 Let us now look at a saturated 2 power 3 factorial design X matrix and so in addition to the intercept we also find A, B, C, A, B, C, A, C and A, B, C that would be 1 plus 7, 8. So all the 8 settings are being used up to determine 8 parameters and when you look at the A column you are having minus 1 plus 1, it is an orthogonal design and so the number of minus 1s would be equal to the number of plus 1s and so the total will be equal to 0. Similarly for B you will again have a total of 0 and so on for C as well. If you look at A, B again you are going to have a combination of plus 1s and minus 1s in equal numbers and so it will become equal to 0 and when you look at A, B, C also we are going to have as 0. So we can say that the first order moments are 0 because we are summing the elements in the X matrix column wise for a given X. In this case X i will be equal to A or i is equal to 8 if you want to put it that way. So the first moments are all 0 and this is the second mixed moment because A is different from B, again that is equal to 0. So it is B, C and AC and when you are looking at the third moment which is also mixed because you are having A and B which is different and also B and C are different, A and C are different and so when you look at the sum it is again equal to 0. So we can say that the third mixed moment is equal to 0 and even A, B squared even though you are having a B squared term you are having A to the power of 1 and hence you are having a mixed moment here and that would be a third mixed moment which is equal to 0. Only when you come to C squared all the elements become 1 and when you total it up you are going to get 8 but anyway you are dividing it by the total design size which is again 8 and so 8 by 8 will give you 1. So this is the way in which the pure and mixed moments are computed for various designs. Now when you look at a central composite design which are now having centre points and also the axial points the moments are carried over up to order 4. So we will take the design matrix for the central composite design and use it to investigate the values of the different moments. So all odd moments through order 4 that is the moments that contain at least one odd power. So i, i to the power of 1, i cube, i to the power of 3, i squared j, i to the power of 2 and j to the power of 1. So it is a odd moment or a mixed moment, i, j, k all of them are different and all of them have odd powers, i cube j. So 3 is also odd and 1 is also odd. The 3 is the exponent for i and 1 is the exponent for j. Both of them are odd numbers and so we can call them as odd moments or mixed moments. And then you also have i squared j, k are 0 for i not equal to j not equal to k. So the tables which are shown next will illustrate this concept. Also when are we going to have the non-zero moments? It appears that many of the moments are 0. Are there any moments which are non-zero? And for the central composite design involving 3 factors the non-zero moments are for k is equal to 3 corresponding to i squared, i squared, j squared, i to the power of 4 for all i not equal to j. So these are the cases where the moments will not be equal to 0 but will the moments be equal to 1. Again this is a very interesting question because n is the size of the run. So n would be the number of experimental settings and for a central composite design it can be shown for 3 factors you have 8 factorial points 2 into 3, 6 axial points so 8 plus 6 that makes it 14 and then you have 14 plus nc center points. But what would be the values taken by the non-zero moments such as i squared, i squared, j squared, i to the power of 4 we will have to compute them. Let us look at the table. So let us look at the moments table for a 2 power 3 design with 6 center points, 8 factorial points that would be again 6 axial points a total of 20 points. So this is 20, 1, 2, 3, 4, 5, 6 so this entire column comprises of 20 elements this is for a and you can see that in addition to the minus 1 and minus 1 plus 1, plus 1, minus 1, minus 1 these correspond to the factorial points. So the number of plus 1 and the number of minus 1 points are matching so that is fine and this minus 1.682 and plus 1.682 refer to the axial points. Similarly b also has its share of equal number of minus 1 and plus 1 points and then it also has 1.682 and minus 1.682 which correspond to the axial points same way we can do for c. Now if I add up the elements in the a column that will correspond to the first moment and that would be equal to 0. So would be the first moment corresponding to b and c but if I look at a squared I am going to get 1, 1, 1, 1 so I will get 1, 2, 3, 4, 5, 6, 7, 8, 1 and then I am also going to get square of minus 1.682 and again square of plus 1.682 which turns out to be 2.828 and 2.828. So when I total the elements of the a squared column vector I am not going to get n as it did for the pure factorial case because now the central composite design we have added the axial points. So the axial point coordinates also will contribute to the moment which is not vanishing. If I look at the a squared vector or the a squared column I am seeing elements which are all 0 or only positive. When I add up the elements I am not going to get a value of n where n is the design size here n is equal to 20. I will get a value different from n because we have 8 ones that could be 8 and then you also have 2.8282 times which is approximately 5.656. So 13.5656 so that value is not equal to n which is 20. I am just illustrating the difference between the central composite design and the regular factorial design. Similarly for b squared you have again a pure second order moment and when you total it up the value will be same as that of a squared. Similarly for c squared when you however look at the second order mixed moments or the odd moments