Paul E. Fishback. Instructors Solutions Manual for Linear and Nonlinear Programming with Maple: An Interactive, Applications-Based Approach

Transcription

1 Paul E. Fishback Instructors Solutions Manual for Linear and Nonlinear Programming with Maple: An Interactive, Applications-Based Approach

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3 Contents I Linear Programming An Introduction to Linear Programming 3. The Basic Linear Programming Problem Formulation Linear Programming: A Graphical Perspective inr Basic Feasible Solutions The Simplex Algorithm The Simplex Algorithm Alternative Optimal/Unbounded Solutions and Degeneracy Excess and Artificial Variables: The Big M Method Duality Sufficient Conditions for Local and Global Optimal Solutions Quadratic Programming iii

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5 Part I Linear Programming

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7 Chapter An Introduction to Linear Programming. The Basic Linear Programming Problem Formulation Linear Programming: A Graphical Perspective inr Basic Feasible Solutions

8 4 Chapter. An Introduction to Linear Programming. The Basic Linear Programming Problem Formulation. Express each LP below in matrix inequality form. Then solve the LP using Maple provided it is feasible and bounded. (a) maximize z=6x + 4x 2 2x + 3x 2 9 x 4 x 2 6 x, x 2, The second constraint may be rewritten as x 4 so that matrix inequality form of the LP is given by (b) maximize z=c x (.) Ax b x, 2 3 where A=, c=[ 6 4 ] 9, b= 4, and x= x. x The solution is given by x=, with z=27. maximize z=3x + 2x 2 x 4 x + 3x 2 5 2x + x 2 = x, x 2. The third constraint can be replaced by the two constraints, 2x + x 2 and 2x x 2. Thus, the matrix inequality form

9 (c).. The Basic Linear Programming Problem Formulation 5 3 of the LP is (.), with A=, c= [ 3 2 ] 4 5, b=, and 2 2 x x=. x 2 The solution is given by x= [] 3, with z=7. 4 maximize z= x + 4x 2 x + x 2 x + 3 x + x 2 5 x, x 2. (d) The third constraint is identical to x x 2 5. The matrix inequality form of the LP is (.), with A=, c=[ 4 ], b= 3, and x= x. x 5 2 [] 3 The solution is x=, with z=3. 4 minimize z= x + 4x 2 x + 3x 2 5 x + x 2 4 x x 2 2 x, x 2. To express the LP in matrix inequality form with the goal of maximization, we set c= [ 4 ] = [ 4 ]. The first two constraints may be rewritten as x 3x 2 5 and

10 6 Chapter. An Introduction to Linear Programming (e) x x 2 4. The matrix inequality form of the LP becomes (.), 3 with A=, c=[ 4 ] 5, b= 4, and x= x. x 2 2 [] 3 The solution is given by x=, with z=. maximize z=2x x 2 x + 3x 2 8 x + x 2 4 x x 2 2 x, x 2. The first and second constraints are identical to x 3x and x x 2 4, respectively. Thus, A=, c=[ 2 ], 8 b= 4, and x= x. x 2 2 (f) The LP is unbounded. minimize z=2x + 3x 2 3x + x 2 x + x 2 6 x 2. Define x = x,+ x,, where x,+ and x, are nonnegative. Then, as a maximization problem, the LP may be rewritten in terms of three decision variables as maximize z= 2(x,+ x, ) 3x 2 3(x,+ x, ) x 2 (x,+ x, )+x 2 6 x,+, x,, x 2.

11 .. The Basic Linear Programming Problem Formulation 7 The matrix inequality form of the LP becomes (.), with A = 3 3, c= [ ] x,+, b=, and x= x 6,. x 2 The solution is given by x,+ = 3 and x, = x 2 =, with z= The LP is given by minimize z=x + 4x 2 x + 2x 2 5 x x 2 2 x, x 2. The constraint involving the absolute value is identical to 2 x x 2 2, which may be written as the two constraints, x x 2 2 and x +x The matrix inequality form of the LP is (.) with A=, c= [ 4 ] 5, b= 2, and x= x. x 2 2 The solution is given by x=, with z=. 3. If x and x 2 denote the number of chairs and number of tables, respectively produced by the company, then z=5x +7x 2 denotes the revenue, which we seek to maximize. The number of square blocks needed to produce x chairs is 2x, and the number of square blocks needed to produce x 2 tables is 2x 2. Since six square blocks are available, we have the constraint, 2x + 2x 2 6. Similar reasoning, applied to the rectangular blocks, leads to the constraint x + 2x 2 8. Along with sign conditions, these results yield the LP. maximize z=5x + 7x 2 2x + 2x 2 6 x + 2x 2 8 x, x 2.

12 8 Chapter. An Introduction to Linear Programming 2 2 For the matrix inequality form, A=, c= [ 5 7 ], b= 2 ] x= [ x x 2. [] 6, and 8 The solution is given by x= [], with z= (a) If x and x 2 denote the number of grams of grass and number of grams of forb, respectively, consumed by the vole on a given day, then the total foraging time is given by z=45.55x x 2. Since the coefficients in this sum have units of minutes per gram, the units of z are minutes. The products.64x and 2.67x 2 represent the amount of digestive capacity corresponding to eating x grams of grass and x 2 grams of forb, respectively. The total digestive capacity is 3.2 gm-wet, which yields the constraint.64x x Observe that the units of the variables, x and x 2, are gm-dry. Each coefficient in this inequality has units of gm-wet per gm-dry, so the sum.64x x 2 has the desired units of gm-wet. Similar reasoning focusing on energy requirements, leads to the constraint 2.x + 2.3x Along with sign conditions, we arrive at minimize z=45.55x x 2.64x x x + 2.3x x, x 2. (b) The solution, obtained using Maple slpsolve command, is given by x = grams of grass and x grams of forb. The total time spent foraging is z 32.7 minutes. Solutions to Waypoints Waypoint... There are of course many possible combinations. Table. summarizes the outcomes for four choices. The fourth combination requires 35 gallons of stock B, so it does not satisfy the

13 .. The Basic Linear Programming Problem Formulation 9 TABLE.: Production combinations (in gallons) Premium Reg. Unleaded Profit ($) listed constraints. Of the three that do, the combination of 7 gallons premium and 5 gallons regular unleaded results in the greatest profit.

14 Chapter. An Introduction to Linear Programming.2 Linear Programming: A Graphical Perspective inr 2. (a) Maximize z=6x + 4x 2 2x + 3x 2 9 x 4 x 2 6 x, x 2, The [ feasible ] region is shown in Figure 2.2. The solution is given by 4.5 x=, with z=27. x 2 z=25 z=27 z=23 x FIGURE.: Feasible region with contours z = 23, z = 25 and z = 27.

15 .2. Linear Programming: A Graphical Perspective inr 2 (b) maximize z=3x + 2x 2 x 4 x + 3x 2 5 2x + x 2 = x, x 2. The feasible region is the [] line segment shown in Figure.2. The 3 solution is given by x=, with z=7. 4 x 2 z=5 z=7 z=3 x FIGURE.2: Feasible region with contours z = 3, z = 5 and z = 7.

