Dr. S. Bourazza Math-473 Jazan University Department of Mathematics

Transcription

1 Dr. Said Bourazza Department of Mathematics Jazan University 1 P a g e

2 Contents: Chapter 0: Modelization 3 Chapter1: Graphical Methods 7 Chapter2: Simplex method 13 Chapter3: Duality 36 Chapter4: Transportation Problem 39 Chapter5: Integer Linear problem 46 Chapter6: Nonlinear programming problem 50 2 P a g e

3 Introduction: Chapter 0: Modelization The operation research (OR), known also as Management Science, System Analysis, Decision Analysis, is the application of the scientific methods, techniques and tools to problem involving the operations of systems so as to provide those in control of the operations with optimum solutions to the problems. During the Second World War, the OR began was to be considered as a separate discipline. In England (Operational Research): British government organized teams of experts to solve perplexing strategic and tactical problems. In USA (Operations Research): Applied OR to the deployment of merchant marine convoys to minimize losses from enemy submarines. After this war, OR extended into industry like Oil refineries, steel, paper mills and to almost all industries, services and large social and urban systems. Mathematical Model: Generally, we look for the solution of the following kind of problem: Max or Min z = c 1 x 1 + c 2 x 2 + c 3 x c n x n (Objective function) Subject to : a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2 (Constraints) Definitions: 3 P a g e a m1 x 1 + a m2 x a mn x n b m all x i 0, i = 1, 2, 3,, n (non-negativity constraints) Linear programming problem: is to optimize the linear function Z subject to certain conditions

4 Objective functions: Is a function of several variables Optimal value: Is a maximum or minimum value of an objective function Constructing the Linear Programming (LP) Problem for optimizing of the Objective Function: Constructing the LP problem requires three steps: Step 1: Define the decision variables. Step 2: Define the objective function. Step 3: Determine the constraints. Example1: Maximization problem A manufacturer produces 2 types of toys A and B. Each toy of type A requires 4Hours of molding and 2 Hours of polishing. Each toy of type B requires 3Hours of molding and 5 Hours of polishing. Molder worker works for 80Hours in a week Polisher worker works 180 Hours in a week Profit on buying a toy A is 3 SAR and on B is 4 SAR Give the linear programming problem of this problem to maximize the profit per week. Answer: This information can be written in tabular form as follows: Molding Polishing Profit A B Time available (Hrs) Step1: Decision variables: x: number of toys of type A per week y: number of toys of type B per week Step 2: Objective function: Maximize the profit Z= 3x+4y Step3: Constraints: Molding total time required 4x+3y 80 Polishing total time required 2x+5y 180 The number of toys produced is a positive integer: x 0, y 0 4 P a g e

5 Summary Maximize Z=3x+4y Subject to: 4x+3y 80 2x+5y 180 x 0, y 0 and integers Example 2: Minimization (Diet Problem) A dietician mixes two kinds of food X, and Y in such way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 12 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1kg of food X and 1 kg of food Y are shown in the following table: 5 P a g e Vitamin A Vitamin B Vitamin C Vitamin D Food X Food Y Kg of food X costs 5 SAR whereas 1Kg of food Y costs 8 SAR. Formulate this text as linear programming problem. Answer: Step1: Decision variables: x: number of kg of food X y: number of kg of food Y Step 2: Objective function: Minimize the cost Z= 5x+8y Step3: Constraints: The word at least in the inequality will be. Vitamin A: x +2y 6 Vitamin B: x + y 7 Vitamin C: x +3y 12 Vitamin D: 2x +y 9 Non negativity of the variables: x 0, y 0 In summary : Min Z= 5x + 8y Subject to: x + 2y 6 x + y 7 x + 3y 12 2x + y 9 x 0, y 0

6 Example 3: Transportation problem There is a factory located at each of the 2 places P and Q. From these places, a certain commodity is delivered to each of 3 depots situated at A, B, and C. The weekly requirements of the depots are respectively 5, 5, and 4 while the production capacity of factories at P, Q are 8 and 6 Units resp. just sufficient for the requirement of the depots. The cost is given by the following table: A B C P Q Formulate the LPP of this transportation problem in order to minimize the transportation cost. Answer: For the Modelization of this problem, we can use the easy way by using six variables or the following way by using only two variables: We obtain the following where x is the quantity of commodity transported from factory P to depot A and y is from P to depot B: Min Z = x - 7y Subject to: x + y 8 x + y 4 0 x 5 0 y 5 6 P a g e

7 Chapter1: Graphical Methods The graphical method is used only for solving LPP with only 2 variables. There are two methods: a) Corner point method b) Iso-profit or Iso-cost method Definitions: Feasible region: The portion determined by all constraints including non-negative constraint of a linear P.P. Feasible solutions: Points within and on the boundary of the feasible region. Optimal solution: Any point in the feasible region gives the optimal value (maximum or minimum) Theorem1: The optimal value of objective function must occur at a corner point (vertex) of the feasible region. Theorem 2: If the feasible region is bounded then the objective function Z has both maximum and minimum values on corner points of the bounded regions. To solve an LP, the graphical method includes two major steps. a) The determination of the solution space that defines the feasible solution. Note that the set of values of the variable x 1, x 2, x 3,...x n which satisfy all the constraints and also the non-negative conditions is called the feasible solution of the LP. b) The determination of the optimal solution from the feasible region. a) To determine the feasible solution of an LP, we have the following steps. Step 1: Since the two decision variable x and y are non-negative, consider only the first quadrant of xy-coordinate plane Step2: each constraint is the form ax+by c or ax+by c. 7 P a g e

8 Draw the line ax + by = c (1) For each constraint, the line (1) divides the first quadrant in to two regions say R 1 and R 2, suppose (x 1, 0) is a point in R 1. If this point satisfies the in equation ax + by c or ( c), then shade the region R 1. If (x 1, 0) does not satisfy the inequality, shade the region R 2. Step 3: Corresponding to each constant, we obtain a shaded region. The intersection of all these shaded regions is the feasible region or feasible solution of the LP. Example1: Let us find the feasible solution for the problem of a decorative item dealer whose LPP is to maximize profit function. Z = 50x + 18y (1) Subject to the constraints: 2x+y 100 x + y 80 x, y 0 Step 1: Since x 0, y 0, we consider only the first quadrant of the xy-plane Step 2: We draw straight lines for the equation 2x+ y = 100 (2) x + y = 80 To determine two points on the straight line 2x + y = 100 Put y = 0, 2x = 100 => x=50 => (50, 0) is a point on the line (2) put x = 0 in (2), y =100 =>(0, 100) is the other point on the line (2) Plotting these two points on the graph paper draw the line which represent the line 2x + y =100. This line divides the 1 st quadrant into two regions, say R 1 and R 2. Choose a point say (1, 0) in R 1. (1, 0) satisfy the inequality 2x + y 100. Therefore R 1 is the required region for the constraint 2x + y 100. Similarly draw the straight line x + y = 80 by joining the point (0, 80) and (80, 0). Find the required region say R 1 ', for the constraint x + y 80. The intersection of both the region R 1 and R 1 ' is the feasible solution of the LPP. Therefore every point in the shaded region OABC is a feasible solution of the LPP, since this point satisfies all the 8 P a g e

