 is the course or optimal control and this course I have just split up into 40 lectures and 20 lectures. Formerly I will cover with the static optimization problem and remaining 20 lecture is a dynamic optimization problems. The books that we will follow as like this one, first is introduction to optimum design. The author is Jesbe S. Aurora, publisher is Elsevier. The second book is an introduction to continuous optimization. The authors are Nicholas Anderson, that Anton Evgrav and Michael Patrickson. This book is Overseas Press Publications Press India Private Limited. These two books mainly I will cover the for what is called the static optimization problems will be covered from these two books. The remaining two books, books three is optimal control systems. Author is D. S. Nidu. It is C. R. C. Press. Fourth book is optimal control and introduction. The author is A. Loka Telly. The publisher is Birkhaus. So, I just mentioned the first two books will be covered the static optimization problems. Third and fourth book is for dynamic optimization problem. Before we start this optimum control problems all this is what do you mean by the optimization of a function that we will discuss is first. So, our first lecture is introduction to optimizations problem and we will explain with some numerical examples or some examples. So, first we will define what is optimization we understand by optimization, then we will consider an example. Optimization is an essential part of design activity in all engineering field. Not only engineering field in other fields also is a design activity in major what is called disciplines. So, we can write optimization is an essential part of design activity. In all major disciplines not restricted to engineering problems. So, it is a process of it is a process of search that seeks to optimize means in other way to maximize or minimize. To optimize bracket either maximization or minimization of a minimization a mathematical function of several variables subject to the constant. The constant may be inequality constant or equality constants. So, a mathematical function of several variables subject to certain constant that constant may be equality constant or inequality constant. The subject of optimization subject of optimizations the subject of optimization is quite general in the sense that it can be looked in different ways depending on the approach. This approach may be a what is called the algebraic approach or geometrical approach. And the nature of variables the nature of variables associated in optimization problems may be real integer or mix of both. And the nature of variables and nature of variables may be real integer or the mix of when it is a real variable this is a continuous variable. When it is a integer variable it is a discrete variable and it can be combination of discrete and continuous variable both used optimization problems. Further this optimization problem can be classified into two groups one is called static optimization another is called dynamic optimization. So, further the optimization problem can be classified into two groups one is static optimization problem and second is dynamic optimization problems. So, static optimization problem basically is a variables which does not change with time. The static optimization problems are the variables that are associated in the optimization problem and it does not change with time and that is called static optimization problems. And static optimization problem can be solved by what is called ordinary calculus. Second the techniques used for solving the optimization static optimization problem is the ordinary calculus then you lagrange multiplier method then third is linear and non-linear programming methods. So, our we just I will write static optimizations what is this it is concerned with design variables that involve in the objective function that back at you can write that involve in the objective function that are not changing that design variables that are not changing are not changing respect to that means, this is a constant variables design variables. The techniques are employed techniques are used to solve this problem that static optimization problem to solve this problem are I just mentioned earlier also it is a ordinary calculus method ordinary calculus second is lagrange method or lagrange multiplier third is linear or non-linear methods programming methods linear or non-linear programming technique. So, this is the static optimization. So, our you can say the static optimization is the is concerned with the design variables that associate in the objective functions and that variables does not change with time that is called static optimization problem. And that optimization problem can be solved based on what is called techniques that techniques are ordinary differential calculus if you know ordinary calculus you can solve it or by using a lagrange what is called multiplier or linear or non-linear programming technique. Similarly, the static optimization we have similarly the dynamic optimization problems optimization you see it is this is dynamic optimization is concerned with the design variable that involved in the objective function again that design variable are a function of time and the time may be involved in the statement of the problem. So, that type of problem is called what is called dynamic optimization problems and dynamic optimization problems are solved the techniques available for solving the dynamic optimization are first is your calculus of variations we can can apply calculus of variations second is your you can apply these dynamic programming third is convex optimization problems. So, first we will see now that what how we will formulate the optimization for static optimization problem based on the statement of the problem then how you formulate. So, I will just write it the definition this is it is concerned with design variables involved in the optimization function involved in the in objective function objective objective function design with design variables are changing respect to time and thus the time is involved the time is involved in the problem statement just I mentioned is the techniques are available to solve such type of problems are three techniques are you can say techniques to solve such problems are one that calculus of variation of variations to the dynamic programming third is your convex optimization problems. So, basically in optimization problem we have two categories static optimization problems and dynamic optimization problems. So, now, I will already explain what