Graphical method. plane. (for max) and down (for min) until it touches the set of feasible solutions. Graphical method


 Clement Benedict Lewis
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1 The graphical method of solving linear programming problems can be applied to models with two decision variables. This method consists of two steps (see also the first lecture): 1 Draw the set of feasible solutions (a feasibility region) on the plane. 2 Draw the objective function for z = 0 and move it parallel up (for max) and down (for min) until it touches the set of feasible solutions.
2 An unique solution A problem may have exactly one optimal solution, which is located in a corner of the feasibility region. max z = 5x 1 + 4x 2 6x 1 + 4x 2 24 x 1 + 2x 2 6 x 2 x 1 1 x 2 2 x 1, x 2 0
3 Infinite number of solutions A problem may have infinite number of optimal solutions located on a segment joining two corner points. max z = 2x 1 + 4x 2 6x 1 + 4x 2 24 x 1 + 2x 2 6 x 2 x 1 1 x 2 2 x 1, x 2 0
4 Unbounded objective function The objective function of a problem may be unbounded. max z = 2x 1 + 2x 2 2x 1 x 2 40 x 2 10 x 1, x 2 0
5 No solution Solving linear programming problems A problem may be infeasible. This happens when the constraints are contradictory. max z = 3x 1 + 2x 2 6x 1 + 4x 2 12 x 1 3 x 2 2
6 Conclusions (2varaiable case) 1 The feasibility region is a convex polyhedron (bounded or unbounded). 2 If a problem has an optimal solution, then at least one optimal solution is located in a corner of the feasibility region. 3 If the objective function is unbounded or the problem is infeasible, then the model is not well defined. In the first case some constraints are missing and in the second case the constraints are contradictory. Perhaps, the model does not describe the real problem correctly.
7 Model in standard form A model is in a standard form if 1 all the constraints are equalities (with the exception of the nonnegativity of the variables), 2 all the right hand sides are nonnegative and 3 all the variables can take only nonnegative values. max (min) z = c 1 x 1 + c 2 x 2 + +c nx n [Obj. function] a 11 x 1 + a 12 x 2 + +a 1n x n = b 1 [Constraint 1] a 21 x 1 + a 22 x 2 + +a 2n x n = b 2 [Constraint 2] a m1 x 1 + a m2 x 2 + +a mnx n = b m [Constraint m] x i 0, i = 1,...,n [Non. constr.] Every linear programming model can be transformed into the standard form.
8 Model in standard form Transform the following model into the standard form: max z = 2x 1 + 3x 2 x 3 x 1 2x 2 5 x 2 3x 3 3 x 1 + x 2 2x 3 = 20 x 1, x 2 0
9 Model in standard form Step 1. Multiply the second constraint by 1. max z = 2x 1 + 3x 2 x 3 x 1 2x 2 5 x 2 + 3x 3 3 x 1 + x 2 2x 3 = 20 x 1, x 2 0 Step 2. Add nonnegative slack variables s 1 and s 2 to the first and second constraint. max z = 2x 1 + 3x 2 x 3 x 1 2x 2 + s 1 = 5 x 2 + 3x 3 s 2 = 3 x 1 + x 2 2x 3 = 20 x 1, x 2, s 1, s 2 0
10 Model in standard form Step 3. Substitute x 3 = u 3 v 3, where u 3 and v 3 are new nonnegative variables. max z = 2x 1 + 3x 2 u 3 + v 3 x 1 2x 2 + s 1 = 5 x 2 + 3u 3 3v 3 s 2 = 3 x 1 + x 2 2u 3 + 2v 3 = 20 x 1, x 2, s 1, s 2, u 3, v 3 0 We get an equivalent model, which is in the standard form.
11 Basic solutions Solving linear programming problems In a set of m equations with n variables (m < n), if we set n m variables equal to 0, then we obtain a system of m linear equations with m variables (called basic variables). If this system has an unique solution, then it is called a basic solution. A basic solution is feasible if all its variables take nonnegative values. In the 2variable case, the set of feasible basic solutions corresponds to the set of corner points of the feasibility region.
