COMPGROUPS.NET | Search | Post Question | Groups | Stream | About | Register

### About the availability and efficiency...

• Email
• Follow

```Hello,

I have corrected an error in my page about the availability and
efficiency, the utilization is equal to the arrival rate divided by
the servive rate, not the inverse..

And here is the part that i have corrected:

As you have noticed, i have finally arrived to the following
mathematical model of our Jackson network that model an Enterprise
website:

T = Ni /A = (Nn + Ns + Nc) /A
= (((A/Ss)/n) / (1 -((A/Ss)/n)) + (A/Sn) / (1 -(A/Sn))
+ (A/Sc) / (1 -(A/Sc))) / A

T: is the average response time..

So read it carefully an you will notice that this mathematical
model also can simulate and predict an intranet website behavior
or an internet website..

First look at the first factor ((A/Ss)/n) / (1 -((A/Ss)/n))
is equal to (A/((Ss x n) - A) in our mathematical model,
as you have noticed as n (number of servers) grows
this will make the denominator grow and this will make
(A/((Ss x n) - A) smaller, that's good for the response time...

Now let's look at the other factors in this mathematical
model:

IF the system is an Intranet, and for example the local
area network Sn is faster than the multiserver, that means:

We have Sn >> Ss, and that means => Sn /A >> Ss /A => that the
Knee in the server side can be reached more quickly than the
Knee in the network node of our Jackson network, so that
the bottleneck becomes the server node.

In the other hand if:

Ss >> Sn => Ss /A >> Sn / A => the Knee in the M/M/1 network
node can be reached more quickly, so that the network node in
the Jackson network becomes the bottleneck.

http://pages.videotron.com/aminer/efficiency_availability.htm

Thank you.
Amine Moulay Ramdane.

```
 0
Reply aminer (401) 6/4/2012 1:43:34 AM

See related articles to this posting

```Hello again,

I have also corrected a typo in my other page about a jackson network
problem...

Here it is:

http://pages.videotron.com/aminer/ParallelGaussSeidel/jackson1.htm

Thank you.
Amine Moulay Ramdane

"aminer" <aminer@videotron.ca> wrote in message
news:jqh0bs\$tv6\$1@dont-email.me...
> Hello,
>
> I have corrected an error in my page about the availability and
> efficiency, the utilization is equal to the arrival rate divided by
> the servive rate, not the inverse..
>
> And here is the part that i have corrected:
>
> As you have noticed, i have finally arrived to the following
> mathematical model of our Jackson network that model an Enterprise
> website:
>
>
> T = Ni /A = (Nn + Ns + Nc) /A
> = (((A/Ss)/n) / (1 -((A/Ss)/n)) + (A/Sn) / (1 -(A/Sn))
> + (A/Sc) / (1 -(A/Sc))) / A
>
>
>
> T: is the average response time..
>
>
>
> So read it carefully an you will notice that this mathematical
> model also can simulate and predict an intranet website behavior
> or an internet website..
>
>
>
> First look at the first factor ((A/Ss)/n) / (1 -((A/Ss)/n))
> is equal to (A/((Ss x n) - A) in our mathematical model,
> as you have noticed as n (number of servers) grows
> this will make the denominator grow and this will make
> (A/((Ss x n) - A) smaller, that's good for the response time...
>
>
> Now let's look at the other factors in this mathematical
> model:
>
>
>
> IF the system is an Intranet, and for example the local
> area network Sn is faster than the multiserver, that means:
>
>
> We have Sn >> Ss, and that means => Sn /A >> Ss /A => that the
> Knee in the server side can be reached more quickly than the
> Knee in the network node of our Jackson network, so that
> the bottleneck becomes the server node.
>
>
> In the other hand if:
>
>
>
> Ss >> Sn => Ss /A >> Sn / A => the Knee in the M/M/1 network
> node can be reached more quickly, so that the network node in
> the Jackson network becomes the bottleneck.
>
>
>
>
> http://pages.videotron.com/aminer/efficiency_availability.htm
>
>
>
> Thank you.
> Amine Moulay Ramdane.
>
>
>
>
>

```
 0
Reply aminer (401) 6/4/2012 1:56:24 AM

1 Replies
30 Views

Similar Articles

12/6/2013 7:49:45 PM
page loaded in 28631 ms. (0)