comprising of ab or ac where a is different from b and a is different from c we see that the total adds up to 0. So the moment values for the second order mixed moments are 0. We can go on for bc, c, ab, c cube, a squared b and abc, b cube, ac squared, ab. So as long as you have odd moments and the moments are carried over up to order 4 we can see as long as there are at least one different term in the product. For example in c squared ab, ab and c squared are different and when you look at ab cube, a and b are different, a squared b, a and b are different. All the cases the moment value becomes 0 when you add up the elements in the column they all become 0 here. So now we are going to the second order models. The scaled prediction variance is normalizing the design size by multiplying by n and removing the variance effect by dividing by sigma squared. And so we simply get n into xm prime x prime x inverse xm. We saw that xm prime is the coordinate point in the exponential design space expanded into the model space. So that xm reflects the model we have chosen to represent the process. And we have also seen that the moment matrix is given by x prime x divided by n. So for a first order factorial design of order k, the moment matrix is the identity matrix of order k plus 1 by k plus 1 that can be easily shown and we have also shown it a few times earlier. So we get using the definition for m, m as x prime x by n. So x prime x inverse would be substituted in terms of m and n. The n cancels out and then we get m inverse here. So this derivation is pretty straightforward. If you want you can try it out. So what we are trying to prove here is the scale the prediction variance at a point x in the exponential design space depends upon the moment matrix. And the scale the prediction variance of x would be xm prime m inverse and xm and you just put the identity matrix here instead of m inverse. So identity matrix inverse is also an identity matrix. So you will have identity matrix here and then it will be each element multiplied with itself in the xm prime and xm. This can be easily verified in our earlier lecture on orthogonal design concepts. We did calculate the scale the prediction variance. The scale the prediction variance we showed in the orthogonal design as 1 plus square of the distance of the coordinate from the exponential design center. You may want to verify that. So what is the importance of the scale the prediction variance in the design space? It is important to have a stable scale the prediction variance. If the stable nature of these SPV is violated then you will have shooting up of the variance at some locations in the exponential design space and the model predictions can be considered to be reliable no longer. So another thing is the quality of the y hat of x should be as uniform as possible throughout the design space. You cannot have certain pockets in the exponential design space where the scale the prediction variance is hitting the roof. Now let us look at the concept of rotatability. This was defined by Box and Hunter in 1957. Rotatable design is 1 for which n into variance y hat of x by sigma square which is nothing but the scale the prediction variance has the same value at any 2 locations which are equidistant from the design center. And for k is equal to 3 n into variance y hat of x by sigma square is constant on spears because any point on the spear would be equidistant from the center. So let us take 2 points in the exponential design space x1 and x2. If you have x1 prime x1 to the power of half is equal to x2 prime into x2 to the power of half then x1 and x2 are said to be equidistant from the origin and the scale the prediction variance is equal at both these points. In other words the predicted values at y hat x1 and y hat x2 should be equally good that is have the same variance. And just because the design is rotatable it does not mean that the SPV is stable everywhere in the design space. But the concept of rotatability also helps us to find the number of center points and also the coordinates of the axial points in the central composite design. So now if you look at the first order models a design is said to be rotatable if and only if the odd moments through order 2 or 0 and the pure moments of order 2 are all equal. So you have i is equal to ij is equal to 0 this can be easily verified for i is equal to 1 2 so on to k and i is not equal to j. And then you also have i squared is equal to lambda 2 where lambda 2 is equal to 1 for 2 power k factorial design with 1- plus or minus 1 settings. So we are looking at the necessary and sufficient conditions for rotatability we are discussing first order design and in the first order design for rotatable conditions to exist the first order moment as well as the second order odd moments or mixed moments should be equal to 0 that means i is equal to ij is equal to 0 for i not equal to j. i squared is equal to lambda 2 where lambda 2 is a constant and it is not equal to 0 and for a 2 power k factorial design alone with plus or minus 1 settings lambda 2 is equal to 1 this also we have seen previously. Again looking at conditions more conditions for rotatability for second order models this is very important all the odd moments through order 4 or 0 and there is a difference between the mixed moment and the odd moments if you have i squared j squared they are considered to be a mixed moments because i is not equal to j. On the other hand you are looking at the power the power is 2 for i as well as j. So you cannot call this an odd moment but it is a mixed moment but if you have i cube j then the power of i is equal to 3 and the power of j is equal to 1 and so here you have odd moments. Now when you are looking at odd moments through order 4 or 0 you have i i cube i squared j j is having a power of 1 ij k all the