End-to-end delay for A to deliver two packets (sent back-to-back) in a packet-sw
ID: 3880737 • Letter: E
Question
End-to-end delay for A to deliver two packets (sent back-to-back) in a packet-switched network to B Each packet is 50 kbits long Queue at R is initially empty Suppose that we put another router R2 in the middle of R-B link, thus forming two routers in the middle (R1, R2), with three links from the sender A to the receiver B. The newly added link R2-B is identical to the original A-R link (now called A-R1 link), with the same link speed of 10Mbps and propagation delay of 10ms. Instead, the middle link R1-R2 is slow, running at 1Mbps, with propagation delay of 90ms. The sender A has 3 packets to transmit to B, each of which is 50kbits. Assume that the queues at each of the routers R1 and R2 are all empty initially and ignore any processing delay at R1 and R2. Assume that A starts transmitting 3 packets back to back at t = 0, when does the first packet arrive to B? a) When does the last 3^rd packet arrive to B (i.e., the total end-to-end delay to transmit all 3 packets back to back to the destination B)? b) Compute the total queueing delay (sum of queueing delays over the path from A to B) for each of the three packets. c) What is the average queueing delay (out of 3 packets)? You need to show clearly all your steps. Suppose that the sender A decides to transmit 3 packets not back-to-back, but with some 'spacing' in the middle. In other words, A starts transmitting packet 1 at t = 0, and once A is done transmitting packet 1, A goes idle, waiting for s seconds, and then start transmitting packet 2 at t = mu + s, where mu is the transmission delay of one packet over the link A-R1. Similarly, there is s-seconds spacing between packet 2 and packet 3. a) What is the minimum value of s such that none of the packets will suffer any queueing delay in the middle? b) Do you think the sender A in reality can find such spacing so that packets won't have to wait in the queue? Here assume there is no other protocol running between A and B (i.e., take the scenario 'as is', or assume UDP is running end-to-end). Carefully write down your logics.Explanation / Answer
First of all to answer this question, the definition of transmission delay and propagation delay must be clear.
Transmission delay- The total time required for a packet to transmit all the packet's bits placed into the link. It is given by the formula, - L/R., where L is packet length(in bits) and R is the link bandwidth.
Propagation Delay- Time required for one bit to travel from one router to the other router. It is given by the formula, - d/S, where d-length of physical link and S- Propagation speed(Speed at which a bit moves through a link It is generally 2 * 108m/s.
So, in the above problem transmission delays are:
i) for A to R1, L = 50Kbits or 50 * 103 bits and R = 10 Mbps or 10 * 106 bps
Therefore Transmission delay 50 * 103 / 10 * 106 = 50 * 10-4 or 5ms(millisecond)
ii)for R1 to R2
L = 50Kbits or 50 * 103 bits and R = 1 Mbps or 106 bps
Therefore Transmission delay 50 * 103 / 106 = 50 * 10-3 or 50ms(millisecond)
iii) for R2 to B
L = 50Kbits or 50 * 103 bits and R = 10 Mbps or 10 * 106 bps
Therefore Transmission delay 50 * 103 / 10 * 106 = 50 * 10-4 or 5ms(millisecond)
Propagation delays for the following links are already given as:
i) For A to R1 - 10 ms
ii) For R1 to R2- 90 ms
iii) For R2 to B - 10 ms
1.) Since there will be no queue delay for first packet, Hence the total time taken by the first packet to reach B is given as : 5 ms + 50 ms + 5ms + 10 ms + 90 ms + 10 ms = 170 ms
2) (b) the queueing delay is given as : dqueue = dtransmission * lqueue(length of the queue)
lqueue is the avg. length of the queue, which depends on the load factor, i.e., the ratio of attempted link transmission rate to the max. link transmission rate. The avg. queue length is typically less than 1(approx. 0.9).
Therefore, the queue delay dqueue for 2nd packet(as 1st packet will not face queue delay, because queue is empty initially) is: dtransmission * lqueue = 170 ms * 0.9 = 153 ms(approx.)
As, the 3rd packet has to wait for both 1st and 2nd packet to get transmitted, hence queue delay for 3rd packet is given as : 2(153 ms ) = 306 ms.
Hence the total queueing delay is : 153 ms + 306 ms = 459 ms.
2) c) Average queue delay is given as : Sum of delays / No. of delays.
Therefore, Averege queue delay is : 459 / 2 = 229.5 ms
2) a) The last 3rd packet will arrive at B at the end, i.e., Sum of Total transmission delays, propagation delays and Queueing delays.
Total Transmission delay for 3 packets is : 3 * 170 = 510 ms
Total Propagation delay for 3 packets is : 3 * (10 + 90 +10) = 330 ms
Total Queueing delay for 3 packets is : 459 ms
Hence, total end-to-end delay is: 510 + 330 + 459 = 1299 ms.
Thank you.
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