b denotes bits and B denotes Bytes (1 Byte = 8 bits) Assume source X is connecte

Question

b denotes bits and B denotes Bytes (1 Byte = 8 bits)

Assume source X is connected to destination Y through 4 store-and-forward routers (there are 5 links on X-to-Y path). Each link has a bandwidth of 100Mbps and a propagation-delay of 10us. The size of each data packets is 20Kb. Assume as soon as a packet arrives at a router the router forwards the packet without any nodal processing-delay or queuing-delay. Assume once a packet arrives at destination Y, the destination immediately replies an acknowledgment packet to X. Assume the size of acknowledgment packet is 60b. It takes some time for the acknowledgment packet to reach source X. The total time it takes for source X to send a data packet and receive the corresponding acknowledgement is: the time it takes for the data packet to reach destination Y plus the time it takes for the acknowledgment to reach source X. In this assignment, you are asked to calculate effective throughput, which is defined as follows: the size of effective data (data reached at destination Y) over the total time spent to deliver data and acknowledge it. Meaning, both data-transmission time and acknowledgement time are considered but only the amount of data arrived at destination is called effective data. The size of acknowledgement replied to the source X is not counted to calculate effective throughput. 4 8 1. 2. 3. How long does it take for a packet to be transmitted from source X to destination Y (in micro-seconds)? How long does it take for an acknowledgement to be transmitted from Y to X (in micro-seconds)? What is the effective throughput (in Mbps) if X sends a message containing 1 data packet? data Total ACK

Explanation / Answer

Solution :-

1. Time for the packet to reach from X to Y

   = Time for X to put the packet on cable +
     Time for routers to pick up and put the packet+
     propagation dely of links+
     Time for Y to pick up the packet

    = (20*1000)/100 * 10^6 + 10 * 10^-6 * 5 + 4 * 2 * (20*1000)/100 * 10^6 + (20*1000)/100 * 10^6
    = 200 * 10^-6 + 50 * 10*-6 + 1600 * 10^-6 + 200 * 10^-6
    = 2050 * 10*-6 seconds
    = 2050 micro seconds

2. Time for ack to reach from Y to X

   = Time for Y to put the ack on cable +
     Time for routers to pick up and put the ack+
     propagation dely of links+
     Time for X to pick up the ack

   = 0.06 * 10^-6 + 50 * 10^-6 + 4 * 2 * 0.06 * 10^-6 + 0.06 * 10^-6
   = (0.06 + 50 + 0.48 + 0.06) * 10^-6
   = 50.6 micro second

3. Effective thtrough put = 20 * 10^3/(2050 + 50.6) * 10^-6 = 9.52 * 10^6 = 9.52 Mbps

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