 Hello, let us have a look at how we estimate the design testing and evaluation or the DTE cost and production cost for military aircraft and as an example, we have taken RQ-4B Block 10 Global Hawk aircraft. At this point, I would like you to ensure that you have watched the video clips of the lecture on estimation of acquisition cost of the aircraft because the example of Global Hawk was shown there and we are going to just take up the details of that particular calculation here. So, like every time we follow the same color scheme in this presentation, the general instructions are going to be given in brown color. The specified values will be in black color. These are the specs of the aircraft available from some reliable source. There are some values which will be assumed and those values will be shown throughout in this light blue color. Wherever there are some calculations to be done, we will have some question marks in red color and there will be this pause symbol. So, whenever you see this symbol, it is a good idea if you pause the video and do the calculations because aircraft conceptual design is learned only by doing the calculations, not by just watching a video and nodding your head. You will have no feel for it unless you do the calculations yourself. The calculated values will be shown in this dark blue color and towards the end, if we do a comparison with any value or parameter published online or in some resource, we are going to show those quoted values in green color so that we can compare with our calculations in blue color. Let us first look at the procedure that we will follow for carrying out the estimation of DTE and production costs for Global Hawk and also some of the baseline assumptions that we will make. So, here also we used a CER based method where CER stands for Cost Estimating Relationships. And in this particular exercise, we are using the same RAND modified DAPCA4 model, but the difference is that now we are going to use it for 1998 formulae given in the textbook by Nikolai and Kariknar. So, let me clarify the other tutorial that talks about the estimation of RDE cost and the production cost of Boeing 787-8 in that the formulae are taken from the RAND modified DAPCA4 model of 2012 as given in the textbook by Daniel Rehmer. But here we are going to look at the same model but 1998 based formulae given by Nikolai and Kariknar. And the costs that we obtain in 1998 either by calculation or by specifications, they will be scaled up to the current values of 2020 using economic inflation factors. So, we are going to make some assumptions of the cost parameters. For example, the cost of avionics is going to be assumed as 12 million dollars per aircraft in 1998. The cost of payloads would be 62.76 million in 1998 and as per the suggestion in the textbook by Nikolai and Kariknar, we should add around 25% for return on investments and initial spares. Let us look at some useful data for RQ-4B Global Hawk aircraft which we will need in our calculations. The aircraft data would be empty weight 5627 kilograms specified in the specifications converted into pounds. The AMPR weight is assumed to be 62% as per standard practice. The maximum speed, so the symbols are a bit different here. The maximum speed is shown by S here and it is specified to be 192 meters per second which converts to 373.21 knots. The maximum thrust per engine is 3447.3 kg, there is only one engine and this becomes 7600 pounds. The turbine inlet temperature is 1540.15 degree Kelvin which becomes 2772.27 degrees Rankine. The maximum Mach number for this aircraft is 0.61. The assumed data related to production is as follows. We assume that 41 aircraft are produced in the production run, 7 aircraft are produced in the RDTNE phase during flight testing and they are not sold eventually and therefore the total number of quantity produced is 48. Let us look at the manoeuvre rates applicable. For this we look at the trends in the hourly rates in the aircraft construction industry. As mentioned in the textbook by Nikola and Karikner, these follow a straight line and there are linear curve fits available for manoeuvre rates versus the year and there are different equations for tooling, engineering, quality control and manufacturing costs. We will use these corresponding equations into our calculations where the letter Y stands for the year for example, 2020. So, it will become 2.883 into 2020 minus 5666 that would give you the value of tooling applicable, tooling manoeuvre applicable in 2020. Let us look at the manoeuvre rates. So, if we use the same equations as mentioned in the previous slide then the costs that the manoeuvre rates which appear are as follows. For tooling it is 157.66 dollars per hour, for engineering it is 145.52 dollars per hour, for quality control it is 140, for manufacturing it is 126. So, we noticed that the highest manoeuvre rates are for the tooling because