Chapter 3: Simplex Method

Transcription

1 Chapter : Simplex Method.1 Standard Problems In Chapter 2 you saw how to solve linear programming problems using the geometric method. While that method works very nicely for problems in which only 2 unknowns are involved, it becomes considerably more difficult to use in problems that have more than 2 unknowns. For example, if the problem has unknowns, a graphical method could be used, but then instead of graphing lines in the plane the problem would involve graphing planes in -dimensional space. If you re good at graphing things in - dimensional space you can see the corner points for the feasible region and determine which planes have intersected at a given corner point. Of course, finding the corner points involves solving equations in unknowns. Another variation on the graphical approach is: 1. Determine all possible intersection points for any combination of three planes (which involves taking every possible combination of three equations and solving for the three unknowns). 2. Check to see which of these points are actually in the feasible region. (i.e. determine which of the intersection points satisfy all of the inequalities.). Substitute the feasible corner points into the objective function and proceed as before. The second technique can be modified for problems with more than three unknowns but, while this method is theoretically possible, practically speaking it quickly becomes unmanageable as the number of equations and variables increases. So instead a technique known as the simplex method is used. The simplex method, like the geometric method, relies on the fact that if a linear function ƒ has a maximum (minimum) on a closed polyhedral region then it occurs at a corner point of the region. However, the simplex method differs from the graphical method in that it s an algebraic method and relies on matrix techniques. As such, it s a method that can be programmed on a computer and used to solve problems involving many variables and many constraint inequalities. In the first part of Section.1, the simplex procedure is outlined. However, even though the simplex method is an algebraic procedure, the underlying ideas are geometric, so in the last part of Section.1, a geometric interpretation of why the simplex method works is presented. When linear programming problems are solved using the simplex method, the problem must first be classified. The two classifications are standard and nonstandard. Standard problems will be dealt with in this section and nonstandard problems will be

2 Section.1: Standard Problems 77 done in Section.2. A linear programming problem has to satisfy three requirements in order to be classified as a standard problem: Standard Problems 1. The objective function is to be maximized. 2. All variables must be nonnegative.. Each constraint inequality must have the form: a 1 x 1 + a 2 x a n x n k, where a 1, a 2,..., a n are constants, x 1, x 2,..., x n are the variables, and k is a positive constant. There are three main steps in the simplex procedure: Let s see how to handle the first step. 1. Setting up the simplex tableau (which is a matrix). 2. Pivoting (this step may need to be done several times).. Reading the final solution. Slack variables and the initial simplex tableau First let s look at several examples to see how to set up a linear programming problem in matrix form. The first example to be set up is a problem which was already solved in Chapter 2 using the geometric method. It s easy to see that this one is a standard problem. (You ll see how to solve the problems a little later). Example.1. A potter is making cups and plates. It takes her 6 minutes to make a cup and minutes to make a plate. Each cup uses /4 lb. of clay and each plate uses 1 lb. of clay. She has 2 hours available for making the cups and plates and has 25 lbs. of clay on hand. She makes a profit of $2 on each cup and $1.5 on each plate. How many cups and how many plates should she make in order to maximize her profit? Solution (setting up the simplex tableau): You saw in Example 2.11 of Chapter 2, that this problem reduces to: Maximize P = 2x + 1.5y.75x + y 25 6x + y 12

3 78 Chapter : Simplex Method where x y x = number of cups produced y = number of plates produced. The graph of the feasible region looks like y x + y = 12 (,25) (12,16).75x + y = 25 (,) 1 (2,) 4 x If (x,y) is some point in the feasible region, then it has to satisfy the constraint inequalities and, in particular, has to satisfy the first two inequalities. However, in order to use matrix methods on linear programming problems, it s necessary to be dealing with equations, not inequalities. That means the constraint inequalities must be converted to equations. Now in this example, depending on the values chosen for x and y, there may be some unused clay and/or some unused time. Letting u be a variable that represents unused clay and v be a variable that represents unused time and using them in the two constraint inequalities converts the inequalities to equations in the following way:.75x + y + u = 25 6x + y + v = 12 Variables like u and v are called slack variables. Practically speaking, a slack variable represents the amount of unused resource in a given constraint inequality. So the first step in the simplex procedure is to introduce slack variables into each of the constraint inequalities (except for the ones which require the variables to be nonnegative those have already been considered in requiring the problem to be a standard problem) to convert the inequalities to equalities. To get the problem in final form for converting to a matrix, the equation for the objective function is rewritten in the following way: 2x 1.5y + P =

4 Section.1: Standard Problems 79 Combining this equation with the other two gives the following set of equations:.75x + y + u = 25 6x + y + v = 12 2x 1.5y + P = with 5 unknowns: x, y, u, v, and P. If the set of equations is now written in the form of an augmented matrix, it becomes: x y u v P This augmented matrix is known as a simplex tableau. Here the first two rows correspond to the equations converted from the constraint inequalities, while the last row is always used for the equation involving the objective function. The columns are also labeled with the variables associated with the coefficients used in the equations. Let s look at a few other examples at this point. For each of them, slack variables will be introduced and the simplex tableau set up. Example.2. Maximize P = x + 5y : 2x + y 12 2x + y 24 x 6 x y Solution (setting up the simplex tableau): Introducing slack variables into the first three constraint inequalities gives the equations: 2x + y + u = 12 2x + y + v = 24 x + w = 6 Putting the equation for the objective function in the form described earlier, gives: so the system of equations is: and the simplex tableau is: x 5y + P = 2x + y + u = 12 2x + y + v = 24 x + w = 6 x 5y + P = x y u v w P

