 class we have discussed that what do you mean by optimizations. We broadly classified the optimization in two classes, one is static optimization another is dynamic optimization. Static optimization is concerned with the design variables, concerned with the design variables that are not changed with time and we have discussed what are the techniques are available to solve such type of static optimization problems. Another class is dynamic optimization problems and that it is concerned with the design variable that changes with time and time is becoming a function of this in the problem statement. So, there also we have discussed what are the techniques are there to solve such type of problems. Then we have taken the what is called that a problem statement is given and this problem statement we have translated into a mathematical form. Then we have seen this mathematical form represents a objective function which we are supposed to optimize either minimize or maximize subject to constraints. Constraint may be equality constraint or that inequality constraint or both and in addition to that there is a side constraints are there. With a suitable example, we have explained how to formulate the statement of the problem into a mathematical form. Then graphically also we have represented for simple example, graphical case also we have seen how to represent the optimization problem in graphical form. Then the next is your what is called that multiple optimization problems today we will just talk about the multiple optimization problems, multiple objective optimization. Let us consider a company has received what is called a production order of 400 units and he has to supply within a week and this company has to produce this unit. There are two options are there two production lines are there one production line is a another production line is b. So, the company has a two production lines two production lines. Let us call this production line is line a line a another production line is line b this line a the regular working hours the regular working hours in a week regular working hours in a week 35 hours 35 hours. Similarly, here also line b also or let us call 30 hours this 30 hours and line b also regular working hours. Thus this unit can be produced through line a or line b in addition to this that line a produces the produces line a produces 6 units per hour and the per hour operation cost per hour operation cost is rupees 400. In addition to this that over time is available up to 35 hours in line a over time is available available up to 35 hours in a week through line a. So, similarly in line b the produces line b produces 5 units and per hour operation charge per hour operation cost or charge is rupees 550 let us call. And similarly, over time available over time is available through line b over time is available up to 35 hours in a week. Now, what is our problem this is the statement of the problem company has got a production order of 40 units and this company has a two lines are there to produce this unit line a. And it has mentioned that in line a the regular working hours in a week is 35 30 hours line b 30 hours, but in one hour the line a produces 6 units and per hour cost is 400 rupees. And it is also mentioned that over time available in line a up to 35 hours in a week. Similarly, the line b produces 5 units per hour and per hour cost or charge is 550 rupees and over time is also available 35 hours in a week. Now, our main objective of this problem is two folds first we have to minimize the production cost the total production cost for 40 units we have to minimize first this is the objective function. Another objective function is at the same time we have to maximize the what is called regular working hours in a week. So, that is the two objectives we have in our hand. So, it is called multi objective more than one objective is a multi objective optimization problems. And we have a from the statement of the problem we have a different constraints are there. So, now our job is to convert this problem into mathematical form and then solve it. So, our main objective if you say the objective are one production objectives are one minimize the overall cost overall production cost for 400 units that we received order from the the next is we have to maximize at the same time at the same time at the same time maximize cost of maximize sorry maximize the utilization utilization of a regular working hours in a week hours in a week. So, this is the our objective there are two objectives are there minimization something and maximization also simultaneously you have to maximize another functions. So, let us this statement of the problem let us write in mathematical form. So, this problem is we can tell more specifically it is two objective optimization problems. So, let us write all the information in tabular form. So, that we can finally, we can write in mathematical expression form. So, we have a units per hour another is total hours for a week total hours for a week then cost in rupees per unit and units per hour can manufacture through two lines. So, line A we have defined the two lines line A and line B and total hours for a week line A and line B cost or the charge operation charge for per each unit or per unit line A and line B line A line B. So, here is our time whether is a regular time or over time. So, regular time the units of per hour through line A produces six unit through line B five units naturally over time also same units will be produced over time the per hour through line A is six per hour through line B is five units total hours for a week line A is 30 hours line B also same time or hours and over time in both the lines is 35 hours cost