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Math Help - attrition rate

  1. #1
    No one in Particular VonNemo19's Avatar
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    attrition rate

    I'm doing a math project and I'm stuck on the last question. OK, so I was told that the rate that a product is produced is given by

    \frac{dN}{dt}=2.5e^{-0.05t}


    so I integrate this and use the initial condition to get the production function. That is, the number of uits produced with respect to the number of days

    N(t)=-50e^{-0.05t}+49.94.

    Now, it would take 95 employees to produce 2500 units

    If there is a 10 percent attrition rate over a period of 25 days, how would I revise the production function to account for this.
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    Re: attrition rate

    Quote Originally Posted by VonNemo19 View Post
    I'm doing a math project and I'm stuck on the last question. OK, so I was told that the rate that a product is produced is given by

    \frac{dN}{dt}=2.5e^{-0.05t}


    so I integrate this and use the initial condition to get the production function. That is, the number of uits produced with respect to the number of days

    N(t)=-50e^{-0.05t}+49.94.

    Now, it would take 95 employees to produce 2500 units

    If there is a 10 percent attrition rate over a period of 25 days, how would I revise the production function to account for this.
    Hi VonNemo19,

    Since the attrition rate is, 10\%\mbox{ per 25 days}\Rightarrow 0.4\%\mbox{ per day}

    Hence the number of items attired per day =\frac{0.4N(t)}{100}

    Therefore your function could be revised as,

    N(t)-\frac{0.4N(t)}{100}=-50e^{-0.05t}+49.94
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Re: attrition rate

    I have a change to make to the last part of my question. It should have read:

    If there is a 10 percent attrition rate over a period of 15 days, how would I revise the production function to account for this.
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    Re: attrition rate

    Quote Originally Posted by VonNemo19 View Post
    I'm doing a math project and I'm stuck on the last question. OK, so I was told that the rate that a product is produced is given by

    \frac{dN}{dt}=2.5e^{-0.05t}


    so I integrate this and use the initial condition to get the production function. That is, the number of uits produced with respect to the number of days

    N(t)=-50e^{-0.05t}+49.94.

    Now, it would take 95 employees to produce 2500 units

    If there is a 10 percent attrition rate over a period of 25 days, how would I revise the production function to account for this.
    If as suggested elsewhere, the loss rate is 0.004N(t) per day, the DE becomes:

    \frac{dN}{dt}=2.5e^{-0.5t}-0.004N(t)

    But that is not the loss rate, what you know is that if you have no production and a initial stock N_0 then:

    \frac{dN}{dt}=-\lambda N(t)

    So: N(t)=N_0 e^{-\lambda t} and you solve for \lambda using N(25)=0.9N_0 which gives \lambda=0.00421..


    Then with production you have:

    \frac{dN}{dt}=2.5e^{-0.5t}-\lambda N(t)

    CB
    Last edited by CaptainBlack; July 12th 2011 at 12:25 AM.
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    Re: attrition rate

    Quote Originally Posted by VonNemo19 View Post
    I have a change to make to the last part of my question. It should have read:

    If there is a 10 percent attrition rate over a period of 15 days, how would I revise the production function to account for this.
    Then you will have to replace 15 instead of 25 and proceed along the same method I had done in my previous post.
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