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I am unable to understand the solution.

(1)Why I/O bus will be used twice?

(2)With what logic, the answer is calculated?

0 votes

1. The bus is used twice because the packet crosses 2 times, one during when it is getting stored and secondly when it is being forwarded from the memory.

It is clearly given in the solution, How you didn't understand this part??

2. Now you want to divide the speed of the bandwidth with twice the(twice because storing and forwarding) * packet size of one packet * forwarding rater per second.

Now just write the SI units, to understand it better per second from the numerator and per second from the denominator is canceled out.

So you are just left with the maximum rate.

Anything beyond this the router will not be able to take the load of this.

It is clearly given in the solution, How you didn't understand this part??

2. Now you want to divide the speed of the bandwidth with twice the(twice because storing and forwarding) * packet size of one packet * forwarding rater per second.

Now just write the SI units, to understand it better per second from the numerator and per second from the denominator is canceled out.

So you are just left with the maximum rate.

Anything beyond this the router will not be able to take the load of this.

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