16 2 Chapter. An Introduction to Linear Programming (c) minimize z= x + 4x 2 x + 3x 2 5 x + x 2 4 x x 2 2 x, x 2. The [ feasible ] region is shown in Figure.3. The solution is given by 3 x=, with z=. x 2 z=5 z=3 z= x FIGURE.3: Feasible region with contours z=5, z=3and z=. (d) maximize z=2x + 6x 2 x + 3x 2 6 x + 2x 2 5 x, x 2.

17 .2. Linear Programming: A Graphical Perspective inr 2 3 The feasible region is shown in Figure.4. The LP has alternative [] 3 optimal solutions that fall on the segment connecting x = to 6 x=. Each such solution has an objective value of z=2, and the parametric representation of the segment is given by 3t+6( t) x= t+( t) 6 3t =, t where t. x 2 z=2 z= z=8 x FIGURE.4: Feasible region with contours z=8, z= and z=2.

18 4 Chapter. An Introduction to Linear Programming (e) minimize z=2x + 3x 2 3x + x 2 x + x 2 6 x 2. The [ feasible ] region is shown in Figure.5. The solution is given by /3 x=, with z= 2 3. x 2 z= 4 3 z= z= 2 3 x FIGURE.5: Feasible region with contours z= 4 3, z=and z= The Foraging Herbivore Model), Exercise 4, from Section. is given by minimize z=45.55x x 2.64x x x + 2.3x x, x 2,

19 .2. Linear Programming: A Graphical Perspective inr 2 5 whose [ feasible ] region is shown in Figure.6. The solution is given by x, with z x 2 z=3 z=32.7 z=2 x FIGURE.6: Feasible region with contours z = 3, z = 2 and z = (a) The feasible region, along with the contours z=5, z=, and z = 5, is shown in Figure.7 (b) If M is an arbitrarily large positive real number, then the set of points falling on the contour z = M satisfies 2x x 2 = M, or, equivalently, x 2 = 2x M. However, as Figure.7 indicates, to use the portion of this line that falls within the feasible region, we must have x x 2 2. Combining this inequality with 2x x 2 = M yields x M 2. Thus, the portion of the contour z=m belonging to the feasible region consists of the ray, {(x, x 2 ) x M 2 and x 2 = 2x M}. (c) Fix M and consider the starting point on the ray from (b). Then

20 6 Chapter. An Introduction to Linear Programming x 2 z=5 z= z=5 x FIGURE.7: Feasible region with contours z = 5, z = and z = 5. x = M 2 and x 2 = 2x M=M 4, in which case all constraints and sign conditions are satisfied if M is large enough. (Actually, M 5suffices.) Since M can be made as large as we like, the LP is unbounded. 4. Suppose f is a linear function of x and x 2, and consider the LP given by maximize z= f (x, x 2 ) (.2) x x 2 x, x 2, (a) The feasible region is shown in Figure.8 (b) If f (x, x 2 )=x 2 x, then f (, )=. For all [] other feasible points, the objective value is negative. Hence, x = is the unique optimal solution. (c) If f (x, x 2 )=x, then the LP is unbounded.

21 .2. Linear Programming: A Graphical Perspective inr 2 7 x 2 x FIGURE.8: Feasible region for LP.2. (d) If f (x, x 2 )=x 2, then the LP has alternative optimal solutions. Solutions to Waypoints Waypoint.2.. The following Maple commands produce the desired feasible region: > restart:with(plots): > constraints:=[x<=8,2*x+x2<=28,3*x+2*x2<=32,x>=,x2>=]: > inequal(constraints,x=..,x2=..6,optionsfeasible=(color=grey), optionsexcluded=(color=white),optionsclosed=(color=black),thickness=); Waypoint.2.2. The following commands produce the feasible region, upon which are superimposed the contours, z = 2, z = 3, and z = 4. > restart:with(plots): > f:=(x,x2)->4*x+3*x2; f := (x, x 2 ) 4x + 3x 2 > constraints:=[x<=8,2*x+x2<=28,3*x+2*x2<=32,x>=,x2>=]: > FeasibleRegion:=inequal(constraints,x=..,x2=..6,optionsfeasible=(color=grey), optionsexcluded=(color=white),optionsclosed=(color=black),thickness=2):

22 8 Chapter. An Introduction to Linear Programming x 2 x FIGURE.9: Feasible region and object contours z = 2, z = 3, and z = 4. > ObjectiveContours:=contourplot(f(x,x2),x=..,x2=..,contours=[2,3,4],thicknes > display(feasibleregion, ObjectiveContours); The resulting graph is shown in Figure.9 Maple does not label the contours, but the contours must increase in value as x and x 2 increase. This fact, along with the graph, suggests that the solution is given by x = 4, x 2 =.

23 .3. Basic Feasible Solutions 9.3 Basic Feasible Solutions. For each of the following LPs, write the LP in the standard matrix form maximize z=c x x [A I m ] = b s x, s. Then determine all basic and basic feasible solutions, expressing each solution in vector form. Label each solution next to its corresponding point on the feasible region graph. (a) Maximize z=6x + 4x 2 2x + 3x 2 9 x 4 x 2 6 x, x 2, Note that this LP is the same as that from Exercise a of Section 2 3., where A=, c=[ 6 4 ] 9, b= 4, and x= x. Since x 6 2 s there are three constraints, s= s 2. In this case, the matrix equation, ( ) 5 2 [A I m ] s 3 [] x s = b (.3) has at most = possible basic solutions. Each is obtained by [] x setting two of the five entries of to zero and solving (.3) for s the remaining three, provided the system is consistent. Of the ten possible cases to consider, two lead to inconsistent systems. They arise from setting x = s 2 = and from setting x 2 = s 3 =. The eight remaining consistent systems yield the following results:

24 2 Chapter. An Introduction to Linear Programming (b) i. x =, x 2 =, s = 9, s 2 = 4, s 3 = 6 (basic, but not basic feasible solution) ii. x =, x 2 = 3, s =, s 2 = 4, s 3 = 3 (basic, but not basic feasible solution) iii. x =, x 2 = 6, s = 9, s 2 = 4, s 3 = (basic, but not basic feasible solution) iv. x = 4.5, x 2 =, s =, s 2 =.5, s 3 = 6 (basic feasible solution) v. x = 4, x 2 =, s =, s 2 =, s 3 = 6 (basic feasible solution) vi. x = 4, x 2 = /3, s =, s 2 =, s 3 = 7/3 (basic feasible solution) vii. x = 4.5, x 2 = 6, s =, s 2 = 8.5, s 3 = (basic, but not basic feasible solution) viii. x = 4, x 2 = 6, s = 7, s 2 =, s 3 = (basic, but not basic feasible solution) The feasible region is that from Exercise a of Section.2, and the solution is x = 4.5 and x 2 = with z=27. minimize z= x + 4x 2 x + 3x 2 5 x + x 2 4 x x 2 2 x, x 2. Note that this LP is the same as that from Exercise d of Section 3., where A=, c=[ 4 ] 5, b= 4, and x= x Since x 2 2 s there are three constraints, s= s 2. s 3 As in the previous exercise, there are ten possible cases to consider. In this situation, all ten systems are consistent. The results are as follows: i. x =, x 2 =, s = 5, s 2 = 4, s 3 = 2 (basic, but not basic feasible solution) ii. x =, x 2 = 5/3, s =, s 2 = 7/3, s 3 = /3 (basic, but not basic feasible solution) iii. x =, x 2 = 4, s = 7, s 2 =, s 3 = 6 (basic feasible solution)

