# attrition rate

• Jul 11th 2011, 07:48 AM
VonNemo19
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

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

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

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

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.
• Jul 11th 2011, 08:14 AM
Sudharaka
Re: attrition rate
Quote:

Originally Posted by VonNemo19
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

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

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

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

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, $\displaystyle 10\%\mbox{ per 25 days}\Rightarrow 0.4\%\mbox{ per day}$

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

Therefore your function could be revised as,

$\displaystyle N(t)-\frac{0.4N(t)}{100}=-50e^{-0.05t}+49.94$
• Jul 11th 2011, 08:37 AM
VonNemo19
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.
• Jul 11th 2011, 11:14 PM
CaptainBlack
Re: attrition rate
Quote:

Originally Posted by VonNemo19
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

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

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

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

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 $\displaystyle 0.004N(t)$ per day, the DE becomes:

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

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

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

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

Then with production you have:

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

CB
• Jul 11th 2011, 11:17 PM
Sudharaka
Re: attrition rate
Quote:

Originally Posted by VonNemo19
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.