9 constraints including the non-negative constraints. b) There are two techniques to find the optimal solution of an LPP. Corner Point Method The optimal solution to a LPP, if it exists, occurs at the corners of the feasible region. The method includes the following steps Step 1: Find the feasible region of the LLP. Step 2: Find the co-ordinates of each vertex of the feasible region. These co-ordinates can be obtained from the graph or by solving the equation of the lines. Step 3: At each vertex (corner point) compute the value of the objective function. Step 4: Identify the corner point at which the value of the objective function is maximum (or minimum depending on the LP) The co-ordinates of this vertex is the optimal solution and the value of Z is the optimal value Corner points Z= 50x + 18y O(0,0) Z O =0 + 0=0 A(0, 80) Z A = =1440 B(20, 60) Z B = =2080 C(50, 0) Z C =2500 Max Z=Z c =2500 for x=50 and y=0. Since our object is to maximize Z and Z has maximum at (50, 0) the optimal solution is x = 50 and y = 0. The optimal value is If an LPP has many constraints, then it may be long and tedious to find all the corners of the feasible region. There is another alternate and more general method to find the optimal solution of an LP, known as 'ISO profit or ISO cost method' ISO- PROFIT (OR ISO-COST) This method of optimization involves the following method. Step 1: Draw the half planes of all the constraints Step 2: Shade the intersection of all the half planes which is the feasible region. Step 3: Since the objective function is Z = ax + by, draw a dotted line for the equation ax + by = k, where k is any constant. Sometimes it is convenient to take k as the Lower Common Multiple of a and b. Step 4: To maximize Z draw a line parallel to ax + by = k and farthest from the origin. This line should contain at least one point of the feasible region. Find the coordinates of this point by solving the equations of the lines on which it lies. To minimize Z draw a line parallel to ax + by = k and nearest to the origin. This line should contain at least one point of the feasible region. Find the co-ordinates of this point by solving the equation of the line on which it lies. Step 5: If (x 1, y 1 ) is the point found in step 4, then x = x 1, y = y 1, is the optimal solution of the LPP and Z = ax 1 + by 1 is the optimal value. Example 2: Solve the following LPP graphically using ISO- profit method. Maximize Z = y. Subject to the constraints: 10x + 5y 80 6x+6y 66 4x+ 8y 24 5x+6y 90 x, y 0 9 P a g e

10 Answer: Since x 0, y 0, consider only the first quadrant of the plane graph and the following straight lines on a graph paper. 10x + 5y = 80 or 2x+y =16 6x + 6y = 66 or x +y =11 4x+ 8y = 24 or x+ 2y = 6 5x + 6y = 90 Identify all the half planes of the constraints. The intersection of all these half planes is the feasible region as shown in the figure. Give a constant value 600 to Z in the objective function, then we have an equation of the line 120x + 100y = 600 (1) or 6x + 5y = 30 (Dividing both sides by 20) P 1 Q 1 is the line corresponding to the equation 6x + 5y = 30. We give a constant 1200 to Z then the P 2 Q 2 represents the line. 120x + 100y = x + 5y = 60 P 2 Q 2 is a line parallel to P 1 Q 1 and has one point 'M' which belongs to feasible region and farthest from the origin. If we take any line P 3 Q 3 parallel to P 2 Q 2 away from the origin, it does not touch any point of the feasible region. The co-ordinates of the point M can be obtained by solving the equation 2x + y = 16 x + y =11 which give x = 5 and y = 6. The optimal solution for the objective function is x = 5 and y = 6. The optimal value of Z : 120 (5) (6) = = P a g e

11 Example 3: Solve the following L.P.P. by using Corner point method: Max Z = 3x + 2y s.t. x + 2y 10; 3x + y 15 x, y 0 Answer: Step 1: The inequalities are converted into equations. We obtain two lines: L1 with the equation x + 2y = 10. Then there are two obvious points belong to it A(0,5) and B(10, 0). L2 with the equation 3x + y = 15. Then there are two obvious points belong to it D(0, 15) and E(5,0). Step2: Draw the lines. Step 3: Determine which side verify the constraint Step 4: Calculate the values of the objective function in all corners of the feasible region and compare them to obtain the optimum. Point X coordinate (X) Y coordinate (Y) Z=3x + 2y O A C E From the table, Max Z=18 for x = 4 and y = P a g e

12 Example4: Solve the following L.P.P. graphically: Answer of example 4: Min and Max Z = x + 2y S.t. x + 2y 100 2x - y 0 2x + y 200 x, y 0 Point X coordinate (x) Y coordinate (y) Z= x + 2y A C D E Max Z= 400 for x = 0 and y=200 and Min Z=100 for x = 20 and y = 40 or x=0 and y= P a g e

13 Introduction Chapter2: Simplex method The Simplex method is an iterative algorithm for efficiently solving LP problems using slack variables, tableaus, and pivot variables as a means to finding the optimal solution of an optimization problem. A linear program is a method of achieving the best outcome given a maximum or minimum equation with linear constraints. Most linear programs can be solved using software, but the Simplex method is a technique for solving linear programs by hand. To solve a linear programming model using the Simplex method the following steps are necessary: Standard form Introducing slack variables Creating the tableau Pivot variables Creating a new tableau Checking for optimality Identify optimal values LP model in equation form: The development of the Simplex method computations is facilated by imposing two requirements on the constraints of the problem: 1. All constraints (with the exception of the nonnegativity of the variables) are equations with nonnegative rift-hand side. 2. All the variables are nonnegative. Converting inequalities into equations with nonnegative right-hand side: To convert a -inequality to an equation, a nonnegative slack S 1 variable is added to the lefthand side of the constraint. For example, 6x 1 + 4x x 1 + 4x 2 +S 1 = 24. To convert a -inequality to an equation, a nonnegative surplus S 2 variable is subtracted to the left-hand side of the constraint. For example, x 1 + x x 1 + x 2 S 2 = 800. If the right-hand side of the resulting equation is nonnegative, we multiply the both sides of the equation by (-1). For example, -x 1 + x x 1 + x 2 +S 3 = -3 ( (-1)) x 1 - x 2 -S 3 = P a g e