25 .3. Basic Feasible Solutions 2 iv. x =, x 2 = 2, s =, s 2 = 6, s 3 = (basic, but not basic feasible solution) v. x = 5, x 2 =, s =, s 2 =, s 3 = 3 (basic, but not basic feasible solution) vi. x = 4, x 2 =, s =, s 2 =, s 3 = 2 (basic, but not basic feasible solution) vii. x = 2, x 2 =, s = 3, s 2 = 2, s 3 = (basic, but not basic feasible solution) viii. x = 7/2, x 2 = /2, s =, s 2 =, s 3 = (basic, but not basic feasible solution) ix. x = /4, x 2 = 3/4, s =, s 2 = /2, s 3 = (basic, but not basic feasible solution) x. x = 3, x 2 =, s =, s 2 =, s 3 = (basic feasible solution) The feasible region is that from Exercise c of Section.2, and the solution is x = 3 and x 2 = with z=. 2. The constraint equations of the standard form of the FuelPro LP are given by x + s = 8 (.4) 2x + 2x 2 + s 2 = 28 3x + 2x 2 + s 3 = 32 If x and s are nonbasic, then both are zero. In this case (.4) is inconsistent. Observe that if x = s =, that the coefficient matrix corresponding to (.4) is given by M= 2, which is not invertible The LP is given by maximize z=3x + 2x 2 x 4 x + 3x 2 5 2x + x 2 = x, x 2. (a) The third constraint is the combination of 2x + x 2 and 2x 2x 2. Thus, the matrix inequality form of the LP has A= and b=

26 22 Chapter. An Introduction to Linear Programming (b) Since there are four constraints, there are four corresponding slack variables, s, s 2, s 3, and s 4, which lead to the matrix equation maximize z=c x [] x [A I m ] = b s x, s, s x s where x= and s= 2. If s x 2 s 3 and s 4 are nonbasic, then both are 3 s 4 zero. In this case, the preceding matrix equation becomes x 3 x 2 = b. 2 s 2 The solution to this matrix equation yields x = 4 t, x 2 = 2+2t, s = t, and s 2 = 5 5t, where t is a free quantity. A simple calculation shows that for x, x 2, s, and s 2 to all remain nonnegative, it must x 4 be the case that t. Note that if t=, then x= = ; x 2 2 [] 3 if t=, then x=. As t increases from to, x varies along the 4 line segment connecting these two points. 4. The FuelPro LP has basic feasible solutions corresponding to the five points [] [] [] x=,,,, and. 4 4 Suppose we add a constraint that does not change the current feasible region but passes through one of these points. While there are countless such examples, a simple choice is to add x 2 4. Now consider the original FuelPro LP, with this new constraint added, and let s [ 4 denote ] the new corresponding slack variable. In the original LP, x = corresponded to a basic feasible solution in which the nonbasic variables 4 were x and s 2. In the new LP, any basic solution is obtained by setting two of the six variables, x, x 2, s, s 2, s 3, and s 4, to zero. If we again choose x and s 2 as nonbasic, then the resulting system of equations yields a basic feasible solution in which s 4 is a basic variable equal to zero. Thus, the LP is degenerate. s 2

27 .3. Basic Feasible Solutions A subset V ofr n is said to be convex is whenever two points belong to V, so does the line segment connecting them. In other words, x, x 2 V, implies that tx + ( t)x 2 V for all t. (a) Suppose that the LP is expressed in matrix inequality form as maximize z=c x (.5) Ax b x. If x and x 2 are feasible then Ax b and Ax 2 b. Thus, if t, then A (tx + ( t)x 2 )=tax + ( t)x 2 tb+( t)b = b. By similar reasoning, if x, x 2, then tx + ( t)x 2 whenever t. This shows that if x and x 2 satisfy the matrix inequality and sign conditions in (.5), then so does tx + ( t)x 2 for t. Hence, the feasible region is convex. (b) If x and x 2 are solutions of LP (.5), then both are feasible. By the previous result, x=tx + ( t)x 2 is also feasible if t. Since, x and x 2 are both solutions of the LP, they have a common objective value z = c x = c x 2. Now consider the objective value at x: c x=c (tx + ( t)x 2 ) = tc x + ( t)c x 2 = tz + ( t)z = z. Thus, x=tx + ( t)x 2 is also yields an objective value of z. Since x is also feasible, it must be optimal as well. 6. Suppose the matrix inequality form of the LP is given by maximize z=c x (.6) Ax b x,

28 24 Chapter. An Introduction to Linear Programming and let x denote an optimal solution. Since x belongs to the convex hull generated by the set of basic feasible solutions, x= p σ i x i, i= where the weightsσ i satisfyσ i for all i and p σ i = and where x, x 2,..., x p are the basic feasible solutions of (.6). Relabeling the basic feasible solutions if necessary, we may assume that c x i c x, for i p. Now consider the objective value corresponding to x. We have p c x=c σ i x i i= p = c (σ i x i ) = i= p σ i (c x i ) i= p σ i (c x ) i= = (c x ) = c x. p σ i i= Thus the objective value corresponding to x is at least as large as that corresponding to the optimal solution, x, which implies that x is also a solution of (.6) i= Solutions to Waypoints Waypoint.3.. The matrix equation [A I 3 ] x = b s has a solution given by x= and s=b. Since the corresponding system has more variables then equations, it must have infinitely many solutions.

