 Hello and welcome. In this last tutorial that is tutorial number 4, we will look at the optimal staging configuration solution using the Lagrange multiplier method and we will understand the solution steps involved in the process. So, let us begin. Let us begin our discussion with the simplified problem of equal stages. So, in this case, let us consider a two stage sounding rocket which has a structural ratio for both the stages as 0.2. Let us try and determine the optimal stage payload ratios that is pi s and the lift of mass m naught for an m star or the payload of 20 kg, if the ideal burnout velocity required in this case is 6000 meters per second and it is burning a propellant of I sp equal to 200 seconds. So, let us start the solution. So, we first calculate the parameter beta which is nothing but V star divided by n g naught I sp where V star is the desired velocity and are the number of stages and I sp are the specific impulses. So, when we substitute these we get beta as 1.5 to 9 0. Here you must keep in mind the fact that as we are going to use exponential functions the number of significant digits have to be sufficient in order for us to capture the accuracy from such manipulations otherwise we may get an erroneous solution. So, in this case we have gone up to 4 decimal places and then we calculate the stage payload ratio that is pi 1 and pi 2 as e to the power minus beta minus epsilon divided by 1 minus epsilon. In the present case the epsilon is 0.2 and when we perform this calculation we find that the two stage payload ratios are 0.0209 fairly small value. Next we talk about the mission payload fraction pi star which is nothing but the stage payload ratio raised to the power n where n is number of stages. So, in this case as there are two stages we get 0.0209 square which is 0.00044 and from this using the payload requirement of 20 kg we find that the lift of mass is 45610 kg or 45.6 tons. What it means is that our payload efficiency in this case is very small. So, just to launch a 20 kg payload you are going to require a 45 tons rocket. Let us now look at problem of a two stage rocket again with the equal stages, but this time we use the payload as the constraint saying that payload is very small. Let us try and find out what is the burnout velocity that we can get but and also we have this time increase the ISP from 200 to 350. So, we have improved the structure from 0.2 to 0.1 so it has become more efficient it can carry more propellant and we have improved the ISP from 200 to 350 and we also have a requirement of a significantly higher payload ratio of 0.1 and we also have a significantly higher requirement of payload of 100 kg. Now indirectly what we are saying is that we do not want the rocket lift of mass to be more than 1 ton as against the 45 tons that we saw in the previous case we would like our rocket mass to be only 1 ton but we are willing to sacrifice the burnout velocity. Let us proceed with the solution. So, in this case the equal stage payload ratio from our formula that we have seen in the lectures is square root of pi square which is square root of 0.1 and we find that the stage payload ratios have to be 0.316. Now with this stage payload ratio we find that B star or the ideal burnout velocity which is minus g0 ISP into number of stages ln of epsilon plus 1 minus epsilon pi if we substitute these numbers we find that we get a velocity of 6561.7 meter second which is a surprise. If we compare these two problems we find that in the previous problem a structural efficiency or ratio of 0.2 and an ISP of 200 when it was changed to such a efficiency of 0.1 and the ISP of 350 we are able to launch a 100 kg payload with 1 ton rocket and which also has a velocity which is higher than 6000 meters per second which obviously means that by small increases in the structural efficiency and small increases in propulsion technology it is possible for us to significantly improve the efficiency of mission and you will find that the optimization techniques that we talk about essentially aim to look at slightly better structure slightly better propulsion in order for us to have significantly better gains in terms of the launch cost which is typically expressed in terms of the payload fraction which is so many kgs of payload per kg of lift of mass. Let us now move over to a problem where we are going to have unequal stages which means the stages are not exactly equal. So, in this case I have taken a three stage rocket which has marginally different ISPs and also marginally different structural efficiencies or structural ratios. You will find that these are typically the numbers that we would be getting in most practical rocket configurations and our objective is to determine the stage wise payload ratios which will maximize the mission payload ratio such that a 122 kg payload is given a velocity of 7800 meters per second. In this context it might be worth noting that the velocity which is specified in this problem corresponds to the spacecraft forming a circular orbit at 200 kilometer altitude. So, this is how the spacecraft mission requirements would appear in the form of rocket configuration design. Let us proceed with the solution. So, first let us look at the basic formulation. So, let us recall the solution for the stage