5 8 Chapter : Simplex Method Example.. Maximize P = 4x + 2y : x + y 1 x + y 12 4x y x y Solution (setting up the simplex tableau): Introducing slack variables gives the equations: x + y + u = 1 x + y + v = 12 4x y + w = Rewriting the equation for the objective function and listing the equations yields so the initial simplex tableau is: x + y + u = 1 x + y + v = 12 4x y + w = 4x 2y + P = x y u v w P Now that you see how to set up the simplex tableau, what do you do with it? Well, if the columns of the simplex tableau are examined you can see they fall into two categories: 1. The column is a unit column which means it has a 1 in one of the positions and s in the other positions. 2. The column is not a unit column. In the initial simplex tableau for Example.1, the unit columns are the u, v and P columns while the x and y columns are nonunit. The variables associated with unit columns are called basic variables and the variables associated with the nonunit columns are called nonbasic variables. Thus, u, v and P are basic variables for this tableau and x and y are nonbasic variables. To start the simplex procedure the value of each of the

6 Section.1: Standard Problems 81 nonbasic variables is set equal to zero. The simplex tableau is then translated into its corresponding system of equations and the basic variables are solved for, assuming the nonbasic variables are zero. Thus, in Example.1, letting x = and y = gives u = 25, v = 12 and P =. (That agrees with the original set of equations since if x = and y = then there are 25 lbs. of unused clay and 12 minutes of unused labor, with a profit of $.) The solution x =, y =, u = 25, v = 12 and P = is called a basic solution. Finding a basic solution 1. Translate the simplex tableau into its corresponding system of equations. 2. Set the nonbasic variables equal to zero. (The nonbasic variables are the variables whose columns are not in unit form).. Solve for the basic variables (the ones associated with the unit columns). If the same procedure is applied to the simplex tableau of Example.2, x and y are found to be nonbasic variables and u, v, w and P are basic variables. Thus in the equations set which gives 2x + y + u = 12 2x + y + v = 24 x + w = 6 x 5y + P = x = and y = u = 12, v = 24, w = 6 and P = as values for the basic variables. The basic solution is x =, y =, u = 12, v = 24, w = 5 and P =. Similarly, using the simplex tableau found in Example., x and y are nonbasic variables, u, v, w and P are basic variables and that gives a basic solution of x =, y =, u = 1, v = 12, w = and P =. Now you re ready for the second step of the simplex procedure, namely pivoting. Pivoting and the complete simplex procedure

7 82 Chapter : Simplex Method Once a basic solution is determined it must be tested to see if it gives an optimal solution. Let s go back to Example.1 again. If x = and y =, then P =. However, it s clear that P = is not the maximum profit because P can be made bigger by simply producing some cups and/or plates (as long as the potter doesn t outstrip her resources or, mathematically speaking, as long as the points (x, y) stay in the feasible region). Since P = 2x + 1.5y, it seems the profit can be improved more by producing cups as opposed to plates, since the coefficient of x is bigger than that of y. To figure that out from the simplex tableau, observe that the number that shows up as the biggest positive coefficient on the right side of the equation P = 2x + 1.5y is 2 and that corresponds to the number that is most negative ( 2) in the last row of the simplex tableau. In the pivoting procedure, the column corresponding to the number in the last row which is most negative is called the pivot column. In this example the pivot column is the x column. Of course, the potter is limited in the number of cups she can produce. Looking at the graph you see that the largest number of cups that can be made and still stay in the feasible region is 2. However, that can be determined from the simplex tableau as well. Look at the pivot column and take each entry in that column (other than the 2 in the last row) and divide it into the constant in the constant column and the same row as that entry. For the.75 entry, the ratio 25/.75 or 1/ is obtained and for the 6 entry, the ratio 2. The number in the pivot column which gives the smallest ratio is called the pivot element. Finding the Pivot Element 1. In the last row, find the negative number that is the largest in absolute value. 2. Divide each positive entry in the pivot column into the number in the column of constants that is in the same row as the entry. The row of the smallest ratio is the pivot row.. The number in the pivot row and pivot column is the pivot element. The next step in the simplex procedure is the pivoting operation.

8 Section.1: Standard Problems 8 Pivoting 1. Divide the pivot row by the pivot element so the pivot element becomes Use the pivot row and row operations to get ʼs in the rest of the pivot column. Continuing with Example.1, box the pivot element in the simplex tableau and pivot on it to get x y u v P x y u v P 1 6 R /2 1/ x y u v P R 1.75R 2 5/8 1 1/ /2 1/6.5 1/ R + 2R 2 This is a new simplex tableau, having as unit columns the x, u and P columns. Thus, there is a new set of basic variables, x, u and P, and a new set of nonbasic variables, y and v. Applying the pivoting procedure converted x to a basic variable and v to a nonbasic variable. Setting the nonbasic variables equal to and finding the values of the basic variables as was done before, gives x = 2, u = 1, y =, v = and P = 4 which is the basic solution at this stage. The new basic solution has improved her profit from P = to P = 4 but is this solution optimal? The answer is no and that can be determined by going to the last row in the simplex tableau. If the last row is converted to its corresponding equation, then x.5y + u + 1/v + P = 4 or P =.5y 1/v + 4. Her profit can be made larger by making y larger since y is currently equal to zero.

