Simplex method - Maximisation Case
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The document discusses the simplex method for solving linear programming problems. It introduces some key terminology used in the simplex method like slack variables, surplus variables, and artificial variables. It then provides an overview of how the simplex method works for maximization problems, including forming the initial simplex table, testing for optimality and feasibility, pivoting to find an optimal solution. Finally, it provides an example application of the simplex method to a sample maximization problem.
Simplex method - Maximisation Case
- 1. Simplex Method - Introduction In the previous chapter, we discussed about the graphical method for solving linear programming problems. Although the graphical method is an invaluable aid to understand the properties of linear programming models, it provides very little help in handling practical problems. In this chapter, we concentrate on the simplex method for solving linear programming problems with a larger number of variables. Many different methods have been proposed to solve linear programming problems, but simplex method has proved to be the most effective. This technique will nurture your insight needed for a sound understanding of several approaches to other programming models, which will be studied in subsequent chapters. Simplex Method is applicable to any problem that can be formulated in terms of linear objective function, subject to a set of linear constraints. Often, this method is termed Dantzig's simplex method, in honour of the mathematician who devised the approach. In the following section, we introduce you to the standard vocabulary of the simplex method. Basic Terminology - Simplex Method Slack variable It is a variable that is added to the left-hand side of a less than or equal to type constraint to convert the constraint into an equality. In economic terms, slack variables represent left- over or unused capacity. Specifically: ai1x1 + ai2x2 + ai3x3 + .........+ ainxn bi can be written as ai1x1 + ai2x2 + ai3x3 + .........+ ainxn + si = bi Where i = 1, 2, ..., m Surplus variable It is a variable subtracted from the left-hand side of a greater than or equal to type constraint to convert the constraint into an equality. It is also known as negative slack variable. In economic terms, surplus variables represent overfulfillment of the requirement. Specifically: ai1x1 + ai2x2 + ai3x3 + .........+ ainxn bi can be written as ai1x1 + ai2x2 + ai3x3 + .........+ ainxn - si = biWhere i = 1, 2, ..., m Artificial variable It is a non negative variable introduced to facilitate the computation of an initial basic feasible solution. In other words, a variable added to the left-hand side of a greater than or equal to type constraint to convert the constraint into an equality is called an artificial variable.
- 2. Simplex Method - Maximization Case Consider the general linear programming problem Maximize z = c1x1 + c2x2 + c3x3 + .........+ cnxn subject to a11x1 + a12x2 + a13x3 + .........+ a1nxn b1 a21x1 + a22x2 + a23x3 + .........+ a2nxn b2 ......................................................................... am1x1 + am2x2 + am3x3 + .........+ amnxn bm x1, x2,....., xn 0 Where: cj (j = 1, 2, ...., n) in the objective function are called the cost or profit coefficients. bi (i = 1, 2, ...., m) are called resources limitaions. aij (i = 1, 2, ...., m; j = 1, 2, ...., n) are called technological coefficients or input-output coefficients. Converting inequalities to equalities Introducing slack variables to convert inequalities to equalities a11x1 + a12x2 + a13x3 + .........+ a1nxn + s1 = b1 a21x1 + a22x2 + a23x3 + .........+ a2nxn + s2 = b2 .............................................................................. am1x1 + am2x2 + am3x3 + .........+ amnxn + sm = bm x1, x2,....., xn 0 s1, s2,....., sm 0 An initial basic feasible solution is obtained by setting x1 = x2 =........ = xn = 0 s1 = b1 s2 = b2 .............. sm = bm The initial simplex table is formed by writing out the coefficients and constraints of a LPP in a systematic tabular form. The following table shows the structure of a simplex table.
- 3. Structure of a simplex table cj c1 c2 c3 --- cn 0 0 0 --- 0 cBi Basic variables B x1 x2 x3 --- xn s1 s2 s3 --- sm Solution values b (=XB) 0 s1 a11 a12 a13 --- a1n 1 0 0 --- 0 b1 0 s2 a21 a22 a23 --- a2n 0 1 0 --- 0 b2 0 s3 a31 a32 a33 --- a3n 0 0 1 --- 0 b3 --- ----- ---- ---- ----- --- ---- --- --- --- --- 0 ----- 0 sm am1 am2 am3 --- amn 0 0 0 --- 1 bm zj-cj -c1 -c2 -c3 --- -cn 0 0 0 --- 0 Z = 0 Where: cj = coefficients of the variables (m + n) in the objective function. cBi = coefficients of the current basic variables in the objective function. zj= ∑aijcBiwhere i = 1,2…..m; for each j= 1,2…..n+m B = basic variables in the basis. XB = solution values of the basic variables. zj-cj = index row. Or Relative Cost factor The rules used for the construction of the initial simplex table are same in both the maximization and the minimization problems.