this is very specialized and quality control is done by the engineers or the quality managers who are actually who have been taken from the production side itself. So, the costs are slightly lower there. We also need to know how to inflate the rates to take care of the consumer price index change per year. So, here is again a graph from the textbook by Nikolay and Kariknar where we have the base value for the consumer price index as 200 for the year 1998. So, if you want to calculate the values in 2020 then we need to just go along this line. So, for the base year the number is 200 that is 89. So, for the year 2020 the number reads from the graph as around 315. So, therefore the inflation ratio to scale up a cost from 1998 to 2020 would be CPI at 2020 upon CPI at 1998. So, at this stage you should just calculate the value. So, the value is going to be 1.575. Remember this number because this number will be used to scale up the costs from 1998 to 2020 in many places in our calculations that follow. So, the inflation fraction, inflation fraction is 1.575. Let us look at first cost of airframe engineering in the DTE phase. The formula given in the book by Nikolay and Kariknar relates the airframe engineering hours with the empty weight in pounds, the maximum speed of the aircraft and the number of aircraft produced or the quantity produced in the DTE phase. So, for our aircraft we know that the empty weight is 12405.4, the maximum speed is 373.21 in knots and that quantity produced in RDT phases 7. So, at this point you should pause the video and do this calculation to get the value of man hours. So, you just have to multiply these quantities which are shown on the screen. The value comes out to be 2.016 into 10 power 6 hours. So, many hours are needed in the airframe engineering and we have just calculated in the previous slide the man hour rate for airframe engineering that is 145.52. This rate is in 2020 and the hours are also there. So, if you multiply both of them you will get the cost of airframe engineering which would be just the multiplication of E, DTE and RE which means it will be 2.016 into 10 power 6 multiplied by 145.52. Please pause the video and calculate this number. You first calculate in million dollars and then you can convert it in billion dollars. So, it turns out to be 2.933 into 10 power 8 or 293.3 million and that will become 0.2933 billion dollars. Approximately 0.3 billion dollars is being estimated as the cost of airframe engineering for the RDT phase. The same formula now applied to the production phase is you have to apply the same formula. The difference is that here we are going to use all the aircraft produced including the 7 aircraft produced in the previous phase in finding out the hours. This is because when you produce 7 aircraft and then after that you start producing the 8 aircraft to 41 more aircraft, the knowledge and experience gained in the first 7 aircraft stays with you. So, therefore the hours that you need is can be obtained by getting the cumulative hours. So, the cumulative hours are going to be 2.759 to 10 power 6. From this we should subtract the hours which were consumed in the first 7 aircraft to be produced which were 2.016 10 power 6. So, the difference is only 0.743 to 10 power 6 hours. So, you notice that for producing 7 aircraft you need 2 into 10 power you know 2 into 10 power 6 and to produce 48 you need only 2.759. So, the difference is not you know something like 7 times the difference is only very small number. This is because of the learning curve. Now the manoeuvre rate is known to us as 145.52 per hour. So, therefore to get the cost of AFM engineering we just have to multiply these 2 numbers 145.52 with 0.743 into 10 power 6. Please do this calculation and you will get the answer as 1.081 into 10 power 8 which is 108.1 million or 0.1081 billion. Let us look at the tooling cost in the DTE phase. The tooling hours again are obtained as a function of the empty weight in pounds maximum speed in knots and the quantity produced during the DTE phase. In our case the empty weight is 12405.4 pounds the maximum speed is 373.21 knots and the quantity produced in the DTE phase is only 7. So, please pause the video and calculate this number it comes out to be 9.343 into 10 power 5 hours. So, this you have to multiply now with the manoeuvre rate of 157.66 per hour and if you do that then you get this expression 9.343 into 10 power 5 into 157.66. Please pause the video and do this calculation and get the value. The CT in DTE phase is 1.473 10 power 8 or 147.3 million or 0.1473 billion. So, so much money is needed in the tooling for only 7 aircraft. Now, let us see what happens in the production phase. In the production phase again the formula remains the same but once again just like in the previous case we use 48 as the quantity we get the cumulative tooling hours and then we subtract the hours required for the