ij and k are having powers of 1 i cube j i squared j k are 0 for i not equal to j not equal to k and very interestingly the ratio of the moments i power 4 by i squared j squared is equal to 3 and i squared is equal to lambda 2. So this is a very interesting condition i power 4 divided by i squared j squared is equal to 3 for a rotatable design and so when you look at i power 4 you will have the elements all 0 or positive you would not have any negative values you will have 0 1 and then you will also have the axial points to the power of 4. So for the factorial points they will all be 1 so the minus 1 will be converted into plus 1's so they will all remain as 1's 0's will of course be 0's so when you total them up that will be equal to the total number of factorial points and then you are going to have 2 alpha to the power of 4 because the axial points for each column would be plus alpha and minus alpha when they are taken to the power of 4 it will become alpha to the power of 4 and alpha to the power of 4 so it will be 2 alpha to the power of 4 and so the fourth order pure moments would be giving a total of f plus 2 alpha to the power of 4 whereas i squared j squared it can be shown very easily will give you only f that is very interesting the fourth order mixed moments i squared j squared is independent of the alpha term whereas the pure fourth order moment is having the alpha term in it. So the ratio of these 2 i to the power of 4 by i squared j squared moments is equal to 3 for a rotatable design and here it can be easily shown that from this relation we are going to have 3f minus f which is 2f so f is equal to alpha to the power of 4 or alpha is equal to fourth root of f and so if you want your central composite design to be rotatable adopting this criteria helps you to find the value of alpha this is what I have been telling about early in the lecture we will as to how the alpha values are going to be set if you want a rotatable design so using this criteria we set the value of alpha as the fourth root of the f where f is the factorial number of points okay f is not the number of factors but the total number of factorial points for a 2 power 2 design f would be equal to 2 power 2 which is 4 for a 2 power 3 factorial design f would be equal to 8 so f refers to the number of factorial points and is independent of the number of center runs again you can see that the alpha value which is recommended to meet the condition of rotatability is independent upon independent of the number of center runs. Please note that the rotatability is achieved if alpha is used as above regardless of the number of center runs. Let us now look at the 2 power 3 central composite design we are having 20 runs how do the 20 runs come about we have 8 factorial points 2 power 3 factorial points which will be 8 and then you have also 6 axial points they are located at minus 1.682 plus 1.682 and that would be 6 so because you are having 3 axis and so each axis has 2 axial points so you have 6 axial points totally 8 plus 6 is 14 and then you also have 6 repeats the repeats are not located in one group you can see that this design has been randomized and so you are having the center points at different locations this is one center point that was one this is the second center point because all the coordinates are 0 0 0 0 that would be 2 so this is 3 and this is 4 5 and 6. So we have 6 center points and you can see that the a squared will not vanish either the terms are 0 or positive but for a b and c you have equal number of negative and positive values and so the total would be equal to 0 a squared and b squared are not going to be 0 a squared is minus 1.68 is squared is 2.83 so these calculations are pretty straight forward so I would not be discussing them any further but when you look at the other elements we are having the fourth order pure moment c power 4 when you total it it adds up to 24 interestingly the axial point has become 8 you can see that the axial point has become 8. So what is 1.68179 according to the conditions of rotatability alpha is equal to fourth root of f where f is the number of factorial points so the fourth root of 8 gives you 1.68179 if the fourth root of 8 is 1.68179 the fourth power of 1.68179 will be equal to 8 and that is the reason why for c power 4 column we are getting 8 here and when you total it up it adds up to a nice 24 and when you look at the fourth order mixed moments i squared j squared the sum is equal to 8 because you are going to have b squared and c squared and you are not having the contribution from the axial point because the axial point in the b column is located at a different place than the axial point in the c column it is very easy to verify here the axial points are here corresponding to the location of the axial point in the b column the corresponding values are both 0 the axial point in the c column for example I was pointing to the wrong column this is the axial point in the b column and corresponding to that the values are 0 okay and the axial points in the c column are located at points where the corresponding points in the b column are both 0 so you are having 0 0 and then you are having these 2 and you are having these 2 and then you are having 0 0 here so when you do b squared this term will of course be positive both will be positive but when you are doing b squared c squared you can see that it will become 1.68179 squared into 0 squared which is 0 so that is why the b squared c squared has 0s even at locations corresponding to the axial points for the b and c columns anyway if you take the ratio of c to the power of 4 and b squared c squared if you take the ratio of c power 4 to b squared c squared you get 24 by 8 which is equal to 3. Now let us see the values of alpha for rotatable central composite design for k is equal to 2 and 4 