Mekelle University College of Business and Economics Department

Mekelle University College of Business and Economics Department of Management Course Operations Research Gebreyohannes G. 12/16/2021

Chapter Two Linear Programming Definition: FLinear programming is a technique applied for the optimizing a linear objective function, subject to linear equality and linear inequality constraints. FIs a particular technique used for economic allocation of ‘scarce’ or ‘limited’ resources. 12/16/2021

Cont… ve. g. labour, material, machine, time, warehouse space, capital, energy, etc. : vseveral competing activities e. g. products, services, jobs, new equipment, projects, etc. on the basis of a given criterion of optimality. 12/16/2021

Components of LPP FThe linear programming model consists of the following components: üDecision variables: mathematical symbols representing levels of activity of an operation. üObjective function: a linear relationship reflecting the objective of an operation. üConstraint: a linear relationship representing a restriction on decision making. üNon-negativity: values of the decision variables are zero or non -zero positives. 12/16/2021

General form of LP Z = c 1 x 1 + c 2 x 2 +…………. +cnxn Subject to restrictions a 11 x 1 + a 12 x 2 + …. . +a 1 nxn (≤ or ≥) b 1 a 21 x 1 + a 22 x 2 + …………. . +a 2 nxn (≤ or ≥) b 2. am 1 x 1 + am 2 x 2 + ………. +amnxn (≤ or ≥) bm & x 1 ≥ 0, x 2 ≥ 0, …, xn ≥ 0 12/16/2021

Cont… Where § Z = value of overall measures of performance § xj = level of activity (for j = 1, 2, . . . , n) § cj = a unit profit/cost increase in Z that would result from each unit increase in the decision variables (j=1, 2, . . . , n) § bi = amount of resource i that is available for allocation to each decision variables (for i = 1, 2, …, m) § aij = amount of resource i consumed by each unit of activity j 12/16/2021

Cont. . . Resource 1 2 Resource usage per unit of activity Activity 1 2 …………. . n a 11 a 12 …………. a 1 n a 21 a 22 …………. a 2 n b 1 b 2 am 1 am 2 …………………amn Bm . . . M Contribution to Z per C 1, c 2 ……………. cn unit of activity 12/16/2021 Amount resource available . . . of

Linear Programming Model Construction v. General steps taken are as under: § Define the decision variables. § Identify the coefficient of each decision variables. § Formulate the objective function. § Identify the coefficients of each decision variables in the obj fun. § Identify the constrained resources or RHS values. § Formulate the suitable mathematical constraints related to each respective resource. § Mention the non-negativity constraints associated with the decision variables. 12/16/2021

Assumptions of Linear Programming v. Proportionality v. Additive v. Divisibility v. Deterministic v. Finiteness v. Optimality v. One objective 12/16/2021

Profit Maximization case Example 1: Problem Definition 1. A company manufactures two products, tables and chairs by using three machines A, B, and C so as to maximize a profit. Machine A has 10 hrs of capacity available during the coming week. Similarly, the available capacity of machine B and C during the coming week is 36 hrs and 30 hrs, respectively. One unit of a table requires 2 hrs of machine A, 3 hrs of machine B and 3 hrs of machine C. Similarly, one unit of a chair requires 1 hr, 12 hrs and 6 hrs of machine A, B, and C respectively. When one unit of table is produced and sold, it yields a profit of Birr 5 and that of a chair is Birr 7. Given the above information, formulate the linear programming model. 12/16/2021

Solution I. Decision variable § how tables and chairs can the managers produce? § the quantities to be produced can be denoted as: X 1: number of tables to be produced X 2: number of chairs to be produced II. The Objective function § It is to max profit (sum of each) § Then total profit = $5 x 1+ 7 x 2 i. e. 12/16/2021

Cont… III. Model Constraints §three resources are used for production are hours of machine A, B, and C respectively. § hours of machine A constraint is 2 x 1+ 1 x 2 ≤ 10 §hours of machine B constraint is 3 x 1 + 12 x 2 ≤ 36 § hours of machine C constraint is 3 x 1 + 6 x 2 ≤ 30 IV. Non-negativity constraint: §x 1 ≥ 0, x 2 ≥ 0. 12/16/2021