14 Dealing with unrestricted variables: If the variable x 1 is unrestricted, we use the following substitution; x 1 = x x x 1 are nonnegative variables, in all equations. with x 1 - and Simplex method (maximization) Example1: Solve the following LPP by simplex method: Maximize Z = 2x 1 + 3x 2 Subject to: 2x 1 + x 2 4 x 1 + 2x 2 5 x 1, x 2 0 Answer: This paragraph breaks down the Simplex method into the above steps and follows the example linear programming model shown below throughout to find the optimal solution. First, we start by converting all inequalities into equations by adding slack variables: 2x 1 + x 2 4-2x 1 + x 2 +S 1 = 4 x 1 + 2x 2 5 x 1 + 2x 2 +S 2 = 5 We add these slack variables into the objective function with coefficient zero: Z = 2x 1 + 3x 2 + 0S 1 +0S 2 - Z - 2x 1-3x 2-0S 1-0S 2 = 0 Second, we construct the initial tableau of Simplex method as follows: Z x 1 x 2 S 1 S 2 Right-Hand side (RHS) Z-row Z S 1 -row S S 2 -row S From the tableau: The basic variables are S 1 = 4, S 2 = 5and the Nonbasic variables x 1 = x 2 = 0. The value of the objective function is 0. Third, choose the entering variable by using the optimality condition; the most negative coefficient in Z-row corresponding to (-3). Thus, the entering variable is x 2. And we choose the leaving variable S 2 by using the feasibility condition; the most nonnegative ratio of the righthand side of the equations by the coefficients under the entering variable x 2. Z x 1 x 2 S 1 S 2 RHS Ratio Z-row Z S 1 -row S /1=4 S 2 -row S /2= P a g e

15 The next tableau is constructed basing on the Gauss-Jordan row operations. It identifies the entering variable column as the pivot column and the leaving variable row as the pivot row. The intersection is called the pivot element. The Gauss-Jordan computations needed to produce the new basic solution include two types. 1) Pivot row: a) Replace the leaving variable S 2 in the basic column with the entering variable x 2. b) New pivot row = Current pivot row pivot element x2-row=s2-row 2 x2-row = (0, 1, 2, 0, 1, 5)/2= (0, ½, 1,0, 1/2, 5/2) 2) All other rows, including Z New Row = (current row) - (its pivot column coefficient) (New pivot row) New Z-row=Z-row (-3) (x2-row) = (1, -2, -3, 0, 0, 0) (-3) (0, ½, 1, 0, 1/2, 5/2) New Z-row=(1, -1/2, 0, 0, 3/2, 15/2) New S 1 -row= S 1 -row (1) (x2-row) = (0, 2, 1, 1, 0, 4) (1) (0, ½, 1, 0, 1/2, 5/2) New S 1 -row= (0, 3/2, 0, 1, -1/2, 3/2) We obtain the following tableau: Z x 1 x 2 S 1 S 2 RHS Z-row Z 1-1/ /2 15/2 S 1 -row S 1 0 3/ /2 3/2 x 2 -row x 2 0 1/ /2 5/2 We cannot terminate now because we still have a negative in Z-row. The solution obtained at this step is S1=3/2, x2=5/2,x1=s2=0, and Z=15/2. Z x 1 x 2 S 1 S 2 RHS Ratio Z-row Z 1-1/ /2 15/ S 1 -row S 1 0 3/ /2 3/2 (3/2) (3/2)=1 x 2 -row x 2 0 1/ /2 5/2 (5/2) (1/2)=5 The entering variable x 1 take the place of the leaving variable S 1 and the pivot is 3/2. We do the following computations: x 1 -row=s 1 -row (3/2) x1-row = (0, 3/2, 0, 1, -1/2, 3/2) (2/3)= (0, 1, 0, 2/3, -1/3, 1) New Z-row=Z-row (-1/2) (x1-row) = (1, -1/2, 0, 0, 3/2, 15/2) (-1/2) (0, 1, 0, 2/3, -1/3, 1) New Z-row=(1, 0, 0, 1/3, 8/6=4/3, 16/2=8) New x 2 -row= x 2 -row (1/2) (x 1 -row) = (0, 1/2, 1, 0, 1/2, 5/2) (1/2) (0, 1, 0, 2/3, -1/3, 1) New x 2 -row= (0, 0, 1, -1/3, 4/6=2/3, 4/2=2) Z x 1 x 2 S 1 S 2 RHS Z-row Z /3 4/3 8 x 1 -row x /3-1/3 1 x 2 -row x /3 2/3 2 At the end Max Z= 8 where x 1 = 1 and x 2 = P a g e

16 Example2: Solve the following LPP by simplex method: Answer: Step 1: Standard Form = s.t.: 5 8,, 0 Standard form for maximization problem is the baseline format for all linear programs before solving for the optimal solution and has two requirements: (1) All linear constraints must be in a less-than-or-equal-to inequality, (2) All variables are non-negative. These requirements can always be satisfied by transforming any given linear program using basic algebra and substitution. Standard form is necessary because it creates an ideal starting point for solving the Simplex method as efficiently as possible as well as other methods of solving optimization problems. Transforming linear constraints from a inequality to a inequality can be by multiplying by -1 on both sides, the inequality can be changed to. 1 ( 5 8) Step 2: Determine Slack Variables Slack variables are additional variables that are introduced into the linear constraints of a linear program to transform them from inequality constraints to equality constraints. If the model is in standard form, the slack variables will always have a +1 coefficient. Slack variables are needed in the constraints to transform them into solvable equalities with one definite answer. Step 3: Setting up the Tableau = = 8,,,, 0 A Simplex tableau is used to perform row operations on the linear programming model as well as to check a solution for optimality. The tableau consists of the coefficient corresponding to the linear constraint variables and the coefficients of the objective function. In the tableau below, 16 P a g e

17 the bolded top row of the tableau states what each column represents. The first row represents the objective function variable coefficients multiply by (-1) and the following two rows represent the linear constraint variable coefficients from the linear programming model = = = 8 Z x1 x2 x3 S1 S2 Right Hand side (RHS) Z-row Z S1-row S S2-row S Once the tableau has been completed, the model can be checked for an optimal solution as shown in Step 4. Step 4: Check Optimality The optimal solution of a maximization linear programming model is the values assigned to the variables in the objective function to give the largest Z value. The optimal solution would exist on the corner points of the graph of the entire model. To check optimality using the tableau, all values in the Z- row must contain values greater than or equal to zero. If a value is less than zero, it means that variable has not reached its optimal value. As seen in the previous tableau, three negative values exist in Z- row indicating that this solution is not optimal. If a tableau is not optimal, the next step is to identify the pivot variable to base a new tableau on, as described in Step 5. Step 5: Identify the entering Variable The entering variable is used in row operations to identify which variable will become the unit value and is a key factor in the conversion of the unit value. The entering variable can be identified by looking at the Z-row of the tableau and the ratio. Assuming that the solution is not optimal, pick the smallest negative value in the Z- row. One of the values lying in the column of this value will be the entering variable. To find the ratio, divide the beta values of the linear constraints by their corresponding values from the column containing the possible entering variable. The intersection of the row with the smallest non-negative ratio and the smallest negative value in the Z- row will become the pivot. In the example shown below, -10 is the smallest negative in the last row. This will designate the x 2 column to contain the entering variable. Solving for the ratio gives us a value of for the first 17 P a g e

