# Network performance (latency)

• posted

Well, I don't know if I should be helping you with your homework, but consider the store & forward switch. Its presence means you're actually sending the data twice. Once from the source to the switch and again from the switch to the destination. That alone would double the latency. However, without looking at the original question, I can't say for certain that's what they're looking for.

• posted

Hi. I'm trying to solve an exercise, I have the solution, but I have some doubts about it. I have to calculate the latency*bandwidth of a network. This network is made up of the source and of the destination, which are linked with a

10Mbps network (10 micro-seconds propagation time) with a store-and-forward switch. The request is to calculate the latency between the first sent bit and the first arrived bit. I solved this way:

latency=2*10micro-seconds+5000bit/10Mbps+1bit/10Mbps

where 5000bits is the packet size. I added 1bit/10Mbps because I need the time to transfer the last bit from the switch to the destination. The solution I have in the book instead, adds another 5000bit/10Mbps so:

latency=2*10micro-seconds+2*5000bit/10Mbps

Why do they add two 5000bit/10Mbps? Is someone able to help me? Thanks for any help :-).

Luca

• posted

James Knott nel messaggio precedente ha scritto:

Yes, this I can see. But what I considered is that the text explicitly said to measure the time from the *first* bit sent to the *first* arrived bit. So, this means, from what I understood, that I need of course two times the propagation time, since as you said I can consider two different wires. But, after this, I have to distinguish. From the source to the switch I need a time t:

t = 5000bit/10Mbps

i. e. the time needed to transfer the packet (5000bits) from the source to the switch. After this time (plus the propagation time), I have

5000bits in the switch. Now, the exercise asks me to calculate the time till the *first* bit arrives. So, I wouldn't say I need other 5000bits/10Mbps, but only 1bit/10Mbps. The first bit needs this time (plus another propagation time of course). If I added other 5000bits/10Mbps I get the time necessary to transfer all the 5000bits of the packet. The exercise, instead, only wants the time till the first bit arrives to the destination. So:

latency = 2*pr latency = 2*propagation_time + 2*5000bits/10Mbps

I would have found the time between the first bit is sent to the last bit is arrived to the destination. Isn't this correct?

The question only asks to calculate the latency on this network measured from the time the first bit is sent by the source to the time the first bit arrives to the destination. Thanks for your answer.

Luca

• posted

Luca nel messaggio precedente ha scritto:

I can answer now: I was right. Thanks.

Luca

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