 we are discussing about the interior point method. That means, if you have a linear programming problem or the convex optimization problem, we can solve by using interior point method. What is this method is there first you have to guess one point inside the feasible region and then we start the process iterative process. So, if you see this we have considered a simple example minimize this function that function is linear function subject to equality and inequality constraints. In this case all are inequality constraints, then if you represent that problem in graphically we will see this is the feasible region. If you see this one is the feasible region and we have taken initial if you take the initial point a here whose coordinates is 1.5 and 1.5 that means, these are the equidistance from the coordinate axis. Another point I have shown it take this is the point B point B whose coordinates are 3 comma 1 this is the off center point. If you start the process with B point off center point, then you will see we are moving towards the one side of the wall, so we are in turn we are on the border of this interior point what is called feasible region. It is not an interior point in the on the border of the feasible point to avoid that one. What we did it if you see this one we have made a one simple transformation that transformation is called affine transformation means linear transformation. So, that in transform coordinate system the coordinates are in the equidistance from the coordinate axis, so that we have shown it how to select the that is transform system matrix how to select it. We have considered the transform system coordinate y 1 0 y 2 0 then what is the initial and original coordinate axis what is the x 1 0 that we are dividing with that element. If you divide by this element similarly here then in transform system that value will be 1 that is what we have shown it. That means in transform system in transform coordinates the our initial interior point is equidistant from the two coordinate axis that we have shown. Now, we will see more general if you see the transform system now you see our point is here B point now transform into a new coordinate axis 1 1 point that is equidistant from y 1 y 1 axis and y 2 axis they are in equidistant 1 1 point. In general now we can we will show it how to proceed that that one. So, first we consider the scaling of a what is called standard LP problem how to do then after doing scaling then we will that means in other words how to select the transformation matrix in new coordinate axis. So, the scaling the standard LP problem, so standard LP problem if you recollect whatever the LP problem is given we have to convert into standard LP problem first. So, our standard LP problem is minimize f of x which is nothing but a c transpose x whose dimension if you consider n cross 1 then subject to subject to a into x is equal to b, b is the number of equality constant even in original problem if there is a inequality constant is there then we know how to convert into a equality constant by introducing slack variable, surplus variable and artificial variable. So, we know these things, so where b is the standard LP problem is b is greater than equal to 0 and x is greater than equal to each element of x is greater than equal to 0. So, this is our standard LP problem let x superscript 0 whose dimension n cross 1 is the initial is the initial feasible interior point feasible interior point point to center this point and this point we are considering it is off centered that means if distance from the coordinate axis are not equal each. So, we consider how to center x 0 defined as to center x 0 we defined as x 0 is equal to d y of 0 y of 0 means that coordinates systems. So, this transformation we have used next question is what should be the choice of d that d will choice will select d like this way x of 0 that if you consider the how many elements of x of 0 are there any elements. So, each element of 0 2 0 dot x n 0, so this is the choice of our transformation matrix which will convert the coordinate x that initial what is called interior point in the feasible region will convert into a what is called transform system equivalent what is called the interior point. So, if you see this one and this is called the scaling matrix or transformation matrix. So, if you look at this one now what is y 0 then you take the transpose of that one agree y 0 y 0 of this one is nothing, but a d inverse you have to take inverse of d d inverse x of 0 and what is d inverse you see d inverse is nothing, but a 1 by x 1 0 because it is a diagonal matrix and inverse of this one reciprocal of each diagonal element. So, this then 1 by x 2 of superscript 0 and this way 1 by x n superscript 0. So, this is our d multiplied by x 0 x 0 is x 1 superscript x 2 superscript 0 and dot dot x n superscript 0. So, if you multiply by this thing you will see in coordinate system the points will be 1 1 1 dot dot last element is 1 this indicates that in coordinate axis that all the point new point in coordinate axis are equidistance from each axis and there in one distance y 1 coordinates is the distance is 1 from this one. So, that point in coordinate axis point is equidistant from each axis. So, it is now centered point in that transform what is called axis. So, you can write it this implies the unit distance unit distance from each coordinate axis from each coordinate axis. So, this transformation is called linear transformation because they are related with a linear expression that sometimes it