Solution#2.33

With circuit switching, at t=s the circuit is set up; at t=s+x/b the last bit is sent; at t=s+x/b+kd the message arrives. With packet switching, the last bit is sent at t=x/b. To get to the final destination, the last packet must be retransmitted k-1 times by intermediate routers, each retransmission taking p/b sec, so the total delay is x/b+(k-1)p/b+kd. The packet switching is faster if s > (k-1)p/b.

Solution#2.34

the total number of packets needed is x/p, so the data + header traffic is (p+h)x/p bits. The source requires (p+h)x/pb sec to transmit these bits. The retransmission of the last packet by the intermediate routers take up a total of (k-1)(p+h)/b sec. Adding up the time for the source to send all the bits and the time for the routers to carry the last packet to the destination, thus clearing the pipeline, we get a total time of (p+h)x/pb + (p+h)(k-1)/b sec. Minimizing this quantity with respect to p we find p=[hx/(k-1)]^1/2.

Solution#2.38

If the switch has n lines, it takes 100n sec to switch each frame . With a frame time of 125 µsec, we get 0.1n=125 or n=1250 lines.

Solution#2.39

The frame buffer must contain 80 x 10 = 800 bits.