Suppose two nodes, A and B, are attached to opposite ends of an 800 m cable, and

Question

Suppose two nodes, A and B, are attached to opposite ends of an 800 m cable, and that they each have one frame of 1 024 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t = 0. Suppose there are five repeaters between A and B, each inserting a 20-bit delay. Assume the transmission rate is 10 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is used. After the first collision, A draws K = 0 and B draws K = 1 in the exponential backoff protocol after sending the 48-bit jam signal.

a. What is the one-way propagation delay (including repeater delays) between A and B in seconds? Assume that the signal propagation speed is 2 * 10^8 m/sec

b. At what time (in seconds) is A’s packet completely delivered at B?

c. Now suppose that only A has a packet to send, and that the repeaters are replaced with switches. Suppose that each switch has a 16-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A’s packet delivered at B?

- Note: you must include all the delays that occurred according to CSMA/CD protocol.

Explanation / Answer

a.Solution

A 's packet is completely delivered at B .

Time = 800 * 10^8 m/sec +4 * 20 bits 100 X 10 ^6 bps

= (4 x 10 ^-6 + 0.8 x 10 ^ -6)sec

= 4 + 0.8

= 4.8 µsec

b.Solution

First note, the transmission time of a single frame is give by 1500/(100Mbps)=15 micro sec, longer than the propagation delay of a bit.

• At time t = 0 , both A and B transmit.

• At time t = 4.8µ sec , both A and B detect a collision, and then abort.

• At time t = 9.6µ sec last bit of B 's aborted transmission arrives at A.

• At time t =14.4µ sec first bit of A 's retransmission frame arrives at B .

• At time t =14.4 µsec + 1500 bits 100 x 10 ^ 6 bps = 29.4 µ sec A 's packet is completely

   delivered at B .

c. Solution

A 's packet is delivered at B .

Time = 800 * 10^8 m/sec +4 * 16 bits 100 X 10 ^6 bps

= (4 x 10 ^-6 + 0.16 x 10 ^ -6)sec

= 4 + 0.16

= 4.16 µsec

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.