Cont. . . §The complete linear programming model for this problem can now be summarized bellow as Max Z = $5 x 1+ 7 x 2 ST: 2 x 1 + 1 x 2 ≤ 10 5 x 1 + 2 x 2 ≤ 36 3 x 1 + 6 x 2 ≤ 30 x 1, x 2 ≥ 0 12/16/2021

Example 2: Problem Definition • Example 1: Assume ‘XY’ company is a small craft operation run by a Native American tribal council. The company employs skilled artisans to produce clay bowls and mugs with authentic Native American designs and colours. The company also employed inspectors who will check the products after finished. The three primary resources used by the company are special pottery clay, skilled labour and inspectors. Given these limited resources, the company desires to know how many bowls and mugs to produce each day in order to max profit. The two products have the following resource requirements for production and profit per item produced (i. e. the model parameters). 12/16/2021

Example 1: Problem Definition. . . Resource requirement Product Labor Inspector Clay Profit Hr/unit Hr/Unit Ib/unit $/unit Bowl 2 1 5 50 Mug 2 4 2 35 There are 40 hours of labour 50 hours of inspector and 60 pounds of clay available each day for production. Required: Formulate the LP model of this problem. 12/16/2021

Example 1: Problem Definition. . . The final complete LP model Max Z = $50 x 1 + 35 x 2 ST: 2 x 1 + 2 x 2 ≤ 40 5 x 1 + 2 x 2 ≤ 50 x 1 + 4 x 2 ≤ 60 x 1 , x 2 ≥ 0 12/16/2021

Exercise 1 A firm manufactures two types of products A and B and sells them at a profit of $2 on type A and $3 on type B. Each product is processed on two machines G and H. Type A requires 1 minute of processing time on G and 2 minutes on H; type B requires 1 minute on G and 1 minute on H. The machine G is available for not more than 6 hours 40 minutes while machine H is available for 10 hours during any working day. Required: a) Formulate the problem as a linear programming problem b) Solve the problem using graphical method Answer for que a) Maximize Z = 2 x 1 + 3 x 2 ST X 1 + x 2 ≤ 400 2 x 1 + 3 x 2 ≤ 600 X 1 ≥ 0, x 2 ≥ 0

Cost Minimization case – Example 1 (sec 1 & 2) A company owns 2 oil mills A and B which have different production capacities for low, high and medium grade oil. The company enters into a contract to supply oil to a firm every week with 12, 8, 24 barrels of each grade respectively. It costs the company $1000 and $800 per day to run the mills A and B. On a day A produces 6, 2, 4 barrels of each grade and B produces 2, 2, 12 barrels of each grade. Required: Develop the LP model. 12/16/2021

Solution Let x 1 be the number of days a week the mill A has to work x 2 be the number of days per week the mill B has to work Grade A B Minimum requirement Low 6 2 12 High 2 2 8 Medium 4 12 24 Cost per day $1000 $800 12/16/2021

Cont… Therefore, the complete LP model is; Minimize Z = 1000 x 1 + 800 x 2 Subject to 6 x 1 + 2 x 2 ≥ 12 2 x 1 + 2 x 2 ≥ 8 4 x 1+12 x 2 ≥ 24 x 1 ≥ 0, x 2 ≥ 0 12/16/2021

Optimal Solution Methods FThere are two approaches to solve linear programming problems: CThe Graphical Algorithm CThe Simplex Algorithm 12/16/2021

Graphical Method FA graphical approach is limited to linear programming problems with only two decision variables. FThe complexity of its application increases as the number of constraint functions increases. FIt is a relatively straightforward method for determining the optimal solution to certain linear programming problems. 12/16/2021

The following are the steps to be followed in solving LP problems graphically: v. Formulate the problem in terms of mathematical constraints and an objective function. v. Convert the inequalities into equation and compute the coordinate points v. Plot the model constraints on the graph v. Identify the feasible region i. e. the area which satisfies all the constraints simultaneously 12/16/2021

