value is determined is called constraints. It can be either a linear equation or inequality.

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1 Contents 1. Index 2. Introduction 3. Syntax 4. Explanation of Simplex Algorithm Index Objective function the function to be maximized or minimized is called objective function. E.g. max z = 2x + 3y Constraints the restriction on the variable by which the objective function value is determined is called constraints. It can be either a linear equation or inequality. Decision variable variables on which our objective function depends is called decision variable. e.g. x, y in this case. URS Variables The variable that can accept both negative as well as positive values example a URS variable x can take values greater than 0 as well as less than 0 values. Non URS The variables that can take only positive values for example a Non URS variable x will have only values greater than 0. Sign Restriction restriction applied on decision variable is called sign restriction. Variables can be restricted in sign in two ways. It may be URS (unrestricted sign) variable i.e. it can assume both positive and negative values including zero or it may be non URS i.e. it should have only non negative values. Feasible region set of all points satisfying all the LP s constraints and sign restrictions. Optimal solution a point in the feasible region which satisfies the objective function.

2 Introduction In everyday life we come across different forms of constraints and problems of maximizing a profit or minimizing a loss. These problems when simplified under any equation and constraints are called linear program. We use some algorithm to find the feasible solution either by graphical method or by simplex method. The graphical method will be efficient only in case of 2 or 3 variables but in the case of more than 3 variables, we will use simplex algorithm. Our application is used to solve any linear program using simplex algorithm. In this program we will be using two methods of displaying the results of any linear program. In the first method we will be required to enter our objective function which has to be either maximized or minimized along with the constraints applied on it. We can enter as many problems we want and at the last line of question we have to add a key word end. In the other method, we will be required to enter the no. of constraints, no. of URS and NON URS variables. URS means un restricted sign variables which can take any value either >= 0 or < 0. Similarly NON URS means which are restricted to take the value which is > 0. Main disadvantage of the second method is that we can get solution of only one Linear Program at a time. And we can use at max 8 constraints and variables for the LPs. The first method is more efficient as it can solve many linear programs simultaneously.

3 But the second method is more user friendly as the user just has to enter the coefficients, of the variables. Syntax In the first method there should be a space before and after every + or in both the objective function as well as the constraints. There should be space after max or min in the objective function. The second method doesn t require any syntactical precautions. Simplex Algorithm Any linear programming problems can be solved either by graphical method (two or three variable) or by using simplex algorithm. So, in order to find an optimal solution to any LP we use simplex algorithm efficiently for more than 2 or 3 variables. Before simplex algorithm is applied to solve LP we first need to convert LP into a problem in which all the constraints or inequality become an equation and all the variable get positive value. This form of LP is called standard form. Consider for an example we have to maximize Z = 4X 1 + 3X 2 Such that X 1 + X 2 <= 40 and 2X 1 + X 2 <=60 and X 1,X 2 >=0 Now for each <= constraint we add a slack variable S i so that S 1 =40 X 1 X 2 and S 2 =60 2X 1 X 1. So any point (X 1, X 2 ) satisfies inequality if slack variable is positive. Similarly for any >= constraint we subtract excess variable (e i >0) so that any point satisfies inequality. So our final standard form of LP is max Z=4X 1 +3X 2 Such that X 1 +X 2 +S 1 =40 2X 1 +X 2 +S 2 =60

4 And X 1,X 2,S 1,S 2 >=0 Now before proceeding to the steps of simplex algorithm we must know the concept of basic and non basic variables. Basic and Non basic variables Suppose we have m linear equation in n variables (n>=m). We make a set of non basic variables which will contain n m variables and set each NBV equal to zero. And we will solve for remaining n (n m) =m variables called BV (basic variables). e.g we have equations x 1 +x 2 =3 x 2 +x 3 = 1 As there are 3 variables and 2 equations so we have to take 3 2=1 NBV. So let s take x 3 as a NBV then our BV will be x 1 and x 2. Now we will set x 3 to 0 which will make this equation to x 1 +x 2 =3 x 2 = 1 Solving these equations we will get the basic solution corresponding to the basic variables. For different set of basic variables we will get different basic solution. But our feasible solutions will be the one which has all the variables containing nonnegative value. Steps to simplex algorithm Step 1 first convert the LP into a standard form. Step 2 obtain a basic feasible solution (bfs) if possible from the standard form Step 3 check if the obtained solution is optimal or not. Step 4 if the current bfs is not optimal then find which non basic variable should become basic variable and vice versa to obtain a better objective function value. Step 5 find new bfs with better objective function value. Go back to step 3. Consider an example where we have to maximize

5 2 2.5 Such that x, y > 0 Using step 1 Proceeding to algorithm steps we first convert the LP into a standard form by adding a slack variable (s i >0) to the constraints. We will form a set of basic and non basic variables by determining n and m here. Here no of equation is 2 (m=2) and variables 3(n=6). So we will get n m = 3 non basic variables Objective function Using step 2 So simplest non basic variables can be NBV={x, y} and in basic variables we will include w as it is not included in any other equations. So BV= {w, s 0, s 1 } We will get this feasible solution w = 0, s 0 = 350, s 1 = 400 by setting NBV equal to 0. Using step 3 We will check if the obtained feasible solution is optimal or not by putting these values in our objective function. i.e. w = 2x + 2.5y. As we get (x, y = 0) so z will become 0. Using step 4 Since our bfs is not optimal we will check for other bfs by forming new set of BV and NBV. If we increase x to 1 leaving y to 0 we will get w increased by 2.similarly if we take y to 1 we will get w increased to 2.5. By above discussion we find that increasing y, we get maximum increase in w. so we will make y a basic variable.

6 In row 1 if we check using NBV as x = 0 => 2y + s 0 = 350 To get s 0 > 0 we should have 350 2y > 0 => y < 175 Similarly in row 2 we will get y + s 1 = 400 => y < 400 So in order to keep all basic variable non negative, the largest value we can make y will be the minimum of the values {175,400} i.e 175. This shows that making y basic variable in row 1 will make NBV positive. This process is also known as determining the entering variable. In short, we can find the row in which we have to make basic variable by the ratio test = right hand side of the row / coefficient of entering variable. Here our entering variable was y. The row which has minimum value of the ratio test is considered to be the winner. We will make coefficient of y in row 1 equal to 1 and all other coefficient of y = 0. So our final BV becomes {w,y,s 1 } and NBV becomes {x,s 0 }. Using elementary row operation on rows we get our new rows as Using step Now check whether objective function is optimal or not Here w = 0.75x 1.25s where BV {w,y,s 1 } and putting NBV = 0,we get w = So checking if we can get other bfs or not. We again find that increasing x will increase w. so using ratio test in row 1 x = 175/.5=350 and in row 2 x = 225/1.5 = 150. Hence x will be a BV in row 2.so using elementary operation on rows again we get new w where BV= {w,y,x} and NBV={s s,s 1 } This will give the value of w = 550

7 As we can see that increasing any of x or y will not increase w so this will fetch feasible solution. So our bfs is x = 150, y = 100 and maximum value of w = 550 Minimization problem In order to maximize any objective function we use two methods Method 1 Minimization can be thought of opposite of maximization i.e. to minimize z = 2x + 3y, we will maximize z = 2x 3y. In this we will solve like the above method for maximization using z as basic variable instead of z. Method 2 In other method,when we find new bf by checking the objective function and taking that variable which has maximum +ve coefficient which will try to decrease our objective function value. In maximization problem we were taking the most negative coefficient variable as bf.