already DTE phase. So, that the number of hours come as 1.5510 power 6 for the total phase or the cumulative phase from there you subtract the value that you got in the previous slide of 0.9.343. So, the answer is 6.157 into 10 power 6 that is the only additional number of hours needed. So, notice to produce 7 aircraft we need 9.343 10 power 5 to produce 41 aircraft after producing 7 aircraft the additional hours are only 6.157 into 10 power 5 and the manoeuvre rate is 157.66 per hour. So, therefore, cost of tooling would be a multiplication of these two quantities and that would be 157.66 into 6.157 into 10 power 5. This number turns out to be 9.707 10 power 7 or 97.07 million or 0.09707 billion dollars. Moving ahead let us look at the manufacturing labour cost for the DTE phase. Here also the formula relates the three parameters as before. We insert the values of those parameters in the formula and please pause the video and do this calculation get the value of L DTE it is 1.025 into 10 power 6 hours. Now, the manufacturing rate is 126.32 dollars per hour as seen earlier. So, therefore, the cost of manufacturing will be a product of these two quantities L RDE L DTE and RM which will be 1.025 10 power 6 into 126.32 pause the video and do this calculation please. This number comes out to be 129.4 million or 0.1294 billion dollars. We repeat the same calculation for the production, but for that we first get the cumulative hours. So, in that the formula remains the same as last one, but the number of aircraft becomes 48 from 7 because the total phase consists of 41 plus 7 48. So, you calculate the total hours needed 3.522 10 power 6 from here, we will subtract the hours needed earlier for 7 aircraft and the balance only is needed in the production phase which is 2.497 into 10 power 6 hours. And the value of the maintenance maneuver rate is the same as obtained earlier. So, when you want to obtain the cost of manufacturing, you have to multiply these two quantities that means 2.947 10 power 6 into 126.32 per hour. Please pause the video here, do this calculation and try to match with our value of 315.4 million or 0.3154 billion. Moving ahead to quality control cost in the DTE phase. It is given in the textbook by character and Nikolai that the quality control hours are just 0.133 times that of the hours required for the manufacturing phase. The only difference being that if the aircraft is a cargo aircraft this ratio is little bit smaller, but ours is not a pure cargo aircraft. So, we simply multiply the hours that we had required for the aircraft production in the DTE phase, multiply those hours that is 1025 into 10 power 6 by 0.133 to get the new hours. And these hours are then multiplied by the maneuver rate. If you multiply you will get 1.363 into 10 power 5 into 40. Pause the video here, do the calculation. The value is 1.908 10 power 7 or 19.08 million or 0.019 1 billion dollars. Let us look at the quality control cost in the production phase. Again the formula is same, the hours needed are only 13.3 percent of the hours spent in production, which were 2.497 10 power 6 as in the couple of slides behind. So, we will just multiply by 0.133 to get the number of hours. Those number comes out to be 3.321 into 10 power 5. Now, the rate is 140 per hour. So, you just have to multiply these two quantities to get the total cost. That cost is going to be 3.321 into 10 power 5 into 140. So, what is the answer? 4649 into 10 power 7 or 46.49 million or 0.04649 billion. Let us look at the manufacturing material cost in DTE and production. Here we will do both together because this cost is not, there are no hours to be calculated. This is just a cost in terms of the number of aircraft. So, to manufacture 7 aircraft you just put 7 and the other parameters are already known to you get the value. So, when you multiply all these numbers 16.39 by 124045.4 to the power 0.921 into 3.21 to the power 0.621 into 7 to the power 0.799, you will get the value as 18.08 million or 0.0180 billion dollars. Similarly, for the manufacturing phase, the formula remains the same. The difference only is that instead of 7 aircraft which were manufactured in that phase, in this phase we will make 41 aircraft. So, it will be as shown on the screen in 1998, then year dollars, but these number has to be multiplied by the inflation fraction to bring it current to 2020 and hence you ultimately get the value as 0.11691 billion dollars. Development support and flight testing cost we are assuming they occur only in the case of the aircraft RDE phase. So, therefore, it will only have formula related to empty weight of the aircraft and the area. This is a theoretical calculation. So, but with that in mind they have obtained the value of CL minimum should be point you know you need a very high CL in this case. And the value of the the value of the cost comes out to be 5.52110 power 7 or 55.21 million dollars