Assume that two computers A and B are connected with a link of 600 km. Assume in

Question

Assume that two computers A and B are connected with a link of 600 km. Assume initially that there are no intermediate packet switches / routers between them (ie a direct link). Data rate on the link is provided to 100Mbps. Signal propagation speed is 2 * 108 meters / second.

a) How long does it take to send a parcel at 1460 bytes from A to B when we only take into account the transmission and propagation delay?

Now assume that there are two packet switches (based on the "store-and-forward") between A and B.

b) How long does it take now to send the packet from A to B?

Assume further that the use of a stop-and-wait protocol based where ACK packets are so small that we can ignore the transmission delay for these.

c) How long does it take now to send a packet from A to B and get a confirmation that this has been received? Otherwise, the conditions of a) and b)

Explanation / Answer

a . data rate = 100 Mbps

message to be sent = 1460*8 = 11680bits

Data rate gives is = 100 * 10^6 bits per sec

so transmission delay = 11680/100*10^6 = 0.00011680 sec = 0.11680 milli seconds

distance = 600 km

Signal propagation speed is 2 * 108 meters / second.

so propogation delay = 600 * 10^2 / 2* 10^8 = 0.3 milliseconds

Time taken to send a parcel at 1460 bytes from A to B when we only take into account the transmission and propagation delay = transmission delay + propogation delay = 0.11680 + 0.3 = 0.41680 milli seconds

Now assume that there are two packet switches (based on the "store-and-forward") between A and B.

b) How long does it take now to send the packet from A to B?

ans > now the propogation delay remains same as packets has to travel 600 km but transmission delay gets added because of switches and given switches store and forward

Tansmission delay = sender delay + switch A forward delay + switch B forward delay

and also all transmission delays are equal , so

Tansmission delay new = 3 * Transmission delay ==>

data rate = 100 Mbps

message to be sent = 1460*8 = 11680bits

Data rate gives is = 100 * 10^6 bits per sec

so transmission delay = 11680/100*10^6 = 0.00011680 sec = 0.11680 milli seconds

Transmission delay new = 3 * 0.11680 = 0.35040 milli seconds

so delay for 1460 bytes is = transmissiondelaynew + propogation delay = 0.35040 + 0.3 = 0.65040milli seconds

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