factorial points the total number of design points would be 4 plus 4 plus nc and alpha is equal to 4th root of 4 which is 1.414 4th power of 4 would be root 2 and that is why you are getting 1.414 and for k is equal to 3 it can be easily shown just now we saw that the alpha is 1.682 for a design involving 4 factors k is equal to 4 we have f is equal to 16 and alpha is coming to a whole 2 and when you look at 5 factors the number of factorial points would be 2 power 5 which is 32 the total number of points would be 32 plus 10 plus nc and alpha is coming out to be 2.378. Now let us see whether the design is spherical for a spherical design you want the points to be equidistant from the origin. So for meeting the condition for spherical design the design points should be located at square root of k where k is the number of factors. The value of alpha for k is equal to 3 is 1.682 for a rotatable design and it is 1.732 for a spherical design coming again in the above table it can be seen that the value of alpha for k is equal to 3 is 1.682. So this is actually meant for a rotatable design but this value of 1.682 is not equal to 1.732 which is corresponding to square root of 3 here k is equal to 3 for a factorial design with 3 factors and hence this alpha value is 4th root of 8 which is 1.682 and not square root of 3 which is 1.732. So all the design points in a rotatable design are not equidistant from the center and the design is not completely spherical. So if you look at central composite designs with k is equal to 2 and k is equal to 4 for k is equal to 2 it is 1.414 for a rotatable design and square root of 2 for a spherical design or a circular design if you want to put it that way. So square root of 2 is equal to 4th root of 2 is equal to 4th root of 2 power 2 the number of factorial points. So both of them are equal to root 2. So the design is both spherical or mean if you want to take it for a 2 dimensional case circular and also rotatable but when you take k is equal to 3 you are getting 4th root of 8 which is 1.682 which is not equal to square root of k which is 1.732. So even though the design rotatable design is not strictly spherical the values of 1.682 and 1.732 are pretty close to each other and so we are getting a nearly spherical design for k is equal to 3. Now coming to k is equal to 4 we are having 16 factorial points and the 4th root of 16 is 2 and the square root of 4 is also equal to 2. So for this case we have both the conditions of rotatability and a spherical design being met. So for both k is equal to 2 and k is equal to 4 the design is both spherical as well as rotatable. The design points are equidistant from the centre and the design also is rotatable. But when you look at a factorial design with 5 factors the condition of both rotatability and sphericity is not met simultaneously. So to reiterate what we said just now for k is equal to 2 we have for a rotatable design 4th root of 2 power 2 which is root 2 and that is equal to root k the condition required for a spherical design. For k is equal to 4 the condition of rotatability stipulates 4th power of 2 power 4 which is 2 which is also equal to root of k is equal to 4 and hence we have both the spherical as well as rotatable characteristics for the central composite design with either 2 factors or 4 factors. When you look at central composite design with 3 factors it is said to be a nearly spherical design because 1.682 and 1.7321 are not very further apart. So central composite designs with k is equal to 2 and k is equal to 4 contain 8 and 24 design points apart from the centre runs that are equidistant from the design centre. For these cases the design is exactly spherical. However for k is equal to 3 the axial points are located at 1.682 and the design is said to be only nearly spherical. For k is equal to 2 and 4 the design points are exactly at root k from the design centre whereas they are only approximately so for a design involving 3 factors. So now we are looking at centre runs. The importance of centre runs we already emphasised as estimation for pure error to detect the curvature effects significance in the model. Now we are going to see another role played by centre runs in the central composite design. For rotatable central composite design it is actually a rotatable central composite design. So the centre runs are important to try and import stability to the scaled prediction variance in the design region. We saw that not necessarily the scaled prediction variance should be high only in the design boundaries or design extremes but it is also possible for the scaled prediction variance to shoot up in the centre of the experimental design space. So to have a check on that we increase the number of centre points. If there are no centre runs in a second order design the matrix X prime X becomes singular and the scaled prediction variance is infinite. So adequate number of centre runs are recommended. Spherical or near spherical designs require 3 to 5 centre runs to avoid severe imbalance of the scaled prediction variance in the experimental design region. So for factors ranging from 2 to 4 it is advisable to have 3 to 5 centre points so that the scaled prediction variance does not shoot up and when you are going in for a rotatable or a nearly rotatable central composite design a few centre runs only is not desirable. So adequate number of centre runs are required for rotatable and nearly rotatable central composite designs. Recommended that you use at least 3 to 5 centre runs to avoid severe imbalance of the scaled prediction variance in the experimental design region. It is not desirable to have very less number of centre runs for the rotatable or nearly rotatable CCD. So we will demonstrate the importance of centre runs with a few examples in the next lecture. So thanks for your attention.