Introduction to Optimum Design phần 4 pdf

206 INTRODUCTION TO OPTIMUM DESIGN TABLE 6-5 Pivot Step to Interchange Basic Variable x 4 with Nonbasic Variable x l for Example 6.5 Initial canonical form. BasicØ x 1 x 2 x 3 x 4 b 1 x 3 -11104 2 x 4 11016 Basic solution: Nonbasic variables: x 1 = 0, x 2 = 0 Basic variables: x 3 = 4, x 4 = 6 To interchange x 1 with x 4 , choose row 2 as the pivot row and column 1 as the pivot column. Perform elimination using a 21 as the pivot element. Result of the pivot operation: second canonical form. BasicØ x 1 x 2 x 3 x 4 b 1 x 3 0 21110 2 x 1 1 101 6 Basic solution: Nonbasic variables: x 2 = 0, x 4 = 0 Basic variables: x 1 = 6, x 3 = 10 Solution. The given canonical form can be written in a tableau as shown in Table 6-5; x 1 and x 2 are nonbasic and x 3 and x 4 are basic, i.e., x 1 = x 2 = 0, x 3 = 4, x 4 .= 6. This corresponds to point A in Fig. 6-2. In the tableau, the basic variables are identified in the leftmost column and the rightmost column gives their values. Also, the basic vari- ables can be identified by examining columns of the tableau. The variables associated with the columns of the identity matrix are basic; e.g., variables x 3 and x 4 in Table 6- 5. Location of the positive unit element in a basic column identifies the row whose right side parameter b i is the current value of the basic variable associated with that column. For example, the basic column x 3 has unit element in the first row, and so x 3 is the basic variable associated with the first row. Similarly, x 4 is the basic variable associated with row 2. To make x 1 basic and x 4 a nonbasic variable, one would like to make a¢ 21 = 1 and a¢ 11 = 0. This will replace x 1 with x 4 as the basic variable and a new canonical form will be obtained. The second row is treated as the pivot row, i.e., a 21 = 1 (p = 2, q = 1) is the pivot element. Performing Gauss-Jordan elimination in the first column with a 21 = 1 as the pivot element, we obtain the second canonical form as shown in Table 6-5. For this canonical form, x 2 = x 4 = 0 are the nonbasic variables and x 1 = 6 and x 3 = 10 are the basic variables. Thus, referring to Fig. 6-2, this pivot step results in a move from the extreme point A(0, 0) to an adjacent extreme point D(6, 0). 6.3.5 Basic Steps of the Simplex Method In this section, we shall illustrate the basic steps of the Simplex method with an example problem. In the next subsection, we shall explain the basis for these steps and summarize them in a step-by-step general algorithm. The method starts with a basic feasible solution, i.e., at a vertex of the convex polyhedron. A move is then made to an adjacent vertex while maintaining feasibility of the new solution as well as reducing the cost function. This is accomplished by replacing a basic variable with a nonbasic variable. In the Simplex method, movements are to the adjacent vertices only. Since there may be several points adjacent to the current vertex, we naturally wish to choose the one that makes the greatest improvement in the cost function f. If adjacent points make identical improvements in f, the choice becomes arbitrary. An improvement at each step ensures no backtracking. Two basic questions now arise: 1. How to choose a current nonbasic variable that should become basic? 