Cont… v. Plot the objective function then move the line out from the origin till it touches the extreme maximum points of the feasible solution area. v. Plot the objective function then move the line down till it touches the extreme minimum point of the feasible solution area. v. Substitute these values into the objective function to find the set of values that result in the maximum profit or the minimum cost. v. Interpreting the results 12/16/2021

Cont. . . Case 1: Graphical Solution Method for Maximization Problem Example 1: Max Z = $50 x 1+ 35 x 2 ST: 2 x 1 + 2 x 2 ≤ 40 5 x 1 + 2 x 2 ≤ 50 x 1 + 4 x 2 ≤ 60 x 1, x 2 ≥ 0 Required: Compute the optimal solution using graphical method. Apply all possible steps. 12/16/2021

Solution: v. First develop the problem into mathematical linear programing if not given. v. The structured model is given so we need to proceed into change the inequalities into equations for computing the coordinate points. 2 x 1 + 2 x 2 = 40 5 x 1 + 2 x 2 = 50 x 1 + 4 x 2 = 60 12/16/2021

Cont. . . 2 x 1 + 2 x 2 = 40 x 1 x 2 0 20 12/16/2021 x 1 + 4 x 2 = 60 20 x 1 0 10 x 1 0 60 0 x 2 25 0 x 2 15 0 When x 1 = 0, x 2 = 20: (0, 20) When x 2 = 0, x 1 = 20: (20, 0) ) 5 x 1 +2 x 2 = 50 When x 1 = 0, x 2 = 25: (0, 25) When x 2 = 0, x 1 = 10 (10, 0) When x 1 = 0, x 2 = 15: (0, 15) When x 2 = 0, x 1 = 60: (60, 0)

Cont… A 12/16/2021

Cont… §In this graph, the FSA is represented by the shaded region. §The feasible region has four corner points i. e. O, A, B, and C. §From these corner points at least one will give us maximum profit §The one that generate the highest profit would be regarded as optimal corner point/s. 12/16/2021

Cont… Corner Coordinate points Points O (0, 0) Z = $50(0)+ 35(0) = $0 A (0, 15) Z = $50(0)+ 35(15) = $525 B (40/9, 125/9) Z = $50(40/9)+ 35(125/9) = $6375/9 C (10, 0) Z = $50(10)+ 35(0) = $500 12/16/2021 Profits

Cont… §To compute the coordinates for B, use simultaneous method as below 5 x 1 + 2 x 2 = 50 x 1 + 4 x 2 = 60 …. (Multiply the equation by 5) 5 x 1 + 2 x 2 = 50 …. (Subtract the equation) 5 x 1 + 20 x 2= -300 -18 X 2 = -250 x 2 = 125/9 , 12/16/2021

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Cont… The optimal solution has been reached at corner point B x 1 = 40/9 Unique Solution x 2 = 125/9 why? Max Z = $6375/9 in profit 12/16/2021

Cont… v. Interpretation §Form this result; we can conclude that the manager has different alternatives to make decisions but since the objective of the problem is to maximize profit, she has to select the best alternative i. e. alternative corner point B at x 1 = 40/9 and x 2 = 125/9 12/16/2021

Case 2: Graphical Solution Methods for Minimization Problem Example 1: Minimize Z = 1000 x 1 + 800 x 2 Subject to 6 x 1 + 2 x 2 ≥ 12 2 x 1 + 2 x 2 ≥ 8 4 x 1 +12 x 2 ≥ 24 X 1 ≥ 0, x 2 ≥ 0 12/16/2021

Cont… Required: A. Sketch the graph and identify the feasible solution area B. How many numbers of products type x 1 and type x 2 the company has to produce to minimize cost? C. What is the z value of the problem? D. Does the problem have unique solution? Why? 12/16/2021

