 Hello and welcome. So, in this lecture, we will explore the possibility of an approximate solution strategy which still provides a reasonably optimal solution, but with smaller computational effort. We will also look at the overall implication of the methodology and its relation to the loss of accuracy in the given solution. So, let us begin. Let us proceed with the approximate staging solution procedure and understand the elements that provide us with this particular simplification. As was mentioned in the previous lecture, it is an alternative to the Lagrange technique which we have seen earlier in which we drop one equation and solve the residual n cross n system of equations. This is the first step, which means that among the n plus 1 equations and n plus 1 unknowns, we drop one equation so that we do not have to drop the or we do not have to actually include the Lagrange variable lambda that we have been doing earlier. We realize that this methodology is going to provide suboptimal solutions because now it is no longer an exact formulation. However, you will find that in many cases, we can use these solutions to initiate a more rigorous design iteration which in many cases require an initial guess of the solution. The question is, how does one decide on the equation to drop? While there can be many options for dropping one equation, this can be achieved exactly if one of the partial derivatives is zero throughout the design space. Now, this is something which is not really very clearly visible at the stage of formulation but if such a thing happened and we were able to pick that particular partial derivative which was zero throughout the design space, then that equation is automatically exactly specified and satisfied at all the places which obviously means that if we drop this partial derivative and the equation corresponding to it, there will be no error committed as far as the solution is concerned because all other points, the remaining ones will together pick a point which is optimal from their perspective and it is optimal also from this particular design variable because anyway this is the minimum value that this design variable has. There are two aspects that become bottlenecked in this case. One, there is no way of knowing upfront which of those equations will result in a zero partial derivative throughout the design space. The second point is the design space may have multiple optima in which case it is possible that we may pick a solution close to a local optimum but may completely miss a global optimum. However, with these two limitations, still the methodology has a reasonable value in trying to reduce the computational effort and still giving a reasonably good initial guess in situations where there is only one optimum in the design space. So, which means if there are functions which are such that there is only one maxima or a minimum in the design space, then these methodologies can still work with reasonable degree of accuracy provided we hit upon the right partial derivative equation to drop. The reason why this is true is that in such a case solution obtained from the remaining equations will automatically become optimal. We must realize that the sensitivity of objective function to design variable is not same for all the partial derivatives particularly in a general context. So, obviously depending upon the equation that we draw, we may be somewhere in the vicinity of the exact optimal point. As I mentioned, these are approximate but in many cases can serve as good starting points for a more rigorous design exercise. Now, in order to ensure that the loss of accuracy is not excessive, we satisfy the constraint exactly which means that the constraint is always exactly satisfied while one of the partial derivatives is probably only approximately satisfied. And this gives us another benefit that the constraint equation can be used to substitute one design variable as a function of the remaining n minus 1 design variables. The benefit that this gives is that the remaining partial derivatives are only n minus 1. So, effectively we need to evaluate only n minus 1 partial derivatives and not n partial derivatives. And these n minus 1 partial derivative equations in terms of n minus 1 design variables can be solved simultaneously to obtain solution for n minus 1 design variables. And once we do that, we substitute this solution back into the one design variable which is a function of the remaining n minus 1 design variables and in the process we get solution for all the n design variables. Let me show you the basic formulation that this strategy is connected to. So, let us assume that we are going to choose pi 1 to be the design variable for which the partial derivative will not be evaluated. So, now what I do is I take this and express pi 1 as a function of the remaining pi s that is from pi 2 to pi n. So, and this is my constraint relation. Here of course, I have used the velocity constraint. So, v star is coming here and then I write my pi star which is product of all the pi i s but because I have removed pi 1 it will be a function of only pi 2 to pi n. And that is why I need to evaluate only those n minus 1 partial derivatives going from j equal to 2 to n. So, these represent only n minus 1 algebraic equations and the nth algebraic equation is my constraint relation in which once I evaluate the n minus 1 design variables pi 2 to pi n I will substitute back into this and evaluate pi 1. When I am talking about the pi star as the constraint then again I use this