12 Basic solutions Solving linear programming problems max z = 2x 1 + 3x 2 2x 1 + x 2 4 x 1 + 2x 2 5 x 1, x 2 0 max z = 2x 1 + 3x 2 2x 1 + x 2 + s 1 = 4 x 1 + 2x 2 + s 2 = 5 x 1, x 2, s 1, s 2 0
13 Basic solutions Solving linear programming problems If a linear programming problem has an optimal solution, then it also has a basic feasible solution which is optimal. A number of basic feasible solutions is finite, so in principle we could enumerate all of them and choose the one which minimizes (maximizes) the objective function. However, this method is very slow and fails when the objective function is unbounded (why?). The simplex algorithm is a procedure which locates the optimal basic feasible solution in an efficient manner.
14 Step 1. Convert the objective function and constraints into the following form: z 2x 1 3x 2 = 0 2x 1 +x 2 +s 1 = 4 x 1 +2x 2 +s 2 = 5 The above problem is in a basis form with respect to the basis (s 1, s 2 ). Notice that s 1 and s 2 do not appear in the first equation which corresponds to the objective function. The problem can also be represented as the following starting simplex tableau: x 1 x 2 s 1 s 2 z s s
15 Remarks: 1 The numbers in the zrow are called optimality coefficients or reduced costs. 2 All the basic variables must have the optimality coefficients equal to 0. 3 For the nonbasic variables two cases are possible: 1 Some variable has a negative optimality coefficient. Then it is possible to increase the value of the objective function by making the value of the variable positive. 2 All the nonbasic variables have nonnegative optimality coefficients, in which case the current basic solution is optimal.
16 Step 2. Perform an iteration (GaussJordan elimination) by inserting x 2 into the basis and removing s 2 from the basis. x 1 x 2 s 1 s 2 z s (4/1) s (5/2) x 1 x 2 s 1 s 2 z s x The nonbasic variable x 2 has the negative coefficient in the zrow. Hence increasing the value of x 2 will increase the value of the objective function z. The maximal value of x 2 is equal to min{4/1, 5/2} = 2.5. After increasing x 2 to 2.5 the value of s 2 falls to 0. So, the variable s 2 leaves the basis. The second basis form is: z 0.5x s 2 = x 1 0.5s 2 +s 1 = x s 2 +x 2 = 2.5
17 Step 3. Perform an iteration (GaussJordan elimination) by inserting x 1 into the basis and removing s 1 from the basis. x 1 x 2 s 1 s 2 z s (1.5/1.5) x (2.5/0.5) x 1 x 2 s 1 s 2 z 0 0 1/3 4/3 8 x /31/3 1 x /3 2/3 2 The nonbasic variable x 1 has the smallest negative coefficient in the zrow. Hence increasing the value of x 1 will increase the value of the objective function z. The maximal value of x 1 is equal to min{2.5/2.5, 2.5/0.5} = 1. After increasing x 1 to 1 the value of s 1 falls to 0. So, the variable s 1 leaves the basis. The second basis form is: z 1/3s 1 +4/3s 2 = 8 2/3s 1 1/3s 2 +x 1 = 1 1/3s 1 +2/3s 2 +x 2 = 2
18 x 1 x 2 s 1 s 2 z 0 0 1/3 4/3 8 x /31/3 1 x /3 2/3 2 The last table represents an optimal solution x 1 = 1, x 2 = 2 with the maximal value of the objective function equal to 8.
