Results 1 to 6 of 6

Math Help - Maximization question involving normal distribution.

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    35
    Thanks
    1

    Maximization question involving normal distribution.

    The cost of keeping a unit of inventory is a.
    The cost of being short a unit is b.
    b>a
    The expected demand is normally distributed with mean m and std. dev. c.

    I'm trying to find a formula for the optimum amount of inventory.

    I'm currently plugging in values for q in the following formula:
    aq+b*Integral of(normal distribution*(x-q)) from q to infinite

    q is the inventory in excess of the expected demand.

    Is this correct? If so, what would the derivative of this formula be so I can find the maximum.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by bob000 View Post
    The cost of keeping a unit of inventory is a.
    The cost of being short a unit is b.
    b>a
    The expected demand is normally distributed with mean m and std. dev. c.

    I'm trying to find a formula for the optimum amount of inventory.

    I'm currently plugging in values for q in the following formula:
    aq+b*Integral of(normal distribution*(x-q)) from q to infinite

    q is the inventory in excess of the expected demand.

    Is this correct? If so, what would the derivative of this formula be so I can find the maximum.
    Let the stock holding he $$ q and the demand $$ u. The the cost is:

    c(q)=aq+b\, p(u>q)

    or:

    \displaystyle c(q)=aq+b\, \left(1-\dfrac{1}{\sqrt{2\pi} c}\int_{-\infty}^q e^{-\frac{(u-m)^2}{2c^2}}\;du\right)

    so:

    c'(q)=a-\dfrac{b}{\sqrt{2\pi}c}e^{-\frac{(q-m)^2}{2c^2}}

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    35
    Thanks
    1
    I don't think this is correct:


    The cost of shortage is b for EACH unit you are short.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack View Post
    Let the stock holding he $$ q and the demand $$ u. The the cost is:

    c(q)=aq+b\, p(u>q)

    or:

    \displaystyle c(q)=aq+b\, \left(1-\dfrac{1}{\sqrt{2\pi} c}\int_{-\infty}^q e^{-\frac{(u-m)^2}{2c^2}}\;du\right)

    so:

    c'(q)=a-\dfrac{b}{\sqrt{2\pi}c}e^{-\frac{(q-m)^2}{2c^2}}

    CB
    Quote Originally Posted by bob000 View Post
    I don't think this is correct:


    The cost of shortage is b for EACH unit you are short.
    OK so that changes $$ c(q) to:

    \displaystyle c(q)=aq+b \int_{u=q}^{\infty}(u-q)p(u)\;du

    ........ \displaystyle =aq+b \int_{u=q}^{\infty}u\, p(u)\;du-b\,q \int_{u=q}^{\infty}p(u)\;du

    So (if I have done the derivative correctly):

    \displaystyle c'(q)=a-b \int_{u=q}^{\infty}p(u)\;du

    and so the stationary point (maximum presumably) is:

    q=P^{-1}_{m,c}(1-\frac{a}{b})

    where $$ P_{m,c}(.) is the cumulative normal distribution function of a normal RV \sim N(m,c^2).

    Then in terms of the standard normal cumulative distribution function:

    q=m+cP^{-1}_{0,1}(1-a/b)

    (Check that to make sure it is correct. for a/b close to zero it will still give a negative $$ q, but that is because with a normal distribution there is always a finite probability of negative sales)

    The above may still have errors you will have to check it, but the key idea is the use of the fundamental theorem of calculus:

    \displaystyle \dfrac{d}{dx} \left( \int_a^x f(\xi)\; d\xi \right)=f(x)

    and/or:

    \displaystyle \dfrac{d}{dx} \left( \int_x^a f(\xi)\; d\xi \right)=-f(x)



    CB
    Last edited by CaptainBlack; August 11th 2010 at 10:25 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    35
    Thanks
    1
    If a>.5b then the value inside of is less than .5 and the value of q is less than 0. But if the cost of inventory is less than the cost of shortage (a<b), we know that optimal inventory I=m+q must be at least the expected demand m. Am I misunderstanding something?

    Thank you for the time you've spent so far btw.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by bob000 View Post
    If a>.5b then the value inside of is less than .5 and the value of q is less than 0. But if the cost of inventory is less than the cost of shortage (a<b), we know that optimal inventory I=m+q must be at least the expected demand m. Am I misunderstanding something?

    Thank you for the time you've spent so far btw.
    Did you read the caveats about checking as there may still be errors in the algebra?

    Think P(.) is the cumulative normal distribution function for a RV \sim N(m,c^2)

    or see the modified earlier post.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Problems involving maximization of Area
    Posted in the Geometry Forum
    Replies: 11
    Last Post: May 24th 2011, 12:31 PM
  2. normal distribution question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: May 22nd 2011, 09:37 AM
  3. Normal distribution question involving samples
    Posted in the Statistics Forum
    Replies: 4
    Last Post: July 18th 2009, 12:38 AM
  4. Normal distribution question
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: December 25th 2008, 02:01 PM
  5. Normal distribution question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 31st 2008, 11:39 PM

Search Tags


/mathhelpforum @mathhelpforum