Suppose that Router A and Router B are connected via a 1 Kbps link, and that the
ID: 3885993 • Letter: S
Question
Suppose that Router A and Router B are connected via a 1 Kbps link, and that the two routers are 10 meters apart. Suppose that Router A has 5 packets in its queue when a new packet arrives at Router A. Assume the following a. Every packet consists of 100 bits. b. Propagation speed of a bit over the link connecting the two routers is 2x 10^2 meter/sec. c. Processing delay of a (whole) packet at a router takes 30 milliseconds (i.e., 0.03 seconds). What is the total delay of the new packet, i.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 four delay components the new packet experiences.Explanation / Answer
Answer is as follows:
Given statements:
Propagation Delay ( pd )= 2 x 102 meter/sec
Processing Delay ( tpd )= 0.03 seconds
Packet Length ( L) = 100 bits
Link Speed ( R ) = 1Kbps = 1000 bps
Calculations:
Transmission Delay (td) = Packet Length/ Link Speed = L / R = 100 / 1000 = 0.1 bps
Quening delay for new nth packet i.e. 6th packet
Queing Delay (qd ) = (n - 1 ) * L / R = (6 - 1) * 0.1 = 5 * 0.1 = 0.5 seconds
There is no mean of using processing delay because it is also covered in queing delay.
So the total delay = propagation delay + transmission delay + Queing Delay
= pd + td + qd
= 2 x 102 + 0.1 + 0.5
= 200 + 0.6 = 200.6
So the total delay of the new packet is 200.6 seconds = 3.343 minutes..
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