is called affine transformation or affine scaling transformation. So, now what is the scaled LP problem original problem is was that one. Now, we have made transformation now we will say what is the our scaled LP problem the scaled LP problem now we are considering the scale LP problem. Now, the scale LP problem is minimize f of x now x I will express in terms of transform coordinates f of y is equal to c transpose original objective function was c transpose x, but x is now you see is related with y new coordinate systems is d y in place of x I have written d y. So, it is nothing, but a c transpose d y which I can write it if you see it is nothing, but a d c whole transpose into y that means we know the properties of product of matrices and whole transpose we can write it in reverse order each matrices transpose. Since, d is diagonal matrix. So, we do not need it for write here d transpose. So, this let us call this d c I denoted by small p and d c if you see it is a vector and d c transpose is a row vector and d c is a column vector. So, that d c this into y so where p is equal to d c and the dimension you can say that this dimension is n row one column and this dimension is n by n. So, this p dimension will be n cross 1 again. So, this is the now transform coordinates system our objective function is a equivalently I can write for original system objective function equivalently in transform domain p transpose y is the transform coordinates and p is what d into c that means scaling matrix multiplied by c objective function that coefficient matrix of objective function. So, that subject to what subject to our a x is equal to b that is standard LP problem. Now, a x I have I have converted into a the transform domain that is our d into y is equal to our b. So, now we can write it a into d y is equal to b then I denoted by a into d is equal to b b into y is equal to b into capital B into y is b and that dimension m cross 1. And y dimension you see the y dimension is n cross 1 here also I can n cross 1. So, and b dimension naturally this dimension is m cross n. So, our b is where our b is equal to a into d and p is equal to d into c. So, keeping this thing in mind not only this our y is all greater than equal to 0 since the x is greater than equal to 0. In this expression if you see that how they are related x is greater than equal to 0 and the scaling matrix agree that each element is positive quantity agree then this if you take inverse this will be a greater than equal to 0 that y this. So, with this one our original feasible interior point our original feasible interior point our original feasible interior point x has now become y of 0 and this coordinates are if you see we have seen 1 1 dot dot last is 1 and this dimension is n cross 1. This indicates we have made in transform coordinate axis the our interior point at the centre means it distance from each coordinate axis are same in this case 1 1 1. So, this is what we have seen it now let us say now our job is in transform coordinate system from y 0 we have to move in such a direction. So, that we can achieve 2 objectives 1 objective is to you have to move in such a direction 1 objective is it should be in the interior point that after movement is made it then that point new point must be in the interior point at the in the feasible region next is that point must be that point should be such that that objective function value should reduce compared to the earlier values. So, 2 objectives we have to meet so to improve y 0 with respect to f of y we need 1 a feasible solution or feasible direction 1 we need 1 a feasible solution or feasible direction 1 first objective second is we need a descent direction. So, this objective we have to form. So, if you see if you do for 2 dimensional case if you say if this is our feasible region let us call this is our feasible. Now, y 0 is this one we have to move from here in such a direction agree then our vector coordinates are of this one is here agree this y of 0 this is y of y of 0 this is y of 1 and it should be in the feasible region this is the feasible region and not only this the function value at y is equal to y 0 on the transform coordinates x is what is the function value and that function value must decrease from this value. So, 2 objectives we have to achieve and for that one we have to see which direction will move it such that both the I mean objectives are satisfied to achieve this one to achieve this one what we are doing it here first we will to achieve above objectives what we did we consider a what is called projection matrix consider a projection matrix with the knowledge of our equality constraints equality constraints in transform domain. An equality constraint is b y is equal to small b with the knowledge of b matrix will form a what is called projection matrix and that projection matrix this projection t is defined as I minus b transpose then b transpose whole inverse into b and this dimension if you see this one that dimension of b is m cross n and if you proceed this one this dimension is t dimension n cross 1. So, the knowledge of b matrix in the equality constraints of our transform coordinate system then we form the what is called projection matrix. This projection matrix indicates that if you multiply it by this by b the result is 0 and this projection matrix in addition to this projective what is called the projection matrix the projection matrix this projection matrix has some properties you one can easily verify that properties. So, one properties is there that t