Two new workers were hired for an assembly line. Jim processed 25 units during first hour and 45

Two new workers were hired for an assembly line. Jim...

Key Concepts

Parameter Estimation in Nonlinear Models

In many applied situations, including production rate analysis, one obtains output data at different time points and uses those observations to estimate unknown parameters in the model. This involves setting up equations based on the observed values and solving for both the rate constant (which dictates how quickly the process approaches its maximum) and the maximum capability itself.

Exponential Saturation Model

This concept refers to a model in which a process starts with a slower rate of output that accelerates quickly but eventually slows down as it asymptotically approaches a maximum limit. In production or human performance tasks, this model captures the idea that there may be a maximum capacity or speed that a worker can eventually reach, regardless of additional time spent on the task.

Asymptotic Behavior

A key idea in analyzing performance models is understanding how a function behaves as time increases. In these models, while early performance (or output) is well below the maximum, the function gradually approaches a horizontal asymptote representing the maximum possible output. Recognizing and estimating this limiting behavior is crucial for determining the worker's maximum processing capability.

Transcript

00:01 As we have derived in equation number 31, p is m plus c times e raised to minus kt, where m is maximum learning, c is an arbitrary constant that arise due to our integration and k is positive constant.

00:14 We are given that gym joins a work and at first his learning is 25 for 1 hour and 45 for 2 hours.

00:24 Similarly for mark it's 35 for 1 hour and 45 for 2 hours.

00:30 Sorry it's 50 for two hours and we need to calculate what is the maximum learning can be for gym and mark we are also provided with a condition that p of zero is zero okay so let's start with the first condition so we have p of zero as zero so at t equals to zero we have p is equal to zero therefore we have 0 is equals to m plus c times e raise to 0 right k times 0 is 0 so we get 0 is m plus c and therefore c is minus m okay so now we can update our equation as a particular solution p is m minus m e raised to minus k t now we'll solve for gym.

01:32 Okay.

01:32 So gym at t equals to one we have at t equals to one we have p as 25.

01:41 So 25 is equals to what is m we don't know.

01:48 We can take m common so this will be one minus e is 2 minus k times 1 and that's a equation for first so i'll write it as e raise 2 minus k okay so if we can we can calculate value of e 2 minus k to substitute in the next step so we'll divide by m so we'll have 25 by m is equals to 1 minus e raise 2 minus k and therefore we get we'll shift these two positions or interchange it so we'll get raised to minus k is 1 minus 25 by m so let this be our first finding now the second condition for jim is at time t equals to 2 is b is 45 okay so therefore 45 is m 1 minus he raised 2 minus k into 2 therefore we have 45 is m 1 minus this is nothing by e minus k square and we have already calculated this value so we can put it here so m times one minus i'll put this value 1 minus 25 by m square okay so if we simplify this this becomes one a minus b whole square is a square minus 2 a b so that is 2 into 25 50 by m plus b square so 25 by m square is 625 by m square or bracket so if you open the bracket we get 1 minus 1 plus 50 by m minus 625 by m square and if you multiply with this m after canceling this we get 50 minus 625 by m right m square and m 1m will get cancelled and here we have 45 so we'll again change this to position.

04:02 So here and this will be brought here.

04:06 So we'll have 625 by m is 50 minus 45.

04:15 So therefore we get 625 by m as 5.

04:22 And interchanging this position, inverting this will get m over 625 is 1 over 625 5 so therefore m will be 625 by 5 and that is 1 to 5 so maximum learning of gym will be 125 on similarly we can solve for mark so what's a condition for mark mark at time t equals to 1 r is maximum working is 35 so that's p and and t equals to 2 p is 50 correct so we have our equation as p is m times 1 minus e raised to minus k t so for 1 r that is 35 m 1 minus e raise to minus k and what will be raised to minus k will be 35 by m and it is subtracted from 1 so this is what we are going to use to substitute in the second equation correct so 50 is equals to that is our second condition we are substituting so 50 is equals to m times 1 minus this is 2k right so i'll first write that value otherwise it's very confusing so 2 okay so m times 1 minus this is nothing but d -res to minus k square...

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