Part 1 Assume a node A is sending a packet to router B who is forwarding the pac
ID: 3887220 • Letter: P
Question
Part 1 Assume a node A is sending a packet to router B who is forwarding the packet to destination C. A B link has transmission rate of 1 Megabit/second while B - C link has transmission rate of 2 Megabit/second. There is only one packet that needs to traverse from A- B- C. The packet size is 1000 Bytes. a) 2] What are the transmission delays over link A - B and link B-C b) [2] Assume nodal processing delay at B is 1 millisecond and queuing delay at B is 3 milliseconds. Then what is the total delay from node A to node C? Assume zero propagation delay.Explanation / Answer
1.
a)
Given,
Packet size = 1000 Bytes= 1000 * 8 bits = 8000 bits (Since, 1 byte= 8 bits)
Transmission rate of link A-B = 1 Megabit/second = 106 bits/second.
Therefore, Transmission delay over link A-B is given by
Transmission delay over link A-B = (Packet Size in bits) / ( Transmission rate of link A-B in bits/second)
= (8000) / (106) seconds
= 8000 micro seconds
= 8 milli seconds (8 msec)
Transmission rate of link B-C = 2 Megabit/second = 2*106 bits/second.
Therefore, Transmission delay over link B-C is given by
Transmission delay over link B-C = (Packet Size in bits) / ( Transmission rate of link B-C in bits/second)
= (8000) / (2*106) seconds
= 4000 micro seconds
= 4 milli seconds (4 msec)
b)
Given,
Processing Delay at B = 1 millisecond
Queuing Delay at B = 3 milliseconds
Therefore, Total Delay for the packet to reach C from A is given by:
Total Delay = Trasmission delay over link A-B + Processing Delay at B + Queuing Delay at B + Transmission delay over link B-C
= 8milliseconds + 1 millisecond + 3 milliseconds + 4 milliseconds
= 16 milliseconds