transpose is equal to t now look at this one if you take the transpose of this one then it will be nothing but a same matrix you will get it is easy to prove from the structure of that t next is that what you will just put the value of t that take the transpose of this one this is one property and second property is t square is nothing but a t how you prove this one proof of second one is easy to check you multiply t into t and t is what i minus b transpose b of b transpose whole inverse into b. So, that you multiply by i minus b transpose b of b transpose b transpose b of b transpose whole inverse whole inverse then multiplied by b. So, if you do you expand this one ultimately you will get it i minus b b of b transpose whole inverse b. So, kindly expand this one and prove yourself this is equal to this. So, there are two properties are there for projection matrix now I told you I have to find out the direction in such a way. So, that our new point from y 0 to y 1 let us call new point that y 1 point must be in the interior of the feasible region interior point. Second the direction what will go get is the new point that new point function value objective function value must be less than the previous objective function value at y is equal to y 0 of y superscript 0. So, this two objective will so our choice of this one let our direction vector choice n cross n is equal to t into w t dimension n cross n w connection n and w is any vector of dimension n cross 1 this is any vector dimension. So, when it is any vector that d is any directions agree. So, now see this one what will be it become to become d to become d as a feasible direction. In other words to be a interior point of the feasible region d then see this one d has to satisfy this condition that b into let us call our new point old plus lambda some scalar which value is greater than 0 multiplied by d. That is our that means if you think of in two dimensional case we are here now y 0 we moved some other d d is in the directions and that quantity is lambda into d and our new coordinates when you new point is that 1 which is your y of this and this must satisfy our equality constraints that is our b. Now let us call b y 0 is equal to lambda is a scalar quantity I can take it anywhere in the second term this equal to b and this since y 0 was the interior point that it must satisfy the our constraints. So, this value is our b. So, this b this we cancelled. So, our things is coming that ultimately this boils down to beta is equal to b into d is equal to 0. Since lambda is greater than 0 any scalar quantity this positive quantity that means from there d direction is there in this direction you just multiply them. So, this condition if it is satisfy b into d if it is 0 agree b into d is 0 it means that our new point belongs to the in the feasible region mean interior point. So, our d choice is here once you selected the our position matrix w can be any matrix because now see why it is any matrix. Now you see b into d b d is what we have selected t into w and w is a vector any direction vector this vector. So, let us call it is nothing but a b into t into w and what is b this and what is t I am writing i minus b t of b b t transpose whole inverse into b into w. So, if you expand that one it will be b minus b multiplied by 0 for any value of vector this results is 0 agree only the condition is there if you see our this that b matrix what we have a and d that must be a full rank that is the only conditions. Then this inversion is exist. So, this is the so our condition is b into d b is formed if it is 0 indicates the direction d should be in such a direction. So, that b into d is equal to 0 then that means we move from initial point to another point which is also in the interior point of this our transform coordinates interior point in the transform coordinates. So, this is our direction we got it now what is the guarantee that you have we have moved in that from y 0 to y 1 in the direction of d that is no doubt it is a we can prove if it is this equal to 0 b into d 0 we proved that it is a interior point, but there is no guarantee that what is called objective function value you will get what is called what is the objective function value you will get less than the previous objective function value at y is equal to y 0. So, in order to ensure that one what we did you see this one now next is your see to make second achievement to make a steepest descent direction to make a steepest which comma we select w vector. Previously we told w is any vector, but it will satisfy that our if you move any direction it will be inside the our interior region but now question is how to get that value of new value new point which will give you the objective function value less than the previous value of our previous point that value of the objective function value. So, if you select w is equal to minus p let us see what will be the our cost function values we know our cost function objective function in what is called coordinate system objective function value is what p transpose what is this y and that value if it is descended that value must be less than equal to p transpose y of 0. Now, let us call p transpose y is we move from y 0 to y 1. So, it is a y 0 plus lambda into d and that objective function value must be less than p transpose y 0. Now, you expand this one y p transpose y 0 plus lambda p transpose d is less than p transpose y 0. So, this this cancel. So, this implies that p transpose d is less must be less than 0. So, if this is the descent