(computer networks) Consider a simple network that is composed of two computers

Question

(computer networks)

Consider a simple network that is composed of two computers (A & B) and a router X in the middle. Computer A is connected to router X by a link that is 10 Mbps and has a propagation delay of 10 msec. Computer B is connected to router X by a link that is 1000 KBps and has a propagation delay of 20 msec.

By what percentage should computer A reduce its traffic in the worst-case to remove the need of a buffer at router X?

In order to increase the carrying capacity (number of bits on the wire) of this simple network, you are asked to compare between two propositions. The first proposition suggested to increase the bandwidth of the link between computer A and router X to 12 Mbps. The second suggested to increase the propagation delay between router X and computer B to 25 msec. Which one of these two propositions would increase the carrying capacity more?

Explanation / Answer

Given bandwidth of the link between A and X is = 10 Mbps

propagation delay between A and X is = 10 msec

We know time taken to send 1 bit is :

1 / B = 1 / 10 Mbps = 1 / 107 = 10-7 secs (1M = 106)

Now in 10 msec we can send 10 msec / 10-7 bits :

10 * 10-3 / 10-7 =

105 bits

Hence through this Link we can send 105 bits of data.

Given bandwidth of the link between B and X is :1000 KBps

propagation delay between B and X is : 20 msec

We know time taken to send 1 bit is :

1 / B = 1 / 1000 KBps = 1 / 8 * 106 = 0.125 * 10-6 secs (1K =103,1B=8bits)

Now in 20 msec we can send 20 msec / 0.125 * 10-7 bits = 16 * 104 bits

Hence through this Link we can send 16 * 104 = 1.6 * 105 bits of data.

Proposal 1 (increase the bandwidth of the link between computer A and router X to 12 Mbps):-

Now bandwidth of the link between A and X is :12 Mbps

propagation delay between A and X is : 10 msec

We know time taken to send 1 bit is = 1 / B = 1 / 12 Mbps = 1 / (12 * 106) = (1/12) * 10-6 secs (1M = 106)

Now in 10 msec we can send 10 msec / (1/12) * 10-6 bits =12 * 104 bits

Hence through this Link we can send 12 * 104 = 1.2 * 105 bits of data.

Proposal 2 (increase the propagation delay between router X and computer B to 25 msec):-

Now bandwidth of the link between B and X is : 1000 KBps

propagation delay between B and X is :25 msec

We know time taken to send 1 bit is = 1 / B = 1 / 1000 KBps = 1 / 8 * 106 = 0.125 * 10-6 secs (1K =103,1B=8bits)

Now in 25 msec we can send 25 msec / 0.125 * 10-7 bits = 2 * 104 bits

Hence through this Link we can send 2 * 104 = 0.2 * 105 bits of data.

Due to proposal 1, 105 bits of data increased to 1.2 * 105 bits of data.

Due to proposal 2, 1.6 * 105 bits of data decreased to 0.2 * 105 bits of data.

Hence proposal 1 is better than 2.

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