2. Which variable from the current basic set should become nonbasic? The Simplex method answers these questions based on some theoretical considerations which shall be discussed in Chapter 7. Here, we consider an example to illustrate the basic steps of the Simplex method that answer the foregoing two questions. Before presentation of the example problem, an important requirement of the Simplex method is discussed. In this method the cost function must always be given in terms of the nonbasic variables only. To accomplish this, the cost function expression c T x = f is written as another linear equation in the Simplex tableau; for example, the (m + l)th row. One then performs the pivot step on the entire set of (m + 1) equations so that x 1 , x 2 , , x m and f are the basic variables. This way the last row of the tableau representing the cost function expression is automatically given in terms of the nonbasic variables after each pivot step. The coefficients in the nonbasic columns of the last row are called the reduced cost coefficients written as c¢ j . Example 6.6 describes the steps of the Simplex method in a systematic way. Linear Programming Methods for Optimum Design 207 EXAMPLE 6.6 Steps of the Simplex Method Solve the following LP problem: maximize Solution. The graphical solution for the problem is given in Fig. 6-3. It can be seen that the problem has an infinite number of solutions along the line C–D (z* = 4) because the objective function is parallel to the second constraint. The Simplex method is illustrated in the following steps: 1. Convert the problem to the standard form. We write the problem in the standard LP form by transforming the maximization of z to minimization of f =-2x 1 - x 2 , and adding slack variables x 3 , x 4 , and x 5 to the constraints. Thus, the problem becomes (a) subject to (b) (c) 24 124 xxx++= 43 12 123 xxx++= minimize fxx=- -2 12 zxx xx xx x x xx=+ +£ +£ +£ ≥2431224240 12 1 2 12 1 2 12 subject to ,,, , 208 INTRODUCTION TO OPTIMUM DESIGN 4 2x 1 + x 2 = 4 4x 1 + 3x 2 = 12 x 1 + 2x 2 = 4 4 3 3 2 2 1 10 A B C D F E Optimum solution line CD x 2 x 1 G z * = 4 z = 2 z = 3 FIGURE 6-3 Graphical solution for the LP problem of Example 6.6. Optimum solution: along line C–D. z* = 4. (d) (e) We use the tableau and notation of Table 6-3 which will be augmented with the cost function expression as the last row. The initial tableau for the problem is shown in Table 6-6, where the cost function expression -2x 1 - x 2 = f is written as the last row. Note also that the cost function is in terms of only the nonbasic variables x 1 and x 2 . This is one of the basic requirements of the Simplex method—that the cost function always be in terms of the nonbasic variables. When the cost function is only in terms of the nonbasic variables, then the cost coefficients in the last row are the reduced cost coefficients, written as c¢ j . 2. Initial basic feasible solution. To initiate the Simplex method, a basic feasible solution is needed. This is already available in Table 6-6 which is given as: basic variables: x 3 = 12, x 4 = 4, x 5 = 4 nonbasic variables: x 1 = 0, x 2 = 0 cost function: f = 0 Note that the cost row gives 0 = f after substituting for x 1 and x 2 . This solution represents point A in Fig. 6-3 where none of the constraints is active except the nonnegativity constraints on the variables. 3. Optimality check. We scan the cost row, which should have nonzero entries only in the nonbasic columns, i.e., x 1 and x 2 . If all the nonzero entries are nonnegative, then we have an optimum solution because the cost function xi i ≥=01; to 5 xxx 125 24++= Linear Programming Methods for Optimum Design 209 cannot be reduced any further and the Simplex method is terminated. There are negative entries in the cost row so the current basic feasible solution is not optimum (see Chapter 7 for further explanation). 