hello everybody and welcome to lesson three linear programming formulation previously in lesson we have discussed about models in operation research we defined what model mean we also discussed about the importance of models and we also talk about the models particularly in operational research today in this lesson we are going to discuss about linear programming formulation thus the objective of this lesson is to understand the linear programming formulation from a real problem first of all we have to have a real problem at hand then we have to formulate that real problem into linear programming formulation so as to get the optimum solution or to find out the solution for the real problem linear programming is a mathematical technique for the optimum allocation of scales or limited resources as you know resources are limited or scarce on the other hand needs are unlimited or demands are unlimited so from these limited resources to achieve maximum satisfaction or to get maximum utility we have to use linear programming technique which is the optimum which is used for the optimum allocation or the efficient allocation of scarce resources to achieve the maximum satisfaction linear programming is applicable in so many areas like production mix to determine the production mix or to examine the data problem or to get the proportion of diet to have media selection for the transportation problem and hc for the linear programming to work effectively there are certain such certain assumptions that should be satisfied the assumptions of the linear programming model is the first one is linearity that means the impact of decision variables is linear in constraints and in the objective functions the relationship between those decision variables in objective function and in constraints are linear and the second is divisibility the divisibility assumption states that none integer or fraction values of the decision variables are also acceptable that means the numbers or the result written in the form of a over b where b is not equal to zero is also accepted and the other the third assumption is certainty that means values of the parameters are known and constant all the values are determined or known and the last one is negativity which means all the values of the decision variables greater than or equal to zero or we can say negative values of the decision variables are not acceptable moving to the components or the structure of linear programming models we have four components for extra for the linear programming the first is constraints these are the limitations in achieving our objective or the resources available on our hand which are used to achieve the objectives and second decision variables these are the variables or the volume which will decide or represent the ultimate solutions for the problem and the other is objective function it is a mathematical representation of the objective of making a decision and represented by z it may be maximization or minimization it will be maximization for the profit and minimization for the cost and the last one is negativity restrictions that means all the decision variable should be greater than or equal to zero or negative value of the decision variable are not accepted these are the components of linear programming models the standard form of linear programming models after formulating the li the real problem looks like this here is a standard form the standard form consists the maximization problem or the objective function the objective function by the way can be minimization or maximization and subjected to the restrictions as you see here are the restrictions the restriction or the constraints used to achieve the objective the limited resources that we are going to use to get the objectives or to achieve the objectives and the last one is the negativity constraints or all the decision variables should be greater than or equal to zero now let's start to formulate the linear programming model before that we have to see the steps clearly the first step in linear programming formulation is list and define the decision variables the decision variables is represented by x1 x2 x3 and so on xn these are by the way the ultimate represents the ultimate solution that we are going to solve or it represents the quantities and the second step is to state the objective functions it includes the decision variable in the model and its contribution to the profit or the cost the profit of the cost and the third step is list the constraints the right hand side constraints that are the limitations or the resources that we are going to achieve our objectives represented by from this equation b1 b2 b3 and bn and the last step for formulating the linear programming model is non-negativity restrictions here we have to state explicitly explicitly the non-negative negativity restrictions of the decision variables let's do an example for better understanding about formulation of linear programming models from a real problem a bakery produce only two types of burger burger type a and burger type b rose burger requires flour and meat only each unit of burger requires two units of flour and three units of meat whereas burger b requires one unit of flour and two units of meat the company kitchen has a total of five units of flour and 30 minutes of meat on each side the company makes a profit of