# Minimizing probability function based on constraint

• Jan 4th 2010, 09:20 PM
bjkrishna
Minimizing probability function based on constraint
Minimize the probability:

$\displaystyle P_b=\left(2^{R-B_1log(1+\frac{P_1}{N_1})}-1\right)\left(\frac{N_2B_2}{P_2}\right)$

with respect to the constraint $\displaystyle P=P_1+P_2$

I am trying to solve this using lagrange multipliers.
Even though i got the $\displaystyle \lambda$ value i am not able to find the $\displaystyle P_1$ and $\displaystyle P_2$ values .

Can anyone please help me to get the $\displaystyle P_1$ and $\displaystyle P_2$ values .

Thanks
• Jan 5th 2010, 03:19 AM
CaptainBlack
Quote:

Originally Posted by bjkrishna
Minimize the probability:

$\displaystyle P_b=\left(2^{R-B_1log(1+\frac{P_1}{N_1})}-1\right)\left(\frac{N_2B_2}{P_2}\right)$

with respect to the constraint $\displaystyle P=P_1+P_2$

I am trying to solve this using lagrange multipliers.
Even though i got the $\displaystyle \lambda$ value i am not able to find the $\displaystyle P_1$ and $\displaystyle P_2$ values .

Can anyone please help me to get the $\displaystyle P_1$ and $\displaystyle P_2$ values .

Thanks

Why are you using Lagrange multiplers? If you are required to do so then fine, but other wise just substitute $\displaystyle P_2=P-P_1$ into the expression from $\displaystyle P_b$ and minimise as usual subject only to $\displaystyle 0 \le P_1\le 1$

Also clarify if the log is natural or common (or base 2).

CB
• Jan 5th 2010, 08:09 AM
bjkrishna
Thanks Bruno

the logarithm is respect to base 2

It is not compulsory to use Lagrange multipliers.
Can you please explain the minimizing procedure in detail.
• Jan 6th 2010, 04:04 AM
CaptainBlack
Quote:

Originally Posted by bjkrishna
Thanks Bruno

the logarithm is respect to base 2

It is not compulsory to use Lagrange multipliers.
Can you please explain the minimizing procedure in detail.

You will need to check the algebra but I have:

$\displaystyle P_b=2^k \left[1+\frac{P_1}{N_1}\right]^{-B} \frac{N_2}{P-P_1}$

Now differentiate this with respect to $\displaystyle P_1$ and set the derivative to zero and solve the resulting equation for $\displaystyle P_1$.

Now compare $\displaystyle P_b$ for $\displaystyle P_1=0,1$ and the solution (if any in $\displaystyle [0,1]$) found above the minimum for $\displaystyle P_b$ is the smallest of these.

CB
• Jan 6th 2010, 07:34 PM
bjkrishna
Thanks a lot for the clue
I am able to get the value of $\displaystyle P_1.$