Ans for A. To sketch the graph, determine the CPs of each constraint by changing the ≥ into = 6 x 1 + 2 x 2 = 12 Let x 1 = 0, x 2 = 6 coordinate points = (0, 6) Let x 2 = 0, x 1 = 2 coordinate points = (2, 0) 2 x 1 + 2 x 2 = 8 Let x 1 = 0, x 2 = 4 coordinate points = (0, 4) Let x 2 = 0, x 1 = 4 coordinate points = (4, 0) 4 x 1 +12 x 2 = 24 • Let x 1 = 0, x 2 = 2 coordinate points = (0, 2) • Let x 2 = 0, x 1 = 6 coordinate points = (6, 0) 12/16/2021

Cont… 12/16/2021

Ans for B. CP Coordinates Profits A (0, 6) Z = $1000(0) + 800(6) = $4800 B (1, 3) Z = $1000(1)+ 800(3) = $3400 C (3, 1) Z = $1000(3)+ 800(1) = $3800 D (6, 0) Z = $1000(6)+ 800(0) = $6000 § Mill A has to work 1 day a week and 12/16/2021 § Mill B has to work 3 days a week

Ans for C. The optimal solution or the minimum possible cost of the problem is $3400. Ans for D The answer is yes because the company has no alternatives days to work more or less for both mill A and mill B without changing the minimum possible cost. 12/16/2021

Graphical Solution of Irregular Types of Linear Programming Problems: FIf a linear programming problem has a unique solution, then it must occur at a corner point of the feasible set. FHowever, there are several special types of typical linear programming problems (linear programming models for which the general rules do not always apply). 12/16/2021

FThese special types of linear programming problems include the following: A. Multiple optimal solutions B. Infeasible problem C. Unbounded problem D. Redundancy 12/16/2021

A. Multiple optimal solutions FIf the objective function is optimized at two adjacent corner points of the feasible area, it is optimized at every point on the line segment joining these corner points, in which case there are infinitely many solutions to the problem. FMultiple optimal solutions can benefit the decision maker since the number of decision options is enlarged. FMultiple optimal solutions provide greater flexibility to the decision maker. 12/16/2021

Con… Example 1: Max Z = 4 x 1 + 3 x 2 Subject to 4 x 1+ 3 x 2 ≤ 24 x 1 ≤ 4. 5 x 2 ≤ 6 x 1 ≥ 0, x 2 ≥ 0 12/16/2021

Solution The first constraint 4 x 1+ 3 x 2 ≤ 24, written in a form of equation 4 x 1+ 3 x 2 = 24 Put x 1 =0, then x 2 = 8 Put x 2 =0, then x 1 = 6 The coordinates are (0, 8) and (6, 0) 12/16/2021

Cont… The second constraint x 1 ≤ 4. 5, written in a form of equation x 1 = 4. 5 The third constraint x 2 ≤ 6, written in a form of equation x 2 = 6 12/16/2021

Cont… 12/16/2021

Cont… The corner points of feasible region are A, B, C and D. So the coordinates for the corner points are § A (0, 6) §B (1. 5, 6) (Solve the two equations 4 x 1+ 3 x 2 = 24 and x 2 = 6 to get the coordinates) §C (4. 5, 2) (Solve the two equations 4 x 1+ 3 x 2 = 24 and x 1 = 4. 5 to get the coordinates) §D (4. 5, 0) 12/16/2021

Cont… We know that Max Z = 4 x 1 + 3 x 2 At A (0, 6) Z = 4(0) + 3(6) = 18 At B (1. 5, 6) Z = 4(1. 5) + 3(6) = 24 At C (4. 5, 2) Z = 4(4. 5) + 3(2) = 24 At D (4. 5, 0) Z = 4(4. 5) + 3(0) = 18 12/16/2021

Cont… Max Z = 24, which is achieved at both B and C corner points. It can be achieved not only at B and C but every point between B and C. Hence the given problem has multiple optimal solutions. 12/16/2021

B. No Optimal Solution/infeasible solution An infeasible solution has no feasible solution area; every possible solution point violates one or more constraints. In other words, if a solution violates at least one constraint, the solution is said to be infeasible. 12/16/2021