9 84 Chapter : Simplex Method (Making v larger would decrease her profit.) But the way to see it directly from the simplex tableau is to note there is still a negative number (.5) in the last row. That negative number is in the y column. So the y column becomes the new pivot column and the pivot element again needs to be determined. Looking at each entry in the pivot column (other than.5) and finding the ratios as was done before yields 1 divided by 5/8 and 2 divided by 1/2. The ratios are 16 and 4 respectively and 16 is the smaller of the two ratios. The entry that gave the smallest ratio is 5/8 so that s the pivot element. Now pivot about 5/8, i.e. multiply that row by 8/5 to convert the pivot element to 1 and then use elementary row operations to convert the pivot column (in this case the y column) to a unit column. The computations are: x y u v P x y u v P 8 5 R 1 5/8 1 1/8 1 1/2 1/6.5 1/ /5 1/ /2 1/6.5 1/ x y u v P R R 1 1 8/5 1/ /5 4/15 4/5 7/ R R 1 Pivoting converted y to a basic variable and u to a nonbasic variable so the basic variables are now x, y and P and the nonbasic variables are u and v. Thus the basic solution is x = 12, y = 16, u =, v = and P = 48. At this point, since u =, the potter has used all her clay and since v =, she has used all her available time. To test to see if this solution is optimal, look at the equation corresponding to the last row of the simplex tableau and solve for P. P = 4/5u 7/v Increasing either u or v decreases P. However, again that can be determined from the simplex tableau by noting that there are no longer any negative numbers in the bottom row. Thus, the optimal solution is: x = 12 y = 16 P = 48 and u = v =.

10 Section.1: Standard Problems 85 Now that one problem has been solved in detail, let s summarize the procedure. Summary: Simplex procedure for standard problems 1. Convert the constraint inequalities to equalities by introducing a slack variable into each inequality. (Remember to exclude the inequalities that require the variables to be nonnegative). Then set up the initial simplex tableau. 2. Determine if the optimal solution has been reached (i.e. check if the bottom row has negative entries). If there are no negative entries, you have the optimal solution: Go directly to step 5. (If there are negative entries then continue with step.). Find the pivot element. 4. Pivot using the pivot element and then go back to step The optimal solution is found by determining the basic solution at this stage. Now let s look at Examples.2 and. again. Since Step 1 has already been done for each example, the examples will begin with Step 2. Example.2 (continued). Maximize P = x + 5y : 2x + y 12 2x + y 24 x 6 x y Solution (continued): The initial simplex tableau was: x y u v w P

11 86 Chapter : Simplex Method Step 2: There are negative numbers in the last row. Step : The most negative is 5, so the pivot column is the y column. Computing ratios gives 12/1 =12 and 24/ = 8. The smallest ratio is 8 so is the pivot element. It s good practice to box the pivot element at this stage so as not to lose track of it. Step 4: Pivoting on this pivot element results in the following row operations: x y u v w P x y u v w P R / 1 1/ R 1 R 2 x y u v w P 4/ 1 1/ 2/ 1 1/ 1 1 1/ 5/ R 4 + 5R 2 Step 2: Since there are no longer any negative numbers in the last row, the basic solution is the optimal solution. Step 5: The optimal solution is: x =, y = 8, u = 4, v =, w = 6 and P = 4. Example. (continued). Maximize P = 4x + 2y x + y 1 x + y 12 4x y x, y Solution (continued): The simplex tableau is

12 Section.1: Standard Problems 87 x y u v w P Step 2: The solution is not optimal since there are negative numbers in the last row. Step : The pivot column is the x column since that is the column associated with the most negative number in the last row. Computing ratios gives 1/1 =1, 12/ = 4 and /4. /4 is the smallest ratio so the pivot element is the entry in the third row and first column. Step 4: Boxing the 4 and pivoting on it yields x y u v w P x y u v w P R /4 1/ /4 9/4 /4 R 1 R x y u v w P R 2 R 7/4 1 1/4 1/4 1 /4 1 /4 1/ /4 9/4 /4. R 4 + 4R The basic solution at this stage is x = /4, y =, u = 7/4, v = 9/4, w = and P =. Step 2: There s still a negative number in the last row so the procedure must be continued. Step : The pivot column is now the y column and computing ratios gives (7/4)/(7/4) = 7/7 and (9/4)/(1/4) =. (Notice when the ratios were computed, the /4 in the third row of the pivot column was ignored since the instructions in step for finding the pivot element specify that the pivot element must be a positive number.) The smallest ratio is

13 88 Chapter : Simplex Method so the pivot element is 1/4, the entry in the second row of the pivot column. Step 4: Pivoting on this element gives x y u v w P 7/4 1 1/4 1/4 1 /4 1 /4 1/ /4 9/4 /4 4 1 R 2 x y u v w P 7/4 1 1/4 1 4/1 /1 1 /4 1/ /4 /4 x y u v w P R R 2 R + 4 R 2 1 7/1 2/1 1 4/1 /1 1 /1 1/1 2/1 2/ R 4 + 5R 2 The basic solution at this stage is x =, y =, u = 4, v =, w = and P = 18. Step 2: However, there s still a negative number in the last row so the basic solution is not the optimal solution. Step : The pivot column is the w column so computing ratios, 4/(2/1) = 26 and /(1/1) = 9. Thus the pivot element is 2/1. Step 4: Pivoting on it, x y u v w P

14 Section.1: Standard Problems /1 2/1 1 4/1 /1 1 /1 1/1 2/1 2/ R 1 x y u v w P 1/2 7/ /1 /1 1 /1 1/1 2/1 2/ x y u v w P R R 1 R 1 1 R 1 1/2 7/2 1 1 /2 1/2 1 1/2 1/ R R 1 Step 5: Since no negative numbers appear in the bottom row, the basic solution is optimal. It is x = 1, y = 9, u =, v =, w = 26 and P = 22. Of course, the simplex method is most useful in situations where it s difficult to use the geometric method, so let s go through one more example, applying the method to a problem which couldn t be solved before. The following example was introduced as Example 2.14 in Chapter 2. Example.4. A manufacturer of outdoor furniture makes wooden couches, tables and chairs, each of which must be fabricated, assembled and finished. Each table requires 1 hour to fabricate, hours to assemble and 4 hours to finish and returns a profit of $1. Each couch requires 6 hours to fabricate, 1 hours to assemble and 8 hours to finish and returns a profit of $12. Each chair requires 1 hour to fabricate, 1 hour to assemble and an hour to finish and returns a profit of $6. The manufacturer has 12, staff-hours available each week for fabrication,, staff-hours for assembling and 4, staff-hours for finishing. How many units of each type of furniture should the manufacturer produce each week in order to maximize his profit?