18 constraint, and a value of for the second constraint. Due to being the smallest non-negative indicator, the pivot value will be in the second row and have a value of 5. Z x1 x2 x3 S1 S2 RHS Ratio Z S / S /5 1.6 Now that the sorting variable has been identified (S2), the new tableau can be created in Step 6 to optimize the variable and find the new possible optimal solution. Step 6: Create the New Tableau The new tableau will be used to identify a new possible optimal solution. Now that the pivot variable has been identified in Step 5, row operations can be performed to optimize the pivot variable while keeping the rest of the tableau equivalent. I. To optimize the pivot variable, it will need to be transformed into a unit value (value of 1). To transform the value, multiply the row containing the pivot variable by the reciprocal of the pivot value. In the example below, the pivot variable is originally 5, so multiply the entire row by. Z x1 x2 x3 S1 S2 RHS Z S1 x2 0 1/5 1 1/5 0 1/5 8/5 II. After the unit value has been determined, the other values in the column containing the unit value will become zero. This is because the x 2 in the second constraint is being optimized, which requires x 2 in the other equations to be zero. Z x1 x2 x3 S1 S2 RHS Z 0 S1 0 x2 0 1/5 1 1/5 0 1/5 8/5 III. In order to keep the tableau equivalent, the other variables not contained in the pivot column or pivot row must be calculated by using the new pivot values. For each new value, multiply the negative of the value in the old pivot column by the value in the new pivot row that corresponds to the value being calculated. Then add this to the old value from the old tableau to produce the new value for the new tableau. New tableau value = (Negative value in old tableau pivot column) x (value in new tableau pivot row) + (Old tableau value) 18 P a g e

19 Old Tableau: Z x1 x2 x3 S1 S2 RHS Z-row Z S1-row S S2-row S New Tableau: Z x1 x2 x3 S1 S2 RHS New Z-row Z New S1-row S1 0 2/5 0 7/5 1-3/5 26/5 x2-row x2 0 1/5 1 1/5 0 1/5 8/5 New Z-row = (10 x2-row) + Z-row New S1-row = ((-3) x2-row) + S1-row Numerical examples are provided below to help explain this concept a little better. Numerical examples: I. To find the s 2 value in S1-row: New tableau value = (Negative value in old tableau pivot column) * (value in new tableau pivot row) + (Old tableau value) New tableau value = (-3) * ( ) + 0 = - II. To find the x 1 variable in Z- row : New tableau value = (Negative value in old tableau pivot column) * (value in new tableau pivot row) + (Old tableau value) New value = (10) * ( ) + -8 = -6 Once the new tableau has been completed, the model can be checked for an optimal solution. Step 7: Check Optimality As explained in Step 4, the optimal solution of a maximization linear programming model are the values assigned to the variables in the objective function to give the largest Z value. Optimality will need to be checked after each new tableau to see if a new entering variable needs to be identified. A solution is considered optimal if all values in the Z- row are greater than or equal to zero. If all values are greater than or equal to zero, the solution is considered optimal and Steps 8 through 11 can be ignored. If negative values exist, the solution is still not optimal and a new pivot point will need to be determined which is demonstrated in Step P a g e

20 Step 8: Identify New Pivot Variable If the solution has been identified as not optimal, a new pivot variable will need to be determined. The pivot variable was introduced in Step 5 and is used in row operations to identify which variable will become the unit value and is a key factor in the conversion of the unit value. The pivot variable can be identified by the intersection of the row with the smallest non-negative indicator and the smallest negative value in the Z- row. Z x1 x2 x3 S1 S2 RHS Ratio Z S1 0 2/5 0 7/5 1-3/5 26/5 (26/5)/(2/5)=13 x2 0 1/5 1 1/5 0 1/5 8/5 (8/5)/(1/5)=8 With the new pivot variable identified, the new tableau can be created in Step 9. Step 9: Create New Tableau After the new pivot variable has been identified, a new tableau will need to be created. Introduced in Step 6, the tableau is used to optimize the pivot variable while keeping the rest of the tableau equivalent. I. Make the pivot variable 1 by multiplying the row containing the pivot variable by the reciprocal of the pivot value. In the tableau below, the pivot value was, so everything is multiplied by 5. Z x1 x2 x3 S1 S2 RHS Z S1 x II. Next, make the other values in the column of the pivot variable zero. This is done by taking the negative of the old value in the pivot column and multiplying it by the new value in the pivot row. That value is then added to the old value that is being replaced. Z x1 x2 x3 S1 S2 RHS Z S x P a g e

21 Step 10: Check Optimality Using the new tableau, check for optimality. Explained in Step 4, an optimal solution appears when all values in the Z- row are greater than or equal to zero. If all values are greater than or equal to zero, skip to Step 11 because optimality has been reached. If negative values still exist, repeat steps 8 and 9 until an optimal solution is obtained. Step 11: Identify Optimal Values Once the tableau is proven optimal the optimal values can be identified. From the first column and the last one, we obtain that: Z =64, x1=8, S1=2, and the others variables are equals to zero. Then, the maximum optimal value is 64 and found at (8, 0, 0) of the objective function. Example3: Solve the following LPP by simplex method: Maximize Z = 5x 1 + 4x 2 Subject to: 6x 1 + 4x 2 24 x 1 + 2x 2 6 -x 1 + x x 2 2 x 1 0 Answer: LP form equation: 6x 1 + 4x x 1 + 4x 2 +S 1 = 24 x 1 + 2x 2 6 x 1 + 2x 2 +S 2 = 6 -x 1 + x 2 1 -x 1 + x 2 +S 3 = 1 x 2 2 x 2 +S 4 = 2 The objective function becomes: Z - 5x 1-4x 2-0 S 1-0S 2-0S 3-0S 4 = 0 The initial tableau of Simplex method will be: entering Basic z X1 X2 S1 S2 S3 S4 RHS Ratio z leaving S1 0 6 (pivot) S S ve S Pivot column The second tableau is: entering Basic z X1 X2 S1 S2 S3 S4 RHS Ratio z 1 0-2/3 5/ x /3 1/ leaving S /3-1/ /4=1.5 S /3 1/ S Pivot column Pivot Row Pivot Row 21 P a g e