is static optimization problem and dynamic optimization problems again next I will that given the statement of the problem how we will form the mathematical model of these optimization problems again. So, let us take one simple example and see how one can formulate a one can get the mathematical model of that optimization problems example. Suppose you have to design a container. So, can is I am just showing it here it is this type of can was height it the height of the can is h which is expressed in terms of centimeter unit is centimeter and the radius of this can this radius are in centimeter this is the can we have to design, but we have a what is our requirement and what is the constraints involved while we will design the can. So, first statement of the problem this can volume should be not more than 200 milliliter that means can contain should not contain more than 200 milliliter of liquid or you can say soda pop because it is restricted that people should not consume or should not drink more than 200 milliliter of soda pop at a time on that basis this can is to be designed in addition to that there are some constraints also involved. Keeping all this thing in mind our job is what is the manufacturing cost is for the can minimum manufacturing cost. So, that we obtain this constraints. So, I will write one by one what is we need it to design before we design this container manufacture this container. So, design a can to hold at most 200 milliliter or 200 cc of liquid or you can say the soda pop that this is restricted because people should not drink more than 200 milliliter of soda pop at a time that is why this the volume of this one is restricted to this one. Second things in addition to this there are some constraints imposed on the physical parameters of this what is called the can that physical parameters is r radius of the can is less than equal to less than equal to 5 centimeter and more greater than equal to what is called 3 centimeter this is the constraint on the radius of this coil or so radius of the can is imposed when you will manufacture the can the radius will be in this range. So, next is imposed that height the height of the can is less than equal to 20 centimeter and greater than equal to 5 centimeter this is the height constraints is given, but for aesthetic reason aesthetic reason means when you will manufacture the can. So, that we can comfortably hold the can again for aesthetic reason there is another constraint is imposed in the design that constraint is the height of the cylinder is greater than equal to 3 r basically you see r and h radius of the can and the height of the can is a side constraints means side constraints means r cannot be more than this or less than this similarly height also cannot be more than this less than this and this constraint is given due to the aesthetic reason in the sense that we can hold the can comfortably with this constraints agree. If the height is more the diameter radius is small then something is not convenient or the diameter is radius is more height is small then it is not convenient to hold the can comfortably. So, this is the constraints now, what is our problem we have to design a can in such a way so that manufacturing cost is minimum. So, our objective is now our objective is to minimize fabrication fabrication cost and assume the cost of the material used to manufacture the can cost of the material used to manufacture or to fabricate the can is rupees c rupees c per centimeter square this is the cost per unit centimeter square of metal sheet or whatever the material you have used to manufacture the can. So, this is our constraint everything is given now I can easily formulate the problems. So, what is the our optimization problems if you look carefully the optimization problem I am just writing given the statement of the problem I am now formulating in mathematical model again form. So, our first problem is optimization problem is let us before that I just considering x 1 is the variable which is denoted by radius of the circle then x 2 is the another variable which is the height of the cylinder in centimeter square. So, our problem is minimize bracket you can say the fabrication fabrication cost. So, our manufacturing fabrication cost or manufacturing cost of can should be minimized and that is generally called objective function that we call generally also objective function or cost function in optimization problems objective. So, what is our objective function that manufacturing cost is involved the circumference of the can which we assume that circular circumference of the can again we are assuming that thickness that cost of the can cost of the material used for fabrication for a fixed thickness it is given that one and with this sheet we are fabricating the can. So, what is our cost involved first is twice pi r circumference of this one multiplied by height. So, this is the material used for covering the circumference of this can plus the bottom and top since it is a circular it is a pi r square again, but there are twice. So, 2 pi r square. So, if you use this what is called that variables now that this 2 pi r is x 1 I have defined that are h is x 2 then 2 pi x 1 square you see this is the centimeter square we know the cost is c rupees c per centimeter square. So, you have to multiply it by c similarly here in terms of variables x 1 x 2 we have to multiply it by c. So, our problem is minimize this manufacturing cost of the can which is expressed by this expression and that expression if you see is a non-linear function and function of x 1 and x 2 radius and height and c is the cost of this material is fixed per centimeter square. So, what is the constraints are involved here next our constraints are here we mentioned it that the can contain at most the 200 milliliter of soda pop or any liquid. So, we can write the minimization subject to. So, what is this one that volume pi r square what is the volume of this one pi r square into h must be equal to 200 milliliter or 200 cc. So, which you can write it pi x 1 square into x 2 is equal to 200 cc. So, let us call this is equation number 1 this is equation number 2. So, this equation if you see this is nothing but a equality constraints. So, I will call this equation is equality constraint and another constraint we have imposed for aesthetic reason. So, that we can hold the can comfortably that constraints is given like this that h is greater than equal to 3 r. In other words we can write it this one that or 3 x 1 r is equal to x 1 minus x 2 is less than equal to 0 this is agree. So, this is