15 9 Chapter : Simplex Method Solution: In Chapter 2 the problem was formulated as: Maximize P = 1x + 12y + 6z x + 6y + z 12, x +1y + z, 4x + 8y + z 4, x, y, z Step 1: Convert the inequalities to equations by introducing slack variables: x + 6y + z + u = 12, x +1y + z + v =, 4x + 8y + z + w = 4, 1x 12y 6z + P = The simplex tableau is: x y z u v w P The basic solution is not optimal so the pivot column and the pivot element are determined and the pivot element is then boxed in the above matrix. Converting the pivot column to a unit column, x y z u v w P 1 6 R 1 1/6 1 1/6 1/ x y z u v w P R 2 1R 1 R 8R 1 1/6 1 1/6 1/6 4/ 2/ 5/ 1 8/ 1/ 4/ R R 1

16 Section.1: Standard Problems 91 The next pivot element is boxed in the above matrix, and the procedure is continued, converting the new pivot column (the x column) to a unit column. 4 R 2 x y z u v w P 1/6 1 1/6 1/6 1 1/2 5/4 /4 8/ 1/ 4/ x y z u v w P R R 2 R 8 R 2 1 1/4 /8 1/8 1 1/2 5/4 / R 4 + 8R 2 Now, in trying to find the next pivot column, there are two choices, since both of the negative numbers in the last row have the same value. Either the z or the u column can be picked. Let s pick the z column. (You might see what happens if the u column is picked your final answer should be the same). The new pivot element is enclosed in a box in the matrix above. Converting the z column to a unit column gives 4R 1 x y z u v w P 4 1 /2 1/2 1 1/2 5/4 /

17 92 Chapter : Simplex Method x y z u v w P R R 1 R R /2 1/ /2 1/2 4 1/2 / R 4 + 8R 1 Since the last row has no negative numbers, the procedure is done. The optimal solution is: x = 9 v = y = w = 1 z = P = 18, u = That means the manufacturer should produce 9 tables, couches, and chairs for a maximum profit of $18,. Since he has 1 excess hours of finishing work, he might also consider reassigning or hiring out a few of the workers assigned to finishing the furniture. (In Section., you ll see how he can best decide what to do). The simplex method looks complicated at first but after going through the procedure on a few problems, you ll quickly get the hang of it. You do have to be careful with your computations and you do need to stick closely to the procedure since deviations from this procedure can mess you up badly. One deviation that is allowed is in picking the pivot column. The procedure calls for choosing a pivot column by looking for the number that is most negative in the last row of the simplex tableau. In fact, you can choose any column that has a negative number in the last row. You might look at Example and see what would happen if you picked y as the pivot column initially. While the simplex method does allow a wide variety of linear programming problems to be solved, there are many technical difficulties that have not been considered that can occur in the process of solving these problems. Some aspects are beyond the scope of this book. However, you should be prepared to deal with a few of the situations that might happen. For example, if the feasible region is unbounded, the objective function may not have a maximum (minimum) on the region. In that case, the simplex method will break down at some point. For instance, when you compute the ratios in the pivot column, you might end up with none of the ratios being positive. Another possible scenario is that there may be more than one solution to the problem (this was discussed when solving problems using the geometric method two adjacent vertices gave the same optimal value for the objective function). In that case, the choice of pivot column may well affect the final solution. See Exercise 25 at the end of this section. One more situation that can arise is the appearance of a zero in the upper part of the

18 Section.1: Standard Problems 9 last column at some point in the simplex procedure. This is associated with a condition called degeneracy. For most problems having only a few variables, the simplex procedure can be followed without difficulty and the optimal solution will then be obtained. However, in some cases degeneracy can lead to a situation known as cycling, in which at some point in the simplex procedure a sequence of simplex tableaux repeats over and over so the objective function stays the same and thus you never arrive at the optimal solution. Cycling is a rare occurrence and there are techniques for dealing with it which are incorporated in computer programs that use the simplex procedure, but they are beyond the scope of this text. Geometric Interpretation of the Simplex Procedure The key to understanding the geometry behind the simplex procedure is recognizing two facts: 1. A basic solution at any stage of the simplex procedure corresponds (in the standard problem) to a vertex of the feasible region. 2. In pivoting the basic solution is exchanged for another one and in the process one nonbasic variable is also exchanged for another one. Geometrically that corresponds to moving from one vertex to an adjacent one along a boundary line. Let s see how to interpret Examples.1 and. geometrically. Example.1 (continued): A potter is making cups and plates. It takes her 6 minutes to make a cup and minutes to make a plate. Each cup uses /4 lb. of clay and each plate uses 1 lb. of clay. She has 2 hours available for making the cups and plates and has 25 lbs. of clay on hand. She makes a profit of $2 on each cup and $1.5 on each plate. How many cups and how many plates should she make in order to maximize her profit? Solution (geometric interpretation): The mathematical statement of the problem was as follows: Maximize P = 2x + 1.5y.75x + y 25 6x + y 12 x y where x = number of cups produced y = number of plates produced.

