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Math Help - Minimizing probability function based on constraint

  1. #1
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    Minimizing probability function based on constraint

    Minimize the probability:

     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 P=P_1+P_2

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

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


    Thanks
    Last edited by CaptainBlack; January 5th 2010 at 03:16 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bjkrishna View Post
    Minimize the probability:

     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 P=P_1+P_2

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

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


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

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

    CB
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by bjkrishna View Post
    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:

    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 P_1 and set the derivative to zero and solve the resulting equation for P_1.

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

    CB
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  5. #5
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    Thanks a lot for the clue
    I am able to get the value of P_1.
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