4. Choice of a nonbasic variable to become basic. We select a nonbasic column having a negative cost coefficient; i.e., -2 in the x 1 column. This identifies a nonbasic variable (x 1 ) that should become basic. Thus, eliminations will be performed in the x 1 column. This answers question 1 posed earlier: “How to choose a current nonbasic variable that should become basic?” Note also that when there is more than one negative entry in the cost row, the variable tapped to become basic is arbitrary among the indicated possibilities. The usual convention is to select a variable associated with the smallest value in the cost row (or, negative element with the largest absolute value). Notation. The boxed negative reduced cost coefficient in Table 6-6 indicates the nonbasic variable associated with that column selected to become basic, a nota- tion that is used throughout. 5. Selection of a basic variable to become nonbasic. To identify which current basic variable should become nonbasic (i.e., to select the pivot row), we take ratios of the right side parameters with the positive elements in the x 1 column as shown in Table 6-7. We identify the row having the smallest positive ratio, i.e., the second row. This will make x 4 nonbasic. The pivot element is a 21 = 2 (the intersection of pivot row and pivot column). This answers the question 2 posed earlier: “Which variable from the current basic set should become nonbasic?” Selection of the row with the smallest ratio as the pivot row maintains feasibility of the new basic solution. This is justified in Chapter 7. TABLE 6-6 Initial Tableau for the LP Problem of Example 6.6 BasicØ x 1 x 2 x 3 x 4 x 5 b 1 x 3 4 310012 2 x 4 2 1010 4 3 x 5 1 2001 4 Cost function -2 -1 000f Notation. The reduced cost coefficients in the nonbasic columns are boldfaced. The selected negative reduced cost coefficient is boxed. TABLE 6-7 Selection of Pivot Column and Pivot Row for Example 6.6 BasicØ x 1 x 2 x 3 x 4 x 5 b Ratio: b i /a i1 ;a i1 > 0 1 x 3 4310012 12 – 4 = 3 2 x 4 21010 4 4 – 2 = 2 ¨ smallest 3 x 5 12001 4 4 – 1 = 4 Cost function -2 -1 000f The selected pivot element is boxed. Selected pivot row and column are shaded. x 1 should become basic (pivot column). x 4 row has the smallest ratio, and so x 4 should become nonbasic. 210 INTRODUCTION TO OPTIMUM DESIGN Notation. The selected pivot element is also boxed, and the pivot column and row are shaded throughout. 6. Pivot operation. We perform eliminations in column x 1 using row 2 as the pivot row and Eqs. (6.15) to (6.17) to eliminate x 1 from rows 1, 3, and the cost row as follows: • divide row 2 by 2, the pivot element • multiply new row 2 by 4 and subtract from row 1 to eliminate x 1 from row 1 • subtract new row 2 from row 3 to eliminate x 1 from row 3 • multiply new row 2 by 2 and add to the cost row to eliminate x 1 As a result of this elimination step, a new tableau is obtained as shown in Table 6-8. The new basic feasible solution is given as basic variables: x 3 = 4, x 1 = 2, x 5 = 2 nonbasic variables: x 2 = 0, x 4 = 0 cost function: 0 = f + 4, f =-4 7. This solution is identified as point D in Fig. 6-3. We see that the cost function has been reduced from 0 to -4. All coefficients in the last row are nonnegative so no further reduction of the cost function is possible. Thus, the foregoing solution is the optimum. Note that for this example, only one iteration of the Simplex method gave the optimum solution. In general, more iterations are needed until all coefficients in the cost row become nonnegative. Note that the cost coefficients corresponding to the nonbasic variable x 2 in the last row is zero in the final tableau. This is an indication of multiple solutions for the problem. In general, when the reduced cost coefficient in the last row corresponding