LINEAR PROGRAMMING math

LINEAR PROGRAMMING A Concise Introduction Thomas S Ferguson Contents Introduction The Standard Maximum and Minimum Problems The Diet Problem The Transportation Problem The Activity Analysis Problem The Optimal Assignment Problem Terminology Duality 10 Dual Linear Programming Problems 10 The Duality Theorem 11 The Equilibrium Theorem 12 Interpretation of the Dual 14 The Pivot Operation 16 The Simplex Method 20 The Simplex Tableau 20 The Pivot Madly Method 21 Pivot Rules for the Simplex Method 23 The Dual Simplex Method 26 Generalized Duality 28 The General Maximum and Minimum Problems 28 Solving General Problems by the Simplex Method 29 Solving Matrix Games by the Simplex Method 30 Cycling 33 A Modification of the Simplex Method That Avoids Cycling 33 Four Problems with Nonlinear Objective Function 36 Constrained Games 36 The General Production Planning Problem 36 Minimizing the Sum of Absolute Values 37 Minimizing the Maximum of Absolute Values 38 Chebyshev Approximation 39 Linear Fractional Programming 39 Activity Analysis to Maximize the Rate of Return 40 The Transportation Problem 42 Finding a Basic Feasible Shipping Schedule 44 Checking for Optimality 45 The Improvement Routine 47 Solutions to the Exercises 50 Related Texts 66 LINEAR PROGRAMMING Introduction A linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to linear constraints The constraints may be equalities or inequalities Here is a simple example Find numbers x1 and x2 that maximize the sum x1 + x2 subject to the constraints x1 ≥ , x2 ≥ , and x1 + 2x2 ≤ 4x1 + 2x2 ≤ 12 −x1 + x2 ≤ In this problem there are two unknowns, and five constraints All the constraints are inequalities and they are all linear in the sense that each involves an inequality in some linear function of the variables The first two constraints, x1 ≥ and x2 ≥ , are special These are called nonnegativity constraints and are often found in linear programming problems The other constraints are then called the main constraints The function to be maximized (or minimized) is called the objective function Here, the objective function is x1 + x2 Since there are only two variables, we can solve this problem by graphing the set of points in the plane that satisfies all the constraints (called the constraint set) and then finding which point of this set maximizes the value of the objective function Each inequality constraint is satisfied by a half-plane of points, and the constraint set is the intersection of all the half-planes In the present example, the constraint set is the fivesided figure shaded in Figure We seek the point (x1 , x2 ), that achieves the maximum of x1 + x2 as (x1 , x2 ) ranges over this constraint set The function x1 + x2 is constant on lines with slope −1 , for example the line x1 + x2 = , and as we move this line further from the origin up and to the right, the value of x1 + x2 increases Therefore, we seek the line of slope −1 that is farthest from the origin and still touches the constraint set This occurs at the intersection of the lines x1 + 2x2 = and 4x1 + 2x2 = 12 , namely, (x1 , x2 ) = (8/3, 2/3) The value of the objective function there is (8/3) + (2/3) = 10/3 Exercises and can be solved as above by graphing the feasible set It is easy to see in general that the objective function, being linear, always takes on its maximum (or minimum) value at a corner point of the constraint set, provided the x2 4x1 + 2x2 = 12 -x1 + x2 = optimal point x1 + 2x2 = x Figure constraint set is bounded Occasionally, the maximum occurs along an entire edge or face of the constraint set, but then the maximum occurs at a corner point as well Not all linear programming problems are so easily solved There may be many variables and many constraints Some variables may be constrained to be nonnegative and others unconstrained Some of the main constraints may be equalities and others inequalities However, two classes of problems, called here the standard maximum problem and the standard minimum problem, play a special role In these problems, all variables are constrained to be nonnegative, and all main constraints are inequalities We are given an m -vector, b = (b1 , , bm )T , an m × n matrix,  a11 a12 · · ·  a21 a22 · · · A=  am1 am2 of real numbers ··· n -vector, c = (c1 , , cn )T , and an  a1n a2n    amn The Standard Maximum Problem: Find an n -vector, x = (x1 , , xn )T , to maximize cTx = c1 x1 + · · · + cn xn subject to the constraints a11 x1 + a12 x2 + · · · + a1n xn ≤ b1 a21 x1 + a22 x2 + · · · + a2n xn ≤ b2 am1 x1 + am2 x2 + · · · + amn xn ≤ bm and x1 ≥ 0, x2 ≥ 0, , xn ≥ (or Ax ≤ b ) (or x ≥ ) The Standard Minimum Problem: Find an m -vector, y = (y1 , , ym ), to minimize y Tb = y1 b1 + · · · + ym bm subject to the constraints y1 a11 + y2 a21 + · · · + ym am1 ≥ c1 y1 a12 + y2 a22 + · · · + ym am2 ≥ c2 (or y TA ≥ cT ) y1 a1n + y2 a2n + · · · + ym amn ≥ cn and y1 ≥ 0, y2 ≥ 0, , ym ≥ (or y ≥ ) Note that the main constraints are written as ≤ for the standard maximum problem and ≥ for the standard minimum problem The introductory example is a standard maximum problem We now present examples of four general linear programming problems Each of these problems has been extensively studied Example The Diet Problem There are m different types of food, F1 , , Fm , that supply varying quantities of the n nutrients, N1 , , Nn , that are essential to good health Let cj be the minimum daily requirement of nutrient, Nj Let bi be the price per unit of food, Fi Let aij be the amount of nutrient Nj contained in one unit of food Fi The problem is to supply the required nutrients at minimum cost Let yi be the number of units of food Fi to be purchased per day The cost per day of such a diet is (1) b1 y1 + b2 y2 + · · · + bm ym The amount of nutrient Nj contained in this diet is a1j y1 + a2j y2 + · · · + amj ym for j = 1, , n We not consider such a diet unless all the minimum daily requirements are met, that is, unless a1j y1 + a2j y2 + · · · + amj ym ≥ cj for j = 1, , n (2) Of course, we cannot purchase a negative amount of food, so we automatically have the constraints (3) y1 ≥ 0, y2 ≥ 0, , ym ≥ Our problem is: minimize (1) subject to (2) and (3) This is exactly the standard minimum problem Example The Transportation Problem There are I ports, or production plants, P1 , , PI , that supply a certain commodity, and there are J markets, M1 , , MJ , to which this commodity must be shipped Port Pi possesses an amount si of the commodity (i = 1, 2, , I ), and market Mj must receive the amount rj of the commodity (j = 1, , J ) Let bij be the cost of transporting one unit of the commodity from port Pi to market Mj The problem is to meet the market requirements at minimum transportation cost Let yij be the quantity of the commodity shipped from port Pi to market Mj The total transportation cost is I J yij bij (4) i=1 j=1 J j=1 The amount sent from port Pi is si , we must have yij and since the amount available at port Pi is J j=1 for i = 1, , I yij ≤ si The amount sent to market Mj is we must have I i=1 yij (5) , and since the amount required there is rj , I i=1 yij ≥ rj for j = 1, , J (6) It is assumed that we cannot send a negative amount from PI to Mj , we have yij ≥ for i = 1, , I and j = 1, , J (7) Our problem is: minimize (4) subject to (5), (6) and (7) Let us put this problem in the form of a standard minimum problem The number of y variables is IJ , so m = IJ But what is n ? It is the total number of main constraints There are n = I + J of them, but some of the constraints are ≥ constraints, and some of them are ≤ constraints In the standard minimum problem, all constraints are ≥ This can be obtained by multiplying the constraints (5) by −1 : J j=1 (−1)yij ≥ −si for i = 1, , I (5 ) The problem “minimize (4) subject to (5 ), (6) and (7)” is now in standard form In Exercise 3, you are asked to write out the matrix A for this problem Example The Activity Analysis Problem There are n activities, A1 , , An , that a company may employ, using the available supply of m resources, R1 , , Rm (labor hours, steel, etc.) Let bi be the available supply of resource Ri Let aij be the amount of resource Ri used in operating activity Aj at unit intensity Let cj be the net value to the company of operating activity Aj at unit intensity The problem is to choose the intensities which the various activities are to be operated to maximize the value of the output to the company subject to the given resources Let xj be the intensity at which Aj is to be operated The value of such an activity allocation is n cj xj (8) j=1 The amount of resource Ri used in this activity allocation must be no greater than the supply, bi ; that is, aij xj ≤ bi for i = 1, , m (9) j=1 It is assumed that we cannot operate an activity at negative intensity; that is, x1 ≥ 0, x2 ≥ 0, , xn ≥ (10) Our problem is: maximize (8) subject to (9) and (10) This is exactly the standard maximum problem Example The Optimal Assignment Problem There are I persons available for J jobs The value of person i working day at job j is aij , for i = 1, , I , and j = 1, , J The problem is to