How to simplify this equation?

**Pan** · January 12th 2012, 06:22 PM

This shouldn't even be difficult for me, but after a good 4 hours and many pages filled, I still cannot get the right answer. I got pretty close, really thought I was going to get it when everything started cancelling, but then my answer is off, although at least the correct format (took me ages to even get that lol, before then I had huge, horribly complex equations that I couldn't simplify).

This is the equation:
$W=\frac {60(\lambda/60)^2}{(120-\lambda)^2 (1+[\frac {\lambda}{60}]+\frac {1}{2}[\frac {\lambda}{60}]^2\frac {120}{120-\lambda})}$

I need it simplified. The correct answer is:

$W = \frac {\lambda^2}{4-\lambda^2}$

After many hours and pages, I managed to get the following result:

$W = \frac {\lambda^2}{864000-60\lambda^2}$

So, it's at least in the right form, but my $\lambda^2$ term in the denominator is out by a factor of 15, and the constant term is out by $15*120^2$ (I put it like that since 120 appears a lot when solving it).

I would really appreciate if someone can show me the correct way to simplify this, I have to be able to do this on an exam (revising at the moment).

**pickslides** · January 12th 2012, 06:28 PM

Post the steps you took.

**Pan** · January 12th 2012, 06:30 PM

Originally Posted by pickslides

Post the steps you took.

OK

This could take a while

**Pan** · January 12th 2012, 06:47 PM

$\frac {60(\lambda/60)^2}{(120-\lambda)^2 (1+[\frac {\lambda}{60}]+\frac {1}{2}[\frac {\lambda}{60}]^2\frac {120}{120-\lambda})}$

first I simplified the numerator:

$60(\frac {\lambda^2}{3600}) = \frac {\lambda^2}{60}$

then I turned this division by 60 of the numerator into a multiplication by 60 on the denominator:

$\frac {\lambda^2}{(120-\lambda)^2 (60+\lambda+60\frac {1}{2}[\frac {\lambda}{60}]^2\frac {120}{120-\lambda})}$

Simplify that a bit:

$\frac {\lambda^2}{(120-\lambda)^2 (60+\lambda+\frac {\lambda^2}{120}\frac {120}{(120-\lambda)})}$

Those 120s cancel:

$\frac {\lambda^2}{(120-\lambda)^2 (60+\lambda+\frac {\lambda^2}{120-\lambda})}$

I turn $(120-\lambda)^2$ into $(120-\lambda)$ by multiplying the whole other term by $(120-\lambda)$ . This cancels with the last fractional term:

$\frac {\lambda^2}{(120-\lambda) [60(120-\lambda)+\lambda(120-\lambda)+\lambda^2]}$

At this point you can probably see how I got my final answer. I multiplied out the terms in the square brackets, and lots of them cancel:

$\frac {\lambda^2}{(120-\lambda) [7200-60\lambda+120\lambda-\lambda^2+\lambda^2]}$

$\frac {\lambda^2}{(120-\lambda) [7200+60\lambda]}$

Then multiply out the two bracket terms:

$\frac {\lambda^2}{864000-60\lambda^2}$

**MarkFL** · January 12th 2012, 07:08 PM

I get the same thing you do, and I fed it into the machine and got the same result as well. Are you certain you are beginning with the correct expression?

**Pan** · January 12th 2012, 07:36 PM

I was pretty sure I was starting with the right thing.

From the start I have this:

$W=\frac{\mu(\lambda/\mu)^k}{(k-1)!(k\mu-\lambda)^2}p_0$

$p_0=\frac {1}{[\sum\limits^{k-1}_{n=0}\frac {1}{n!}(\frac{\lambda}{\mu})^n]+[\frac{1}{k!} (\frac{\lambda}{\mu})^k \frac{k\mu}{k\mu-\lambda}]}$

for me $k=2$ and $\mu=60$

therefore

$W=\frac{60(\lambda/60)^2}{1*(120-\lambda)^2}p_0$

and

$p_0=\frac{1}{[1+(\frac{\lambda}{\mu})]+[\frac{1}{2}(\frac{\lambda}{\mu})^2 \frac{2\mu}{2\mu-\lambda}]} = \frac{1}{1+(\frac{\lambda}{60})+\frac{1}{2} (\frac{\lambda}{60})^2 \frac{120}{120-\lambda}}$

so

$W=\frac{60(\lambda/60)^2}{(120-\lambda)^2} \times \frac{1}{1+(\frac{\lambda}{60})+\frac{1}{2} (\frac{\lambda}{60})^2 \frac{120}{120-\lambda}}$

$W=\frac{60(\lambda/60)^2}{(120-\lambda)^2 [1+(\frac{\lambda}{60})+\frac{1}{2} (\frac{\lambda}{60})^2 \frac{120}{120-\lambda}]}$

which is where I started in the first post

**Pan** · January 12th 2012, 08:01 PM

It is the right answer.

I was using the wrong $\mu$ . It should have been 1 instead of 60. I was in fact looking at the wrong question entirely!

Sorry everybody! Doh

**MarkFL** · January 12th 2012, 08:04 PM

Check $p_0$ ; it appears the first term in the denominator should be $\frac{1}{60}$ rather than 1. However, I still do not get the intended result...I get:

$W=\frac{\lambda^2}{(120-\lambda)(119\lambda+120)}$

**Pan** · January 12th 2012, 08:15 PM

Sorry, I have put more brackets in to make that clearer. It wasn't clear what the power was for. When n=0

$\frac{1}{0} (\frac{\lambda}{\mu})^0$

This must be 1 since anything to the power 0 is 1. You are right if the power of n was on the lambda.

**MarkFL** · January 12th 2012, 08:55 PM

Glad you figured it out!

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