- 4. .Simplex Algorithm Unquestionably, the simplex technique has proved to be the most effective in solving linear programming problems. In the simplex method, we first find an initial basic solution (extreme point). Then, we proceed to an adjacent extreme point. We continue this process until we reach an optimal solution Steps (Simplex Method - Maximization Problem) 1. Formulate the Problem Formulate the mathematical model of the given linear programming problem. If the objective function is given in minimization form then convert it into maximization form in the following way: Min z = - Max (-z) Any minimization problem can be converted into an equivalent maximization problem by multiplying the objective function with (-1) Convert every inequality constraint in the L.P. problem into an equality constraint by adding a slack variable to each constraint. 2. Find out the Initial Solution Calculate the initial basic feasible solution by assigning zero value to the decision variables. This solution is shown in the initial simplex table. 3. Test for Optimality Calculate the values of zj – cj If the values of zj– cj are positive, the current basic feasible solution is the optimal solution. If there are one or more negative values, choose the variable corresponding to which the value of zj– cj is least (most negative) as this is likely to increase the profit most. 4. Test for Feasibility Divide the values under XB column by the corresponding positive coefficient (aij) in the key column, and compare the ratios. The row that indicates the minimum ratio is called the key row. However, division by zero or negative coefficients in the key column is not allowed. In the case of a tie, break the tie arbitrarily. 5. Identify the Pivot Element (Key Element) The number that lies at the intersection of the key column and key row of a given table is called the key element. It is always a non-zero positive number. 6. Determine the New Solution
- 5. The numbers in the replacing row may be obtained by dividing the key row elements by the pivot element. The numbers in the remaining rows may be calculated by using the following formula: New number= old number- (corresponding no. of key row) X (corresponding no. of key column) ------------------------------------------------------------------------------ pivot element 7. Revise the Solution Go to step 3 and repeat the procedure until all the values of zj– cj are either zero or positive. For the time being we assume that a feasible solution exists and the optimal value of the objective function is finite. Later in the chapter, we will remove these restrictive assumptions and consider several special cases like unbounded solution, multiple optimum solution, no feasible solution, and degeneracy. Simplex Method Examples - Maximization Problems Get ready for a few examples of simplex method. In this section we will take maximization problems only. Do you know how to divide, multiply, add, and subtract? Yes. Then there is a good news for you. About 50% of this technique you already know. Example 1 Example 2 Now, open the door and windows of your mind and concentrate on the following example.
- 6. Simplex Method: Example 1 Maximize z = 3x1 + 2x2 subject to -x1 + 2x2 4 3x1 + 2x2 14 x1 – x2 3 x1, x2 0 Solution. First, convert every inequality constraints in the LPP into an equality constraint, so that the problem can be written in a standard from. This can be accomplished by adding a slack variable to each constraint. Slack variables are always added to the less than type constraints. Converting inequalities to equalities -x1 + 2x2 + x3 = 4 3x1 + 2x2 + x4 = 14 x1 – x2 + x5 = 3 x1, x2, x3, x4, x5 0 Where x3, x4 and x5 are slack variables. Since slack variables represent unused resources, their contribution in the objective function is zero. Including these slack variables in the objective function, we get Maximize z = 3x1 + 2x2 + 0x3 + 0x4 + 0x5 Initial basic feasible solution Now we assume that nothing can be produced. Therefore, the values of the decision variables are zero. x1 = 0, x2 = 0, z = 0 When we are not producing anything, obviously we are left with unused capacity x3 = 4, x4 = 14, x5 = 3 We note that the current solution has three variables (slack variables x3, x4 and x5) with non- zero solution values and two variables (decision variables x1 and x2) with zero values. Variables with non-zero values are called basic variables. Variables with zero values are called non-basic variables.