29 .3. Basic Feasible Solutions 25 Waypoint.3.2. To solve the given matrix equation, we row reduce the augmented matrix The result is given by 4 3/2. 4 If we let s 2 and s 3 denote the free variables, then we may write s 2 = t, s 3 = s, where t, s R. With this notation, x = 4+t s, x 2 = 3 2t+s, and s = 4 t+s. We obtain the five extreme points as follows: [] 8. t=28 and s=32 yield x= and s= [] 8 2. t=2 and s=8yield x= and s= 2 8 [] 8 3. t=4and s= yield x= and s= t=s=yields x= and s= 8 5. t=and s=4 yield x= and s= 4 4 These extreme points may then be labeled on the feasible region. Waypoint The feasible region and various contours are shown in Figure?? [] 3/2 6/7 The extreme points are given by the x =, x 2 =, x 3 =, 2/7 [] and x 4 = x s 2. If c=[ 5, 2], A=, b=, x=, and s=, then the x 2 s 2 standard matrix form is given by

30 26 Chapter. An Introduction to Linear Programming x 2 z= z= 3 z= 5 z= 7.5 x FIGURE.: Feasible region with contours z =, z = 3, z = 5 and z= 7.5. maximize z=c x x [A I 2 ] = b s x, s, 3. The two constraint equations can be expressed as 3x + 2x 2 = 6 and 8x + 3x 2 = 2. Substituting x = x 2 = into this equations yields the basic solution x =, x 2 =, s = 6 and s 2 = 6. Performing the same operations using the other three extreme points from () yields the remaining three basic feasible solutions: (a) x =.5, x 2 =, s =.5 and s 2 =

31 .3. Basic Feasible Solutions 27 (b) x = 6 7, x 2= 2 7, s = and s 2 = (c) x =, x 2 = 3, s = and s 2 = 6 4. The contours in Figure?? [ indicate ] the optimal solution occurs at the.5 basic feasible solution x=, where z= 7.5.

32 28 Chapter. An Introduction to Linear Programming

33 Chapter 2 The Simplex Algorithm 2. The Simplex Algorithm Alternative Optimal/Unbounded Solutions and Degeneracy Excess and Artificial Variables: The Big M Method Duality Sufficient Conditions for Local and Global Optimal Solutions Quadratic Programming

34 3 Chapter 2. The Simplex Algorithm 2. The Simplex Algorithm. For each LP, we indicate the tableau after each iteration and highlight the pivot entry used to perform the next iteration. (a) TABLE 2.: Initial tableau z x x 2 s s 2 s 3 RHS TABLE 2.2: Tableau after first iteration z x x 2 s s 2 s 3 RHS TABLE 2.3: Tableau after second iteration z x x 2 s s 2 s 3 RHS (b) The solution is x = 3 and x 2 = 4, with z=7.

35 2.. The Simplex Algorithm 3 TABLE 2.4: Final tableau z x x 2 s s 2 s 3 RHS /5 7/5 7 -/5 3/5 3 /5-3/5 2/5 -/5 4 TABLE 2.5: Initial tableau z x x 2 s s 2 s 3 RHS The solution is x = 4 and x 2 = 5, with z=3. (c) The solution is x = 2, x 2 =, and x 3 = 74, with z= 3 3. (d) This is a minimization problem, so we terminate the algorithm when all coefficients in the top row that correspond to nonbasic variables are nonpositive. The solution is x = and x 2 = 4, with z= If we rewrite the first and second constraints as x 3x 2 8 and x x 2 4 and incorporate slack variables as usual, our initial tableau becomes that shown in Table 2.3 The difficulty we face when attempting to perform the simplex algorithm, centers on the fact that the origin, (x, x 2 )=(, ) yields negative values of s and 2. In other words, the origin corresponds to a basic, but not basic feasible, solution. We cannot read off the initial basic feasible solution as we could for Exercise (d). 3. We list those constraints that are binding for each LP. (a) Constraints 2 and 3

36 32 Chapter 2. The Simplex Algorithm TABLE 2.6: Tableau after first iteration z x x 2 s s 2 s 3 RHS TABLE 2.7: Final tableau z x x 2 s s 2 s 3 RHS 5/2 3/ /2 -/2 5 -/2 /2 5 (b) Constraints and 3 (c) Constraints and 2 (d) Constraint 2 Solutions to Waypoints Waypoint 2... We first state the LP in terms of maximization. If we set z= z, then maximize z=5x 2x 2 (2.) x 2 3x + 2x 2 6 8x + 3x 2 2 x, x 2, where z= z. The initial tableau corresponding to LP (2.) is given by Since we are now solving a maximization problem, we pivot on the highlighted entry in this tableau, which yields the following: All entries in the top row corresponding to nonbasic variables are positive. Hence x = 3 2 and x 2 = is a solution of the LP. However, the objective value corresponding to this optimal solution, for the original LP, is given by z= z= 5 2.

37 2.. The Simplex Algorithm 33 TABLE 2.8: Initial tableau z x x 2 x 3 s s 2 s 3 RHS TABLE 2.9: Tableau after first iteration z x x 2 x 3 s s 2 s 3 RHS -2/3 2/3 5/3 7/3 2/3 /3 /3 4/ TABLE 2.: Final tableau z x x 2 x 3 s s 2 s 3 RHS 3/2 /6 74/3 /2 -/6 /3 /2 -/4 /4 2 2 /2 -/2 TABLE 2.: Initial tableau z x x 2 s s 2 RHS TABLE 2.2: Final tableau z x x 2 s s 2 RHS TABLE 2.3: Initial tableau z x x 2 s s 2 s 3 RHS

38 34 Chapter 2. The Simplex Algorithm TABLE 2.4: Initial tableau z x x 2 s s 2 s 3 RHS TABLE 2.5: Final tableau z x x 2 s s 2 s 3 RHS 3/8 5/8 5/2-3/8 -/8 /2 7/8-3/8 3/2 3/8 /8 3/2

39 2.2. Alternative Optimal/Unbounded Solutions and Degeneracy Alternative Optimal/Unbounded Solutions and Degeneracy. Exercise The initial tableau is shown in Table TABLE 2.6: Initial tableau for Exercise z x x 2 s s 2 RHS After performing two iterations of the simplex algorithm, we obtain the following: TABLE 2.7: Tableau after second iteration for Exercise z x x 2 s s 2 RHS At this stage, the basic variables are x = 3 and x 2 =, with a corresponding objective value z=2. The nonbasic variable, s 2, has a coefficient of zero in the top row. Thus, letting s 2 become basic will not change the value of z. In fact, if we pivot on the highlighted entry in Table 2.7, x 2 replaces s 2 as a nonbasic variable, x = 6, and the objective value is unchanged. 2. Exercise 2 For an LP minimization problem to be unbounded, there must exist, at some stage of the simplex algorithm, a basic feasible solution in which a nonbasic variable has a negative coefficient in the top row and nonpositive coefficients in the remaining rows. 3. Exercise 3 (a) For the current basic solution to not be optimal, it must be the case that a<. However, if the LP is bounded, then the ratio test forces b>. Thus, we may pivot on row and column containing b, in which case, we obtain the following tableau:

