 Welcome back. So, in the last lecture we had discussed the operation of a multistage rockets is performance and we derived an expression for the velocity increment for a multistage rocket. What we have shown is that the overall increment of the velocity of a multistage rocket is the algebraic sum of the increment attained by individual stages, where subscript i represents the stage and n is the total number of stages. And since we know that for a given stage the velocity increment is function of equivalent velocity and the mass ratio given like this. Therefore, the total velocity increment can be written like this and now the equivalent velocity we have shown again and again is the function of specific impulse of that stage. So, therefore, this is equal to i 1 to n i s p i g e l n r i, where i s p i represents the specific impulse of individual stages. We have shown this last time apart from that we had shown some relationships which we said will be useful in optimizing the performance. So, before we go to optimization let us recap what we had discussed and list the relationships that we had obtained. So, one such relationship was relating r i to the payload factor lambda i and the structural coefficient epsilon i for different stages. We had shown this before where lambda i for a multistage rocket is the payload for ith stage which is equal to the initial mass for the i plus 1th stage divided by all the mass of the i th stage except the payload. So, it is given like this. So, this is the payload factor for i th stage. Similarly, we had looked at the structural coefficient for the i th stage which is defined as m s i divided by m naught i minus m naught i plus 1. So, this is the structural mass of the i th stage, the initial mass of the i th stage minus the payload for the i th stage. So, this is the structural coefficient. We had also defined the payload mass fraction for i th stage as m l i divided by m naught i, where m l i is the payload mass for the i th stage and m naught i is the total mass for the initial mass for the i th stage. So, once again since this is the initial mass for the i plus 1th stage, we can write this as m naught i plus 1 by m naught i. So, we had defined these parameters and we had proved that r i is equal to 1 upon l i 1 minus epsilon i plus epsilon i. So, these are the things that we had discussed in the last class. Now, let us continue from here. What we want to do today is optimization of multi stage rocket. Today's topic is multi stage optimization. Now, what is our goal when we are designing a rocket? The goal is to get to the most economical design, which attains a given mission. So, first of all to attain a given mission is the most important thing. We want to attain it in the most economical way. So, this is most economical design. We have gone to get the most economical design to attain a given mission. So, this is our mission objective or design objective. So, when we are doing optimization, these are the things that we have to keep in mind. So, the given mission becomes our constraint and the economical design is our goal. So, therefore, what is the overall goal for the design effort? First of all again for a given mission. Now, what is a given mission? Important to know what is the given mission? What is the program director has given a designer? The given mission is to put a given payload into a given orbit. Now, that payload when it is in a particular orbit, in order to remain in that orbit, it must be delivered with the given velocity only then it will take that orbit and remain there. So, therefore, the given mission as far as the designers point of view is considered can be considered to be a given velocity increment delta u and the given payload ml. So, this is my given mission is overall delta u and overall ml, the velocity increment and the payload. So, this is the given mission. The overall goal is that what should be the relative sizes of these stages in order to minimize the initial mass. Let me explain what I mean by this. This is the design objective. We have to attain this mission. We have to get a certain delta u and certain payload we have to deliver, but we want to do it in such a manner that our initial mass is minimum. If we have to carry the initial mass, the lowest initial mass, therefore, if I look at the mass ratio, overall mass ratio or this factor r, this factor r is going to be higher. So, therefore, we get higher delta v or if I reduce this, it is more economical operation. So, therefore, if I have to if I can reduce the initial mass, we can now reduction in initial mass keeping the payload constant means either we reduce the structural mass or the propellant mass. If you are reducing the structural mass, we have to carry less. If you are reducing the propellant mass, we have to carry less propellant at the same time, we are saving on the propellant cost. So, both of them taken together essentially improves the rather economy of the operation. Now, how do we attain that without compromising