- 7. Simplex Method: Table 1 cj 3 2 0 0 0 cB Basic variables B x1 x2 x3 x4 x5 Solution values b (=XB) 0 x3 -1 2 1 0 0 4 0 x4 3 2 0 1 0 14 0 x5 1 -1 0 0 1 3 zj-cj -3 -2 0 0 0 Z=0 a11 = -1, a12 = 2, a13 = 1, a14 = 0, a15 = 0, b1 = 4 a21 = 3, a22 = 2, a23 = 0, a24 = 1, a25 = 0, b2 = 14 a31= 1, a32 = -1, a33 = 0, a34 = 0, a35 = 1, b3 = 3 Calculating values for the index row (zj– cj) z1 – c1 = (0 X (-1) + 0 X 3 + 0 X 1) - 3 = -3 z2 – c2 = (0 X 2 + 0 X 2 + 0 X (-1)) - 2 = -2 z3 – c3 = (0 X 1 + 0 X 0 + 0 X 0) - 0 = 0 z4 – c4 = (0 X 0 + 0 X 1 + 0 X 0) - 0 = 0 z5 – c5 = (0 X 0 + 0 X 0 + 0 X 1) – 0 = 0 Choose the smallest negative value from zj – cj(i.e., – 3). So column under x1 is the key column. Now find out the minimum positive value Minimum (14/3, 3/1) = 3 So row x5 is the key row. Here, the pivot (key) element = 1 (the value at the point of intersection). Therefore, x5 departs and x1 enters. We obtain the elements of the next table using the following rules: 1. If the values of zj– cj are positive, the inclusion of any basic variable will not increase the value of the objective function. Hence, the present solution maximizes the objective function. If there are more than one negative values, we choose the variable as a basic variable corresponding to which the value of zj– cj is least (most negative) as this will maximize the profit. 2. The numbers in the replacing row may be obtained by dividing the key row elements by the pivot element and the numbers in the other two rows may be calculated by using the formula: New number= old number- (corresponding no. of key row) x (corresponding no. of key column) ------------------------------------------------------------------------------ pivot element
- 8. Calculating values for table 2 x3 row a11 = -1 – 1 X ((-1)/1) = 0 a12 = 2 – (-1) X ((-1)/1) = 1 a13 = 1 – 0 X ((-1)/1) = 1 a14 = 0 – 0 X ((-1)/1) = 0 a15 = 0 – 1 X ((-1)/1) = 1 b1 = 4 – 3 X ((-1)/1) = 7 x4 row a21 = 3 – 1 X (3/1) = 0 a22 = 2 – (-1) X (3/1) = 5 a23 = 0 – 0 X (3/1) = 0 a24 = 1 – 0 X (3/1) = 1 a25 = 0 – 1 X (3/1) = -3 b2 = 14 – 3 X (3/1) = 5 x1 row a31 = 1/1 = 1 a32 = -1/1 = -1 a33 = 0/1 = 0 a34 = 0/1 = 0 a35 = 1/1 = 1 b3 = 3/1 = 3 Table 2 cj 3 2 0 0 0 cB Basic variables B x1 x2 x3 x4 x5 Solution values b (= XB) 0 x3 0 1 1 0 1 7 0 x4 0 5 0 1 -3 5 3 x1 1 -1 0 0 1 3 zj-cj 0 -5 0 0 3 Calculating values for the index row (zj– cj) z1 – c1 = (0 X 0 + 0 X 0 + 3 X 1) - 3 = 0 z2 – c2 = (0 X 1 + 0 X 5 + 3 X (-1)) – 2 = -5 z3 – c3 = (0 X 1 + 0 X 0 + 3 X 0) - 0 = 0
- 9. z4 – c4 = (0 X 0 + 0 X 1 + 3 X 0) - 0 = 0 z5 – c5 = (0 X 1 + 0 X (-3) + 3 X 1) – 0 = 3 Key column = x2 column Minimum (7/1, 5/5) = 1 Key row = x4 row Pivot element = 5 x4 departs and x2 enters. Calculating values for table 3 x3 row a11 = 0 – 0 X (1/5) = 0 a12 = 1 – 5 X (1/5) = 0 a13 = 1 – 0 X (1/5) = 1 a14 = 0 – 1 X (1/5) = -1/5 a15 = 1 – (-3) X (1/5) = 8/5 b1 = 7 – 5 X (1/5) = 6 x2 row a21 = 0/5 = 0 a22 = 5/5 = 1 a23 = 0/5 = 0 a24 = 1/5 a25 = -3/5 b2 = 5/5 = 1 x1 row a31 = 1 – 0 X (-1/5) = 1 a32 = -1 – 5 X (-1/5) = 0 a33 = 0 – 0 X (-1/5) = 0 a34 = 0 – 1 X (-1/5) = 1/5 a35 = 1 – (-3) X (-1/5) = 2/5 b3 = 3 – 5 X (-1/5) = 4 Don't convert the fractions into decimals, because many fractions cancel out during the process while the conversion into decimals will cause unnecessary complications.