per 8 per unit of household and per 7 per unit of b sold so this is a per unit profit of each burger before formulating the linear programming model from the problem it is advisable to put the data in the table there are two types of burger burger type a and burger type b and burger type any these two units of flour and three units of meat and we can get from burger from burger assault we can get eight beer as a per unit profit and burger b needs one unit of flour and two units of meat and we can make a profit of seven by selling a single unit of burger b and we have a total of 5 units of flour and 12 mils of meat on our kitchen thus the first step to formulate the linear programming is we have to define the decision variables our decision variables are two that is the number of units of burger a and the number of units of burger b that we are going to make that we are going to prepare that's the number of units of burger a is x1 and lets the number of units of burger b is x2 this is the first step in linear programming formulation and the second step is to define or to state the objective function our objective function here is to maximize the profit to maximize the profit obtained from both burgers so the total profit is represented by z and the profit is the total number of units of a and b produced multiplied by its per unit profit that is br eight and percent respectively so this is the maximization object function maximize z equals 8 times x 1 8 means the per unit profit obtained from a and x 1 is the number of a burger plus seven seven is a per unit profit of b type burger and x two means the number of b type burgers that we are going to prepare so this is objective function and the third step is to list down the constraints to restrain the consequence the company will try to produce as many units of end b to maximize the profit so this as a number of a and b units increases we can increase the profit we can increase the profit but the results of floor and meat are available in limited amount this is a constraint each unit of a and b requires two and one units of floor respectively a requires two and b requires one unit of floor respectively the total amount of floor available is 5 units so to represent this mathematically we can state like this 2 x 1 plus x 2 less than or equal to 5. we can go up to 5 but not more than five because we only have five units of floor and for each end of a and b requires three units and units of meter respectively the total amount of meter available is 12 units so we can state mathematically three x one plus two x two less than or equal to we can go up to two but not more than twelve because we only have two units of meat and the fourth step is our decision variables that are x1 greater x1 and x2 are greater than or equal to zero both x1 and x2 are greater than or equal to zero so after finishing the problem formulation we can put like this standard we can put as a standard form of linear programming model that is maximization that equals 8x1 plus 7x2 subjected to the constraints to x one plus x two less than equal to five five is the units available the floor units available and three x one plus 2 x 2 less than equal to 2 and 12 is the available meet the available mid and the other is the negative restriction that that is x 1 and x 2 are greater than or equal to 0 this is what we call formulating the real-world problem in dramatic mathematical model after this the next step is to find out the optimal the feasible and optimal solution let's add more problems to better understand how to formulate the lead the real time problem into linear programming model the second problem is a workshop manufacturers three types of products a1 a2 and s3 each product uses cutting drilling and polishing machine the process time per unit of a1 on cutting is 10 hours drilling machine is 8 hours and polishing machine is nine hertz and the process time per unit of a2 on cutting is 15 hour drilling machine is general and polishing machine is 12 power the process time per unit of s3 on the cutting is 10 hours drilling tiravar and polishing is 15 hours but the maximum number of hours available per week on cutting drilling and polishing machines are 80 70 and 60 hours respectively these are the resources the limited resources that we have also the profit per unit of a1 a2 and s3 are 25 35 and 40 dollars respectively thus formulate the linear programming model of the product such that the total profit is maximized so we are formulating the maximization problem before formulating the linear programming model it is to put all the it is better to put all the data in the table so as to better understand also is the formulation so it is presented in step like this so the machines cutting machine drilling machine and polishing machines are listed here and product a1 requires 10 hour from cutting machine eight hour from drilling machine nr from polishing machine and we can make a profit of 25 by selling a single product of a1 and for product a2 it requires 15 10 and 12 hours from cutting drilling and polishing respectively and we can make 35 dollar or that vapor by selling a single item of a2 and the same is for product s3 and the house or the limited resources are the limited resource are the time available for each machines the cutting machining machine is a polishing machine for the cutting mat we