equation number 3. So, this is equal to you see this is the inequality constraint this is the inequality constraint. So, we have a objective function we can from the statement of the problem we can write it the what is the objective function for this particular example the objective function is the minimization of manufacturing cost of a can. Then we have a equality constraints from the statement of the problem we can write it equality constraint another another constraint is there which is inequality constraint from the statement of the problem we have written. There are some other constraints are there we call what is called side constraints. So, this side constraints are this one that we have written that radius of this can cannot be more than is less than equal to x less than equal to 5 and greater than equal to 3. Another constraint we have given a height of the can will be less than equal to 22 centimeter and greater than equal to 5 centimeter. So, these constraints are called the side constraints and x 1 x 2 you see is the only design variables which will play role for minimization of the cost manufacturing cost. So, these variables side constraints is given to you agree. So, now one can solve this problem by keeping in mind we have a objective function any optimization problem we have an objective function. Then we have a constraint equality constraint and equal inequality constraint may be there or both is there equality and inequality constraint then we have a side constraints. So, next is this side constraints are necessary just you can say the side constraints are necessary part of the solution techniques. This will tell you about the acceptable region of design variables. So, let us see graphically this problem we can represent this problem graphical representation. So, now see this one x x is I have considered x 1 the radius of the can and y x is x 2 which is the height of the can this is you can say radius of the can. Then with based on the side constraints I am drawing this one you see 1 2 3 4 5 0 1 2 3 4 5 this negative cannot be because radius cannot be negative that the high dimension. So, this part is not there in the another is your height varies from what is called lower limit is 5 and upper limit is 22. Let us call this is 5 10 15 20 and this is 25. So, I know with this constraint side constraint where our design variables may lie. So, this is 3 x 1 varies from 3 to 5 and x 2 varies from 5 to 22 this is 20 let us call this is 22 is somewhere here. So, this is our region where our design variables x 1 x 2 must lie to obtain the minimum fabrication cost somewhere here not only this there is a another equality constraints are there further the search region we can just compress it. What is this constraints are there another constraint if you see this one that it is given that h is greater than 3 r which in turn we can write it which in turn we can write it that 3 x 1 minus x 2 less than equal to 3 because our constraint if you recollect our constraint h is greater than is equal to 3 r. So, 3 r x r is a x r is we have defined as a x 1 variable. So, 3 x 1 minus x 2 is equal to this constraint. So, this inequality constraint I can show in this graph where it is. So, first I will draw 3 x minus x 2 is equal to 0. So, 3 x 3 x 1 minus x 2 is equal to 0 it is a equation of a straight line and in this case this equation of straight line whose slope is 3. That means, this angle will be you can just draw it from this one this is x is equal to 1 that will be x 2 will be 3 somewhere here x 3 is equal to 2 that will be 8. So, if you just draw it this one this is our line. So, this line represents 3 x 1 minus x 2 is equal to 0, but our constraint is given 3 x 1 minus x 2 less than equal to 0. Then which side of this straight line which side of this straight line satisfy this constraints. One can easily say the upper portion of this line upper portion of this line satisfy this equation. So, I am just writing in this side arrow. So, this is the this part is 3 x 1 minus x 2 is less than equal to 0. You can check by any value in this one in this region that this inequality satisfied. Now, see previously we have shown it this whole shaded area. Now, this has become the search area now becoming a smaller with this constraints. So, our if you see this one this is our the search region with the black one this. So, in order to optimize the objective function that our design variable lies in this zone in this region. So, now question is I know this in this region we have a design variables of this one. If it is in this region some value of x 1 and x 2 the our what is called the manufacturing cost will be minimum, but we do not know what should be the value of x 1 and x 2, but we know the region is in this region. So, this is the graphical representation to understand some optimization problems. If you look in this span this optimization problems may be two categories. I am talking about all our static optimization problem because x 1 and x 2 does not vary with time this is fixed this one. Now, you see this this optimization problem just I have written it here. So, this you see f the minimization of fabrication cost if you consider this f f 1 is function of x 1. So, it is a non-linear function this is a non-linear objective function whereas, see equation number 2 this is also non-linear inequality constraints because of it is a product of x 1 square and x 2 and here is product of x 1 square and also x 1 and x 2 and the third equation which is inequality constraint it is a linear equations. So, this problem is called non-linear static optimization problems we have in our hand. So, there are two types of optimization, static optimization may be linear optimization problems and non-linear optimization problems. We will call the linear optimization problem when the objective function and all the constraints are linear then we call it a linear optimization problems. And non-linear optimization problem we will call if any one of this either objective or any inequality constraint may be more than one equality constraints is there in the problem statement or more than one inequality constraint is there in the problem statement. If any one of the constraint is non-linear either objective function or constraints either inequality constraint or inequality constraint then we will call it is a non-linear static optimization problems. Anyone either objective function or constraints any constraint or more than one constraints are non-linear function in variables then we will call non-linear optimization problems. So, we have a two now if