Cont. . . Example 1: Solve graphically Max Z = 3 x 1 + 2 x 2 Subject to x 1 + x 2 ≤ 1 x 1 + x 2 ≥ 3 x 1 ≥ 0 , x 2 ≥ 0 12/16/2021

Cont… Solution v. The first constraint x 1+ x 2 ≤ 1 is written in the form of x 1+ x 2 = 1 Put x 1 =0, then x 2 = 1 Put x 2 =0, then x 1 = 1 The coordinates are (0, 1) and (1, 0) 12/16/2021

Cont… v. The second constraint x 1+ x 2 ≥ 3, written in a form of equation x 1 + x 2 = 3 Put x 1 =0, then x 2 = 3 Put x 2 =0, then x 1 = 3 The coordinates are (0, 3) and (3, 0) 12/16/2021

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Cont… There is no common feasible region generated by two constraints together i. e. we cannot identify even a single point satisfying the constraints. Hence there is no optimal solution. 12/16/2021

C. Unbounded Solution FIn an unbounded problem, the feasible solution area formed by the model constraints is not closed. FIn unbounded problem the objective function can increase indefinitely without reaching a maximum value. FUnlimited profits are not possible in the real world; and an unbounded solution, like an infeasible solution, typically reflects an error in defining the problem or in formulating the model. 12/16/2021

Cont… Example 1: Solve by graphical method Max Z = 3 x 1 + 5 x 2 Subject to 2 x 1+ x 2 ≥ 7 x 1 + x 2 ≥ 6 x 1+ 3 x 2 ≥ 9 x 1 ≥ 0 , x 2 ≥ 0 12/16/2021

Cont… Solution v. The first constraint 2 x 1+ x 2 ≥ 7, written in a form of equation 2 x 1+ x 2 = 7 Put x 1 =0, then x 2 = 7 Put x 2 =0, then x 1 = 3. 5 The coordinates are (0, 7) and (3. 5, 0) 12/16/2021

Cont… v. The second constraint x 1+ x 2 ≥ 6, written in a form of equation x 1 + x 2 = 6 Put x 1 =0, then x 2 = 6 Put x 2 =0, then x 1 = 6 The coordinates are (0, 6) and (6, 0) 12/16/2021

Cont… v. The third constraint x 1+ 3 x 2 ≥ 9, written in a form of equation x 1+ 3 x 2 = 9 Put x 1 =0, then x 2 = 3 Put x 2 =0, then x 1 = 9 The coordinates are (0, 3) and (9, 0) 12/16/2021

Cont… 12/16/2021

Cont… The corner points of feasible region are A, B, C and D. So the coordinates for the corner points are: A (0, 7) B (1, 5) (Solve the two equations 2 x 1+ x 2 = 7 and x 1+ x 2 = 6 to get the coordinates) C (4. 5, 1. 5) (Solve the two equations x 1+ x 2 = 6 and x 1+ 3 x 2 = 9 to get the coordinates) D (9, 0) 12/16/2021

D. Redundancy FA redundant constraint is simply one that does not affect the feasible solution region. FOne constraint may be more binding or restrictive than the other and thereby negate its need to be considered. 12/16/2021

Con… Example 1: Maximize Z = X + 2 Y Subject to: X + Y ≤ 12 2 X + Y ≤ 30 X ≤ 25 X, Y ≥ 0 12/16/2021

2. Simplex Solution Methods v. This technique is used to alleviate the limitation of the graphical solution method. v. In the simplex method, the search usually starts at the origin and moves to that adjacent corner that increases (for maximization) or decreases (for minimization) the value of the objective function. v. The process continues until no further improvement is possible. In each of the iteration, the objective function improves. 12/16/2021

Basic Terms Involved in Simplex Procedure 1. Standard Form: when all constraints are written as equalities it is called standard form 2. Utilization of Resources I. Slack Variable: A Variable added to the left-hand side of a ‘less-than or equal to’ constraint to convert the constraint into an equality is called a slack variable. II. Surplus variable: A variable subtracted from the left hand side of the “greater than or equal to” constraints to convert the constraints into equality is called a surplus variable. 12/16/2021