on this? That is the crux of this optimization of design. We do not want to compromise on this, but we want to have the minimum initial mass. We can attain that by distributing this masses, relative sizes of different stages. We can distribute the structural mass and the propellant mass in such a way among different stages that the final overall mass of the rocket is minimum. So, that is the design of objective. Equivalently, we can restate this that this is equivalent to essentially finding L i's. L i is our structural sorry payload mass fraction. So, it is equivalent to finding the L i's such that delta u is maximum for a given overall L. So, this statement here that we want to minimize the initial mass is actually equivalent to finding the payload mass fraction. For different stages, the optimum payload mass fraction for different stages such that the velocity increment is maximum for the given overall payload ratio. So, essentially here what we are saying is that we keep this constant and M L is constant. We can reduce M naught so that we get the best possible performance. Here what we are saying that this is equivalent to saying that we have the given M L, we have the given initial mass M naught. What will be the distribution that will give us the maximum delta u? Now, I am claiming that these two are equivalent statement. Let us now see whether it is truly correct or not are these two equivalent statement. For that what I am going to do is I will take an example. So, to show this equivalent of statement let me consider a two stage rocket. I am considering a two stage rocket. Now, for this rocket since it is a two stage rocket the overall payload mass fraction is given as the product of the payload mass fraction for the first and second stage. This we have proved in the last lecture. Now, for this rocket let us say what we do is let us say for a given value of L 1 what we do is we consider a value of L different values of L. So, let us say we consider L equal to C 1 or L equal to C 2 L equal to C 3 etcetera. So, let us say for a given value of L for example, let us consider L equal to C 1 is a constant value of L. So, overall mass fraction is constant what we do is we vary L 1 and then estimate the corresponding delta u and plot it. So, we see that the variation will look something like this. This is for L equal to C 1. Let us now increase L say C 1 is less than C 2 is less than C 3. So, now we increase L and again repeat the same thing that is we vary L 1 and estimate L 2 for this given L and plot the variation in delta u. So, this will be like this and like this. This is L equal to C 2 L equal to C 3. Now, this is our increasing L. Let us see now on this curve. What is happening? First of all let us look at the variation of for a given L 1. Let us say let us look at the variation of L. What we are seeing here is that as we are increasing L as we are increasing L delta u is decreasing. So, as L increases delta u decreases. This is seen for all the values of L. Now, how we are increasing L? How do we increase L? If you have to keep our the final payload mass M L to be given we do not want to change this. Then the only way we can increase L is by reducing M naught. So, here what we are seeing is that if we have to increase L, if we increase L delta u decreases or in other words if M naught is increased keeping M L constant delta u decreases coming to this then this statement. Here we want to keep this constant. We want to minimize the initial mass delta u M naught. Therefore, delta u will decrease. Now, the question is but we want to keep this same. How can we do that? By we are decreasing it at the same time we want to keep this same. Now, that is possible if we work along the horizontal line like this. If I work here let us along this horizontal line. At this point this is L equal to C 1. From here and to if I come to this line I have reduced L in increased L which means I have reduced M naught initial mass I have reduced, but I have kept the same delta u. So, if I come from here to here to let us say this point I have achieved this. See attaining that delta u was by goal M L was given to me which I have to keep constant. So, I have achieved this statement by moving like this. Now, this is one of the conditions of the first statement that we have looked at. Now, this value here then which we are getting here is now if I look at this point this point is optimum L 1 for given delta u. Now, if I look at this delta u value and compare it with L 1 what we see is that if let us say if we are on this curve this value of delta u L 1 is here, but as I go up there is one point here at which L 1 there is a value of L 1 for which delta u is maximum. If I increase L 1 beyond that delta u starts to drop similarly for this L up to this point if I increase L 1 delta u increases is reaches this maximum then delta u starts to drop same here. So, therefore, if I look at this point then for this value of delta u this is the optimum delta L 1. So, for any value