Network performance (latency)

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Archived from groups: comp.dcom.lans.ethernet (More info?)

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

 

[Guest]

Archived from groups: comp.dcom.lans.ethernet (More info?)

Luca wrote:

> 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 .

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.

 

[Guest]

Archived from groups: comp.dcom.lans.ethernet (More info?)

James Knott nel messaggio precedente ha scritto:

> 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.

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*propagation_time + 5000bits/10Mbps + 1bit/10Mbps

so, 5000bits/10Mbps is the transmission time for 5000bits from the
source to the switch. Then 1bit/10Mbps is the transmission time for the
first bit to go from the switch where it is to the destination. If I
said:

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?

> However, without looking at the original question, I can't say for
> certain that's what they're looking for.

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

 

[Guest]

Archived from groups: comp.dcom.lans.ethernet (More info?)

Luca nel messaggio precedente ha scritto:

> Isn't this correct?

I can answer now: I was right.
Thanks.

Luca