 let us introduce this notion of system compensation. So, negative feedback system often tries to maintain the goal in presence of some other exogenous factor either a constant inflow or a constant outflow rate over which the feedback system exerts no control. So, in the presence of such an external exogenous variable, how does this system behave? So, that is what we will try to understand today. What effect will this constant exogenous rate have? So, negative feedback systems will compensate for the additional inflow or flow by attending an equilibrium which will be different from the desired one. So, though we will have a goal, the system because of the constant exogenous variable will actually come to an equilibrium which will be different from the goal as we will see in using the example. So, before we go into the analytical version of it, let us take our laptops and expand the labour model that we did yesterday in the phase of a constant exogenous outflow. So, let us look at what we want to do in this model. Consider same labour example suppose, so in the labour example there was a desired labour that a company desired and we had the current level and the gap was adjusted based on some adjustment time I think we the base case assume 6 months of adjustment time. So, we had built that. So, in that consider labour example suppose people leave the company at some arbitrary constant rate over which you have no control people may leave the company at any point they choose. Now, we want to update the model reflect this scenario and what effect does this constant exogenous rate have on the system? We can start with parameter desired labour 100 initial labour 100 and the leaving rate as step we will come to that what it means. So, let us go to Vensim and try to model this. The first one is this one. Suppose people leave the company at some arbitrary constant rate we need to update the model to reflect this scenario. So, people are leaving that has to be modeled. So, let us go to Vensim. So, this is the model we used yesterday. I am just going to save as a new version. So, let us before we change the model. So, let us just quickly click equations button f of x click that and just check if adjustment time is 6 months net hiring rate is labour gap divided by adjustment time units is person per month labour gap is desired labour minus labour units is person desired labour is 100 desired labour 100 percent. Now, let us take labour that is the stock I just opened the stock labour the initial value it will say 0. So, let us just change that to 100. So, now the goal as well as actual stock is 100. So, we do not expect any new people joining the company. Now, click rate we need to model people leaving the company. So, let us click rate click inside the labour. So, just include a variable called leaving rate. So, there is since it is exogenous we do not need to show anything coming into this it is already given as an exogenous. Let us go to the equations. As soon as you click equations both these things will become kind of on the black background. So, let us click labour first and you will see the integration equation has changed net hiring rate minus leaving rate has automatically did that. So, which is what we want we want the stock to change the hiring rate minus leaving rate. So, all we have to do is ok we are done nothing to change there come to leaving rate. Let us just put a value 0 and what should be units for this persons per month it is a rate this always be per time unit we use the same time unit as the inputs ok. So, initial value of labour is 100 desired labour is also 100. So, then we can expect net hiring rate should be 0 and leaving rate also is 0. So, when we actually simulate the model the labour does not change it is always at 100, leaving rate is 0, net hiring rate is 0. This is what we mean by dynamic equilibrium. The model starting a dynamic equilibrium I think is changed. Now, so we just included this one extra variable within the model. Now, let us go to our model say that here it says leaving rate is step of 4 comma 5 people per month or person per month I guess. Find out what step command does let us do that before I change the equation. See there will be many such commands it is a pretty decent help. So, you just click help click when some manual it gets installed by default you do not need internet for it it is including a local drive. So, you should open up a screen like this search step submit if a various things in mine it comes as option 5, mine it comes as option 5 to click step. So, step command is a step you can use step function. So, the it takes a value of height at the step time until then it is 0. So, the current time is greater than the step time then it takes a value of height else it continues to remain 0. So, that is what the step function does. So, there are various such functions that is available like you know in other courses in most of you are either in the fourth year of undergrad or already master students you may stress in other courses on how to give the various inputs in the systems. So, we can have various types you know pulse inputs or a ramp input or a step functions and stuff. So, there are specific functions for each. So, this is just a simple step function. So, we want a step function of height comma time. So, pretty much what we want to model it as this input. So, this is your height and this is the time. So, value 0 until then before the time and after that it takes a value of height. So, in our example suppose step is say 4 comma 5 that means a time you can have the axis 1, 2, 6. So, step of 4 comma 5 means at time 5 I am going to get a start getting a step input of height 4. So, this is 4 units and this is going to go on. So, it can take a positive value or a negative value the height you can put a minus 4 also it will come a pulse in the downward direction. So, you can use multiple step functions to construct whatever functions you want. There are other functions like step called ramp, pulse, pulse strain etcetera which you can use to model variety of inputs in the system as required. So, let us go to the leaving rate and in the equation we need to type step. So, list of all the functions that is available is that when some supports is listed here you can see a drop down called as function and common. So, these are the common functions that is listed here, but if you want all the functions you just select all then every single function that when some supports is put here. So, you can look for step. So, once you click step the basic key also comes there. So, step of height comma start time S time. So, we can pretty much guess it. So, height is what we want is 4 and start time we want is 5 that is it as this model step function into the system. Now, let us save it and let us run the model and see what happens. Remember desired labor is 100, current labor is 100. As soon as people leave let us see what happens. Let us just click play click run let us then touch click labor click here causes strip. So, I can get all the graphs at one shot ok. So, let us see the leaving rate became a step function because of time resolution you can get you know time step is 1. So, after that you get a increase in the next time unit it looks like a RAM, but it is actually step function. So, reduce the time unit you can make a more clear step function. So, this is a leaving rate. So, as soon as that happened the same goal seeking behavior is observed where. So, to compensate for people leaving that is 4 people are leaving. So, for system to reach again equilibrium system reach equilibrium when inflow is equal to your outflow right it will be steady state again inflow is