Cont… 3. Basic variable: for m simultaneous linear equations in n variables(n>m), a solution obtained by setting (n-m) variables equal to zero and solving for the remaining m variables is called a basic variables 4. Basic feasible solution: a basic feasible solution is a basic solution for which the variables have a value of greater than or equal zero. 12/16/2021

Cont… 5. Simplex Tableau: is a table used to keep track of the calculation that made at each of the iteration when the simplex solution method is employed. 7. Zj Row: The numbers in this row under each variable represents the total contribution of outgoing profit when one unit of a non-basic variable is introduced into this in place of a basic variable. 12/16/2021

Cont… 8. Cj – Zj Row: The row containing the net profit that will result from introducing one unit of the variable indicated in that column in the solution numbers in index rows are also known as shadow prices. 9. Pivot column: The column with the largest positive number in the net evaluation row of maximization problem, or the largest negative number in the net evaluation row in the cost minimization problem(Cj-Zj) 12/16/2021

Cont… Incoming variables § An incoming variable is currently a non-basic variable (the current value is zero) and will be changed to a basic variable (introduced into the solution). 12/16/2021

Cont… 10. Pivot row: the row corresponding to the variable that will leave the basis in order to make room for the entering variable. Outgoing Variable § To determine the outgoing variable, compute the ratio of the Quantity to the coefficient of the incoming variable for each basis row. § For both the maximization and minimization problems, the outgoing variable is the basic variable with the smallest ratio. § The coefficient of the incoming variable in the outgoing row is called the pivot element. 12/16/2021

Cont… 11. Pivot Number: The element at the intersection of the pivot row and pivot column 12. Iteration: An iteration of the simplex method consists of the sequence of steps performed in moving from one basic feasible solution to another. 12/16/2021

Cont… The procedures of Simplex Method I. Standardize the problem. II. Generate an initial Solution. III. Test for optimality if in case the solution is optimal Otherwise, go to Step 4. IV. Identify the incoming and outgoing variables. V. Generate an improved solution. Go to Step 3. VI. Check for other optimal solutions. 12/16/2021

Case 1: Simplex Method for the Maximization Problem Example 1: Max z= 40 x 1 +30 x 2 S. T x 1 + 2 x 2 ≤ 40 hour of labour 4 x 1 + 3 x 2 ≤ 120 Ib of clay X 1, x 2 ≥ 0 Required: § Standardize the model § Generate an initial solution and the initial simplex tableau § Determine the value of x 1 and x 2 at the optimal simplex tableau § What is the z-value at the optimal solution? § Does the problem have a multiple solution? Why? 12/16/2021

A. Solution: Convert the model into standard form by adding slack variables to each constraint as follows. N. B: slack variables are added to ≤ constraints and represent unused resources. Max Z = Max z= 40 x 1 +30 x 2+0 s 1 + 0 s 2 S. T x 1 + 2 x 2 +s 1 +0 s 2 = 40 4 x 1 + 3 x 2 + 0 s 1+ s 2 =120 X 1, x 2, s 1, s 2 ≥ 0 12/16/2021

Cont… B. At the origin where nothing is being produced, the values of all the decision variables are zero which implies the value of the slack variable equals to the right hand side value. 1(0) + 2(0) +s 1+ 0 S 2 = 40 S 1= 40 ------------------------- 1 4(0) + 3(0) + 0 S 1 + s 2 = 120 S 2 = 120 ------------------------ 2 N. B: at the origin all decision variables are non-basic whereas all slack are basic. 12/16/2021

Cont… § The objective is to maximize profit with a maximum available of 40 hours of labour and 120 pound of clay. Since all resources are idle, a zero profit is made. 12/16/2021 Therefore, the solution is not optimal

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Cont… The optimal solution has been reached because all values of Cj-zj row are less than and 12/16/2021

