B: $$

c.

Explanation of Solution

Given that:

There are “N” number of nodes in which the node “A” has retransmission probability “2p” whereas the other nodes has retransmission probability “p”.

To find: Average throughputs of node “A” and other nodes...