per unit through line A is 400 rupees per unit and through line B is 550 when it will go for over time then it is a 500 rupees instead of 400 500 and that is a 600 rupees. So, this table we can we have written from the statement of the problems. So, let us consider our design variables the objective function that is what we will define that is the function of our design variables first X A R this indicates production of units production of units during a regular working hours per week working hours per week through line A that is the unit that means X A R unit is produced during the week through regular hours working hours then X B B stand per line B similarly production of units during the regular working hours through line B per week. Now, X A suffix 0 line through line A during the over time working hours that is the production of units production of units during over time through line A and similarly B line through line B over time production of units during over time through line B in a week. So, this is the design variables we have defined. Now, immediately we can say that what is the cost that is overall cost is for the for manufacturing 40 400 units what will be the total cost will be for the 40 units. So, our problem is minimize the total cost minimize the total cost. So, we know if you see the from the table we know the production unit during the regular working hours through line is A R this number of units through line A is produced or manufactured then per unit cost is 400 rupees. So, 400 multiplied by A R is the total cost through line A during the working hours regular working hours. So, it will be a this total cost minimization total cost is a function of design variables X A R X B R X A 0 over time number of units and X B 0 and which equal to 400 into X A R plus 400 rupees. When you will manufacture the unit through line A during the what is called working hours the cost per unit is through line A is 400 rupees we have a X A R number of units produced through the through line A. Similarly, through line B is X B number of units produced through line B during the working hours regular working hours and the cost of this one is if you see from the table that is given 550. So, 550 into this one plus over time through line A is number of unit produced in through line A during the over time is X A 0 and the cost is your 500 rupees over time through line A. So, it is multiplied by 500 plus number of units produced during and number of units produced through line B during the over time hours is X B 0 and each unit cost is 600 rupees each unit cost is 600 rupees. So, it is a 600. So, this is the total cost. So, this cost we have to minimize that objective function I told you last class is an scalar function and it is a function of design variables our design variables are 4 A R B X B R X A 0 X B 0. So, you have to minimize this cost total cost. So, in addition to this we have a another objective function that is the maximization of regular working hours. So, another objective function is the maximize the utilization of regular working. So, that is let us call we write that is the function of what X A R X B R regular working hours. In the regular working hours we have produced X A R units and through line A how many we have produced X B R through line B. Then what is the time required to produce that X A R units through line A during the regular hours. So, that is we have seen from the table or from the statement of the problem that 6 units is what is called 6 units is manufactured in 1 hour. So, it will take how many hours is there X A R by 6 plus X B R by 5. So, this is the regular hours through both the lines. So, that time we have to maximize. So, there are two objective functions as there and this objective function one is maximization this maximization another is minimization. So, subject to our constraints then what is the constraints the subject to the following constraint and that constraints we are obtained from the statement of the problem constraints subject to. So, what is this that number of units we have to produce through regular working hours through line A and B as well as number of units we have to produce through over time in line A and line B. So, what is the total units we can write it one that is one conditions that X A R plus X B R plus X A 0 over time number of units produce through line A during the over time hours is X A 0 plus X B 0 is equal to 400 units that is as per from the statement of the problem. Second condition is given constraints are given that if you see the total hours through line A maximum hours in a week is 30 line B 30 and the over time through line A 35 line B is 35. So, we have a this constraint. So, we have a X A R is the number of units produced through line A during the regular working hours. So, how many hours is required then six units is produced in one hour then this much of unit will produced in this much of hours. So, that must be less than equal to 30 hours as per statement of the problems again. So, another things are there that is X B R by 5 also less than equal to 30 X B R is the number of units produced through line B during the working regular working hours and through line B 5 units that time required per hour it can produce 5 units. So, total hours is this much it must be less than 30 as per specific statement of the problem. And also through over time the third X A 0 divided by 6 time is must be less over time must be less than 35 and fourth constraints is X B 0 number of units produced through line B during the over time working hours is X B. Since it is required 5 in one hour through line B 5 units can produced then this unit how many times how many hours is required that we