22 We obtain the optimal tableau: Basic z X1 X2 S1 S2 S3 S4 RHS z /4 1/ x /4-1/ x /8 3/ /2 S /8-5/ /2 S /8-3/ /2 Based on the optimality condition, none of the z-row coefficients associated with the nonbasic variables, S1 and S2, are negative. Hence the last tableau is optimal. Thus, max Z = 21 where x2=3, x2=3/2, S3=5/2, S4=1/2, S1=S2=0 The solution also gives the status of the resources. A resource is designated as scarce if the corresponding slack variable is zero. Otherwise the resource is abundant. The following table classifies the constraints of the model: Resource Value of slack Status Constraint 1 S1=0 scarce Constraint 2 S2=0 scarce Constraint 3 S3=5/2 abundant Constraint 4 S4=1/2 abundant Summary of the Simplex Method So far we have dealt with the maximization case. In minimization problems, the optimality conditions calls for selecting the entering variable as the nonbasic variable with the most positive objective coefficient in the objective equation. This follows because Max Z = - Min ( -Z). As for the feasibility condition for selecting the leaving variable, the rule remains unchanged. Optimality condition: the entering variable in maximization (minimization) problem is the nonbasic variable having the most negative (positive) coefficient in the Z-row. Ties are broken arbitrarily. The optimum is reached at the iteration where all the Z-row coefficient of the nonbasic variables are nonnegative (nonpositive). Feasibility condition: for both the maximization and minimization problems, the leaving variable is the basic variable associated with the smallest nonnegative ratio. Ties are broken arbitrarily. Gauss- Jordan row operations: 1) Pivot row: a. Replace the leaving variable S 2 in the basic column with the entering variable x 2. b. New pivot row = Current pivot row pivot element 2) All other rows, including Z New Row = (current row) - (its pivot column coefficient) (New pivot row) 22 P a g e

23 The steps of the Simplex method are: Step1. Determine a starting basic feasible solution. Step2. Select an entering variable using the optimality condition. Stop if there is no entering variable; the last solution is optimal. Else, go to step3. Step3. Select a leaving variable using the feasibility condition. Step4. Determine the new basic solution by using the appropriate Gauss-Jordan computations. Go to step2. Artificial starting solution We said that the LPs is ill behaved if the constraints are (=) or ( ). We will use artificial variables that play the role of slacks at the first iteration. Two methods are introduced here: the M-method and two-phase method. M-method (Big M-method): The artificial variable Ri is added to form a starting solution and they are assigned high penality in the objective function: Given M, a sufficiently large positive value. M, in maximization problems Artificial variable objective coefficient = +M, in minimization problems Example 4: Solve the following LPP by using M-method: Min z = + s.t.: + = + +, Answer: Using x 3 as a surplus variable in the second constraint and x 4 as a slack in the third constraint, and we add R in the first equation and R in the second constraint, the equation form of the problem is given as: Min z = + + = s.t.: + +, Min z = = s.t.: + = + + =,,, Min z = = s.t.: + + = + + =,,,,, 23 P a g e

24 Using M=100, the starting simplex tableau is given as follows (for convenience, the z-column is eliminated because it does not change in all the iterations): basic X1 X2 X3 X4 R1 R2 RHS z R R X Before proceeding with the Simplex method computations, we need to make the z-row consistent with the rest of the tableau. From the tableau, we z =0 for R1=3, R2=6,X4=4,and X1=X2=X3=0. But, z =100*3+100*6=900 not 0. We can eliminate this inconsistency by: New z-row = Old z-row+ (100 R1-row R2-row) The modified tableau thus becomes: basic X1 X2 X3 X4 R1 R2 RHS Ratio z R /3=1 R /4=1.5 X /1=4 Because it is a minimization problem we start by choosing the most positive number in z-row to obtain the entering variable and we obtain the optimum if the entire coefficient in this row is 0. We obtain the following tableau: basic X1 X2 X3 X4 R1 R2 RHS z x1 1 1/ /3 0 1 R2 0 5/ /3 1 2 X4 0 5/ /3 0 3 The last tableau shows that x2 and R2 are the entering and the leaving variables, respectively. Continuing with the simplex computations, two more iterations are needed to reach the optimum: x 1 =2/5, x 2 =9/5, and z=17/5. basic X1 X2 X3 X4 R1 R2 RHS z / / /5 3 3/5 Ratio X /5 0 3/5-1/5 3/5 3 X /5 0-4/5 3/5 1 1/5-2 X P a g e

25 basic X1 X2 X3 X4 R1 R2 RHS z /5-98 3/ /5 X /5 2/5 0 2/5 X /5-1/ /5 X Remark: If we obtain in the last tableau that an artificial variable has a value not zero then the LP does not have a feasible solution. Exercise: Consider the problem Maximize Z = 2x 1 +4x 2 +4x 3-3x 4 Subject to: x 1 + x 2 + x 3 = 4 x 1 + 4x 2 + x 4 = 8 x 1, x 2, x 3, x 4 0 The problem shows that x 3 and x 4 can play the role of slacks for two equations. They differ from slacks in that they have nonzero coefficients in the objective function. We can use x 3 and x 4 as starting variable, but as in the case of artificial variables, they must be substituted out in the objective function before the simplex iterations are carried out. Solve the problem with x 3 and x 4 as the starting basic variables and without using any artificial variables. Two-phase method: As the name suggests, the method solves the LP in two phases: Phase I attempts to find a starting basic feasible solution by using always minimize the sum of artificial variables as objective function, and, if one is found the minimum value of the objective function is zero, Phase II is invoked to solve the original problem. We will use the same example 4 given in the M-method. Phase I: Min r = + s.t.: + + = + + = + + =,,,,, The associated tableau is given as 25 P a g e

26 Basic X1 X2 X3 X4 R1 R2 RHS r R R X As in the M-method, r1 and R2 are substituted out in the r-row by using the following computations: New r-row = Old r-row + (1 R1-row+1 R2-row) Basic X1 X2 X3 X4 R1 R2 RHS r R R X Basic X1 X2 X3 X4 R1 R2 RHS r 0 1 2/ /3 0 2 ratio X1 1 1/ / R / / X / / The optimum tableau of the first Phase is: Basic X1 X2 X3 X4 R1 R2 RHS r x /5 0 3/5-1/5 3/5 x /5 0-4/5 3/5 6/5 X Because minimum r=0, Phase I produces the feasible solution x1=3/5, x2=6/5, and x4=1. At this point we eliminated the R1 and R2 columns from the tableau and move on to phase II. Phase II After deleting the artificial columns, we substitute the r-row by the original objective function. Min z = 4x 1 + x 2 26 P a g e

