Consider a simple network that is composed of two computers (A & B) and a router

Question

Consider a simple network that is composed of two computers (A & B) and a router X in the middle. Computer A is connected to X by a link that is 1 Mbps and has a propagation delay of 1 msec. Computer B is connected to X by a link that is 2 Mbps and has a propagation delay of 2 msec. Suppose that computer A generates two packets (destined to B) that are sent back-to-back via router X. Assume that each packet is 1000 Bytes in size.

(a) When would the first packet arrive fully at X?

(b) When would the first packet arrive fully at B?

Explanation / Answer

Since it is mentioned that this is a simple network and also other parameters like link length are not given, we are only going to consider transmission time as the total time for the packet to reach the destination.

Transmission time is nothing but the time taken to place the packet onto the link. It mainly depends on the link capacity (nothing but bandwidth) and the amount of data that needs to be placed (packet size here).

So, transmission time for a packet = packet size / link bandwidth;

Transmission time for packet 1 to reach Router X is

TAX = (1000 bytes) / (1 x 1000 bytes/sec) { where bandwidth = 1Mbps = 1 X 1000 bytes /sec}

TAX = 1 sec

Similarly, transmission time to reach packet 1 from X to B is

TXB = packet size / bandwidth of XB link

TXB = 1000 bytes / (2 X 1000 bytes /sec) { since bandwidth between X and B link is 2 Mbps = 2 x 1000 bytes/sec}

TXB = 0.5 sec

Therefore packet A reaches X at 1sec and

reaches B at 1+0.5 sec = 1.5 sec

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