in 1989. So, then this cost has to be escalated by 1.575 and brought to the current levels which is 850, 86.95 million dollars or 0.0806 billion dollars. The next calculation for the flight test cost, this also is a fixed number. Remember that these coefficients change from the formula if given in rember to the formula given here only because the formulae in the remers book are for 2012. So, this number will be appropriately maybe higher and in our case here the costs are given in 1989 terms. So, these values are tend to be lower. Again we plug in the value of WES and Q for the flight test there are only 7 aircraft the initial prototype. So, the cost comes out to be 5.429 10 power 7 or 52.29 million but this cost has to be scaled up to the year 2020 by multiplying by a scale coefficient 1.575. So, if you apply this coefficient to multiply you will get the value as 0.0855 billion dollars or 85.5 million dollars. Now, we go to the engine cost. So, there is a graph in the textbook by Nikolay and Karikner which talks about the unit production cost of various types of power plants as a function of the maximum sea level static thrust. And there are two lines here the one on the top is basically for aircraft with after burners and the one on the bottom most of the aircraft are like our aircraft Benign, relaxed aircraft going at mark number 0.61 maximum but it should be it should be able to provide some connectivity. So, we know that the thrust max of this engine is 7600 pounds. So, going on the screen this gives you an indication this gives you a global average. So, the global average says that the engine cost should be somewhere between 7.5 to a billion dollars. So, we have taken the minimum value as cut by this line and the unit production cost per engine comes out to be 0.75 million dollars in 1989 1998 and turns out to be 0.75 now that is actually a very low value. So, even if you scale it up to 2020 it is only 1.8 million dollars it is actually a unrealistically low value for the cost of an engine. So, therefore, we do not use this graph we use the formula given by Nikolay and Karekna this is the formula which has been given in the book applicable for transport aircraft. So, we use this formula and so there was a requirement there was a requirement to know a few things we will see how it goes ahead. Now, in this expression the term 7600 stands for the maximum thrust in pounds at sea level 0.61 stands for the maximum Mach number and 2772 0.27 stands for the value of thrust required. So, therefore, if you if you plug in these values you get the value as 2.152 into 10 power 6 or 2.15 that seems reasonable in 1998 around 2 million dollars instead of 0.75 million dollars and then when you scale it up from 98 to 2000 and 20 the factor you know is 1.575. So, therefore, the cost in 2020 would be into 1.575 or 3.3894 million dollars. Now, there are 7 engines needed for 7 aircraft because of single engine aircraft. So, in the DTE phase it will cost us 7 times this number 3.3894 per million per year. So, we will get 23 million dollars to be spent every year in making this fleet run. Now, only two costs are remaining the cost of Avionics and the cost of payloads. As far as Avionics cost is concerned we have assumed the cost to be 12 million dollars per aircraft in 1998. So, in 1998 the value will come out to be 7 into 12 please calculate the number I know it is 84 million dollars and then we have to multiply by the factor to upscale that cost to 9 from 1998 to 2020. So, you multiply by 1.575 to get 132.5 million dollars this is the money that will be spent in the RDT phase on the engine cost. And then the next thing would be what will be the cost of the payloads which will go on the aircraft. So, each of them carries 41 point each of them is going to so in there will be a cost of 41 into 12 that is 492 million dollars because there are 41 aircraft which are going to carry the Avionics and this number turns out to be 779.4 million dollars for the production aircraft where we have 1100 aircraft for those where we have 7 aircraft it comes out to be much smaller only 492 million. But here also there is a question how do you get such a large amount of money 492,000 is not a small amount to save or to obtain. Let us also look at the payload cost we have assumed that the payload cost is 67 point so the 2.76 power 6 dollars per aircraft and therefore for 7 aircraft we are going to get 7 into 62.76 million which is 439.32 million. Same way you can get the cost in escalated fashion for 90 for 2020 as 691.9 million dollars. Now if you want to calculate the cost of payload for the production aircraft we are using the full production quantity here we are assuming that these 7 kits which are there in the aircraft they actually have to be kept as a standby so we have to do afresh and many changes would be made so we may have to go afresh in the calculation. So we assume that a small sub