19 94 Chapter : Simplex Method y x + y = 12 (,25) (12,16).75x + y = 25 (,) 1 (2,) 4 x In order to understand the geometric ideas behind the simplex procedure, it s important to realize that for the graph of the feasible region, each of the boundary lines can be described by setting one of the original variables, x or y, or one of the slack variables, u or v, equal to. Thus, the 4 corner points can be represented as the intersection of lines in which two of the variables are set equal to. (x = and y = in the case of A, y = and v = in the case of B, u = and v = in the case of C and x = and u = in the case of D). y 4 D(,25) or x =, u = 2 1 6x + y = 12 or v = C(12,16) or u =, v =.75x + y = 25 or u = A(,) or x =, y = 1 B(2,) or 4 y =, v = x The first step in the simplex procedure was to set up the simplex tableau. x y u v P

20 Section.1: Standard Problems The basic solution at this point is x =, y =, u = 25, v = 12 and P =. The values of the nonbasic variables are x = and y = which correspond to the vertex A. The next step is to find the pivot element which was done by finding the pivot column and computing ratios to find the pivot element. The pivot column was the x column and the ratios were 25/.75 = 1/ and 12/6 = 2. The ratios represent the intersection of the x axis (y = ) with the line.75x + y = 25 (u = ) and the line 6x + y = 12 (v = ) respectively. The first point (1/, ) is not in the feasible region while the second point (2, ) is in the feasible region. Check this out on the graph above. Thus, the smallest ratio represents a limiting point for the feasible region. The next step was to pivot about the pivot element 6 which gave x y u v P x y u v P 1 6 R /2 1/ x y u v P R 1.75R 2 5/8 1 1/ /2 1/6.5 1/ R + 2R 2 The basic solution (which is not optimal) at this point is x = 2, y =, u = 1, v = and P = 4. The basic solution corresponds to the vertex x = 2, y = which is the intersection of the boundary lines y = and v =. So, in pivoting, the nonbasic variable x was changed to basic and the basic variable v was changed to a nonbasic variable. Geometrically, the basic solution moved from the vertex A (x =, y = ) along the boundary line y = to the vertex B (y =, v = ). Since the entry in the last row of the y column is negative, y should be increased and v should be kept equal to zero. Moving along the line v =, observe that it intersects the line u = at the point (12, 16) and the line x = at the point (, 4). Clearly the next vertex to move to is (12, 16) since it s in the feasible region while (, 4) is not. The corresponding step in the simplex procedure is choosing a pivot column and computing ratios to determine the pivot

21 96 Chapter : Simplex Method element. The pivot column is the y column and the ratios are 1/(5/8) = 16 and 2/(1/2) = 4. Again, choosing the smallest ratio in the pivoting step guarantees the solution stays in the feasible region. The next step algebraically was to pivot about 5/8, obtaining x y u v P x y u v P 8 5 R 1 5/8 1 1/ /2 1/6.5 1/ /5 1/ /2 1/6.5 1/ x y u v P R R 1 1 8/5 1/ /5 4/15 4/5 7/ R R 1 The basic (and optimal) solution at this point is x = 12, y = 16, u =, v = and P = 48. The basic solution corresponds to the vertex x = 12, y = 16 which is the intersection of the two boundary lines u = and v =. In pivoting, the basic solution moved from the vertex B (y =, v = ) to the vertex C (u =, v = ) along the boundary line v =. Example. (continued). Maximize P = 4x + 2y : x + y 1 x + y 12 4x y x, y Solution (geometric interpretation):

22 Section.1: Standard Problems 97 (,1) or x =, u = y x + y = 12 or v = (1,9) or v =, u = 8 6 4x y = or w = 4 2 (,) or v =, w = x + y = 1 or u = (,) The simplex tableau is (/4,) or y =, w = x y u v w P x The initial basic solution is x =, y =, u = 1, v = 12, w = and P =. The vertex corresponding to this is the origin. The pivot column is the x column and the ratios are 1/1 = 1, 12/ = 4, and /4. Those ratios correspond to the intersection of the boundary line y = with the boundary lines u =, v = and w = respectively. Picking the smallest ratio of /4 keeps the solution in the feasible region. Pivoting about 4 gives x y u v w P x y u v w P R /4 1/ /4 9/4 /4 x y u v w P

23 98 Chapter : Simplex Method R 1 R R 2 R 7/4 1 1/4 1/4 1 /4 1 /4 1/ /4 9/4 /4 R 4 + 4R The basic solution is now x = /4, y =, u = 7/4, v = 9/4, w = and P = which corresponds to the vertex x = /4, y = which is the intersection of the boundary lines y = and w =. Next the y column is picked as the pivot column and ratios are computed, ignoring the /4. (Can you think of a geometric reason why /4 is ignored?) The ratios are 7/7 and which represent the intersection of the w = line with u = and v = respectively. Picking the smaller ratio and pivoting on 1/4 gives x y u v w P 7/4 1 1/4 1/4 1 /4 1 /4 1/ /4 9/4 /4 4 1 R 2 x y u v w P 7/4 1 1/4 1 4/1 /1 1 /4 1/ /4 /4 x y u v w P R R 2 R + 4 R 2 1 7/1 2/1 1 4/1 /1 1 /1 1/1 2/1 2/ R 4 + 5R 2

24 Section.1: Standard Problems 99 The basic solution is x =, y =, u = 4, v =, w = and P = 18. In pivoting the basic solution has moved to the vertex x = and y = along the boundary line w =. This vertex is the intersection of the boundary lines v = and w =. The w column now becomes the pivot column. Computing ratios gives 26 and 9. What do these numbers correspond to on the graph? The pivot element is 2/1 and again pivoting, x y u v w P 1 7/1 2/1 1 4/1 /1 1 /1 1/1 2/1 2/ R 1 x y u v w P 1/2 7/ /1 /1 1 /1 1/1 2/1 2/ x y u v w P R R 1 R 1 1 R 1 1/2 7/2 1 1 /2 1/2 1 1/2 1/ R R 1 The basic (and optimal) solution is x = 1, y = 9, u =, v =, w = 26 and P = 22. The vertex which gives this result is x = 1, y = 9 and represents the intersection of the boundary lines u = and v =. Notice that in the simplex procedure the basic solution started out at the origin and, after pivoting, it moved along the boundary line y = to the adjacent vertex (/4, ). After another pivoting operation it moved along the boundary line w = to the vertex (, ) and the final pivoting operation moved it along the boundary line v = to the vertex (1, 9) which is the vertex corresponding to the optimal solution.