direction means objective function value will decrease if you move from y 0 to new point y 1 that objective function value will decrease provided this condition is satisfied. Now, since lambda is greater than 0. So, this condition is satisfied when if you choose w is equal to minus p then it is obvious that p transpose d will be less than 0 p and d we have if you choose d is what if you see t w d of direction of d we have now I am telling if you select the w instead of any direction now I am choosing w is equal to minus w is I am selected minus of p. So, w I am selecting minus of p you see here minus p if you selected this that minus p this one. So, it will be coming now p transpose t w minus p and minus is I take it minus here then p just now exploiting the stretch what is this called properties of t. So, this t I can write t square because t square is equal to t into t into p. So, I am written now I am writing this one t into p there also t into p whole transpose minus. So, this and this is what if you see this one this is our vector of dimension n row 1 column this will be a whole thing will be 1 row n column. So, it is nothing but a a clean norm of the vector p into t square and this quantity this quantity is always greater than equal to 0. This is facility with minus. So, this indicate with the choice of p is equal to what is called w is equal to minus p that we are getting with the choice of with the with the choice of d is equal to t minus p d is equal to t minus p we are achieving the two objective simultaneous. One is feasible direction we are moving another is the objective function new point objective function value at the new point value is less than the objective function value at y is equal to the that earlier point. That means in the y of 0 is less that ensures. So, therefore, the direction vector d is equal to minus t p is a feasible direction and also and a descent direction direction direction as well. So, what is our coming in short that means if the original problem is given if it is of center d is there you transform this into a new coordinate axis by selecting the what is called the scaling matrix after you reach to the what is called coordinate. So, what is a new coordinate system then you select from your starting point of y 0 you move in such a direction and the direction is d is equal to minus t p and p is the nothing but a coefficient of transform coordinate system objective function coefficient matrix or coefficient vector this one p and t is form with the knowledge of v matrix which is called projection matrix. If you select this one d is equal to minus t p we will simultaneously achieve both the objectives. So, this is now let us take one simple example and see now before that we will see what is what should the choice of our lambda what should be our choice of choice of that means in d direction how far we will move from y of 0 to what is what length in the direction of d we will move it. So, lambda will decide the what length so how to choice of d now look at this expression y is new point is we will get it with the knowledge of initial point plus lambda into d where lambda is greater than equal to 0 and our d where our choice of d is equal to minus t into p. And this choice will satisfy both the objectives will satisfy what will satisfy equality constant in transform coordinate axis b y is equal to small b this will satisfy one another condition is what will satisfy p transpose y will be less than equal to p transpose y 0 sorry superscript 0 this both the condition will satisfy this one. Now, let us say what is mean this one if you see this one y is equal to y 0 of lambda d and in our what is called transform coordinate system our l p problems that means scaled l p problem you will see the y values will be greater than equal to 0 y value is greater than equal to 0. Since y 0 is greater than equal to 0 lambda is positive quantity. So, we have a two choice on l if d d elements of d if it is a positive then the let us call ith element of d is positive then it ensure the ith element of new point also positive. Now, question is if the ith element of d is negative then we may get the difference even though y 0 is positive lambda is positive but ith element of d is negative if it is negative the this two difference of this two we can get negative, but that depends on the choice of lambda and we can make the choice of lambda. So, that in that situation the ith element of y must be positive or greater than equal to 0. So, let us see that since you can write it since y 0 sorry y of 0 is feasible that means y j jth component of y j is greater than equal to 0 for all j this implies feasible mean this implies. Now, there are two situation I have considered first situation is if d j of 0 is all are greater than equal to 0 this then y j will be greater than equal to 0 since lambda is greater than 0 this is straight forward. There are two second situation if d j 0 is less than equal to 0 then y j since lambda is positive y j jth element of y j can be negative or positive depending upon the choice of lambda. So, we are now keeping the choice of what is called lambda should be make. So, that we can get that y jth component of y is positive then you can write it y j y j 0 plus lambda j d j 0 what we want to do is we want to greater than 0 if this is less than 0 this we want greater than 0. Now, both side you have multiplied by minus that you know y j 0 value is what positive and that value after scaling that transformation and in what is called in transform coordinate axis this