40 36 Chapter 2. The Simplex Algorithm TABLE 2.8: Tableau obtained after pivoting on row and column containing b z x x 2 s s 2 RHS a b 3 a b 2 a b b b 3+ 2 b b b 2 b Thus, x = 3+ 2 b, x 2=, and z= 2 a b. (b) If the given basic solution was optimal, then a. However, if the LP has alternative optimal solutions, then a=. In this case, pivoting on the same entry in the original tableau as we did in (b), we would obtain the result in Table 2.8, but with a=. Thus, x = 3+ 2 b, x 2=, and z=. (c) If the LP is unbounded, then a<and b. 4. Exercise 4 (a) The coefficient of the nonbasic variable, s, in the top row of the tableau is negative. Since the coefficient of s in the remaining two rows is also negative, we see that the LP is unbounded. (b) So long as s 2 remains non basic, the equations relating s to each of x and x 2 are given by x 2s = 4 and x 2 s = 5. Thus, for each unit of increase in s, x increases by 2 units; for each unit of increase in s, x 2 increases by unit. (c) Eliminating s from the equations x 2s = 4 and x 2 s = 5, we obtain x 2 = 2 x + 3. (d) When s =, x = 4 and x 2 = 5. Since s, x 4 and x 2 5. Thus,we sketch that portion of the line x 2 = 2 x + 3 for which x 4 and x 2 5. (e) Since z=,, z=x + 3x 2, and x 2 = 2 x + 3, we have a system 5. Exercise 5 of equations whose solution is given by x = and x 2 = 6 5. (a) The LP has 2 decision variables and 4 constraints. Inspection of the

41 2.2. Alternative Optimal/Unbounded Solutions and Degeneracy 37 graph shows that the constraints are given by 2x + x 2 x + x x + x x + x 2 2. By introducing slack variables, s, s 2, s 3, and s 4, we can convert the preceding list of inequalities to a system of equations in six variables, x, x 2, s, s 2, s 3, and s 4. Basic feasible solutions are then most easily determined by substituting each of the extreme points into the system of equations and solving for the slack variables. The results are as follows: 5 [] x = = s 6 4 3/2 2 9/ = = = [] 4 Observe that the extreme point x= yields slack variable values 2 for which s = s 2 = s 3 =. This means that when any two of these three variables are chosen to be nonbasic, the third variable, which is basic, must be zero. Hence, the LP is degenerate. From a graphical perspective, the LP is degenerate because the boundaries of three 4 of the constraints intersect at the single extreme point, x= 2 (b) If the origin is the initial basic feasible solution, at least two simplex algorithm iterations are required before a degenerate basic feasible solution results. This occurs when the objective coefficient of x is larger [] than that of x 2. In this case, the intermediate iteration yields 5 x=. If the objective coefficient of x 2 is positive and larger than that of x, then at most three iterations are required.

42 38 Chapter 2. The Simplex Algorithm Solutions to Waypoints Waypoint The current basic feasible solution consists of x = a, s = 2, and x 2 = s 2 =. The LP has alternative optimal solutions if a nonbasic variable has a coefficient of zero in the top row of the tableau. This occurs when a= or a=2. 2. The LP is unbounded if both a< and a 3<, i.e., if <a 3. The LP is also unbounded if both 2 a< and a 4, i.e., if 2<a 4.

43 2.3. Excess and Artificial Variables: The Big M Method Excess and Artificial Variables: The Big M Method. (a) We let e and a denote the excess and artificial variables, respectively, that correspond to the first constraint. Similarly, we let e 2 and a 2 correspond to the second constraint. Using M=, our objective becomes one of minimizing z= x + 4x 2 + a + a 2. If s 3 is the slack variable corresponding to the third constraint our initial tableau is as follows: TABLE 2.9: Initial tableau z x x 2 e a e 2 a 2 s 3 RHS To obtain an initial basic feasible solution, we pivot on each of the highlighted entries in this tableau. The result is the following: TABLE 2.2: Tableau indicting initial basic feasible solution z x x 2 e a e 2 a 2 s 3 RHS Three iterations of the simplex algorithm are required before all coefficients in the top row of the tableau that correspond to nonbasic variables are nonpositive. The final tableau is given by TABLE 2.2: Final tableau z x x 2 e a e 2 a 2 s 3 RHS -3/4-397/4 - -7/4 5/2 -/4 /4 -/4 3/2 -/4 /4 3/4 7/2 -/2 /2 - /2 The solution is x = 7 2 and x 2= 3 2, with x 2= 5 2.

44 4 Chapter 2. The Simplex Algorithm (b) We let e and a denote the excess and artificial variables, respectively, that correspond to the first constraint. Similarly, we let e 3 and a 3 correspond to the third constraint. Using M=, our objective becomes one of minimizing z=x + x 2 + a + a 3. If s 2 is the slack variable corresponding to the second constraint our initial tableau is as follows: TABLE 2.22: Initial tableau z x x 2 e a s 2 e 3 a 3 RHS To obtain an initial basic feasible solution, we pivot on each of the highlighted entries in this tableau. The result is the following: TABLE 2.23: Tableau indicting initial basic feasible solution z x x 2 e a s 2 e 3 a 3 RHS Three iterations of the simplex algorithm lead to a final tableau given by TABLE 2.24: Final tableau z x x 2 e a s 2 e 3 a 3 RHS -3/7-697/7 -/7-2 -5/7 5/7 3/7-6 -/7 /7 2/7 6-2/7 2/7-3/7 6 The solution is x = 6 and x 2 = 6, with x 2 = 2. (c) We let a denote the artificial variable corresponding to the first (equality) constraint, and we let e 2 and a 2 denote the excess and artificial variables, respectively, that correspond to the second constraint. Using M =, our objective becomes one of maximizing z=x + 5x 2 + 6x 3 a a 2. The initial tableau is given by To obtain an initial basic feasible solution, we pivot on each of the highlighted entries in this tableau. The result is the following:

45 2.3. Excess and Artificial Variables: The Big M Method 4 TABLE 2.25: Initial tableau z x x 2 x 3 a e 2 a 2 RHS TABLE 2.26: Tableau indicting initial basic feasible solution z x x 2 x 3 a e 2 a 2 RHS Three iterations of the simplex algorithm are required before all coefficients in the top row of the tableau that correspond to nonbasic variables are nonnegative. The final tableau is given by TABLE 2.27: Final tableau z x x 2 x 3 a e 2 a 2 RHS /2 2 /2 25 The solution is x = x 2 = and x 3 = 25, with z=5. Solutions to Waypoints Waypoint The LP is given by where A= ] s= [ s s 2. maximize z=c x [] x [A I 2 ] = b s x, s, , c= [ ] 3.2, b=, x= [ x x 2 ], and

46 42 Chapter 2. The Simplex Algorithm Waypoint Let s denote the slack variable for the first constraint, and let e 2 and a 2 denote the excess and artificial variables, respectively, for the second constraint. Using M =, our objective becomes one of minimizing z=45.55x x 2 + a 2. The initial tableau is as follows: TABLE 2.28: Initial tableau z x x 2 s e 2 a 2 RHS To obtain an initial basic feasible solution, we pivot on the highlighted entry in this tableau. The result is the following: TABLE 2.29: Tableau indicting initial basic feasible solution z x x 2 s e 2 a 2 RHS Two iteration of the simplex algorithm yield the final tableau: TABLE 2.3: Final tableau z x x 2 s e 2 a 2 RHS Thus x is nonbasic so that the total foraging time is minimized when x = grams of grass and x grams of forb, in which case z 32.7 minutes

47 2.4. Duality Duality. At each stage of the simplex algorithm, the decision variables in the primal LP correspond to a basic feasible solutions. Values of the decision variables in the dual LP are recorded by the entries of the vector y. Only at the final iteration of the algorithm, do the dual variables satisfy both ya c and y, i.e., only at the final iteration do the entries of y constitute a basic feasible solution of the dual. 2. The given LP can be restated as maximize z=3x + 4x 2 x 4 x + 3x 2 5 x + 2x 2 5 x x 2 9 x + x 2 6 x x 2 6 x, x 2. In matrix inequality form, this becomes maximize z=c x Ax b x, 4 3 where A=, c= [ 3 4 ] 5 5, b=, and x= The corresponding dual is given by minimize w=y b ya c y, [ x x 2 ].