27 Basic X1 X2 X3 X4 RHS z x /5 0 3/5 x /5 0 6/5 X Because the variables x1 and x2 have nonzero coefficients in the z-row, they must be substituted out, using the following computations. New z-row = Old z-row + (4 x1-row+1 x2-row) The initial tableau of the phase II is thus given as Basic X1 X2 X3 X4 RHS z 0 0 1/5 0 18/5 x /5 0 3/5 x /5 0 6/5 X Because we are minimizing x3 must enter the solution. Application of the Simplex method produces the optimum in one iteration. Basic X1 X2 X3 X4 RHS Z 0 0 1/ /5 Ratio X /5 0 3/5 3 X / /5-2 X Basic X1 X2 X3 X4 RHS Z /5 3 2/5 X /5 2/5 X /5 1 4/5 X P a g e

28 Special cases: This section considers 4 special cases that arise in the use of the Simplex method: 1. Degeneracy In the application of the feasibility condition, a tie for the minimum ratio occur and can be broken arbitrary when this happens at least one basic variable will be zero in the next iteration and the new solution is said to be degenerate. From the practical standpoint, the condition reveals that the model has at least one redundant constraint. Example 5 : max z= 3x 1 +9x 2 + s.t.: +, Given the slack variables x3 and x4, the following tableaus provide the simplex iterations of the problem: Iteration Basic X1 X2 X3 X4 RHS 0 Z X2 enters X X3 leaves X X1 enters X4 leaves 2 optimum Z -3/4 0 9/ X2 1/4 1 1/4 1/4 2 X4 1/2 0-1/2 1 0 Z 0 0 3/2 3/2 18 X2 0 1 ½ -1/2 2 X Alternative optima When the objective function is parallel to a constraint that satisfied as an equation at the optimal solution (non redundant binding constraint), the objective function can assume the same optimal value at more than one solution point, thus giving rise to alternative optima. Example 6: max z= 2x 1 +4x 2 + s.t.: +, 28 P a g e

29 The iterations of the model are given by the following tableaus. Iteration Basic X1 X2 X3 X4 RHS 0 Z X2 enters X X3 leaves X X1 enters X4 leaves 2 Alternative optimum Z X2 1/2 1 1/2 0 5/2 X4 1/2 0-1/2 1 3/2 Z X X Iteration 1 gives the optimum solution x1=0, x2=5/2 and z=10. look at the z-row coefficients of the nonbasic variables in iteration1. The coefficient of x1 is zero indicating that x1 can enter the basic solution without changing the value of z. 3. Unbounded solutions In some LP models the value of the variables may be increased indefinitely without violating the constraint. That means that the solution space is unbounded. Example 7: max z= 2x 1 +3x 2 s.t.:, The initial simplex table is: Basic X1 X2 X3 X4 RHS Z X X We cannot choose the leaving variable. 4. Non existing (or infeasible ) solutions LP models with inconsistent constraints have no feasible solution. This situation can never occur if all the constraints are of the type with non-negative right-hand sides. For others types of constraints, we use artificial variables if one of them in optimum point has the nonzero value then the problem has infeasible solution. Example 8: max z= 3x 1 +2x 2 + s.t.: +, Using the penalty M=100 for the artificial variable R, the following tableau provide the simplex iteration of the model. 29 P a g e

30 iteration basic X1 X2 X3 X4 R RHS 0 z X2 enters X X3 leaves R Pseudo optimum z X R Optimum iteration 1 show that the artificial variable R is positive (=4), which indicates that the problem is infeasible. Sensitivity analysis 1. Sensitivity analysis, which deals with determining the conditions that will keep the current solution unchanged. 2. Post-optimal analysis, which deals with finding a new optimal solution when the data of the model are changed. In sensitivity Analysis two cases will be considered: 1. Sensitivity of the optimum solution to change in the availability of the resources (righthand side of the constraints) 2. Sensitivity of the optimum to changes in unit profit or unit cost (coefficients of the objective function) Example9: A factory produces two products on two machines. A unit of product 1 requires 2 hours on machine 1 and 1 hour on machine 2. For product 2, a unit requires 1 hour on machine 1 and 3 hours on machine 2. The revenues per unit of product 1 and 2 are 30 SAR and 20 SAR, respectively. The total daily processing time available for each machine is 8 hours. Letting x 1 and x 2 represent the daily number of units of products 1 and 2 respectively the LP model is given as Maximize Z=30x x 2 Subject to : 2x 1 +x 2 8 (machine1) x 1 +3x 2 8 (machine 2) x 1 and x 2 0 Change in the right-hand side constraint: If we increase the right-side of the first constraint from 8 hours to 9 hours we will obtain the following graph: 30 P a g e

31 Dual or Shadow price= = 14/h The change in the optimal objective value per unit change in the availability of the resource represents the unit worth of a resource (in SAR/hr). This means that a unit increase (decrease) in machine 1 capacity will increase (decrease) revenue by 14 SAR. The dual price of 14 SAR/hr remains valid for changes in machine 1 on any point of the line segment BF. Where B(0,2.67) and F(8,0). Regarding the constraint 1, we can conclude that the dual price of 14 remains valid for the range : 2.67 Capacity of machine1 16hr Some calculations can be done for the machine 2. Dual price =2SAR /hr and the range is on the line segment DE where D(4, 0) and E(0, 8). So the Dual price remains valid for the range is 4 capacity of machine 2 24 Changes in the Objective Coefficients: The optimum occurs at point C (x1=3.2, x2=1.6, Z C =128). Changes in revenues units will change the slope of Z. However, as we can see from the figure the optimum solution will remain at the point C so long as the objective function lies between line BF and DE, the two constraints that define the optimum point. This means that there is a range for the coefficients of the objective function that will keep the optimum solution unchanged at C. 31 P a g e

32 We can write the objective function in the general format: Maximize Z= c 1 x 1 +c 2 x 2. Imagine now that the line Z is pivoted at C and that it can rotate clockwise and counterclockwise. The optimum solution remains at the point C as long as z=c 1 x 1 +c 2 x 2 lies between the two lines x 1 +3x 2 =8 and 2x 1 +x 2 =8. This means that the ratio c 1 /c 2 can vary between 1/3 and 2/1 which yield the following condition: c 1 /c 2 2. The new objective function is maximize Z= 35 x1 +25 x2. The solution at C will remain optimal because c1/c2=35/25=1.4 remains within the optimality range (0.33,2). When the ratio falls outside this range, additional calculations are needed to find the new optimum. Notice that although the values of the variables at the optimum point C remain unchanged, the optimum value of Z changes to 35*3.2+25*1.6=152. Algebraic sensitivity Analysis Changes in the right hand side Example 10: Consider the following LPP: Max Z= 3x 1 +2x 2 +5x 3 Subject to : x 1 + 2x 2 +x mn (operation 1) 3x 1 + 2x mn (operation2) x 1 + 4x mn (operation3) x 1, x 2, x 3 0 Using x4, x5, and x6 as the slack variables for the constraints of operations 1, 2,and 3, respectively, the optimum tableau is Basic X1 X2 X3 X4 X5 X6 solution Z X2-1/ /2-1/ X3 3/ / X Max Z= 1350 when x2=100, x3=230, x6=20, and x1=x4=x5=0 1) Give the shadow or dual price for each constraint. The Z-row in the optimal tableau yields directly the dual prices, as the following table shows: 32 P a g e