to a nonbasic variable is zero, the problem may have multiple solutions. We shall discuss this point later in more detail. Let us see what happens if we do not select a row with the least ratio as the pivot row. Let a 31 = 1 in the third row be the pivot element in Table 6-6. This will inter- change nonbasic variable x 1 with the basic variable x 5 . Performing the elimination steps in the first column as explained earlier, we obtain the new tableau given in Table 6-9. From the tableau, we have basic variables: x 3 =-4, x 4 =-4, x 1 = 4 nonbasic variables: x 2 = 0, x 5 = 0 cost function: 0 = f + 8, f =-8 TABLE 6-8 Second Tableau for Example 6.6 Making x 1 a Basic Variable BasicØ x 1 x 2 x 3 x 4 x 5 b 1 x 3 01 1 -204 2 x 1 1 0.5 0 0.5 0 2 3 x 5 0 1.5 0 -0.5 1 2 Cost function 0 0 0 1 0 f + 4 The cost coefficient in nonbasic columns are nonnegative; the tableau gives the optimum solution. Linear Programming Methods for Optimum Design 211 The foregoing solution corresponds to point G in Fig. 6-3. We see that this basic solution is not feasible because x 3 and x 4 have negative values. Thus, we conclude that if a row with the smallest ratio (of right sides with positive elements in the pivot column) is not selected, the new basic solution is not feasible. Note that a spreadsheet program, such as Excel, can be used to carry out the pivot step. Such a program can facilitate learning of the Simplex method without getting bogged down with the manual elimination process. TABLE 6-9 Result of Improper Pivoting in Simplex Method for LP Problem of Example 6.6 BasicØ x 1 x 2 x 3 x 4 x 5 b 1 x 3 0 -51 0 -4 -4 2 x 4 0 -30 1 -2 -4 3 x 1 1200 14 Cost function 0 3 00 4 f + 8 The pivot step making x l basic and x 5 nonbasic in Table 6-6 gives a basic solution that is not feasible. 6.3.6 Simplex Algorithm In the previous subsection, basic steps of the Simplex method are explained and illustrated with an example problem. In this subsection, the underlying principles for these steps are summarized in two theorems, called basic theorems of linear programming. We have seen that in general the reduced cost coefficients c¢ j of the nonbasic variables may be positive, neg- ative, or zero. Let one of c¢ j be negative, then if a positive value is assigned to the associated nonbasic variable (i.e., it is made basic), the value of f will decrease. If more than one neg- ative c¢ j is present, a widely used rule of thumb is to choose the nonbasic variable associated with the smallest c¢ j (i.e., the most negative c¢ j ) to become basic. Thus, if any c¢ j for (m + 1) £ j £ n (for nonbasic variables) is negative, then it is possible to find a new basic feasible solu- tion (if one exists) that will further reduce the cost function. If a c¢ j is zero, then the associ- ated nonbasic variable can be made basic without affecting the cost function value. If all c¢ j are nonnegative, then it is not possible to reduce the cost function any further, and the current basic feasible solution is optimum. These ideas are summarized in the following Theorems 6.3 and 6.4. Theorem 6.3 Improvement of Basic Feasible Solution Given a nondegenerate basic fea- sible solution with the corresponding cost function f 0 , suppose that c¢ j < 0 for some j. Then, there is a feasible solution with f < f 0 . If the jth nonbasic column associated with c¢ j can be substituted for some column in the original basis, the new basic feasible solution will have f < f 0 . If the jth column cannot be substituted to yield a basic feasible solution (i.e., there is no positive element in the jth column), then the feasible set is unbounded and the cost func- tion can be made