- 10. Simplex Method: Final Optimal Table cj 3 2 0 0 0 cB Basic variables B x1 x2 x3 x4 x5 Solution values b (= XB) 0 x3 0 0 1 -1/5 8/5 6 2 x2 0 1 0 1/5 -3/5 1 3 x1 1 0 0 1/5 2/5 4 zj-cj 0 0 0 1 0 Since all the values of zj – cj are positive, this is the optimal solution. x1= 4, x2 = 1 z = 3 X 4 + 2 X 1 = 14. The largest profit of Rs.14 is obtained, when 1 unit of x2 and 4 units of x1 are produced. The above solution also indicates that 6 units are still unutilized, as shown by the slack variable x3 in the XB column. Real life complex applications usually involve hundreds of constraints and thousands of variables. So virtually these problems can not be solved manually. For solving such problems, you will have to rely on employing an electronic computer
- 11. Simplex Method - Maximization Example Now, let us solve the following problem using Simplex Method. Maximization Problem: Example 2 Luminous Lamps produces three types of lamps - A, B, and C. These lamps are processed on three machines - X, Y, and Z. The full technology and input restrictions are given in the following table. Product Machine Profit per unitX Y Z A 10 7 2 12 B 2 3 4 3 C 1 2 1 1 Available Time 100 77 80 Find out a suitable product mix so as to maximize the profit. Solution. The decision problem can be formulated as Maximize z = 12x1 + 3x2 + x3 subject to 10x1 + 2x2 + x3 100 7x1 + 3x2 + 2x3 77 2x1+ 4x2 + x3 80 x1, x2, x3 0 Converting inequalities to equalities 10x1 + 2x2 + x3 + x4 100 7x1 + 3x2 + 2x3 + x5 77 2x1+ 4x2 + x3 + x6 80 x1, x2, x3, x4, x5, x6 0 Where x4, x5 and x6 are slack variables. Including these slack variables in the objective function, we get Maximize z = 12x1 + 3x2 + x3 + 0x4 + 0x5 + 0x6 Initial basic feasible solution
- 12. x1 = 0, x2 = 0, x3 = 0, z = 0 x4 = 100, x5 = 77, x6 = 80 Simplex Method: Table 1 cj 12 3 1 0 0 0 cB Basic variables B x1 x2 x3 x4 x5 x6 Solution values b (=XB) 0 x4 10 2 1 1 0 0 100 0 x5 7 3 2 0 1 0 77 0 x6 2 4 1 0 0 1 80 zj-cj -12 -3 -1 0 0 0 Key column = x1 column. Minimum (100/10, 77/7, 80/2) = 10 Key row = x4 row Pivot element = 10 x4 departs and x1 enters Simplex Method: Table 2 cj 12 3 1 0 0 0 cB Basic variables B x1 x2 x3 x4 x5 x6 Solution values b (= XB) 12 x1 1 1/5 1/10 1/10 0 0 10 0 x5 0 8/5 13/10 -7/10 1 0 7 0 x6 0 18/5 4/5 -1/5 0 1 60 zj-cj 0 -3/5 1/5 6/5 0 0 Simplex Method: Final Optimal Table cj 12 3 1 0 0 0 cB Basic variables B x1 x2 x3 x4 x5 x6 Solution values b (= XB) 12 x1 1 0 -1/16 3/16 -1/8 0 73/8 3 x2 0 1 13/16 -7/16 5/8 0 35/8 0 x6 0 0 -17/8 11/8 -9/4 1 177/4 zj-cj 0 0 11/16 15/16 3/8 0 An optimal policy is x1 =73/8, x2 = 35/8, x3 = 0. The associated optimal value of the objective function is z = 12 X (73/8) + 3 X (35/8) + 1 X 0 = 981/8.