can find out that must be less than equal to 35. So, this is line A through line A line B you can say line A line B line A and line B these are the constraints is given to you in addition to that we have a side constraints are there what is the side constraints all the design variables this design variables X A R X B R X A 0 B 0 is greater than equal to 0 it cannot be negative quantity. So, another side constraints are the side constraints X A R X B R X A 0 A O X B O is all is greater than equal to 0 non negative this is called non negative design variables and this is called side constraints. So, this is the mathematical form these are starting from let us call this is equation number 1 this is equation number 2 and this equation number 2 is a equality constraints. Then you can say equation number 3 4 5 6 all are inequality constraints this all these things are inequality constraints constraints and these are the side constraints. So, this is the mathematical form of mathematical model with 2 objective functions agree. So, it is called multi objective optimization problems. So, we will discuss later how to solve such type of problems. So, we have now come across what is called single objective function multiple objective function in this case. So, next is your classification of optimization. So, if you see this the 2 example we have considered one for single optimization problems from the statement of the problem we have formed the mathematical formulation of optimization problems agree. Then multiple objective optimization problems we have seen and we now we will discuss the classification of optimization problems. So, first is level 1 level 1 is the statement of the problem is given that means general problem. Next is next problem is next level is your cost or objective function classification. So, next level is cost to objective function classification. What is this means whether is a single objective function optimization problem or it is a multiple optimization problems multiple objective optimization problems it is either this is a single objective objective optimization problem or multiple multi objective optimization problem. So, this is classification second level third level is problems classification. So, problem classification means whether it is a unconstrained optimization or it is a constrained optimizations that you have to classify. So, far we have discussed the 2 problems both the problems is constrained optimization it involves inequality constraint and equality constraints. Some problems may be there is no constraints involved in the problem agree. So, we have to maximize or minimize or optimize the function agree which is a function of design variables only agree there is no constraint. So, we have a problem we have to say whether this optimization problem is unconstrained unconstrained or constrained optimization problem level 3. Level 4 is variable classification variable classification means that the design variables what we have considering here the design variables are real or integer or mix of both. So, depending upon it will be called that classification of variable we are dealing with the continuous variables or discrete variables agree that not discrete what is called integer variables or mix of both real and integer variables. So, next is variable classification variable design variable you write design variable design variable classifications that is real variables are real or integer or mixed mixed mix of both real and this when you have to call real variable it is a continuous integer means discrete form. So, this and last level is your level 5 function classification that means function classification is this optimization problem whether it is a linear or non linear optimization problems or convex optimization problems or non convex optimization problems or it is a this problem that optimization problem is differentiable or non differentiable. So, we can do this one if you see the example 1 we have considered that example 1 for designing a can. So, that is a non linear optimization problems because in objective function was non linear and other constraints you can see it that mainly objective function is a non linear functions is there that case whereas, in the what is called the multi objective optimization problems. If you see the carefully the equations see this equation objective function is a linear equation the constraint this is the linear equation then this is equalities all are linear equation. So, this multi objective optimization problem is a linear optimization problems. So, function classification means linear the optimization problems is linear non linear or non linear non linear convex or non convex or differentiable differentiable or non differentiable. So, basically the general problem the problem of optimization can be labeled into 5 one is general problem statement is there from second level is your cost and optimization cost and objective function classification. Third is problem statement problem classification whether it is unconstrained or constant optimization problems. Then fourth is variable classification means whether the design variables are real integer or mix up of both and last level 5 is a function classification whether it is optimization problem is linear non linear convex or non convex or the this objective function all these things are differentiable or non differentiable class if we have can classify. So, this is that what the about the details about the different types of class what is called optimization problems. The before that we drive the conditions for optimal to get the optimal value of the function all these things we must know something about the what is called