arbitrarily small (toward negative infinity). Theorem 6.4 Optimum Solution for LP Problems If a basic feasible solution has reduced cost coefficients c¢ j ≥ 0 for all j, then it is optimum. According to Theorem 6.3, the basic procedure of the Simplex method is to start with an initial basic feasible solution, i.e., at the vertex of the convex polyhedron. If this solution is not optimum according to Theorem 6.4, then a move is made to an adjacent vertex to reduce the cost function. The procedure is continued until the optimum is reached. The steps of the Simplex method illustrated in the previous subsection in Example 6.6 are summarized as follows assuming that the initial tableau has been set up as described earlier: Step 1. Initial Basic Feasible Solution This is readily obtained if all constraints are “£ type” because the slack variables can be selected as basic and the real variables as nonbasic. If there are equality or “≥ type” constraints, then the two-phase simplex procedure explained in the next section must be used. Step 2. The Cost Function Must be in Terms of Only the Nonbasic Variables This is readily available when there are only “£ type” constraints and slack variables are added into them to convert the inequalities to equalities. The slack variables are basic, and they do not appear in the cost function. In subsequent iterations, eliminations must also be performed in the cost row. Step 3. If All the Reduced Cost Coefficients for Nonbasic Variables Are Nonnegative (≥0), We Have the Optimum Solution Otherwise, there is a possibility of improving the cost function. We need to select a nonbasic variable that should become basic. We identify a column having negative reduced cost coefficient because the nonbasic variable associated with this column can become basic to reduce the cost function from its current value. This is called the pivot column. Step 4. If All Elements in the Pivot Column Are Negative, Then We Have an Unbounded Problem Design problem formulation should be examined to correct the situation. If there are positive elements in the pivot column, then we take ratios of the right side parameters with the positive elements in the pivot column and identify a row with the smallest positive ratio. In the case of a tie, any row among the tying ratios can be selected. The basic variable associated with this row should become nonbasic (i.e., become zero). The selected row is called the pivot row, and its intersection with the pivot column identifies the pivot element. Step 5. Complete the Pivot Step Use the Gauss-Jordan elimination procedure and the pivot row identified in Step 4. Elimination must also be performed in the cost function row so that it is only in terms of nonbasic variables in the next tableau. This step eliminates the nonbasic variable identified in Step 3 from all the rows except the pivot row. Step 6. Identify Basic and Nonbasic Variables, and Their Values Identify the cost func- tion value and go to Step 3. Note that when all the reduced cost coefficient c¢ j in the nonbasic columns are strictly pos- itive, the optimum solution is unique. If at least one c¢ j is zero in a nonbasic column, then there is a possibility of an alternate optimum. If the nonbasic variable associated with a zero reduced cost coefficient can be made basic by using the foregoing procedure, the extreme point (vertex) corresponding to an alternate optimum is obtained. Since the reduced cost coef- ficient is zero, the optimum cost function value will not change. Any point on the line segment joining the optimum extreme points also corresponds to an optimum. Note that all these optima are global as opposed to local, although there is no distinct global optimum. Geo- metrically, multiple optima for an LP problem imply that the cost function hyperplane is par- allel to one of the constraint hyperplanes. Example 6.7 shows how to obtain a solution for an LP problem using the Simplex method. 