value is if you see is nothing but 1. So, y j both side I am multiplying by minus. So, this is a lambda j d j of 0 is less than equal to 0. So, now you can write it that one minus lambda j d j superscript 0 is equal to y j 0. So, lambda j is equal to less than equal to y j 0 minus d j 0. Now, you see the d j 0 is negative and negative positive this is already positive quantity and it is greater than 0. So, this quantity lambda is positive quantity. Now, we have a how many components of y and d j we have we have a n components j is equal to 1 2 3 this when any one of the component of d j is positive we ignore we do not need it because the sum of y 0 sum of this one will be sum of this one will be always positive if d j is 0 we have positive then this will be always 0. So, when d j is positive quantity then we do not consider ignore this if it is a negative then see this one. Now, I am choosing that what is the value of lambda then lambda max is equal to lambda max is equal to minimum of lambda j superscript 0 less than 0 for negative value of lambda. See this ratio will be minimum which case it will be minimum and that will consider lambda max because this value is positive and this value is d j value is negative negative positive. So, this ratio that d j component let us call in d vector there are 5 components are there is negative out of this out of this which one is minimum this ratio because this minus this ratio will be positive which one is minimum that will consider as a lambda max and this will ensure this will ensure that y of that is y of y j will be always greater than equal to 0. This will ensure this will be greater than equal to 0. So, now the remarks I am writing can you tell me when it will be the 0 y y j will be 0 when because I am taking the minimum of this one which case it is the minimum let us call fifth position of d j j is equal to 5 this is I got it minimum and corresponding that y j element will be 0 because it will be cancelling that one lambda j value is this one multiplied by d j jth component 5 jth component 5 denominator cancelled. This will be cancelled then this what is called it is same as your y j 0 that means it will be cancelled ultimately it will be 0. So, our remarks is that if we choose lambda is equal to lambda max as minimum of delta j superscript 0 less than 0 that is negative for negative quantity only jth component of d is equal to y j 0 y j 0 divided by minus d j 0 and this value is 1 in case this. Then one at least one will be 0 in y j component at least one will be 0 in y j component at least one it may be more than one 1 or 2 variables will be 0. If it is 0 our new point our new point will no longer be at interior point because it is boundary on the feasible region it is it will be boundary on the feasible region, but it is not in the interior point of the feasible region. So, you can write it and the new point and the new point will no longer be an interior point interior point. So, that is why because I told you it will be 0 because if you take lambda is in this way lambda max as like this way. So, what to avoid that problems because at least one will be 0 when one coordinate axis value will be coordinates of y will be 0 that means it is on the boundary then y is 0 means it is on the boundary. Then what to avoid that one what you did it we whatever the lambda max will get it to avoid that situation you multiplied by this thing by a 0.9 to 0.95 within this range you make it if you multiplied by lambda max with 0.9 to 0.95 then corresponding that element will not become 0 that is our main aim before it reaching to the optimal point it will not be 0. So, to avoid that one we select lambda is equal to lambda max as lambda max is equal to or you see we select lambda ultimately lambda as a as lambda is equal to 0.9 to 0.95 multiplied by lambda max. What is the lambda max we got it here lambda max as this one that you multiplied by 0.9 to 0.95 and then you consider that is the our step length in the direction of d will move and that will overcome that problem that means that no component of y new point y will be 0 that means it will not reach to the boundary point of this region. So, if you look at this one our major competition burden to compute that one i minus b b into b transpose whole inverse b this is the major competition involved in this process. Let us call we solve the same problem what we have considered earlier. I am briefly and quickly I will solve this problems refer our earlier problem we have considered minimize f of x is equal to minus x 1 plus x 2 and subject to this is equal to minus 1 minus 1 1 that x 1 x 2 and our subject to subject to x 1 plus x 2 is less than equal to 5 then x 2 is equal to less than equal to 4 and x 1 x 2 is greater than equal to. So, it is our LP problem, but it is not a standard LP problem. So, you have to convert first the standard LP problem this. So, solution convert to into convert into standard LP problem. So, without much discussion it is a we have discussed. So, standard LP problem minimize minus x 1 plus x 2 and it is of course, minus 1 if you say 1 x 1 and x 2 and subject to we have to convert into a equality constant. So, this quantity is less than 5 that means, we have to add in a what is called slack variables. Let us call we added slack variables is x 3 is equal to 5 and in this case x 4 x 2 plus x 4 is equal to 4. So, our x 1 x 2 x 3 and x 4 is greater than equal to 0. So, this has now converted into our standard LP problem. So, now you identify which one is our A B and C. So, if you see