Q4. Suppose that Router A and Router B are connected via a 1 Kbps link, and that

Question

Q4. Suppose that Router A and Router B are connected via a 1 Kbps link, and that the two routers are 4 meters apart. Suppose that Router A is in the process of transmitting a packet, and has another 4 packets in its queue, when a new packet arrives at Router A. Assume the following ) Even-packet consists of.100bits. b) 50 bits of the currently-being-transmitted packet have been transmitted by Router A. d Propagation speed of a bit over the link comecting the two routers is 2x 10 c) Propagation speed of a bit over the link connecting the two routers is 2x 10 meter d) Processing delay of a (whole) packet at a router takes 0.03 seconds What is the total delay of the new packet that arrives, .e., how long does it take the new packet to arrive at Router B from the time it arrived at Router A? In your solution, clearly identify all (10 points) four delay components the new packet experiences.

Explanation / Answer

Router A has to send
1. 50 bits of the remaining packet
2. Process 4 packets
3. Send 4 packets

Then it can send the new packet after processing it.

time to put 50 bits on cable= 50/1000 s
Propagation delay = 4/ (2 * 10^2) = 2 * 10^-2 s
= 0.05 + 0.02 = 0.07 s

For 4 packets of 100 bits

For one packet:

time to put 100 bits on cable= 100/1000 s
Processing delay for each packet = 0.03 s
propagation delay = 4/ (2 * 10^2) = 2 * 10^-2 s
Total delay for one packet = 0.1 + 0.03 + 0.02 = 0.15 s

Total dealy with 4 packets = 4 * 0.15 = 0.6 s

Total delay for new packet to reach B = finish sending the remaing bits of the packet being currently sent + time to send 4 packets + Processing delay for this packet + Time to send this packet

Total delay for new packet to reach B = 0.07 + 0.6 + 0.03 + 0.15 = 0.85 s

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