payload ratio in the present case which is minus epsilon i divided by 1 minus epsilon i into 1 plus lambda G naught ISPI where lambda is the Lagrange multiplier and using the parameters that have been specified in the previous slide. Let us now write down pi 1, pi 2 and pi 3 in terms of lambda. So, pi 1 is minus 0.142 divided by 0.856 into 1 plus 2658.5 into lambda which is nothing but this is the G naught ISP. Similarly, for pi 2 we get minus 0.157 divided by 0.843 into 1 plus 2580 lambda and pi 3 is minus 0.134 by 0.866 into 1 plus 2589.8 lambda. The next step is to bring in the velocity constraint relation. So, here I suggest that you go and look up in the lecture the velocity constraint relation which is given as V star equal to minus G naught sum over ISPI into ln of epsilon i lambda G naught ISPI divided by 1 plus lambda G naught ISPI. This is the expression that you can verify from the lecture. What we do is we substitute the applicable parameters into that expression and we do one more step. We divide the left hand side that is the velocity with minus G naught and also divide by the ISP 1 that is 271. So, if we do that on the left hand side we are going to get minus 2.937 and then on the right hand side what we are going to get will be the ratios of the two ISPs with the ISP 1 and that resulting ratio of the two stages is 0.945. This you can independently verify and then of course, we have epsilon 1 G naught ISP 1, epsilon 2 G naught ISP 2 and epsilon 3 G naught ISP 3 as the numerator and 1 plus G naught ISP lambda for first stage 1 plus G naught ISP 2 lambda and 1 plus G naught ISP 3 lambda. So, this is now the constrained relation that we must first solve to generate the value of the Lagrange multiplier lambda. Let us now look at how we are going to get this. So, what we do is first of all we remove the natural logarithm from right hand side by taking the exponential on the left hand side that gives us 0.0532 and then of course, we perform all those multiplications. So, we start getting large numbers. Here again I must emphasize the need to retain all the significant digits because as large numbers are involved any truncation of number here is going to lead to large errors. So, you see that when we expand the denominator we are retaining the terms of the almost up to eighth or ninth decimal place so that the significant digits are all retained in the expansion. This results in the following cubic algebraic equation in lambda that is 1.681754 into the power 10 lambda q plus 204 to 5597.3 lambda square plus 7828.3 lambda plus 1 equal to 0. Now, any standard numerical solver can be used including MATLAB to extract the roots. Please note that as this is the cubic equation there are going to be three roots of which only one of them will be the valid or the feasible solution root. So, in this case we find that the one real root lambda is minus 0.6145 into 10 to the power minus 3 while the other two roots appear in the form of a complex conjugate and hence they are invalid. What we now do is take this value of lambda and go back to these three expressions of pi 1, pi 2 and pi 3. Now, you can clearly see that lambda is a negative quantity. So, when you multiply this you are going to get a minus number which is greater than 1 that can be confirmed. So, this denominator will become negative this negative will cancel the numerator and we will get a positive value number for pi 1, pi 2 and pi 3 as shown below. So, pi 1 turns out to be 0.262, pi 2 turns out to be 0.318 and pi 3 turns out to be 0.262. We multiply all these three to get pi star which is 0.0218 and from this we can calculate the lift of mass for 122 kg payload as about 5600 kg. Hi. So, in this problem we have seen that unequal stages problem where both epsilon and ISPs are different in different stages. The numerical effort is primarily in setting up the equation for the Lagrange multiplier lambda and you will find that the order of that equation will be equal to the number of stages and that setting up that problem and numerical solution will require some effort which is significantly larger than when we have equal stages. Now, I will make a suggestion to all of you. Take problem number 3 the way we have defined and look at the structural ratio same for all the three stages as the geometric mean of the three values given in this example which means we have given epsilon 1, epsilon 2, epsilon 3. You multiply all the three take cube root that is the geometric mean and use this in all the three stages. Similarly, take the three ISP values and take their arithmetic mean that is some all of these and divide by 3. Please note these actions carefully and now assuming that these are to be used as constants for all the three stages generate the solution of the stages in terms of the bias for all the three stages and check that value with respect to the geometric mean of the three pi values that we have obtained in this problem. I am sure you will get some surprising results. I hope you can go on to this journey and discover that how the simplified formulations and solution techniques that we have developed for optimal configuration can provide realistic and practical rocket configurations with not much computational effort. With that we come to the end of this tutorial. So, bye and thank you.