preliminaries of vector calculus and the quadratic functions what is quadratic function all these things. So, the preliminaries of vector calculus and quadratic function preliminaries of vector calculus and quadratic function most of the optimization problems involve the function of several variables if it is a most of the optimization problem the function of several variables. So, it is necessary to introduce vector calculus. So, what is vector calculus we will see so gradient of a function. Suppose we have a function of a suppose we have a function f of x. Now, this x is I will define as a vector of dimension n cross 1 that means x is a vector this elements of this vectors I denoted by x 1 x 2 x 3 x 4 dot dot x n that is why I have given the dimension is n cross 1. So, this is in more details one can write it is a function of x 1 x 2 dot dot x n which in compact form I can write function of x whose dimension is n cross 1 or you can say where x is equal to x 1 x 2 dot dot x n. So, I am written a dimension of this n cross 1 and here also you can write it n cross 1. So, we have a function that function is a that f is a function of several variables x 1 x 2 dot dot x n. So, you know already how to find out the gradient of that function. So, gradient of function is denoted by symbol f of x this symbol will read as a grad of f x and this grad of f x we have written as f of x. We are finding out the value of this gradient of the function at x is equal to x star some value is given that vector x 1 x 2 value numerical value is given at this point what is the gradient of this function find out gradient of the function is nothing, but a partial derivative of the function with respect to different variables x 1 x 2 dot dot x n. So, this we can write it symbol del f del x divided by del x same thing by written mathematically is like this way x is equal to x star where x star dimension is n cross 1 similarly x n dimension is this. So, you compute this one and that will give you the information of the gradient of this function. So, how do you compute that one. So, I told you just is a partial differentiation of function with respect to x 1 then partial differentiation of the function with respect to x 2 keeping all other variables are fixed. So, this equal to you can say this equal to or gradient of f of x equal to partial differentiation of the function x which is a function x means x 1 x 2 dot dot x n differentiate this with respect to x 1. So, this function you differentiate with respect to x 1 keeping other variables x 2 x 3 dot dot x n constant. So, similarly again you differentiate f of x with respect to second variable x 2 keeping x 1 x 3 dot dot x n remain constant in the function and then differentiate and this way you proceed and find up to x n. So, once you got it this one you put the value of x is equal to x star. So, this indicates gradient of the function which a function that f is a function of x 1 x 2 dot dot x n and compute the gradient of the function at x is equal to x star. So, this you can compute like this way or one can write it this is nothing but this if you see look the x is a vector. So, it other way you can say or you can write x 1 is equal to x 1 star put it in this x 1 x 2 is equal to x 2 star and dot dot x n is equal to x n star. This one can write it like this way will show the transpose form del f of del x 1 del f of del x 2 dot dot del f of del x 1 del f of del x del x 2 dot dot del f of del x del x and this whole transpose transpose of a matrix transpose of a vector this. So, gradient of a gradient of a function which is a scalar if you differentiate with respect to a vector the results is a column vector. So, this I can write into this row vector take the transpose also in this form. So, these are all are computed at the given point all partial derivatives all are computed you can write it here x is equal to x star at the vector. Given point x is equal to x star whose dimension is n cross n. So, gradient of a function function that function is a scalar function is a is nothing but a you do the partial derivative of f with respect to different variables write into a vector form. So, that is the gradient of this one. So, next is if you different if you do that derivative of a gradient of a function once again then you will get a matrix what is it mean if you differentiate a vector with respect to a vector then you will get a matrix. So, before that I just I will take you one example and tell you the how to compute this one example. Calculate the gradient for the function f of x is equal to x is a vector in your case is dimension is 3 cross 1. That means, it has a x 1 x 2 and x 3. So, this function is written x 1 square plus twice x 1 x 2 plus 3 x 2 square plus 4 x 2 square x 1 x 2 x 3 plus x 3 cube. Calculate the gradient of this gradient vector for the function this at x is equal to x dimension is 3 x 1 x 2 x 3 that is why I am written in suffix this and that value is 1 2 0 and I have written is transpose. That means the row vector transpose is become a column vector. So, this you have to so you can find out this gradient of this vector like this way. So, first you find out f of x with respect to x 1 you calculate this expression is twice x 1 if you see the expression that one you have to differentiate this one with respect to x 1 keeping other variables constant. So, if you do this one it will be 2 x 1 2 x 2 plus 4 x 2 x 3. Similarly, del f of x del x 2 will come 2 x 1 plus 6 x 2 plus 4 x 2 plus 4 x 2 plus 4 x 2 plus 4 x 2 plus x 1 x 3. I told you I repeat once again then when you differentiate partial differential rule with respect to x 2 keeping x 1 and x 3 constant and del f x then del x 3 is equal to 4 x 1 x 2 plus 3 x 3 x 3 