19 Exercise Solve the Reddy Mikks model (first lecture) using the simplex algorithm. First transform the model into the standard form. As a result we get the first basis form: z 5x 1 4x 2 = 0 6x 1 + 4x 2 +s 1 = 24 x 1 + 2x 2 +s 2 = 6 x 1 + x 2 +s 3 = 1 x 2 +s 4 = 2 x 1, x 2, s 1, s 2, s 3, s 4 0
20 x 1 x 2 s 1 s 2 s 3 s 4 z s (24/6) s (6/1) s () s () x 1 x 2 s 1 s 2 s 3 s 4 z 02/3 5/ x 1 1 2/3 1/ (4*3/2) s 2 0 4/31/ (2*3/4) s 3 0 5/3 1/ (5*3/5) s (2/1)
21 x 1 x 2 s 1 s 2 s 3 s 4 z /4 1/ x /41/ x /8 3/ /2 s /85/ /2 s /83/ /2
22 If the objective function should be minimized, we can multiply it by 1 and then maximize. Before the simplex algorithm starts we should represent the problem in a basic form. This is easy when all the constraints have the form of with nonnegative right hand sides. In this case the additional slack variables form the first basis. Otherwise and additional procedure should be used.
23 Artificial base method Consider the following example: max z = 4x 1 3x 2 3x 1 + x 2 = 3 4x 1 + 3x 2 6 x 1 + 2x 2 4 x 1, x 2 0 Step 1. Convert the problem to the standard form: max z = 4x 1 3x 2 3x 1 + x 2 = 3 4x 1 + 3x 2 s 1 = 6 x 1 + 2x 2 + s 2 = 4 x 1, x 2, s 1, s 2 0
24 Artificial base method Step 2. Add additional variables r 1 and r 2 and change the objective function. max z = 4x 1 3x 2 100r 1 100r 2 3x 1 + x 2 + r 1 = 3 4x 1 + 3x 2 s 1 + r 2 = 6 x 1 + 2x 2 + s 2 = 4 x 1, x 2, s 1, s 2, r 1, r 2 0 Remark: We should multiply the artificial variables with a suitably large constant so that their values in every optimal solution are 0. Step 3. Obtain a basic form by eliminating r 1 and r 2 from the objective function. z 696x 1 399x s 1 = 900 3x 1 + x 2 +r 1 = 3 4x 1 + 3x 2 s 1 +r 2 = 6 x 1 + 2x 2 + +s 2 = 4 Step 4. Apply the simplex algorithm to solve the problem.
25 Artificial base method If the value of some artificial variable in the optimal solution is positive, then the original problem is infeasible. In commercial packages, typically a different method of obtaining the first base is used. This method is called a twophase method and its description can be found in the literature.
26 Alternative optima x 1 x 2 s 1 s 2 z x x x 1 x 2 s 1 s 2 z x 2 1/2 1 1/ s 2 1/2 01/
27 Alternative optima If in the last (optimal) simplex tableu, some nonbasic variable has the optimality coefficient equal to 0, then we can obtain an alternative optimum by performing additional iteration in which we insert this variable into the basis. If (x 1,...,x n) and (y 1,...,y n) are two different optimal solutions, then each solution of the form (λx 1 +(1 λ)y 1,...,λx n +(1 λ)y n), λ [0, 1] is also optimal. Hence in this case, the number of optimal solutions is infinite.
28 Unbounded objective function x 1 x 2 s 1 s 2 z s () s ()
29 Unbounded objective function If in some simplex tableu, there exists a variable with a negative optimality coefficient and all the constraint coefficients in this variable s column are negative or zero then the objective function is unbounded. If the objective function is unbounded, then the model is not well defined. Perhaps, some constraints of the model are missing.
30 Degenerate solutions Consider the following problem: max z = 3x 1 + 9x 2 x 1 + 4x 2 8 x 1 + 2x 2 4 x 1, x 2 0 The second simplex tableu is the following: x 1 x 2 s 1 s 2 z 3/4 0 9/ x 2 1/4 1 1/4 0 2 (2*4) s 2 1/2 01/2 1 0 (0*2) In the basis solution, the basic variable s 2 = 0. Such a solution is called degenerate. After performing the next iteration, the value of the objective function will not increase.
31 Degenerate solutions The last simplex tableu is the following: x 1 x 2 s 1 s 2 z 0 0 3/2 3/2 18 x /21/2 2 x It is possible (in very rare situations) for the simplex method to enter a repetitive sequence of iterations never improving the objective value and never satisfying the optimality conditions (it is called cycling). However, there exist some technical modifications of the simplex method which allow us to avoid such a bad behavior (see the literature).
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