program is launched to get the things going and that will cost us by inflation it will cost us this number as shown on the screen the escalation factor into the year into the or divide by year into the cost per day. So the payload cost comes out to be 48 into 62 2.76 million dollars but this number is in 1998 3 billion dollars so you have to multiply by 1.525 and get 4.749 billion dollars so large amount of money is required by the payload. So let us look at the summary of all the items in one shot the first one was the AFRAM engineering where the costs were 293.3 and 108.1 respectively similarly tooling similarly manufacturing similarly quality control manufacturing material development support flight testing engine production and avionics these are the costs which will be incurred on the military aircraft and all these have cost some money so we have to recover this money from the from the sale of the remaining aircraft and payloads are very expensive the largest component of the cost as you can see is the payload 692 million dollars in this entire landscape and 47,449 million dollars is in this particular scale. So lot of work is going on all over the world in reducing the cost of the avionics so that these numbers do not add up like this but we have no information on anything working right now so we go for ROI of 25 percent we assume that the person who does the coating will only take maybe around 10 to 12 percent and the remaining will be available as the additional cost of some of the subsystem that you may have to buy. So the grand total comes out to be approximately 2 billion dollars for the DTE phase and approximately 7.9 or 8 billion dollars for the production phase together it comes to around it is at these numbers they come to nearly 1 million I do not mind my spreadsheet can justify the using 1 million because I have not done any rounding off errors. So here is a breakdown of the DTE cost we can see that the largest component 34 percent or nearly one third happens to be the cost of payloads the remaining major components are 20 percent and 14 percent for airframe engineering and the ROI and spares. ROI and the spares is also a large value and everything else comes in this small portion of the pie chart. When we move to the production cost again we know now that the largest component would be the cost of the payload because this aircraft is nothing but a carrier for a payload. So payload is very very expensive and that is the main thing notice the other two major costs are the ROI and spares here the ROI is basically driving this number high and the cost of avionics remaining everything engineering tooling etc it becomes a very small component of the total cost. So now we want to estimate the cost of fly away so we have just now seen that the RDTNE cost or the DTNE cost is 2.048 billions this cost has to be recovered from the 41 aircraft which are sold and to produce the aircraft we need to spend around 8 billion dollars 7.983 to be precise. So the total cost is around 10.031 billion dollars and if you want to calculate the fly away cost you just divide this number converting it into millions divide by 41 because only 41 aircraft are sold get this value please this value is 244.67 million dollars or your answer will be actually 0.24467. Finally let us compare our calculations with the quoted price in some source and literature the cost of the development testing and engineering RDTNE the quoted value in 2012 is 3971.8 but the scale value is 4635 but the value that we estimate is much lower is 2047.8 because we have been little liberal in assuming that the material that are also conventional and the aircraft also conventional actually the aircraft is highly unconventional therefore there will be larger cost in RDTNE phase but we have no way of estimating it. Secondly the production cost you can notice that the quoted value is around 5914 including inflation and here the scaled value is 6901. So there is also there is a difference around 10 lakhs that is because of the years the differences in the years also the total program cost is 10 021 million dollars which is spent on a market set of 41 aircraft so if all of them are sold. Now when you look at the data from the GAO office which is given in the bottom of the screen the quantity produces 45 but actually we have seen that the quantity produced is 41 plus 7 as per our assumption. So we get a flyaway cost using that so what we do is 10031 divided by 41 is 244.67 and 536 divided by 28 or whatever is the time 28 is when he is running to go back that the number difference comes out to be very less. In fact, the percentage difference is only around 5 percent we are under predicting by around 5 percent. Thank you very much. Before I close I need to thank Lillian, Nikolai and Grant Karikner for their excellent updated textbook with large database and Namanuddin for help in creating this tutorial. Thank you.