outflow. So, system is going to keep hiring people until the net hiring rate is equal to 4 when it balances out the net leaving rate right. So, that many how many people I am going to hire and that will have a nice goal seeking approach that we see here. Now, let us see what happens to our actual labor. You can see that the labor does not remain at the desired level of 100 instead it now finds a new equilibrium at a point about 70, 75, 78 I guess is where it is attaining its new equilibrium which is right here which is different from the desired goal that we achieve here. So, in presence of a constant exogenous variable for example, this outflow system will come to a equilibrium, but it is different than what the desired state will be. So, that is how system is going to behave because system will have equilibrium at that point that hiring will be equal to this the system when it reaches steady state hiring will equal your people who are leaving. So, system will reach steady state only at that point that that point system becomes balanced correct, but you are not considering any extra information in this system right. Every time only you look at the current labor and based on that you are making a decision to hire. You do not account for this information in making a decision at all within the model right. So, you looked at every time you look at the labor and say for people are short. So, I am going to hire some for people and I keep doing that then I find that I am achieving a legally very which is lower than that. I can show you analytically in a minute. So, let us go back to the graphs here the labor gap is at every time only looking at the desired labor and the current level of labor and taking their difference only and based on adjustment time trying to do the hires and system will already reach equilibrium when as soon as inflows equal to your outflows system reaches equilibrium. So, net hiring keeps increasing until it reaches 4 when it becomes equal to your net leaving rate. So, I am not taking the leaving rate information in my labor gap. So, system will achieve a different equilibrium than your desired labor though your desired is 100 based on the system will actually achieve equilibrium at a much lower level because of this constant exogenous flow. So, going back to that causes strip. So, this part here until time 5 when everything is constant there is no change. So, that phase is called as dynamic equilibrium after that say until time 28 or. So, system is what we call as in transient phase and after that it reaches steady state. So, most of our analysis. So, just to expand on that let me just increase the model setting I am just going to model setting and increasing the final time to 50. Simulating it again so that we can view it nicely oops. So, about say from time 5 to time say approximately 25 or 30 system is in transient phase and after 30 system is in steady state. So, most of our analysis especially analytical computations we end up focusing only on the steady state systems when does it be steady state or what is the value of the steady state and system and then operate during the steady state. But using simulation we need to spend sufficient time in analyzing the transient phase also saying how the transient shape is happening because that is what is causing the dynamics in the system because in reality there is hardly any system which is operating in steady state all the systems are in transient state eternally. So, we need to focus on transient phase also. So, we have dynamic equilibrium from 0 to 5 then we have transient phase and then we have steady state. At steady state you can observe that the leaving rate equal to net hiring rate or your inflows equal to your outflow rate then system is steady because then stock won't change right. Stock change will happen only if this inflows and outflows are different if they are equal change in stock is 0 because stock level saturates there and the new equilibrium value we wanted to see that is I just opened the tables to see the exact value. The new equilibrium value seems to be about 76 instead of 100 it has saturated at 76 as a new equilibrium value or the labour the desired labour saturated at 76. So, let us see why that happens. So, steady state value of net hiring rate became 4 percent per month and value of labour saturated at 76 percent. So, let us look at this behaviour of steady state system or simple linear systems analytically to better understand it. Pretender is going to flip the direction. Suppose there is a constant inflow and I have a negative feedback loop in the outflow. So, in phase of a constant inflow system is correcting the discrepancy's from the goal in the state of system to as per fraction per time f it corrects it. So, for this system let us try to compute it then you can use the same analogy to do the math for the example that you just saw. So, let us assume the same system that is one inflow and an outflow. So, outflow is a negative feedback system inflow is a exogenous. So, we have a constant k. So, let us write the equations we get d s by d t is equal to rate 1 minus rate 2 rate 1 is k rate 2 we are going to adjust the discrepancy. So, let us call it f into discrepancy f is a fraction per time that is f I am getting f from here. So, rate 2 is fraction per time into discrepancy k minus f into s minus s star. So, let us say that f is a fraction per time so this is the system that I am simulating. When 2 flows affect a stock when you did graphical integration what did we do? We took the net rate when 2 flows affect a stock we just take the net rate and if stock does not need to change the net rate has to be 0. So, at steady state the net rate is going to be 0 2 flows affect stock net rate here net rate net rate is same as this. So, whatever you see here at steady state or equilibrium net rate will be equal to 0 because in balancing inflow is equal to your odd flows that means k minus f into s minus s star is going to balance itself out going to be equal to 0 means I want to write it in terms of s what is the new value of stock at which I am going to balance this out. So, if I rewrite it I can get s is equal to s star plus k by f. So, this is the new value in which system is going to achieve its equilibrium. So, yes. So, this is the value of stock at equilibrium or steady state this is your goal case your constant inflow f is a fraction per time in which I am adjusting the discrepancy. When yesterday we saw another expression adjustment time is equal to 1 by f. So, in that case I can rewrite my expression as s star plus k into adjustment time same thing. You can try it out for the example that we did there you might you will find that s is s star minus of the exogenous. So, the new level will be less than your desired goal by that in proportion to your inflow size. In this particular example. So, suppose initial value is 200 and s star is 100 and f is 0.2 what will be equilibrium value there is value of stock in steady state when k equal to 5 units per time when k is 0 what will be equilibrium steady state value of stock will be 100. Yes, it will reach s star when k is equal to 5, 125 yeah it will saturate at 125 itself that is at a value higher than the goal in case of there is a constant input. So, the input is coming with you are trying to get rid of it, but it just keeps coming all the time. So, that accumulates over time which results in not saturating at 105, but stops at 125 in proportion to the rate at which I am able to adjust the discrepancy because average time to adjust the discrepancy is 1 over f which is 5 and 5 times k which becomes 25. So, new steady state value becomes 100 plus 25 is 127.