48 44 Chapter 2. The Simplex Algorithm where y= [ y y 2 y 3 y 4 y 5 ] y 6. In expanded form this is identical to minimize w=4y + 5y 2 5y 3 + 9y 4 + 6(y 5 y 6 ) y + y 2 + y 3 y 4 + (y 5 y 6 ) 3 3y 2 y 3 + y 4 + (y 5 y 6 ) y, y 2, y 3, y 4, y 5, y 6. Now define the new decision variable ỹ= y 5 y 6, which is unrestricted in sign due to the fact y 5 and y 6 are nonnegative. Since y 5 and y 6, together, correspond to the equality constraint in the primal LP, we see that ỹ is a dual variable unrestricted in sign that corresponds to an equality constraint in the primal. 3. The standard maximization LP can be written in matrix inequality form as maximize z=c t x Ax b x, where x belongs tor n, c is a column vector inr n, b belongs tor m, and A is an m-by-n matrix. The dual LP is then expressed as minimize w=y t b y t A c t y, where we have assumed the dual variable vector, y, is a column vector inr m. But, by transpose properties, this is equivalent to maximize w= b t y A t y c y. The preceding LP is a maximization problem, whose dual is given by

49 2.4. Duality 45 minimize z=x t ( c) x t ( A t ) b t x. Using transpose properties again, we may rewrite this final LP as which is the original LP. maximize z=c t x Ax b x, 4. (a) The original LP has four decision variables and three constraints. Thus, the dual LP has three decision variables and four constraints. (b) The current primal solution is not optimal, as the nonbasic variable, s, has a negative coefficient in the top row of the tableau. By the ratio test, we conclude that s will replace x as a basic variable and that s = 4 in the updated basic feasible solution. Since the coefficient of s in the top row is 2, the objective value in the primal will increase to 28. By weak duality, we can conclude that the objective value in the solution of the dual LP is no less than The dual of the given LP is which is infeasible. minimize w= y 2y 2 y y 2 2 y + y 2 y, y 2, 6. (a) The current tableau can also be expressed in partitioned matrix form as z x y yb [, ] [,, ] MA M Mb Comparing this matrix to that given in the problem, we see that

50 46 Chapter 2. The Simplex Algorithm /3 M= /2. Thus, /3 b=m (Mb) 3 6 = 3/2 3 8 = 2, 6 and A=M (MA) 3 2/3 = 3/2 3 4/3 2 3 = 2 3/2. 4 Using A and b, we obtain the original LP: maximize z=2x + 3x 2 2x + 3x 2 8 2x x 2 2 4x + x 2 6 x, x 2. (b) The coefficients of x and x 2 in the top row of the tableau are given by c+ya= [2, 3]+ [,, ]A = [ 4, ] The unknown objective value is z=yb 8 = [,, ] 2 6 = 8. (c) The coefficient of x in the top row of the tableau is -4, so an additional iteration of the simplex algorithm is required. Performing

51 2.4. Duality 47 the ratio test, we see that x replaces s 2 as a nonbasic variable. The updated decision variables become x = and x 2 = 2 3, with a corresponding objective value of z = 22. The slack variable coefficients in the top row of the tableau are s = 3, s 2= 4 3, and s 3=, which indicates that the current solution is optimal. Therefore the /3 dual solution is given by y= 4/3 with objective value w= Since the given LP has two decision variables and four constraints, the corresponding dual has four decision variables, y, y 2, y 3, and y 4 along with two constraints. In expanded form, it is given by minimize z=4y + 6y y y 4 y+2y 3 + 2y 4 y + y 2 + 3y 3 + y 4 5 y, y 2, y 3, y 4. To solve the dual LP using the simplex algorithm, we subtract excess variables, e and e 2, from the first and second constraints, respectively, and add artificial variables, a and a 2. The Big M method with M= yields an initial tableau given by TABLE 2.3: Initial tableau w y y 2 y 3 y 4 e a e 2 a 2 RHS The final tableau is TABLE 2.32: Final tableau w y y 2 y 3 y 4 e a e 2 a 2 RHS -/2-3 -5/2-985/ /2 -/2 -/2 /2 /2 5/2-2 3/2-3/2-7/2. The excess variable coefficients in the top row of this tableau are the

52 48 Chapter 2. The Simplex Algorithm additive inverses of the decision variable values in the solution of the primal. Hence, the optimal solution of the primal is given by x = 5 2 and x 2 = 6, with corresponding objective value z= By the complementary slackness property, x is the optimal solution of y y the LP provided we can find y = 2 inr y 4 such that 2 y 4 [ y A c ] i [x ] i = i 6 and We have [y ] j [b Ax ] j = j 4. b Ax =, 6/7 so that in a dual solution it must be the case that y 3 =. Furthermore, when y 3 =, y + 7y 2 + 7y 4 5 2y 5y 2 + 8y 4 4y y A c= + 2y 2 3y 4. 3y + y 2 + 5y 4 4 4y + 2y 2 + 2y 4 6y + 3y 2 + 4y 4 2 By complementary slackness again, the first, third and fourth components of this vector must equal zero, whence we have a system of 3 linear equations, whose solution is given by y = 3 24, y 2 = 37 24, and y 3 = If we define y = 3/24 37/24 47/2, then w=y b = 53 7 = cx so that x is the optimal solution of the LP by Strong Duality.