212 INTRODUCTION TO OPTIMUM DESIGN Linear Programming Methods for Optimum Design 213 EXAMPLE 6.7 Solution by the Simplex Method Using the Simplex method, find the optimum (if one exists) for the LP problem of Example 6.3: (a) subject to (b) (c) (d) Solution. Writing the problem in the Simplex tableau, we obtain the initial canoni- cal form as shown in Table 6-10. From the initial tableau, the basic feasible solution is Note that the cost function in the last row is in terms of only nonbasic variables x 1 and x 2 . Thus, coefficients in the x 1 and x 2 columns and the last row are the reduced cost function: f = 0 from the last row of the tableau nonbasic variables: xx 12 0== basic variables: xx 34 46==, xi i ≥=01; to 4 xxx 124 6++= -+ + =xxx 123 4 minimize fxx=- -45 12 TABLE 6-10 Solution of Example 6.7 by the Simplex Method Initial tableau: x 3 is identified to be replaced with x 2 in the basic set. BasicØ x 1 x 2 x 3 x 4 b Ratio: b i /a iq x 3 -1110 4 4 – 1 = 4 ¨ smallest x 4 1101 6 6 – 1 = 6 Cost -4 -5 00 f Second tableau: x 4 is identified to be replaced with x 1 in the basic set. BasicØ x 1 x 2 x 3 x 4 b Ratio: b i /a iq x 2 -1 1 1 0 4 Negative x 4 2 0 -11 2 2 – 2 = 1 Cost -9 0 5 0 f + 20 Third tableau: Reduced cost coefficients in nonbasic columns are nonnegative; the tableau gives optimum point. BasicØ x 1 x 2 x 3 x 4 b Ratio: b i /a iq x 2 01 1 – 2 1 – 2 5 Not needed x 1 10- 1 – 2 1 – 2 1 Not needed Cost 0 0 1 – 2 9 – 2 f + 29 Example 6.8 demonstrates the solution of the profit maximization problem by the Simplex method. 214 INTRODUCTION TO OPTIMUM DESIGN cost coefficients c¢ j . Scanning the last row, we observe that there are negative coeffi- cients. Therefore, the current basic solution is not optimum. In the last row, the most negative coefficient of -5 corresponds to the second column. Therefore, we select x 2 to become a basic variable, i.e., elimination should be performed in the x 2 column. This fixes the column index q to 2 in Eq. (6.15). Now taking the ratios of the right side parameters with positive coefficients in the second column b i /a i2 , we obtain a minimum ratio for the first row as 4. This identifies the first row as the pivot row according to Step 4. Therefore, the current basic variable associated with the first row, x 3 , should become nonbasic. Now performing the pivot step on column 2 with a 12 as the pivot element, we obtain the second canonical form (tableau) as shown in Table 6-10. For this canonical form the basic feasible solution is The cost function is f =-20 (0 = f + 20), which is an improvement from f = 0. Thus, this pivot step results in a move from (0, 0) to (0, 4) on the convex polyhedron of Fig. 6-2. The reduced cost coefficient corresponding to the nonbasic column x 1 is negative. Therefore, the cost function can be reduced further. Repeating the above-mentioned process for the second tableau, we obtain a 21 = 2 as the pivot element, implying that x 1 should become basic and x 4 should become nonbasic. The third canonical form is shown in Table 6-10. For this tableau, all the reduced cost coefficients c¢ j (corre- sponding to the nonbasic variables) in the last row are ≥0. Therefore, the tableau yields the optimum solution as x 1 = 1, x 2 = 5, x 3 = 0, x 4 = 0, f =-29 (f + 29 = 0), which cor- responds to the point C (1,5) in Fig. 6-2. nonbasic variables: xx 13 0== basic