of now instead of plotting like this if I take any value of delta u for that value of delta u there is a value of L 1 which will with the optimum for a given value of delta u and therefore, this is the value we are trying to find. What we are getting seeing here first of all that when we are increasing L when sorry when we are decreasing L delta u is increasing right when we are increasing L delta u is increasing sorry when we are decreasing L delta u is increasing and that increase is that because how do we increase L by reducing just a second from here to here L is decreasing. When we are decreasing L for this constant we are actually increasing M naught and then that increase essentially is because of increase in propellant mass. So, we are having more propellant here. So, if you are burning more propellant we get higher velocity right. So, therefore, this is obvious that increasing the propellant mass will give us higher velocity on the other hand decreasing L will reduce to reduction in will lead to reduction in propellant mass therefore, delta u will decrease. So, now from this curve we see that there is a point here this point this point here corresponds to both this statement right. First since we are moving horizontally here it satisfies the first statement if I look at this statement we are finding a value of L i this is this optimum L i L 1 right. Now, from this plot if I can get L 1 and my L is fixed here L is equal to C 2 therefore, L 2 is equal. So, L 1 is obtained from here L is equal to C 2 according to this curve. So, therefore, L 2 is equal to L by L 1 which is equal to C 2 by L 1. So, L 2 is already fixed once we have chosen this point right. So, therefore, what we see is that this point here also corresponds to a distribution of L i that is L 1 and L 2 such that delta u is maximum at this point for this given value of L. So, therefore, this discussion shows that both this statement are equivalent and that is what we want to use because minimizing initial mass essentially is something that is difficult to work with because then our expression for velocity increment it includes initial mass because that becomes more complex whereas, the expression for delta u is pretty straight forward right. So, if we say that our objective function is maximizing delta u it is a straight forward approach. So, therefore, we will look at this statement that we will find out different distribution for which delta u is maximum. So, this discussion is for a two stage rocket if we generalize it for n stage then we will have an n minus one dimensional plot this is a one dimensional plot if we increase it to n stages there will be n minus one dimensional plot similar plot can be obtained. Now, with this discussion where we have shown that the multi stage optimization essentially leads to fulfilling this requirement our objective function then is maximizing delta u and the constraints are either this or the overall because m l is known. So, having the overall payload fraction these are the constraints. So, we have now defined the problem let us look at the mathematical way to optimize this problem for that we will consider some specific cases. So, first and foremost let me consider a case where we have say n stages with certain conditions. So, first let us look at case once in case one what we will say is that let us say all the stages all n stages all have same equivalent velocity which essentially means that u equivalent i is equal to u equivalent and we know that equivalent velocity essentially is ISP which implies that ISP for all the stages are same specific impulse for all the stages are same. So, for this then we will have another condition that we have same structural coefficient for all the stages which essentially means that epsilon i is equal to epsilon all the stages have same structural coefficient. Then we want to get the distribution of L i or lambda i which will give us the maximum delta u. So, for this delta u i which is the velocity increment for ith stage is equal to u equivalent l n 1 plus lambda i upon lambda i plus epsilon right this is something that we have shown before. Now, if that is the case then for the n stages for the all the stages the velocity increment total velocity increment is delta u which is equal to sigma i equal to 1 to n u equivalent l n 1 plus lambda i upon lambda i plus epsilon right epsilon is epsilon i is constant therefore, we write it as epsilon only. Now, in this equation u equivalent is constant. So, we can take it out of the summation sign and then take it to the left hand side then we can write delta u by u equivalent is equal to sigma i equal to 1 to n l n 1 plus lambda i lambda i plus epsilon. So, in this expression the only dependence on the stage comes through lambda i right every other parameter is independent of the stage. This then is essentially a function of lambda i nothing else. So, we can write this then as sigma i equal to 1 to n a function of lambda i let me call this first of all let me call this