square. So, these things I have just written from this functions partial differentiation of the function with respect to x 1 x 2 x 3 and now the gradient of that vectors is nothing but a grad of this function gradients of the function this I have to compute x is equal to x star and this star means that what value x is given. So, this is or you can say x 1 is given what value of this one is given x 1 is equal to x 1 star whose values is 1 x 2 is equal to x 2 star whose values is 2 x 3 is equal to x 3 star whose values is 0. So, we put these values this is nothing but a this mathematically you can write del f of del x of del x x is a vector of dimension 3. So, this put x is equal to x star and these values are like this way if you put these values are here 2 x 1 plus 2 x 2 I am writing from this expression del f del x 1 plus 4 x 2 x 3 then del f del x 2 is your 2 x 1 plus 6 x 2 plus 4 x 1 x 3 then from this one del f del x 3 is equal to 4 x 1 x 2 plus 3 x 3 whole square and put this value x is equal to x star and this values you know x 1 is equal to 1 x 2 is equal to x 2 star is 2 and 3. So, these values are nothing but a if you see this one is nothing but a 1 2 0 this if you put these values this this shows the value is 16 6 14 8. So, this is mathematically geometrically what does it mean geometrically it indicates geometrically the gradient of a function the gradient of a function is normal is normal to the tangent plane at the point x is equal to x star. This is the now the second derivative of second derivative of a real valued function how to compute real valued function. So, second derivative of a real value function means the function is given first you derivative that function with respect to x. So, that is called gradient of that function then gradient of that function once again you differentiate with respect to a x and x is a vector. So, first thing we are while we are finding out the gradient of a vector or gradient of a function function that function is a scalar quantity if it is then we will get a vector again what we are doing we are differentiating that vector with respect to a vector then we will get a matrix. So, let us see what we are doing it here. So, the second derivative of real function is denoted by f of x that is and finding out the value x is equal to x star double star by definition I told you first you take the gradient of this function f of x then del of x and then once again you differentiate this with respect to this one with respect to del x. So, this is nothing but a and then you put the value of x is equal to x star where the variables are n cross 1 let us call in general. So, this and this we can write it del square of x del x square the second partial derivative you have to do it and put the value x is equal to x star that is. So, let us see what is done this one del of del x. So, first we have done the differentiation of a gradient of f with respect to x that we know what is this expression. So, I will write it this expression here what is this expression this expression is put the value x is equal to x star del del x as it is I have written then this value I have written this value is what del f del x del x 1 del f del x del x 2 del f del x dot dot dot x n then whole transpose this should be a column vector. So, I can write a draw vector transpose it does not matter this one. So, this thing gradient of the function is this one then you can you will differentiate once again. So, how you do it this is a scalar quantity this you see when you differentiate f with respect to or you will get a scalar quantity again you are differentiating this scalar quantity with respect to vector this again you are differentiating this scalar quantity with respect to vector. So, ultimately you will get a matrix. So, this will become now del square f of x x is equal to x star the matrix will be look like this way del square f of x del 1 square then del square f of x del x 1 del x 2 and dot dot dot you will get del square f of x is equal to del x 1 del x n. Second just you differentiate these things you will get this results del square f of x del x 1 del x 2 actually it will be del x 2 del x 1, but these are two values are same. So, next is del square f del x 1 x 2 square x 2 square dot dot dot del square f del x 2 into del x n differentiate them and in this way if you continue the last row you will be del square x del x this is the del x n x 1 del x n then del square f del x del x 2 del x n and so on last element of this matrix will be del square f del x n square. So, this is a matrix and you have to find out the value of this matrix elements at x is equal to x 1. So, this is called that second derivative of real functions how to find out and if you look this matrix it is a symmetric matrix and that matrix name is called Hessian matrix in short it is called Hessian. So, this is a symmetric matrix. So, in short I can tell you if you differentiate second if you take the second derivative of a real valued function of f of x which is a scalar quantity then the result is a matrix that matrix is a symmetric matrix and each element of this matrix is nothing but a the partial derivative of x second partial derivative of f with respect to x 1 x 1 and x 2 all these things. So, one can compute this one and that matrix is called Hessian matrix simply it is Hessian which is a symmetric matrix this is a symmetric matrix symmetric matrix and you know the definition of symmetric matrix the in general if a is a matrix and elements is a i j a i j is equal to a j i i is not equal to j. So, today I will stop it here and I will continue next class with an example that how to find out the second derivative of a function considering the same example. Thank you.