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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.

Please read more on my page:

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

Thank you.

Amine Moulay Ramdane.

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Hello again,

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

Here it ...

Thank you.

Amine Moulay Ramdane

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