Question

Two nodes A and B are connected via switch S.

*About packet switching Latency

bandwidth : 4Mbps

number of packets : 10^3

Propagation time : 2msec

size of a packet: 8000 bit (there for there are 8M bit data, 8000 * 10 * 3)

switching time : 1msec

Queuing time : 0

If a header must be added to the data size (8000 bits), I wonder what is the length condition of the header that will take less time than 2024 msec.

I think it will be solved by pipelining, but I cannot solve. Can you help me?

Answer 1

Total delay = Propagation delay + Transmission delay + processing delay + queuing delay

Propagation delay is the time taken to cover physical distance from A to B.

Propagation delay = 2ms

Queuing delay is the time that the packet has to wait in the queue.

Queuing delay = 0ms

Processing delay is the time taken to process packets.

processing delay = switching time = 1ms

Transmission delay is the time taken to put all the packets in the wire between A to B.

Transmission delay = Packet size * Number of packets / bandwidth

Let the size of header that can be added as x. So, now size of packet will be 8000 + x

Transmission delay = (8000 + x) * 1000 / 4000 ms = (8000 + x)/4 ms

Total delay = Propagation delay + Transmission delay + processing delay + queuing delay

= 2ms + 0ms + 1ms + (8000+x)/4

Total delay should be less than 2024ms. So,

2ms + 0ms + 1ms + (8000 + x)/4 <2024

3 + (8000 +x)/4 < 2024

3 + 2000 + x/4 < 2024

x/4 + 2003 < 2024

x/4 <21

x < 84

Hence, length of header should be less than 84 bits to take less than 2024 ms.