variables: xx 24 42==, EXAMPLE 6.8 Solution of Profit Maximization Problem by the Simplex Method Use the Simplex method to find the optimum solution for the profit maximization problem of Example 6.2. Solution. Introducing slack variables in the constraints of Eqs. (c) through (e) in Example 6.2, we get the LP problem in the standard form: (a) subject to (b) (c) 1 28 1 14 1 124 xxx++= xxx 123 16++= minimize fxx=- -400 600 12 Linear Programming Methods for Optimum Design 215 (d) (e) Now writing the problem in the standard Simplex tableau, we obtain the initial canonical form as shown in Table 6-11. Thus the initial basic feasible solution is x 1 = 0, x 2 = 0, x 3 = 16, x 4 = x 5 = 1, f = 0, which corresponds to point A in Fig. 6-1. The initial cost function is zero, and x 3 , x 4 , and x 5 are the basic variables. Using the Simplex procedure, we note thata 22 = 1 – 14 is the pivot element. This implies that x 4 should be replaced by x 2 in the basic set. Carrying out the pivot operation using the second row as the pivot row, we obtain the second tableau (canonical form) shown in Table 6-11. At this point the basic feasible solution is x 1 = 0, x 2 = 14, x 3 = 2, x 4 = 0, x 5 = 5 – 12 , which corresponds to point E in Fig. 6-1. The cost function is reduced to -8400. The pivot element for the next step is a 11 , implying that x 3 should be replaced by x 1 in the basic set. Carrying out the pivot operation, we obtain the third canonical form shown in Table 6-11. At this point all reduced cost coefficients (corresponding to nonbasic variables) are nonnegative, so according to Theorem 6.4, we have the optimum solution: x 1 = 4, x 2 = 12, x 3 = 0, x 5 = 3 – 14 . This corresponds to the D in xi i ≥=01; to 5 1 14 1 24 1 125 xxx++= TABLE 6-11 Solution of Example 6.8 by the Simplex Method Initial tableau: x 4 is identified to be replaced with x 2 in the basic set. BasicØ x 1 x 2 x 3 x 4 x 5 b Ratio: b i /a iq x 3 1110016 16 – 1 = 16 x 4 1 – 28 1 – 14 01 01 1 — 1/14 = 14 ¨ smallest x 5 1 – 14 1 – 24 00 11 1 — 1/24 = 24 Cost -400 -600 00 0f - 0 Second tableau: x 3 is identified to be replaced with x 1 in the basic set. BasicØ x 1 x 2 x 3 x 4 x 5 b Ratio: b i /a iq x 3 1 – 2 01-14 0 2 2 — 1/2 = 4 ¨ smallest x 2 1 – 2 1 0 14 0 14 14 — 1/2 = 28 x 5 17 — 336 00 - 7 – 12 1 5 – 12 5/12 — 17/336 = 140 — 17 Cost -100 008400 0 f + 8400 Third tableau: Reduced cost coefficients in the nonbasic columns are nonnegative; the tableau gives optimum solution BasicØ x 1 x 2 x 3 x 4 x 5 b Ratio: b i /a iq x 3 10 2-28 0 4 Not needed x 2 01 -1 28 0 12 Not needed x 5 00- 17 — 168 5 – 6 1 3 – 14 Not needed Cost 0 0 200 5600 0 f + 8800 [...]... be replaced with x2 in the basic set to obtain another optimum point Basicỉ x1 x2 x3 x4 b Ratio: bi /aiq x3 0 2 1 -1 4 4 = 2 ă smallest 2 x1 1 1 2 0 1 2 4 Cost 0 0 0 1 2 f +4 4 = 8 1/2 Third tableau: Second optimum point Basicỉ x1 x2 x3 x2 0 1 1 2 1 4 x4 b Ratio: bi /aiq 1 - 2 2 Not needed 3 4 Not needed x1 1 0 - 3 Cost 0 0 0 1 2 f +4 This solution corresponds to point C on Fig 3-7 Note that any... 6. 14 EXAMPLE 6. 14 Implications of Degenerate Basic Feasible Solution Solve the following LP problem by the Simplex method: maximize z = x1 + 4x2 subject to x1 + 2x2 Ê 5, 2x1 + x2 Ê 4, 2x1 + x2 4, x1 - x2 1, x1, x2 0 Solution 226 The problem is transcribed into the standard LP form as follows: INTRODUCTION TO OPTIMUM DESIGN minimize f = - x1 - 4 x 2 (a) x1 + 2 x 2 + x3 = 5 (b) 2 x1 + x 2 + x 4 = 4. .. = 9 to both sides, as -5 + 9 Ê D1 Ê + 9, or 4 Ê b1 Ê (c) For the second constraint (k = 2), x4 is the slack variable Therefore, we will use elements in column x4 of the nal tableau (a 4, j = 4) in the inequalities of Eq (6. 