33 Resource Slack variable Optimal z-equation coefficient Dual price of slack variables Operation1 X4 1 1 SAR/min Operation2 X5 2 2 SAR/min Operation 3 X6 0 0 SAR/min The zero dual price for opeartion3 means that there is no economic advantage in allocating more production time to this operation. 2) Give the feasibility conditions of any change at right-hand sides of the constraints. Assume that D i is the change in the right hand side of the i th constraint. Basic solution D1 D2 D3 Z X /2-1/4 0 X /2 0 X From the last tableau, we can easily obtain the feasibility conditions: X2= 100+(1/2)D1+(-1/4)D2 0 X3=230 +(1/2)D2 0 X6=20+(-2)D1 +D2+ D3 0 3) Give the optimality conditions of any change in the coefficients of decision variables in the objective function. Assume that d i is the change in the coefficient of the variable x i. d1 d2 d Basic X1 X2 X3 X4 X5 X6 solution 1 Z d2 X2-1/ /2-1/ d3 X3 3/ / X The optimality conditions are: 1*4+ (-1/4)d2+ (3/2)d3 +2*0 - d1 0 1+(1/2)d2 0 2 (1/4)d2 +(1/2)d3 0 4) If we change the right-hand side of the constraints to the values 400, 448, and 430 respectively. What will be the optimal solution? Since D= new value old value, then D1= =-30, D2= =-12, and D3= =10. Substituting these values in the feasibility conditions, we obtain: X2= 100+ (1/2)(-30)+ (-1/4)*(-12)=103-15=88 >0 X3=230 + (1/2)(-12)=224 > 0 X6=20+ (-2)(-30) +(-12)+ 10 =78 > 0 The new feasible solution is x2=88, x3=224, and x6=78 with Z= (0)+2(88)+5(224)= P a g e

34 Notice that optimum objective value can also be computed as Z= (-30) + 2(- 12)= ) Give the optimal solution if Z is substituted by Z =2x1 + x2 + 6x3. The old objective function was Z= 3x1 +2x2 +5x3 Since d = new value old value. Then d1= 2-3=-1, d2=1-2=-1, and d3=6-5=1. Substitution in the optimality conditions yields 1*4+ (-1/4)d2+ (3/2)d3 +2*0 - d1 =6.75 >0 1+(1/2)d2 =0.5 >0 2 (1/4)d2 +(1/2)d3 =2.75 >0 The results shows that the proposed changes will keep the current solution (x1=0, x2=100,x3=230) optimal. Hence no further calculations are needed except that the objective value will change to Z= d d3 =1478. If any of the conditions is not satisfied, anew solution must be determined by using the Simplex method. Remark: The computation above show the dual prices are determined from the optimal tableau for constraints. For constraints, the same idea remains applicable except that the dual price will assume the opposite sign of that associated with constraint. In the case of the constraint is an equation, the determination of the dual price from the optimal simplex table require somewhat involved calculation. If the change doesn t satisfy the feasibility or optimality conditions. 34 P a g e

35 Chapter3: Duality -With every LP maximization problem, there is an associated minimization problem and vice versa. -The original problem is called the primal and the other is called the dual. Properties of duality 1. If the primal is a maximization problem, the dual is a minimization problem, and vice versa. 2. An optimal solution to the dual exists only when the primal has an optimal solution, and vice versa. 3. Both the primal and dual problems have the same optimal value of the objective function. 4. The dual of the dual is the primal. 5. The solution of the dual problem can be obtained from the solution of the primal problem, and vice versa. 6. The dual variables may assume negative values. Formulation of the dual Objective If the primal is a maximization problem, then the dual is a minimization problem, and vice versa. Decision variables For each constraint in the primal (not nonnegativity constraints), there is one decision variable in the dual. Objective function The coefficient of each decision variable in the objective function of the dual is equal to the right-hand side of the corresponding constraint in the primal. Constraints -Before structuring the dual constraints, all primal constraints should be transformed to equalities. -For each decision variable in the primal, there is a corresponding constraint in the dual. -The right-hand side of the dual constraints is the same as the corresponding coefficients of the objective function in the primal. -In the dual problem, all constraints should be in maximization and in minimization. Example1: Find the dual problem for the following LPP: max z= 2x 1 +4x 2 + 4x s.t.: + =,, Answer: First of all, make the primal problem in the equation form without using the artificial variable just slack and surplus variables. 35 P a g e

36 Primal Primal in equation form Dual variables max z= 2x 1 +4x 2 + 4x 3 s.t.: =,, max z= 2x 1 +4x 2 + 4x 3 + 0x = s.t.: + + =,,, y 1 y 2 The dual problem is: Minimize w=10 y y 2 s.t.: + +, + (, Example2: Find the dual problem for the following LPP: Min z= 15x 1 +12x 2 + s.t.:, Answer: Primal Primal in equation form Dual variables Min z= 15x 1 +12x 2 Min z= 15x 1 +12x 2 + 0x 3 + 0x = y 1 s.t.: s.t.: + + = y 2,,,, The dual problem is: Maximize w=10 y y 2 s.t.:, + (, ) 36 P a g e

37 Example3: Find the dual problem for the following LPP: max z= 5x 1 +6x 2 + = s.t.: + +, Answer: Primal Primal in equation form Dual Max z= 5x 1 +6x 2 s.t.: + = + +, Substitute = Max z= x 2 + 0x 3 + 0x = s.t.: = =,,,, variables y 1 y 2 y 3 The dual problem is: = : + +,,, = =..: + +, 37 P a g e