25 1 Chapter : Simplex Method Problems In Exercises 1-4, determine if the stated problem is standard or nonstandard. If it is nonstandard, explain why. If it is standard, set up the initial simplex tableau. 1. Maximize P = 2x + y 2. Minimize P = x + y 4x + 2y 9 2x + 5y 14 x + y 12 x + y 2 x, y x 2y 5 x, y. Maximize P = x + 2y + z 4. Maximize P = 2x y + 4z x + y + z 15 x + 2y + 4z 12 x + 6 2z x + y 4 x, y, z y + 2z 5 x, y, z In Exercises 5-8, find the basic solution for the simplex tableau and determine if the solution is optimal. If it is not optimal find the pivot element, go through one complete pivoting step and state the new basic solution. 5. x y u v P x y u v w P x y u v w P x y z u v w P / / /2 1 1/1 1/ In Exercises 9-12, introduce slack variables to convert the constraint inequalities to equations. Then graph the feasible region, labeling all boundary lines and vertices, using x =, y =, etc. Then solve the problem using the simplex procedure. After each complete pivoting step, state the basic solution and find the corresponding vertex on the graph. Check your answer using the geometric method. 9. Maximize P = 2x + y 1. Maximize P = 2x + y

26 Section.1: Standard Problems 11 x + 2y x + y 1 x + y x + y 4 x, y x + 2y 5 x, y 11. Maximize P = x + 2y 12. Maximize P = x + y the constraints given in x + y 2 Exercise 11 x + y 6 x y 2 x, x, y In Exercises 1-2, solve the problems using the simplex procedure 1. Maximize P = x + 4y 14. Maximize P = x + 4y x + y 5 2x + y 1 y 9 x + y 1 x, y x, y 15. Maximize P = 2x + y 16. Maximize P = 4x + 2y + z x + y 9 2x y + z 6x + y 12 x + y + z 5 x, y x + y + z 2 x, y, z 17. Maximize P = x + 6y + 8z 18. Maximize P = x + 2y + z 4x + 2y + z 6 x + 2y + z 11 2x + y + 2z 6 x y + 2z 12 x, y, z x, y, z 19. Maximize P = 6x + 6y + 5z 2. Maximize P = 7x + 12y + 8z x + y + z 12 x + 2y + z 144 x + y + z 5x + 2z 216 x, y, z x + 8y + 6z 768 x, y, z 21. Maximize P = 5x + 4y + z 22. Maximize P = 1 x + 2y + 12z 1/2 x + 1/ y + 1/2 z 12 x + 2y 15 1/2 x + 1/ y 8 x + 2z 15 1/ y + 1/2 z 1 2x + z 8 x, y, z 2x + y + z 225 x, y, z

27 12 Chapter : Simplex Method 2. Maximize P = 25x + 5y + 8z + 2q x 15 y 1 z 1 q x + 2y + 2z + 4q 4 x, y, z, q 24. Try to solve the following problem using the simplex method. Then graph the feasible region to see what s happening. Maximize P = x + 2y x y x 2y 2 x, y 25. Solve the following problem using the simplex method by (a) choosing the x column as the first pivot column (b) choosing the y column as the first pivot column Then solve the problem using the graphical method. Explain your results. Maximize P = x + 2y x + 2y 6 x + 2y 4 x, y 26. A farmer has a 2 acre farm on which she plants two crops: corn and soybeans. For each acre of corn planted, her expenses are $5 and for each acre of soybeans planted, her expenses are $1. Each acre of corn requires 1 bushels of storage and yields a profit of $6; each acre of soybeans requires 4 bushels of storage and yields a profit of $9. If the total amount of storage space available is 19,2 bushels and the farmer has only $2, on hand, how many acres of each crop should she plant in order to maximize her profit? What will her profit be if she follows this strategy? 27. A company manufactures checker sets and chess sets. Suppose each day the company has available 19 boards (which can be used for both games) and 8, units of wood for making pieces. Each checker set uses 2 units of wood and each chess set uses 8 units of wood. The distributors the company sells to can take up to 125 checker sets per day and up to 75 chess sets per day. The company makes a profit of $1 on each checker set and $1.25 on each chess set. How many checker sets and how many chess sets should the company make each day in order to

28 Section.1: Standard Problems 1 maximize its profits? What is the profit per day using this strategy? 28. A developer is planning to build a new subdivision consisting of townhouses, singlestory detached houses and two-story detached houses. On one acre he can put 6 townhouses or 4 single-story houses or 2 two-story houses and he has 6 acres available. It costs him $4, to build each townhouse, $5, to build each single-story house and $6, to build each two-story house. He makes a profit of $15, on each townhouse, $18, on each single-story house and $2, on each two-story house and he has $2,88, of capital available. Townhouses require 25 hours of labor, single-story houses require hours of labor and twostory houses require 4 hours of labor and he has 24, hours of labor available. How many houses of each type should he construct in order to maximize his profit? 29. A company makes two types of sofas, regular and long, at two locations, one in Hickory and one in Lenoir. The plant in Hickory has a daily operating budget of $45, and can produce at most sofas daily in any combination. It costs $15 to make a regular sofa and $2 to make a long sofa at the Hickory plant. The Lenoir plant has a daily operating budget of $6,, can produce at most 25 sofas daily in any combination and makes a regular sofa for $15 and a long sofa for $18. The company wants to limit production to a maximum of 25 regular sofas and 5 long sofas each day. If the company makes a profit of $5 on each regular sofa and $7 on each long sofa, how many of each type should be made at each plant in order to maximize profit? What is the maximum profit?.2 Nonstandard Problems In Section.1 standard linear programming problems were considered. That meant that the problem had to satisfy three conditions: 1. The objective function had to be maximized. 2. All the variables (including the slack variables) had to be nonnegative.. All of the constraint equations (other than the ones implied in condition 2) had to be written as: a 1 x 1 + a 2 x a n x n k, where k is a positive constant. If any of these three conditions are violated, the problem is nonstandard. This section will deal with handling certain types of nonstandard problems. For that type of nonstandard problem, procedures will be applied to its simplex tableau which will get the simplex tableau of that problem into standard form (no negative numbers in the last column, except possibly in the last row), at which point the method of Section.1 can take over. The first type of nonstandard problem that is going to be considered is one in which conditions 2 and hold but the objective function is to be minimized rather than maximized. Since condition 1 is violated, the problem is thus nonstandard. That situation is easily handled by making the following observation: Minimizing the objective function ƒ is equivalent to maximizing the function g = ƒ.