our A matrix from this 2 equation I can write it A matrix quickly I am writing that A coefficient of x 1 is a sorry this 0 coefficient of x 4 is 0 there and then this is 0 1 coefficient of x 2 is 1 and coefficient of x 3 is 0. Then this is 1 this is A matrix this is B matrix is 5 4 and our C matrix is what you see this our C matrix x 1 x 2 coefficient x 3 4 coefficient is 0 x 4 coefficient is 0. So, we will write it minus 1 1 that is I am writing 0 0. So, this is our C matrix. So, our basic step if you recollect this one that first you convert into a what is called our what is our initial coordinates was there if you see the our earlier problem we have considered the point B is our coordinates sorry point B coordinates in x and x 1 and x 2 coordinates of the original problem B coordinates was 3 1 and it is off centered. So, what you do then iteration before starting iterations what you have to see it the our initial interior point our how many coordinates are points are there we have a 4 coordinates are there x 1 x 2 x 3 x 4. So, initial interior point given as or you have to find out as now I know 3 1 x is x superscript 3 is 3 1 what is x 3 and x 4 we have to select x 3 and x 4 in such a way it satisfies the equality constants agree. So, in order to check the equality constant your C a into x of superscript is equal to our case is 5 and 4. So, first equation you see here the first equation x 1 is minus 3 and x 1 is your sorry x 1 is 3 x 2 is 1 x 3 must be 1 then only it will be 5 x 4 0. So, our things is x 1 is x 2 is x 3 that is 1 for the time I can make it x x 4 0, but here you see come to this point x 2 is what 1, but I have to get it right inside 4 then x 4 must be 3 then only 1 plus 3 will be 4. So, our this will be 3. So, this is our the initial guess which satisfies our equality constant that you have to check first agree. Once you get it that one then your find first see what is the objective function value at x is equal to x interior point of that one. This is our interior point we have verified if you consider how you have selected this one just we have mentioned it this one. So, this equal to this value if you put it C transpose into x of superscript 0 if you put it the value of C minus 1 1 0 0 and x 0 is 3 1 1 3 if you multiply it by this it is a minus 2. So, our at this interior point the objective function value is minus 2. So, what you have to do it next we have to transform and our new transformation is now transformations coordinate of transformations then our x of 0 u 3 is equal to d y of 0 and you know what is y of 0 y of 0 is our d inverse x of 0 what is our d if you remember our d is what 3 1 1 3 this all other elements this d matrix is 4 by 4 all other elements is d. So, d inverse will be 1 by 3 1 1 1 by 3. So, you multiplied by this value will be all 1 1 1 1. So, it is now centered in new coordinate axis it is in centered means that point is that new point y 0 is equidistance from all coordinate axis. So, now see this one the scaled problem the scaled LP problem now what is this scaled LP problem f of y is equal to if you we have done it this things d c whole transpose into y d c is p transpose y and subject to if you say a into d y is equal to b and this is I consider b b into y is equal to b. Now, let us compute what is p p is d into c and d you know it is a 3 1 1 3 4 3 4 3 multiplied by c c is what minus 1 1 0 0 if you multiply it by this is a diagonal matrix other elements are 0 if you multiply it by this I will get it minus 3 minus 3 then this this 1 this and this if you multiply it by this you will get 0 you multiply it by this you will get it 0. So, our p matrix is minus 3 1 0 0 again because I need the information of p in order to get the what is called directional vector and b b is what d into c. So, you put this value of d sorry this is a into d. So, a matrix is what 1 1 1 0 0 1 0 1 and d matrix is 3 1 1 3. So, all other elements is 0 if you multiply it by this I will get the matrix 2 row 4 columns 3 1 1 0 then 0 1 0 1 0 1 0 1 0 1 1 3 just multiply this one. So, this is our matrix b matrix is this one which is in transform coordinate axis b matrix equality constraint we will get it this one. So, now, what is our y y coordinates y coordinate is d inverse that is we have calculated or no I think you have calculated here we have calculated that y inverse is we have calculated we need not to do once again in this. So, now let us see in the transform coordinate axis what is the function value f of y is nothing but a p transpose of y p value just now we have calculated value of p. If you see the value of p what we calculated minus 3 p transpose p is this one is a minus 3 1 0 0 then y values what we got it the y value we got it if you say y of 0 y of 0 value what we got it 1 1 1 1. So, we will see in the transform coordinate this value is minus 2 it is same as our original problem we get it. So, once you just you check the original way now what is our p sorry our directional vector direction will be d if you choose minus t into p then this direction it will satisfy simultaneously two condition one is the feasible direction it will go and not only that the function value that objective function value will be less than the previous objective function value. So, this what is t is nothing but a if you see i minus b transpose b into b transpose i minus b transpose b into b transpose whole inverse b this is our t into p. So, you put the values of d b you know p you know you will get the directional vectors for this one. So, I will continue the next class after this.