as equation 1 let me call this equation 2. Here what is f lambda i f lambda i is equal to only this term 1 plus lambda i upon lambda i plus epsilon. So, here f lambda i is a function of only lambda i given as 1 plus lambda i upon lambda i plus epsilon. Now, before we progress proceed further I would like to derive another expression relating l and lambda i why we want to do that because we have a constraint on l like we have discussed here the product of all l i is equal to l. So, we have a constraint at l. So, therefore let us first derive an expression relating l and lambda i because here our objective function is in terms of lambda i. So, let us first define get an expression relating l and lambda i for that let me first see what is lambda i lambda i lambda by definition is m l upon m naught minus m l that is how we have defined lambda. If I add 1 to this then this is equal to m l minus plus m l upon m naught minus m l. So, m l will cancel off right. So, therefore, 1 plus lambda is equal to m naught upon m naught minus m l. Then if I do now lambda upon 1 plus lambda then that will be equal to m l upon m naught minus m l divided by m naught upon m naught minus m l. You can see that the function numerator denominator for both of this will cancel off. Then what we will be left with is equal to m l upon m naught and that is equal to l the payload my fraction. Therefore, this shows that l is equal to lambda upon 1 plus lambda which is now applicable to any stage. Therefore, I can write that l i equal to lambda i upon 1 plus lambda i. So, this relationship is going to be useful when we try to maximize this delta u. So, now let me look at the expression for l overall l. We have proved the relationship between l and lambda i. Let us now look for l. l is our overall payload fraction. Now, this is something as far as our optimization is concerned is the given quantity. So, this is equal to m l upon m naught. This is given to us. We know that we have proved that l is equal to the product of the payload mass fraction for all individual i stages. Therefore, l is equal to pi i equal to 1 to n l i. Now, if I put this relationship into this equation then this is equal to pi i equal to 1 to m lambda i upon 1 plus lambda i. Now, this is now our constraint because we have said that this relationship is always valid for any multi stage rocket. Therefore, this must be satisfied at the same time we are saying that l is given to us. So, then this becomes a constraint. So, this relationship now what we are saying is that we want to maximize this delta u, but this maximization must be according to this rule. We cannot violate this at the same time we have to maximize delta u. So, how do we achieve that? What we do is we will form a different function. Before doing that I will simplify this expression little more. What I will do is I will take logarithm of l l n l equal to now this is a product. So, logarithm of product is essentially sum of logarithm of individual variables. Therefore, l n l will be equal to sigma i equal to 1 to l l n lambda i plus 1 upon lambda i. So, now what we get here is l n l is equal to the sum of all of this function lambda i upon 1 plus lambda i. Let me say that we define another function g of lambda i which is nothing, but lambda i 1 plus lambda i. We are defining this as a function. Then this constraint that we have now can be written as l n l equal to sigma i equal to 1 to n. So, now what we are having here is our what we want to maximize is delta u which is nothing but sigma i equal to 1 to n f lambda i where f is a function of lambda i. Here this is our constraint that l n l equal to i equal to 1 to m g lambda i where g is again a function of only lambda i. So, now what we can do is we can define a new function which includes both of this. So, we will define this because essentially what we want to do is we are trying to maximize this. This is what we are trying to maximize with this constraint. So, we will define a new function let us call that l which is also a function of lambda i which includes both this maximization variable as well as the constraint. So, if I define a new variable here or the sorry a new function here we can write it as l lambda i. So, this is just an algebraic sum of f lambda i plus some fraction alpha times g lambda i. What we can see is that there is a linear relationship relating this to get a new function l. Now, this equation again let me call this as 3 and let me call this equation as equation 4. Here this parameter alpha is called Lagrange multiplier. This is an undetermined constant. This is an undetermined constant Lagrange's multiplier. Now, when we combine all this we get this equation as we see that is a function of only lambda i. There is a function l lambda i which is the sum of f lambda i plus alpha g lambda i. So, alpha is a variable or a parameter which will actually dictate what is going to be the maximum value for this rather the what will be the maximum combination will be dictated by alpha. So, we