29 14 Chapter : Simplex Method For example, if you had a set of numbers S, where then the minimum of S is 5. But S = { 1,, 15, 5, 7, 9} S = {1,, 15, 5, 7, 9} so the maximum of S is 5. Thus, if an objective function C is to be minimized, the problem can be changed so that C is the objective function and the goal is to maximize C. Once the maximum of C is found then solving for C gives the minimum of C. Let s look at an example to see how this works in practice. Example.5. Minimize C = x 4y x + y 6 2x + y 16 x 4 x y Solution: Let s solve the problem: Maximize C = x + 4y x + y 6 2x + y 16 x 4 x y.1. Since this is a standard problem, it can be solved using the techniques of Section x y u v w C x y u v w C R / 1 1/ / 4 x y u v w C

30 Section.2: Nonstandard Problems 15 R 1 R 2 1/ 1 1/ 2/ 1 1/ 1 1 1/ 4/ 1 2/ 16/ 4 64/ R 4 + 4R 2 x y u v w C R / 1 1/ 1 1 1/ 4/ / 4 64/ x y u v w C R 2 2 R 1 R R R R 1 The solution is x = 2, y = 4, u =, v =, w = 2, with the maximum of C = 22 so the minimum of C is C = 22. A more common (and also more complicated) situation arises when condition is violated. (Condition is the one that requires the constraint inequalities to be written as: a 1 x 1 + a 2 x a n x n k, where k is a positive number.) Consider the problem: Maximize P = x + 4y 2x + y 12 2x y 4 x y. All of the constraint inequalities can be written in the form: ax + by k but if that is done, then the second inequality becomes

31 16 Chapter : Simplex Method 2x + y 4 and the constant term on the right is now negative, which is a violation of condition. Let s overlook that difficulty for now and find the initial simplex tableau for the problem which is now stated as: Maximize P = x + 4y 2x + y 12 2x + y 4 x, y As before, introduce slack variables u and v to change the inequalities to equalities. 2x + y + u = 12 2x + y + v = 4 x 4y + P = Then write the initial simplex tableau. x y u v P The basic solution associated with this simplex tableau is: x = y = u = 12 v = 4 P = However v = 4 is a violation of condition 2 which requires that all variables, including slack variables, have to be nonnegative. Let s see what s happening geometrically in this problem, since this problem can be solved graphically. y 8 (1) 6 2x + y = -4 or v= 4 2 (,2) or u=, v= 2x + y = 12 or u= (2,) or y=, v= (6,) or y=, u= 1 x

32 Section.2: Nonstandard Problems 17 Looking at the feasible region for this problem, it s obvious that the initial basic solution, which says x = and y =, is not in the region, so the basic solution is, in fact, not feasible. However, even though the origin is not in the feasible region, it s still the intersection of the lines x = and y = which are the boundary lines for the constraint inequalities x and y. If you think about all the standard problems which were done in Section.1, you ll realize that in every case, the initial basic solution (which was the origin) was always a feasible solution. Geometrically, it meant that the origin was always in the feasible region. Of course, it generally wasn t the optimal solution so, in using the simplex method, the basic solution moved along a boundary line to an adjacent vertex in the feasible region and the new vertex was tested to see if it gave the optimal solution. The part of the simplex method that allowed the basic solution to move to that adjacent vertex was the pivoting operation. It can be shown that the pivoting operation could be used in the above problem to move the basic solution from the origin along a boundary line to an adjacent vertex which will be in the feasible region. The procedure is as follows: Nonstandard Linear Programming Problems 1. Get all constraint inequalities (other than the ones that require the variables to be nonnegative) in the form a 1 x 1 + a 2 x a n x n k, (k may be negative). Then introduce slack variables and write the initial simplex tableau. 2. Go to the row which has a negative number in the constant column. (If thereʼs more than one such row, pick any one of them.). Pick any negative number in that row that is to the left of the negative constant. (If thereʼs more than one negative number in the row, pick any one of them.) The column containing that entry will then be the pivot column. Divide each entry in the pivot column (except for the one in the last row) into the number that is in the constant column and the same row as the entry. The entry which gives the smallest positive ratio is the pivot element. Pivot about that entry. 4. Continue with the procedure until there are no negative numbers in the last column (with the possible exception of the entry in the last row and column). At this point the problem is a standard problem so the procedure of Section