38 Chapter4: Transportation Problem Let S i be the supplier or factory, D i destination or depot and C ij is the cost of transportation from S i to D j. If the supply and the demand are equal the problem is called balanced ( = ). D1 D2 D3 D4 D5 Supply S1 C11 C12 C13 C14 C15 a1 S2 C21 C22 C23 C24 C25 a2 S3 C31 C32 C33 C34 C35 a3 S4 C41 C42 C43 C44 C45 a4 S5 C51 C52 C53 C54 C55 a5 Demand b1 b2 b3 b4 b5 We are focalizing on: North-West Corner Method Vogel s approximation method Check the optimality Example1: The following table is the cost of transportation problem with 3 factories and 4 warehouses: 1. Find the basic feasible solution by the North-West Corner Method (NWCM) and its cost. 2. Find the basic feasible solution by the Vogel s approximation method (VAM) and its cost. 3. Check the optimality of the solution obtained by Vogel s approximation method. Answer: 1. We copy the table without the cost of transportation Cij and we start at the cell C11 (North West of the table). We calculate the min (a1,b1). Put the results on this cell and subtract it from the a1 and b1 and write the result. 38 P a g e

39 W1 W2 W3 W4 Supply F1 Min(11,6)=6 11-6=5 F2 13 F3 19 Demand 6-6= We cannot go vertically, but horizontally is possible from 6 W1 W2 W3 W4 Supply F1 6 Min(5,10)=5 5-5=0 F2 13 F3 19 Demand = W1 W2 W3 W4 Supply F F2 Min(5,13)=5 13-5=8 F3 19 Demand 0 5-5= W1 W2 W3 W4 Supply F F2 5 Min(8,12)=8 8-8=0 F3 19 Demand =4 15 W1 W2 W3 W4 Supply F F F3 Min(4,19)=4 19-4=15 Demand =0 15 W1 W2 W3 W4 Supply F F F3 4 Min(15,15)=15 15 Demand P a g e

40 We can do it in one table and the basic feasible solution obtained by NWCM is: x 11 =6,x 12 =5,x 22 =5, x 23 =8, x 33 =4, x 34 =15. The cost of this solution is: = The VAM method also takes costs into account in allocation. Five steps are involved in applying this heuristic: Step 1: Determine the difference between the lowest two cells in all rows and columns, including dummies. Step 2: Identify the row or column with the largest difference. Ties may be broken arbitrarily. Step 3: Allocate as much as possible to the lowest-cost cell in the row or column with the highest difference. If two or more differences are equal, allocate as much as possible to the lowest-cost cell in these rows or columns. Step 4: Stop the process if all row and column requirements are met. If not, go to the next step. Step 5: Recalculate the differences between the two lowest cells remaining in all rows and columns. Any row and column with zero supply or demand should not be used in calculating further differences. Then go to Step 2. The Vogel's approximation method (VAM) usually produces an optimal or near- optimal starting solution. One study found that VAM yields an optimum solution in 80 percent of the sample problems tested. We use all values of the table here. We add one column, on the right side, containing the difference between the two smallest value in the each row of Cij. And add one row on the bottom of the table containing the difference between the two smallest value of Cij in each column. W1 W2 W3 W4 Supply Difference F F =3 F =9 Demand Difference 21-17= = = =10 The big difference is 10 on w4 column. We choose in this column the smallest Cij corresponding to 13. W1 W2 W3 W4 Supply F // 13 0 F F Demand P a g e

41 W1 W2 W3 W4 Supply Difference F // = =3 F =9 Demand =0 Difference 32-17= = = =18 W1 W2 W3 Supply Difference F2 6// = =3 F =9 Demand 6-6= Difference 32-17= = =4 W2 W3 Supply Difference F2 3// = =4 F =9 Demand 10-3=7 12 Difference 27-18= =4 W2 W3 Supply Difference F3 7//27 12// = =9 Demand 7-7=0 12 Difference The solution is: x 14 =11, x 21 =6, x 22 =3, x 24 =4, x 32 =7, x 33 =12. W1 W2 W3 W4 Supply F // F2 6 // 17 3 // // F // // Demand The cost is = st step: consider only the nonempty boxes and the row or the column having the greatest number of solutions. 41 P a g e

42 Remark: Consider the second row and put r2 = 0. r2 +c1=17 => c1=17, r2 + c2=18 =>c2= 18, r2 +c4=23 =>c4=23 r3 + c2=27 => r3= 9, r3 + c3=18 => c3= 9, r1 + c4=13 => r1= nd step: Consider only the empty boxes and calculate Eij = ri + cj- Cij E11=r1+c1-21= E12= E23= E31= E34 = For all empty boxes, Eij is negative, Then the solution obtained is optimal. In process, if we find one positive Eij for empty boxes. We stop and then the solution is not optimal. Assignment problem The assignment problem is a special type of transportation problem, where the objective is to minimize the cost or time of completing a number of jobs by a number of persons. In other words, when the problem involves the allocation of n different facilities to n different tasks. The assignment problem also encompasses an important sub-class of so-called shortest- (or longest-) route models. The assignment model is useful in solving problems such as, assignment of machines to jobs, assignment of salesmen to sales territories, travelling salesman problem, etc. Hungarian Method 1. Identify the minimum element in each row and subtract it from every element of that row. 2. Identify the minimum element in each column and subtract it from every element of that column. 3. Make the assignments for the reduced matrix obtained from steps 1 and 2 in the following way: i. For each row or column with a single zero value cell that has not be assigned or eliminated, box that zero value as an assigned cell. ii. For every zero that becomes assigned, cross out (X) all other zeros in the same row and the same column. 42 P a g e

43 iii. If for a row and a column, there are two or more zeros and one cannot be chosen by inspection, then you are at liberty to choose the cell arbitrarily for assignment. iv. The above process may be continued until every zero cell is either assigned or crossed (X). 4. An optimal assignment is found, if the number of assigned cells equals the number of rows (and columns). In case you have chosen a zero cell arbitrarily, there may be alternate optimal solutions. If no optimal solution is found, go to step Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix obtained from step 3 by adopting the following procedure: 1. Mark all the rows that do not have assignments. 2. Mark all the columns (not already marked) which have zeros in the marked rows. iii. Mark all the rows (not already marked) that have assignments in marked columns. iv. Repeat steps 5 (i) to (iii) until no more rows or columns can be marked. v. Draw straight lines through all unmarked rows and marked columns. You can also draw the minimum number of lines by inspection. 5. Select the smallest element from all the uncovered elements. Subtract this smallest element from all the uncovered elements and add it to the elements, which lie at the intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment. 6. Go to step 3 and repeat the procedure until you arrive at an optimal assignment. Example: The Funny Toys Company has four men available for work on four separate jobs. Only one man can work on any one job. The cost of assigning each man to each job is given in the following table. The objective is to assign men to jobs in such a way that the total cost of assignment is minimum. Answer: 43 P a g e

44 The total cost of assignment = A1 + B4 + C2 + D3 Substituting values from original table: = SAR. 78. Since the number of assignments is equal to the number of rows (& columns), this is the optimal solution. 44 P a g e