can maximize this equation. Once we maximize this we will find a relationship between alpha and lambda i. So, then for a given value of lambda i we get the alpha. So, that is how once we get the alpha we had the corresponding best possible combination of lambda i. That is what is the optimization all about. So, essentially now this becomes our objective function l lambda i which we want to maximize. So, for a maximum what we do is we will differential l with respect to lambda i. Now, as we can see that is a linear function this and this are adding together the linear operator. Therefore, d l d lambda i is nothing but d f d lambda i plus alpha d g d lambda i. So, in order to maximize then what we do is we differentiate this expression with respect to lambda i and then we equate it to 0 that gives us the maximum. So, now what we have is essentially d f d lambda i plus alpha d g d lambda i equal to 0 where our f is given by this and our g is given by this. So, now let us look at the first term here d f d lambda i. This is equal to nothing but d d lambda i of this term 1 plus lambda i upon lambda i plus epsilon. So, now we can differentiate this with respect to lambda i this will be equal to 1 upon 1 plus lambda i minus 1 upon epsilon plus lambda i. So, if I differentiate this is what I am going to get. So, now the first term here is taken care of let us look at the second term. The second term here is this g lambda i. Multiplied with an alpha. So, first of all what we do is let us get d g d lambda i. So, in this expression let me first write g lambda i then we will do d g d lambda i. So, g lambda i is equal to lambda i 1 plus lambda i sorry here one thing is missing l and z is equal to l n sign. I forgot to write the l n here there is an l n that is why we get this expansion. Sorry this thing is missing from this expression there will be a normal coming here in this expression. So, this will have an l n term. So, now here also we have l n d lambda i l n lambda i 1 plus lambda i. Now, if I do the if I differentiate this expression with respect to lambda i I will get this is equal to sorry d g d lambda i equal to d d lambda i of l n lambda i 1 plus lambda i. So, this is equal to then 1 upon lambda i minus 1 upon 1 plus lambda i. So, after differentiating we get this expression. Now, let us take this and this and put it back into this expression for maximization. Then what we get is 1 upon 1 plus lambda i minus 1 upon epsilon plus lambda i plus alpha upon lambda i minus alpha 1 plus lambda i is equal to 0. Let me call this equation 5. We got it by putting this and that expression back into this equation. Now, we have this expression here what we see is that is a function of essentially relationship between lambda i and alpha for constant value of epsilon. So, this can be reduced to then after some algebraic manipulation to alpha epsilon 1 minus alpha minus epsilon. Let me call this equation 6. So, the optimum value of alpha will be obtained for like this. So, now if lambda i is given we can get it like this. So, this expression then gives us the value of alpha or rather if alpha is known we get lambda i. If lambda i is known we get alpha and I will come to that how do we get this two different parameters little later. So, just to conclude what we discussed today is that this is our expression for the final velocity that we get which in terms of lambda i and epsilon is given like this. What we say is that this term here within the summation sign is a function of lambda i. We defined it as f lambda i equal to l n 1 plus lambda i upon epsilon plus lambda i. Therefore, if you maximize this we maximize delta u as well. Now, this is our objective function and then we had the condition that the product of l i is equal to l overall payload fraction. From there we got relationship relating g and lambda i. Then what we say is this is our constraint. So, the objective function and the constraint together multiplied by a Lagrange multiplier gives us a new objective function which we want to maximize. So, in order to maximize this we differentiate this expression with respect to lambda i. Well, lambda i is our variable differentiate this and set it equal to 0. Once we do that we get this expression which after simplifying gives us lambda i as a function of epsilon alpha and epsilon. Therefore, this is the optimum combination that we are looking for because now the objective function is optimized with the constraint. So, this gives us this is the method for optimization and here what we have considered is a specific case where equivalent velocity was constant for all the stages. That means, we have the same specific impulse for all the stages and the structural coefficient was considered to be same for all the stages. So, under this assumptions this is the optimization. So, let us stop this lecture right now. In the next lecture we will continue from here and we will discuss little more on the multi stage optimization. We will look at some other cases as well. Thank you.