have only 80 hour drilling machines 70 hours pushing machine 60 hours so using the previous steps determine we have to determine the production volume of a1 a2 and s3 so us to maximize our objectives that is profit maximization profit maximization keeping the constraints one let step one is we have to define the distribution variable x1 x2 and x3 be the production volumes of products a1 a2 s3 respectively and step two is we have to state the objective function that is maximization optimization z equals 25 times x1 25 means the per unit profit of a1 and x1 is the number of production volumes of a1 35 is approved plus 35 times x2 35 is a number of no 35 is a per unit profit of a2 whereas x2 is the number of product a2 and plus 40 40 is a profit per unit whereas x3 is the number of s3 so this is the profit the profit after stating the objective function the steps is we have to mention the constraints we have to mention the constraints the constraints for the cutting machine is 80 we only have 80 hours for cutting machine so 10 x 1 plus 15 x 2 plus 10 x 3 less than or equal to 80 all all the products used up a time of cutting machine not more than h not more than it because the maximum available time for cutting machine is 80 and the same is true for the remaining same step for the remaining constraints and the last is step 4 where x1 x2 and x3 are greater than or equal to zero that is the negativity restrictions after formulating the linear programming model from the real problems we have to put it in a standard form of linear parameters that's maximization z equals 25 times x1 plus 35 times x2 plus 40 times x3 the objective function that's maximization subjected to the constraints 10 x 1 plus 15 x 2 plus 10 x 3 less than or equal to 80 that is available house for cutting eight x one plus ten x two plus twelve x three less than equals seventy n x one plus twelve x two plus fifteen x three less than or equal to sixty and the last one is negativity restrictions let me add one more last example from minimization you are on a special diet and know that your daily requirements of four nutrients is 60 mg of vitamin c 1000 milligrams of calcium 18 milligrams of iron and 360 milligram milligrams of magnesium you have two supplements to choose from vega vita and haphaels vega vita con costs 20 cents per tablet whereas half health costs certain sense per tablet the gravita contains 20 milligrams of vitamin c 500 milligrams of calcium 9 milligrams of iron and 60 milligrams of magnesium happy health contains 30 milligrams of vitamin c 20 250 milligrams of calcium and 2 milligrams of iron and 90 milligrams of magnesium how many of each tablet should you take each day to meet your minimum requirements while spending the least amount of money so here our objective is our objective is to minimize the cost to minimize the cost by achieving the minimum requirements the minimum requirements as i said before before moving to the solution it's better to put the data in the table to ease our analysis to is our formulation so the items or the nutrients are vitamin c calcium iron and magnesium and the minimum requirements are 60 1018 and 360 respectively and this is the content of the content of the nutrients of for vega beta and this is the content of the nutrients for happy health appeals and their correspondent costs are 0.2 dollar and 0.3 dollars for vega vita and half hills respectively by the way even for the minimization the step is the same so the first step is to define the decision variables that is x 1 and x 2 the number of vega beta and half l respectively and step 2 is a minimization because here is a cost so this is the total cost the turquoise is expressed with 0.2 times this is the cost per unit times the number of x1 that is the number of vega beta plus 0.3 is a cost per unit times x to the number of half health tablet so this is the total cost that is minimization and step three is the constraints this is the requirements the minimum requirement for vitamin c is at least 60 milligram at most it can be more than 60 milligram but at least this is a minimum requirement so we have we can express like 20 times x1 plus 30 times x2 greater than or equal to 60 because this at least at most we can't go more than 60. at least 1000 milligrams of calcium is required so 500 times x1 plus 250 times x2 greater than or equal to 1000 that is the minimum requirement and the third constraint is at least 80 milligram of iron is necessary so 9 times x1 and mx1 plus 2 times x2 greater than or equal to the minimum requirement and the last one is at least 300 60 milligram of magnesium is required so 60 x 1 plus 90 x 2 greater than or equal to 360. these are the constraints the constraint after stating the constraint the last is all the decision variables should be greater than or equal to zero that is the negativity restrictions and this is the standard form of the linear programming model that's minimization z equals 0.2 times x1 plus 0.3 times x2 subjected to the constraint 70 x1 plus 30 x 2 greater than or equal to 60 500 x 1 plus 250 x 2 greater than or equal to 1009 x 1 plus 2 x 2 greater than or equal to 18 60 x 1 plus 90 x 2 greater than or equal to 360 and x 1 and x 2 both should be greater than or equal to 0. this is all what i have for today next time we will meet with another lesson till then have a good time bye thank you