expression that is ln pi 1 equal to ln pi star minus this. And this is the equation which now I am going to substitute in my objective function for pi star and the objective function which is my v star. So, v star will contain only pi 2 to pi n and then I differentiate this again with j going from 2 to n. So, I only have n minus 1 partial derivatives. So, I solve for these n minus 1 partial derivatives to solve for pi 2 to pi n and substitute back into this constraint relation to obtain pi 1 value. Let us demonstrate this idea through the example that we have seen in the previous lecture. So, again let us consider a two-stage rocket with the same epsilon that is 0.15 and the same ISP of 240 seconds and I would like the velocity to be 4000 meters per second. You recall that we had done this exercise for a two-stage sounding rocket and we had obtained a solution. Please keep those solutions in mind because now we are going to see what this particular methodology gives as far as those solutions are concerned. So, the solution is as follows. We first express pi 1 in terms of pi 2 because that is the only other unknown. So, I can solve for pi 1 from this. So, the following is the expression for pi 1 that is 0.2152 divided by 0.15 plus 0.85 pi 2 minus 0.1765 that is the solution of pi 1 in terms of pi 2. My ln pi star is ln of pi 1 into pi 2. So, this ln pi star is function of only pi 2 because it is only function of a single variable. I need only one partial derivative of pi star with respect to pi 2 which I evaluate and then I get a constraint relation saying that my pi star is this expression and with that expression I get a quadratic equation for pi 2. Again, you will realize that because your pi star derivative is going to give us the quadratic equation, only one of the pi 2 solutions will be consistent, other pi 2 solution will be inconsistent. I have not mentioned the inconsistent solution but I am mentioning the consistent solution and you immediately see that the pi 2 turns out to be 0.327 and correspondingly if I take this pi 2 and substitute into my constraint solution of pi 1, I get pi 1 as 0.326 and you realize that I have got an exact solution. The exact solution for this problem through the Lagrange multiplier method was 0.327 for both the stages. Here, you can see there is a marginal leakage possibly because of the truncation error in the decimal places. Let us now do the same exercise for unequal stages. So, we have the problem of Angara 1.2 and let us redesign this to have a burnout velocity of 8338 with the fixed parameters. So, it is a velocity constraint problem. Let us determine the approximate stage-wise payload ratios. So, the solution in this case is as follows. So, again I take the pi 1 and substitute into the constraint relation with the velocity of 8338 meters per second and get an expression of pi 1 in terms of pi 2 as 0.0694 divided by 0.089 plus 0.911 pi 2 to the power 1.1048 minus 0.0776. We define pi star as product of these two, differentiate this with this 2 pi 2. I have skipped those intermediate steps but my suggestion is that you verify those steps. And once you verify those steps, you can obtain the solution for pi 2 as 0.19 and if you take this 0.19 as pi 2 and substitute into pi 1 relation, you can show that pi 1 will be 0.22. Now, how do these compare with the actual values? I leave this exercise to you but you will find that these numbers are not exactly equal to the pi 1 and pi 2 numbers that we had obtained for this problem using the Lagrange multiplier method. Now, option 1 and option 2 as you can see are by interchanging the variables and you will see that there is now a marginal difference between these two solutions. In one case, if you get pi 2 in terms of pi 1 as an x constraint and then differentiate with respect to pi 1, you are actually ignoring the partial derivative corresponding to pi 2. Whereas in the other case, you are ignoring partial derivative with respect to pi 1. You will realize that these two provide two different solutions as we had anticipated that as a sensitivity of the partial derivative for different design variables is going to be different. You are going to get different solutions depending upon which particular partial derivative you drop. So, to summarize, the approximate staging solution method simplifies the solution steps that is clear to us. However, we also see that it results in loss of accuracy and moreover, that loss of accuracy is also dependent on the equation which is ignored. In this lecture, we have seen the methodology for obtaining the approximate solution by dropping one partial derivative equation and we realize that while we get a solution which is in the vicinity of the actual optimal solution, it is not really the optimal solution and that it varies depending upon which particular design variable you choose for dropping the partial derivative. However, we still note that within the context of the objective for this exercise, we still get a reasonably good solution at the starting point for any rigorous design exercise which would ultimately give us the exact optimal solutions. With this, we have completed our discussion on the optimal solutions for staging. In the next lecture, we will look at a related concept called the concept of variant of a launch vehicle which is an extremely important practical concept which is commonly practiced by most space agencies to minimize their cost of inventory as well as cost of launch vehicle development. So, bye. See you in the next lecture and thank you.