Cont… E. § Multiple optimal solutions on a simplex tableau can be determined from Cj- Zj or Zi-Cj row value. An alternative optimal solutions have the same z value but different variable values. § The answer for the question is yes because optimal simplex tableau is said to be multiple if at least one non-basic variable has zero value in the Cj-Zi or Zj-Cj row. § Now you can see the above table, x 2 is non-basic variable but its value corresponds to Cj-zj row is zero. § Alternative optimal solution is determined by selecting the nonbasic variable with Cj-zj = 0 as the entering or incoming variable. 12/16/2021

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Case 2: Simplex Tableau for minimization problem Example 1: St: Min Z = 45 x 1 + 12 x 2 x 1 +x 2 ≥ 300 x 1 ≥ 250 x 1 , x 2 ≥ 0 Required: A. Standardize the model B. Generate an initial solution and the initial simplex tableau. C. Determine the value of x 1 and x 2 at the optimal simplex tableau D. What is the z-value at the optimal solution? E. Does the problem have a multiple solution? Why? 12/16/2021

Solution A Surplus variable: A variable subtracted from the left hand side of the greater than or equal to constraints to convert the constraints into equality is called a surplus variable. Min Z = 45 x 1 + 12 x 2 St: x 1 +x 2 – s 1 = 300 x 1 - s 2 = 250 x 1 , x 2 , s 1 , s 2 ≥ 0 12/16/2021

Cont… Surplus variable represents the extra amount of that exceeds the minimum requirement specified in the constraints. However, the simplex method requires that the initial basic feasible solution be at the origin, where x 1= 0 and x 2 = 0. Testing these solution values, we have x 1 +x 2 – s 1= 300 s 1 = -300 ---------------------------------- 1 x 1 - s 2 = 250 s 2 = -250 ---------------------------------- 2 12/16/2021

Cont… The negative sign is illogical and violates the nonnegative restriction of the linear programming. In order to alleviate this difficulty and get a solution at the origin, we need to add an artificial variable (Ai) to the constraint equations; x 1 +x 2 – s 1 +A 1 = 300 x 1 - s 2 + A 2 = 250 12/16/2021

Cont… The artificial variables do not have a meaning as slack or surplus variable does. It is inserted into the equation simply to give a positive solution at the origin; we artificially creating a solution. x 1 +x 2 – s 1 +A 1 = 300 x 1 - s 2 + A 2 = 250 12/16/2021

Cont… § The purpose of the artificial variables is to get us off ground; but once we get started, it has no real use and thus is discarded. § The artificial solution helps get the simplex process started, but we do not want it to end up in the optimal solution, since it has no real meaning. § Like a slack variable, a surplus variable has no effect on the objective function in terms of increasing or decreasing cost. Thus, a coefficient of 0 is assigned to each surplus variable in the objective function. 12/16/2021

Cont… § If the artificial variable appeared in the solution, it would render the final solution meaningless. Therefore, we must ensure that an artificial variable is not in the final solution. § Artificial variable are assigned a large cost in the objective function to eliminate them from the final solution. 12/16/2021

Cont… The completely transformed minimization model can now be summarized as; Min Z = 45 x 1 + 12 x 2 +0 s 1 +0 s 2 +MA 1 + MA 2 St: x 1 +x 2 – s 1+ A 1 = 300 X 1 - s 2 + A 2 = 250 X 1, x 2, s 1, s 2, A 1, A 2 ≥ 0 12/16/2021

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Cont… C. §Apply same procedure as you did in the case of maximization problem. §The variable which corresponds to the highest positive value in Zj - Cj row is called the incoming variable. Hence, in our case x 1 is the entering or incoming variable. 12/16/2021

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Cont… v. Simplex tablea for minimization problem is said to be optimal when § All value of Zj - Cj row becomes less than or equal to zero. § All artificial variables are eliminated from the tableau. § In our case, the simplex tableau does not satisfy all the conditions. Thus, the optimal solution has not been reached. 12/16/2021

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Cont… § The optimal solution has been reached because all zj-cj row values are less than or equal to zero and § All artificial variables are also eliminated from the simplex tableau but this is not necessarily a condition for a problem to be reached at optimal. 12/16/2021

Cont… D. 12/16/2021