attrition rate

**VonNemo19** · July 11th 2011, 07:48 AM

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.

**Sudharaka** · July 11th 2011, 08:14 AM

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

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

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,

$N(t)-\frac{0.4N(t)}{100}=-50e^{-0.05t}+49.94$

**VonNemo19** · July 11th 2011, 08:37 AM

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.

**CaptainBlack** · July 11th 2011, 11:14 PM

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

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

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

**Sudharaka** · July 11th 2011, 11:17 PM

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.

Thread: attrition rate

LinkBack

Thread Tools

Display

attrition rate

Re: attrition rate

Re: attrition rate

Re: attrition rate

Re: attrition rate

Similar Math Help Forum Discussions

Communication channel: baud rate, bit rate, etc.

Relationship rate of change of diagonal and rate of change of area of rectangle

Having trouble with continuous growth rate and growth rate

Average rate and instantaneous rate?

Tax rate

Search Tags