53 2.4. Duality Player s optimal mixed strategy is given by the solution of the LP maximize z Ax ze e t x= x, and Player 2 s by the solution of the corresponding dual LP, minimize w ya we t y e= y. [] Here e=. The solution of the first LP is given by x =.5 solution of the second by x =.5.7 and the.3. The corresponding objective value for each LP is z=w=.5, thereby indicating that the game is biased in Player s favor.. (a) Since A is an m-by-n matrix, b belongs tor m, and c is a row vector inr n, the matrix M has m+n rows and m+n [ columns. ] Furthermore, the transpose of a partitioned matrix,, is given A B C D A t C by t, which can be used to show M t = M. B t D t y t x (b) Define vecw =. Then w belongs tor m+n+ and has nonnegative components since x and y are primal feasible and dual fea- sible vectors belonging tor n andr m, respectively. Furthermore, [w ] m+n+ = >. We have m m A b y t Mw = A t m n c t x b t c Ax + b = A t y t c t b t y t + cx

54 5 Chapter 2. The Simplex Algorithm since the primal feasibility of x dictates Ax b, the dual feasibility of y and transpose properties imply A t y t c t, and strong duality forces b t y t = cx. (c) Lettingκ, x, and y be defined as in the hint, we have Mw m m A b κy t = A t m n c t κx b t c κ m m A b y =κ A t m n c t t b t x c Ax + b =κ A t y t c t b t y t + cx Sinceκ >, we conclude that Ax b, A t y t ct, and cx b t y t. Transpose properties permit us to rewrite the second of these matrix inequalities as y A c. Thus x and y are primal feasible and dual feasible, respectively. Moreover, the third inequality is identical to cx y b if we combine the facts b t y t = y b, each of these two quantities is a scalar, and the transpose of a scalar is itself. But the Weak Duality Theorem also dictates cx y b, which implies cx = y b. From this result, we conclude x and y constitute the optimal solutions of the (4.) and (4.3), respectively. Solutions to Waypoints Waypoint In matrix inequality form, the given LP can be written as where vecx= given by x x 2 x 3 maximize z=c x Ax b x, 2 2, c=[3, 5, 2], A=, and b= 3 5 = The dual LP is

55 2.4. Duality 5 minimize w=y b ya c y, where y= [ y y 2 y 3 ] y 4. In expanded form, this is equivalent to minimize w=2y + 2y 2 + 4y 3 + 3y 4 2y + y 2 + 3y 3 y 4 3 y + 2y 2 + y 3 + y 4 5 y + y 2 + 5y 3 + y 4 2 y, y 2, y 3, y 4. 2/3 The original LP has its solution given by x = 2/3 with z = 6 3. The solution of the dual LP is y = [ /3 7/3 ] with w = 6 3. Waypoint If Steve always chooses row one and Ed chooses columns one, two, and three with respective probabilities, x, x 2, and x 3, then Ed s expected winnings are given by f (x, x 2, x 3 )= [ ] A = x x 2 + 2x 3, which coincides with the first entry of the matrix-vector product, Ax. By similar reasoning, f 2 (x, x 2, x 3 )= [ ] A x x 2 and = 2x + 4x 2 x 3 f 3 (x, x 2, x 3 )= [ ] A = 2x + 2x 3, which equal the second and third entries of Ax, respectively. x x 2 x 3 x 3 x x 2 x 3

56 52 Chapter 2. The Simplex Algorithm 2. If Ed always chooses column one and Steve chooses rows one, two, and three with respective probabilities, y, y 2, and y 3, then Steve s expected winnings are given by g (y, y 2, y 3 )= [ y y 2 y 3 ] A = y + 2y 2 2y 3, which coincides with the first entry of the product, ya. By similar reasoning, g 2 (y, y 2, y 3 )= [ y y 2 y 3 ] A = y + 4y 2 and g 3 (y, y 2, y 3 )= [ y y 2 y 3 ] A = 2y y 2 + 2y 3, which equal the second and third entries of ya, respectively. Waypoint To assist in the formulation of the dual, we view the primal objective, z, as the difference of two nonnegative decision variables, z and z 2. Thus the primal LP can be expressed in matrix inequality form as maximize [ z 3 ] z 2 x e e A z 3 e t z 2 e t x x and z, z 2. If the dual LP has its vector of decision variables denoted by [ y w w 2 ], where y is a row vector inr 3 with nonnegative components, and both w and w 2 are nonnegative as well, then the dual is given by

57 2.4. Duality 53 minimize [ 3 y w w 2 ] [ e e A y w w 2 ] e t 3 e t y 3 and w, w 2. If we write w=w w 2, then this formulation to the dual simplifies to be minimize w ya we t y e= y.

58 54 Chapter 2. The Simplex Algorithm 2.5 Sufficient Conditions for Local and Global Optimal Solutions. If f (x, x 2 )=e (x2 +x2 2), then direct computations lead to the following results: f (x)= 2x f (x, x2) 2x 2 f (x, x 2 ) and [ ( 2+4x 2 H f (x)= ) f (x, x2) 4x ] x 2 f (x, x 2 ) 4x x 2 f (x, x 2 ) ( 2+4x 2 ) f (x, x2) 2 so that f (x ) and H f (x ) Thus, Q(x)= f (x )+ f (x ) t (x x )+ 2 (x x ) t H f (x )(x x ) x.895x x x x x 2 The quadratic approximation [ together ] with f are graphed in Figure The solution is given by x=, with z= Calculations establish that Hessian of f is given by H f (x) = φ(x)a, whereφ(x)= and where (74+.5x + x 2 + x 3 ) / A=

59 2.5. Sufficient Conditions for Local and Global Optimal Solutions 55 FIGURE 2.: Plot of f with quadratic approximation Observe thatφ(x)>for x in S and that A is positive semidefinite. (Its eigenvalues are.2,.423, and zero.) Now suppose that x belongs to S and thatλ is an eigenvalue of H f (x ) with corresponding eigenvector v. Then λ v=h f (x )v =φ(x )Av, implying λ φ(x ) is an eigenvalue of A. Sinceφ(x )> and each eigenvalue of A is nonnegative, thenλ must be nonnegative as well. Hence, H f (x ) is positive semidefinite, and f is convex.

60 56 Chapter 2. The Simplex Algorithm 2x + x 3. (a) We have f (x) = 2, which yields the critical point, x + 4x 2 4 [] 2 x =. Since H f (x)= has positive eigenvalues,λ=3± 2, 4 we see that f is strictly convex onr 2 so that x is a global minimum. 2x x (b) We have f (x)= 2 + 4, which yields the critical point, x + 4x 2 8 /3 x =. Since H /3 f (x) = [ 2 4 ] has mixed-sign eigenvalues,λ=8±2 34, x is a saddle point. 4x + 6x (c) We have f (x)= 2 6, which yields the critical point, 6x x x =. Since H f (x) = has negative eigenvalues, 6 λ= 7±3 5, we see that f is strictly concave onr 2 so that x is a global maximum. [ 4x 3 (d) We have f (x)= 24x2 + 48x ] 32, which yields a single 8x critical point at x =. The Hessian simplifies to H /2 f (x) = 2(x 2) 2 and has eigenvalues of 2(x 8 2) 2 and 8. Thus, f is concave onr 2 and x is a global maximum. cos(x ) cos(x (e) We have f (x) = 2 ), which yields four critical sin(x ) sin(x 2 ) π/2 points: x =± and x =±. The Hessian is π/2 sin(x ) cos(x H f (x)= 2 ) cos(x ) sin(x 2 ), cos(x ) sin(x 2 ) sin(x ) cos(x 2 ) π/2 which yields a repeated eigenvalue,λ=, when x =, a π/2 repeated eigenvalueλ=, when x =, and eigenvalues π/2 λ=±, when x =. Thus, x ±π/2 = is a local maximum, π/2 x = is a local minimum, and x = are saddle points. ±π/2 [ 2x /(x (f) We have f (x)= 2 + ) ], which yields a single critical point x 2