I write it in general statement of this one you will see our problem is like this way. So, now we have a we can say it is a linear optimization problem. We have at an linear optimization problem one then another is non-linear optimization. I mentioned you the linear optimization problem will call if the objective function and including all the constraint either equality constraint or non-inequity inequality constraint all are linear then it is a linear optimization problem linear programming problem. In any one of this objective function or inequality constraint either inequality or inequality constraint is non-linear one or more than one is non-linear then it is called non-linear optimization problems. So, in general now I will write the suppose we have a any problems optimization problem is given statement of the problem is given to you. Next with the with the with the statement of the problems I can formulate the mathematical model of the optimization problems. In general if you have a design variables more than two we have consider two design variables one is for this example one is radius of this can and height of the can. In general it may be a n variables. So, we can write in case of n variables n variables means design variables and design variables what are the those are involved in the optimization problems or statement of the optimization problems associated design variables. So, we have a n variables are there and let us say these n variables are in case of n variables bracket you can write say we have a variable x 1 x 2 x 3 dot dot x n these are n variables are there all these variables are not changing with time that is called static optimization problem these design variables. So, we have a n variables are there these n variables are there. So, we can now in general we can write the problem optimization problems when you will call optimization problems the problem may be a maximization of a function objective function or minimization of a objective function previous example it was a minimization of a function cost minimization fabrication of the can will be minimized. So, in general now I am writing f is a function of x 1 x 2 x 3 dot dot x n. So, this is the objective function or it is called the cost function or problem is minimize this function which is called the objective function or this is called objective function or it is called cost function minimize this objective function then subject to conditions are there subject to in general we have shown it here that in general we can write it h i x 1 x 2 dot dot x n equal to 0. So, this is our equality constraints h i how many equality constraints are there I am writing i is equal to now 1 2 dot dot p. So, we have a and any objective function is a scalar function but in this case only we have considered a single objective function we will show with an example that we may have an in optimization problem more than one objective functions simultaneously you have to optimize more than one objective functions at this moment I am just considering one objective function and subject to there are p equality constraints and g j which is a function of design variables x 1 x 2 dot dot x n this x n x n is less than equal to 0 for j is equal to 1 2 dot dot m. So, we have a m equality constraint there is a sorry p equality constraint and there is a m inequality constraints are there in addition to that I told you there is a sum what is called side constraints are there what is this side constraints x i is less than equal to x i u less than equal to x i l this superscript u indicates the upper value of this design variable x i and superscript l means the lower range or value of x i design variables and i varies from what you which have i we have considered there are n variables are there. So, we can write i is equal to 1 2 dot dot n variables. So, this is our more general statement of our optimization problems we have a objective function which is a scalar and which is a function of n design variables x 1 x 2 dot dot x n subject to equality constraints how many equality constraints are there p constraints are there and how many inequality constraints are there there are m inequality constraints and this constraints are called side constraints this is called side constraints. So, this is our statement of the problem in general. So, if this function this optimization function will be a linear optimization problem or non-linear optimization problems if all the objective function equality constraints inequality constraints are linear then it will be called linear optimization problem or linear programming problems if any one of the either objective function or any one of this function is a non-linear then it is a non-linear optimization problems. One can write this equation more compact form by assigning that say let x is a vector whose dimension is n cross 1. I am writing this one x 1 x 2 dot dot x n is the dimension x 1 x 2 the elements of vector x agree how many elements any elements. So, dimension is n cross 1 in the column vector form I have denoted. So, our problem statement now I can write it minimize f of x subject to h i of x equal to 0 i is equal to 1 2 dot dot p and g j of x is less than equal to 0 and j is equal to 1 2 dot dot m and our side constraints are we can write it this one our side constraints is x i x i u and this is x i l and i is in this variables how many variables are and variables are there. So, it is a general what is called the statement of the optimization problems. So, this is only one optimization problems are there only one single objective function is there single objective function is there we are minimizing or maximizing agree. So, next is your what is called multi objective optimization that means in this case we have a more than one optimization problems are more than one objective function is there the same as this one only our objective function is more than one and we have simultaneously we have a subject to equality constraint and inequality constraint and we have also side constraints are there agree. So, this type of problems are called multi optimization problems more than one optimization more than one objective function to be optimized that means maybe one objective function is maximization another objective function may be minimizations. Such type of problems we have and how to handle such type of problems we will discuss first we will how to formulate that type of problems from the statement of the problems. So, I will stop here today. So, next class I will discuss what is multi objective optimization problems. Thank you.