33 18 Chapter : Simplex Method.1 can take over. (Algebraically, the basic solution at this stage is a feasible solution. Geometrically, what has happened is the basic solution has moved to a vertex which is in the feasible region.) Let s now apply the above procedure to the initial simplex tableau (1). At the same time, look at the graph of the problem to see what s happening geometrically. x y u v P The only negative entry in the last column is 4 which is in the second row. The only entry in that row to the left of 4 is 2 which is in the first column. So the pivot column is the x column. Computing ratios gives x y u v P ratios /2 = 6 4/ 2 = 2 The pivot element is 2. (Just as in Section.1, notice that the ratios correspond to the intersection of the line y = with the line u = (for the ratio 6) and the line v = (for the ratio 2). In this case, both points are in the feasible region, but 2 in the first column is chosen as the pivot element because the ratio 2 is less than the other ratio which is 6. Note that taking the smallest ratio moves the basic solution along the boundary line to the nearest vertex. In the case of a standard problem that guarantees that the solution stays in the feasible region. In the case of a nonstandard problem it at least gives a systematic way to tackle the vertices). Pivoting about the pivot element, x y u v P 1 2 R /2 1/ x y u v P

34 Section.2: Nonstandard Problems 19 R 1 2R /2 1/2 11/2 / R + R 2 At this stage, the basic solution is x = 2, y =, u = 8, v = and P = 6. This solution is feasible since there are no negative entries in the last column. So the simplex tableau is now standard. (And, as you can see, the point (2,) is in the feasible region). Continuing, using the procedure for standard simplex problems, x y u v P x y u v P 1 4 R /2 1/2 11/2 / /4 1/ /2 1/2 11/2 / x y u v P R R 1 1 1/4 1/4 1 1/8 /8 11/8 1/ R R 1 x y u v P 4R /8 /8 11/8 1/ x y u v P

35 11 Chapter : Simplex Method R R /2 1/2 1/2 / R R 1 The solution is: x = 6, y =, u =, v = 8 and P = 18. Check it using the graphical method. Let s do a few more examples to make sure you understand the technique and how to apply it. In problems where it s possible, a graph will also be provided. It would be worthwhile for you to look at the graph as the simplex procedure is applied to see what s happening geometrically. Example.6. Maximize P = x + y x + 2y 18 2x + y 21 x + y 1 x y Solution: y 2 (,21) 2x + y = 21 or v= (,1) (2,8) or u=, w= x + y = 1 or w = (8,5) or u=, v= x + 2y = 18 or u= 1 2 x Following the procedure outlined above, multiply the first and third inequalities by 1 to get them in the correct form. Introduce slack variables u, v, and w to get the following equalities: x 2y + u = 18

36 Section.2: Nonstandard Problems 111 The initial simplex tableau is 2x + y + v = 21 x y + w = 1. x y u v w P Going to step 2 in the procedure for nonstandard problems, either the first or third rows can be chosen. Suppose the first row is chosen. There are two negative entries in the first row and either may be picked. If the 1 is chosen then that means the first column is the pivot column. Computing ratios gives 18/ 1 = 18, 21/2 = 1.5 and 1/ 1 = 1. The smallest ratio is 1 so the pivot element is the entry in the third row and the first column. Pivoting about this entry, x y u v w P ratios R x y u v w P / 1 = 18 21/2 = 1.5 1/ 1 = R 1 + R x y u v w P R 2 2R R 4 + R The basic solution corresponding to this tableau is x = 1, y =, u = 8, v = 1, w = and P =. (What has the pivoting operation accomplished geometrically? At which vertex is the solution? Is it in the feasible region?) Since there s still a negative number in the last column, the procedure is continued.

37 112 Chapter : Simplex Method There are two negative entries in the first row. Let s pick the one in the y column. So the pivot column is the y column and computing ratios yields: 8/ 1 = 8, 1/ 1 = 1 and 1/1 = 1. The smallest positive ratio is 8 so the pivot element is the entry in the first row and second column. Pivoting about this entry, x y u v w P x y u v w P R x y u v w P R 2 + R 1 R R R 4 2R 1 The basic solution is x = 2, y = 8, u =, v = 9, w = and P = 14. Geometrically, the pivoting operation has moved the basic solution along the boundary line w = to the vertex (2, 8) which is the intersection of the boundary lines w = and u =. Now as can be seen from the graph the basic solution is feasible but that can also be determined from the simplex tableau since there are no negative entries in the last column. So the problem is now standard, but of course it s not finished since the solution isn t optimal there s have a negative number in the bottom row. Continuing with the procedure, x y u v w P x y u v w P R / 1/ x y u v w P

38 Section.2: Nonstandard Problems 11 R 1 R 2 R + 2R 2 1 2/ 1/ 1/ 1/ 1 1 1/ 2/ 1/ 5/ R 4 + 5R 2 The optimal solution is x = 8, y = 5, u =, v =, w = and P = 29. Example.7. Maximize P = 4x + y + 2z 2x + 4y + z 16 2x + 5y z 1 x, y, z Solution: The initial simplex tableau is: x y z u v P There is one negative number in the last column and that s in the second row. Either the first or second entry in that row can be picked. Choosing the first entry means the first column is the pivot column and after computing ratios, it s clear that the pivot element is 2, the entry in the second row and first column. Pivoting on that element, x y z u v P x y z u v P 1 2 R /2 1/2 1/ x y z u v P

39 114 Chapter : Simplex Method R 1 2R /2 1/2 1/ R + 4R 2 This is a standard simplex tableau so, proceeding as usual, the final simplex tableau becomes: 1 2 R 1 x y z u v P 1/2 1 1/2 1/2 1 5/2 1/2 1/ x y z u v P R R 1 1/2 1 1/2 1/2 1 9/4 1/4 1/ /2 2 R + 4R 1 The optimal solution is x = 1/2, y =, z =, u =, v = and P = 2. Example.8. Minimize C = 2x y x + y 6 x + 2y 6 x y Solution: To solve this problem, it s necessary to use the technique employed in Example.5 as well as the method used in the other examples. Thus, the following problem will be solved: Maximize C = 2x + y