Factoring response time equation

• Aug 8th 2007, 02:39 PM
Heather
Factoring response time equation
Remember the calculation of response time in the lecture?

R = SQ + S(2)

Based on one Queuing Theorem:

Q = a * R(3)

Manipulating equations (2) and (3) using Factoring method, we obtain:

S
R = ---------- (4)
1 – a S

Show how you manipulate the two equations (2) and (3) in order to get (4).

The lecture notes remind us that Q=a(SQ=S) if that helps anyone out there. Thanks so much!
• Aug 8th 2007, 06:42 PM
topsquark
Quote:

Originally Posted by Heather
Remember the calculation of response time in the lecture?

R = SQ + S(2)

Based on one Queuing Theorem:

Q = a * R(3)

Manipulating equations (2) and (3) using Factoring method, we obtain:

S
R = ---------- (4)
1 – a S

Show how you manipulate the two equations (2) and (3) in order to get (4).

The lecture notes remind us that Q=a(SQ=S) if that helps anyone out there. Thanks so much!

I doubt most of us are taking this class with you! ;)

Can you give us some background in regard to what you are talking about?

-Dan
• Aug 9th 2007, 07:18 AM
Heather
Most of what we are studying right now is factoring and finding GCF. Somehow with this problem you are supposed to be doing factoring, but the only way i see it possible is by starting with substitution.
• Aug 9th 2007, 10:49 AM
topsquark
Quote:

Originally Posted by Heather
Remember the calculation of response time in the lecture?

R = SQ + S(2)

Based on one Queuing Theorem:

Q = a * R(3)

Manipulating equations (2) and (3) using Factoring method, we obtain:

S
R = ---------- (4)
1 – a S

Show how you manipulate the two equations (2) and (3) in order to get (4).

The lecture notes remind us that Q=a(SQ=S) if that helps anyone out there. Thanks so much!

Oh okay, from the way you phrased the question I thought there was more to it.

So we have:
$\displaystyle R = SQ + S$

and

$\displaystyle Q = aR$

So put the bottom equation into the top:
$\displaystyle R = S(aR) + S = aSR + R$

Now we want to solve for R:
$\displaystyle R - aSR = S$

There is a common R to both terms on the left, so factor it:
$\displaystyle (1 - aS)R = S$

Now divide:
$\displaystyle R = \frac{S}{1 - aS}$

-Dan
• Aug 10th 2007, 10:02 AM
Heather
Thank you soooooo much this has been driving me crazy! But now i feel dumb because this was very simple:rolleyes:
• Aug 10th 2007, 10:08 AM
topsquark
Quote:

Originally Posted by Heather
Thank you soooooo much this has been driving me crazy! But now i feel dumb because this was very simple:rolleyes:

Don't feel dumb. There have been times I've forgotten how to add. (Seriously!) Everyone has their moments. ;)

-Dan