61 2.5. Sufficient Conditions for Local and Global Optimal Solutions 57 at the origin. Since f is differentiable and nonnegative onr 2, we conclude immediately that the origin is a global minimum. In addition, H f (x)= 2( x ) 2 (x 2 +)2 The eigenvalues of this matrix are positive when <x <. Thus f is is strictly convex its given domain. ( ) (g) If f (x, x 2 )=e x 2+x x, then f (x)= f (x) so that f has a single critical point at the origin. The Hessian simplifies to H f (x)= x 2 x x 2 f (x) x x 2, which has eigenvalues in terms of x given by x 2 2 f (x) λ= (x 2 +. Since f (x)>and x x2 ) f (x) 2 + x2 <, we have that 2 2 f is strictly concave on its given domain and the origin is a global minimum. 4. We have f (x)=ax b, which is the zero vector precisely when x = A b. The Hessian of f is simply A, so f is strictly convex (resp. strictly concave) onr n when A is positive definite (resp. negative definite). Thus, x is the unique global minimum (resp. unique global maximum) of f. When A= 2, b= 3, and c=6, x 4 =. x 2 7/5. The eigenvalues of 9/5 A areλ= 5± 5, so x is the global minimum. 2 n 5. Write n= n 2. Since n, we lose no generality by assuming n 3. n 3 Solving n t x=d for x 3 yields x 3 as a function of x and x 2 : f :R 2 Rwhere f (x, x 2 )=x 2 + x2 2 + (d n x n 2 x 2 ) 2. The critical point of f is most easily computed using Maple and simplifies to x = [ n d/ n 2, n d/ n 2] The eigenvalues of the Hessian are λ = 2 andλ 2 = 2 n 2 so that f is strictly convex and x n 2 is the global 3 minimum. The corresponding shortest distance is d n n 2 3

62 58 Chapter 2. The Simplex Algorithm /4 For the plane x + 2x 2 + x 3 =, n= 2 and d=. In this case, x = /7 3 3/4 with a corresponding distance of. 4 4x 3 6. The gradient of f is f (x)=, which vanishes at the origin. Since f 2x 2 is nonnegative and f ()=, we see that x = is the global minimum. The Hessian of f at the origin is H f ()= [ 2 ], which is positive semidefinite. The function f (x, x 2 )= x 4 + x2 has a saddle point at the origin because 2 f (x, )= x 4 has a minimum at x = and f (, x 2 )=x 2 has a maximum 2 at x 2 =. The Hessian of f at the origin is again H f ()= [ 2 ]. 7. The function f (x)=x 4 + x4 2 has a single critical point at x =, which is a global minimum since f is nonnegative and f ()=. The Hessian evaluated at x is the zero matrix, whose only eigenvalue is zero. Similarly, f (x)= x 4 x4 2 has a global maximum at x =, where its Hessian is also the zero matrix. 8. (a) Player s optimal mixed strategy is the solution of maximize = z 2x 3x 2 z x + 4x 2 z x + x 2 = x, x 2, 7/ which is given by x =, with corresponding objective value 3/ z = 2. Player 2 s optimal mixed strategy is the solution of the corresponding dual LP, minimize = w 2y y 2 w 3y + 4y 2 w y + y = y, y 2,

63 2.5. Sufficient Conditions for Local and Global Optimal Solutions 59 /2 which is y =, with corresponding objective value w /2 = 2. (b) We have t ] y f (x, y )= x A[ y x = x y 5x 7y + 4, whose gradient, f (x, y ) vanishes at x = 7, y =. The Hessian 2 of f is a constant matrix having eigenvaluesλ=± so that this ( 7 critical point is a saddle point. The game value is f, ) = 2 2. (c) The only solution of f (x, y )= 2 is simply ( 7, 2 (d) A game value 2% higher than 2 is 3 5. The equation f (x, y )= 3 5 simplifies to x y 5x 7y =. The graph of this relation is shown in Figure??. 9. Calculations, which are most easily performed with Maple, establish that f (x, y ) = x t ] y A[ has a saddle point at x y x = d b a+d b c, y = f ( d c a+d b c. Since d b a+d b c, d c a+d b c ) = ). ad bc a+d b c, the game is fair provided det A, or, in other words, if A is noninvertible.

64 6 Chapter 2. The Simplex Algorithm y x FIGURE 2.2: The relation x y 5x 7y =.

65 2.6. Quadratic Programming Quadratic Programming 4. (a) The Hessian of f is 6. If p= 2, A=, and b=, then the problem is given as 3 minimize f (x)= 2 xt Qx+p t x Ax=b. (b) The eigenvalues of Q areλ = (repeated) andλ = 2, so Q is positive definite and the problem is convex. Thus, [ µ x ] m m A b = A t Q p 44/9 65/9 /3. = 4/3 /3 The problem then has its solution given by x = 4/3. The corresponding objective value is f (/3, 4/3)= (a) The [ matrix ] Q is indefinite. However, the partitioned matrix, m m A A t is still invertible so that Q [ µ x ] m m A b = A t Q p 259/3 78/3 = 9/3. /3 42/3

66 62 Chapter 2. The Simplex Algorithm 9/3 Thus, the problem has a unique KKT point, x = /3, with 42/3 259/3 corresponding multiplier vector,µ =. The bordered 78/3 Hessian becomes 2 4 B= In this case there are n=3 decision variables and m=2 equality constraints. Since the determinants of both B and its leading principal minor of order 4 are positive, we see that x is the solution. 6 (b) The problem has a unique KKT point given by x = 3/4 with /4 5/2 corresponding multiplier vector,λ = 3. At x only the first two constraints of Cx d are binding. Thus, we set C= and d=, in which case B= There are n = 3 decision variables, m = equality constraints, and k = 2 binding inequality constraints. Since the determinant of B and its leading principal minor of order 4 are positive, we see that x is the solution. 7/4 / (c) The are two KKT points. The first is given by x = with 7/8.445 corresponding multiplier vectors,λ andµ

67 2.6. Quadratic Programming 63 The second constraint is binding at x, so we set C= [ 3 ] and d= [ ], in which case, A B= 3 C A t C t Q = There are n = 4 decision variables, m = 3 equality constraints, and k= binding inequality constraints. Since n m k=, we only need to check that the determinant of B itself has the same sign as ( ) m+k =, which it does. Hence, x = whereas f 7/4 / 7/8 is a solution The second KKT point is given by x with corresponding multiplier vectors,λ = andµ [] In this case neither inequality constraint is binding. Performing an analysis simi-.67 lar to that done for the first KKT point leads to a situation in which the bordered Hessian test is inconclusive. In fact, f (x ) 2., 7/4 / 7/ To show that m m A S A t = S AQ Q Q A t S Q (Q+A t S A)Q, (2.2) m m A S we will verify that the product of A t and S AQ Q Q A t S Q (Q+A t S A)Q