24) i The ratios of the right side parameters with the elements in column 4, ri of Eq (6. 24) , are calculated as ri = - 4 13 á bi ẽ 1 ,,= è = {2.5, -20.0, -9.286} ai 4 ể -0 .4 0.2... + x 4 = 8 (c) xi 0; i = 1 to 4 (d) subject to Table 6-12 contains iterations of the Simplex method The optimum point is reached in just one iteration because all the reduced cost coefcients are nonnegative in the second canonical form (second tableau) The solution is given as basic variables: x1 = 4, x3 = 4 nonbasic variables: x2 = x4 = 0 optimum cost function: f = -4 The solution corresponds to point... gives optimum point Basicỉ x1 x2 x3 x4 x5 b x2 0 1 0.2 -0 .4 0 1 x1 1 0 0 .4 0.2 0 4 x5 0 0 0.8 1 .4 1 13 Cost 0 0 1.6 1.8 0 f + 18 (c ) 1 (c ) 2 (c ) 3 (c ) 4 (c ) 5 x3, x4, and x5 are slack variables Linear Programming Methods for Optimum Design 231 1 For 2x1 + x2 Ê 9: 2 For x1 - 2x2 Ê 2: 3 For -3x1 + 2x2 Ê 3: y1 = 1.6 (c in column x3) 3 y2 = 1.8 (c in column x4) 4 y3 = 0 (c in column x5) 5 Therefore,... INTRODUCTION TO OPTIMUM DESIGN x2 4 2x 1 + x2 =4 3 F 2 1 x1 f = 6 f = 5 f = 4 A 0 x2 = 1 x1 E B 1 +2 x2 =5 Optimum point C 2 D 3 4 x1 5 FIGURE 6-8 Constraints for Example 6.16 Feasible region: line E-C x1 - x 2 - x 4 + x 6 = 1 (d) xi 0; i = 1 to 6 (e) where x3 is a slack variable, x4 is a surplus variable, and x5 and x6 are articial variables The problem, solved in Table 6-19, takes just two iterations to reach... the equality constraint is changed to 5 from 4, the cost function f = (-x1 - 4x2) changes by 5 5 Df = - y2 De2 = - (5 - 4) = 3 3 5 That is, the cost function will decrease by , from -13 to -6 (z = 6) 3 3 2 34 INTRODUCTION TO OPTIMUM DESIGN (h) 6.5.2 Ranging Right Side Parameters When the right side of a constraint is changed, the constraint boundary moves parallel to itself, changing the feasible region... because it gives the designer options; any suitable point on the straight line joining the two optimum designs can be selected to better suit the needs of the designer Note that all optimum design points are global solutions as opposed to local solutions Example 6.10 demonstrates how to recognize an unbounded feasible set (solution) for a problem Linear Programming Methods for Optimum Design 217 EXAMPLE... 1.8 á j =è , = {4, 9}; - Ê Dc1 Ê min {4, 9}; or - Ê Dc1 Ê 4 a2 j ể 0 .4 0.2  (a) The range for c1 is obtained by adding the current value of c1 = -5 to both sides of the above inequality, - Ê Dc1 Ê - 1 (b) Thus, if c1 changes from -5 to -4, the new cost function value is given as * * fnew = f * + Dc1 x1 = -18 + ( -4 - (-5)) (4) = - 14 (c) That is, the cost function will increase by 4 For the second... + 4x2 subject to x1 + 2x2 Ê 5, 2x1 + x2 = 4, x1 - x2 1, x1, x2 0 Solution Constraints for the problem are plotted in Fig 6-8 It can be seen that line EC is the feasible region for the problem and point E gives the optimum solution Converting the problem to standard Simplex form, we obtain: minimize f = - x1 - 4 x 2 (a) x1 + 2 x 2 + x3 = 5 (b) 2 x1 + x 2 + x 5 = 4 (c) subject to 232 INTRODUCTION TO . xx=+ +£ +£ +£ ≥ 243 12 242 40 12 1 2 12 1 2 12 subject to ,,, , 208 INTRODUCTION TO OPTIMUM DESIGN 4 2x 1 + x 2 = 4 4x 1 + 3x 2 = 12 x 1 + 2x 2 = 4 4 3 3 2 2 1 10 A B C D F E Optimum solution. b i /a iq x 3 1110016 16 – 1 = 16 x 4 1 – 28 1 – 14 01 01 1 — 1/ 14 = 14 ¨ smallest x 5 1 – 14 1 – 24 00 11 1 — 1/ 24 = 24 Cost -40 0 -600 00 0f - 0 Second tableau: x 3 is identified to be replaced with x 1 in. x 1 x 2 x 3 x 4 b Ratio: b i /a iq x 3 -1110 4 4 – 1 = 4 ¨ smallest x 4 1101 6 6 – 1 = 6 Cost -4 -5 00 f Second tableau: x 4 is identified to be replaced with x 1 in the basic set. BasicØ x 1 x 2 x 3 x 4 b