choose an assignment of persons to jobs to maximize the total value xij An assignment is a choice of numbers, xij , for i = 1, , I , and j = 1, , J , where represents the proportion of person i’s time that is to be spent on job j Thus, J j=1 xij ≤ for i = 1, , I (11) xij ≤ for j = 1, , J (12) I i=1 and xij ≥ for i = 1, , I and j = 1, , J (13) Equation (11) reflects the fact that a person cannot spend more than 100% of his time working, (12) means that only one person is allowed on a job at a time, and (13) says that no one can work a negative amount of time on any job Subject to (11), (12) and (13), we wish to maximize the total value, I J aij xij i=1 j=1 (14) This is a standard maximum problem with m = I + J and n = IJ Terminology The function to be maximized or minimized is called the objective function A vector, x for the standard maximum problem or y for the standard minimum problem, is said to be feasible if it satisfies the corresponding constraints The set of feasible vectors is called the constraint set A linear programming problem is said to be feasible if the constraint set is not empty; otherwise it is said to be infeasible A feasible maximum (resp minimum) problem is said to be unbounded if the objective function can assume arbitrarily large positive (resp negative) values at feasible vectors; otherwise, it is said to be bounded Thus there are three possibilities for a linear programming problem It may be bounded feasible, it may be unbounded feasible, and it may be infeasible The value of a bounded feasible maximum (resp, minimum) problem is the maximum (resp minimum) value of the objective function as the variables range over the constraint set A feasible vector at which the objective function achieves the value is called optimal All Linear Programming Problems Can be Converted to Standard Form A linear programming problem was defined as maximizing or minimizing a linear function subject to linear constraints All such problems can be converted into the form of a standard maximum problem by the following techniques A minimum problem can be changed to a maximum problem by multiplying the n objective function by −1 Similarly, constraints of the form j=1 aij xj ≥ bi can be n changed into the form j=1 (−aij )xj ≤ −bi Two other problems arise n (1) Some constraints may be equalities An equality constraint j=1 aij xj = bi may be removed, by solving this constraint for some xj for which aij = and substituting this solution into the other constraints and into the objective function wherever xj appears This removes one constraint and one variable from the problem (2) Some variables may not be restricted to be nonnegative An unrestricted variable, xj , may be replaced by the difference of two nonnegative variables, xj = uj − vj , where uj ≥ and vj ≥ This adds one variable and two nonnegativity constraints to the problem Any theory derived for problems in standard form is therefore applicable to general problems However, from a computational point of view, the enlargement of the number of variables and constraints in (2) is undesirable and, as will be seen later, can be avoided Exercises Consider the linear programming problem: Find y1 and y2 to minimize y1 + y2 subject to the constraints, y1 + 2y2 ≥ 2y1 + y2 ≥ y2 ≥ Graph the constraint set and solve Find x1 and x2 to maximize ax1 + x2 subject to the constraints in the numerical example of Figure Find the value as a function of a Write out the matrix A for the transportation problem in standard form Put the following linear programming problem into standard form Find x1 , x2 , x3 , x4 to maximize x1 + 2x2 + 3x3 + 4x4 + subject to the constraints, 4x1 + 3x2 + 2x3 + x4 ≤ 10 x1 − x3 + 2x4 = x1 + x2 + x3 + x4 ≥ , and x1 ≥ 0, x3 ≥ 0, x4 ≥ Duality To every linear program there is a dual linear program with which it is intimately connected We first state this duality for the standard programs As in Section 1, c and x are n -vectors, b and y are m -vectors, and A is an m × n matrix We assume m ≥ and n ≥ Definition The dual of the standard maximum problem maximize cTx subject to the constraints Ax ≤ b and x ≥ (1) is defined to be the standard minimum problem minimize y Tb subject to the constraints y TA ≥ cT and y ≥ (2) Let us reconsider the numerical example of the previous section: Find x1 and x2 to maximize x1 + x2 subject to the constraints x1 ≥ , x2 ≥ , and x1 + 2x2 ≤ 4x1 + 2x2 ≤ 12 −x1 + x2 ≤ (3) The dual of this standard maximum problem is therefore the standard minimum problem: Find y1 , y2 , and y3 to minimize 4y1 + 12y2 + y3 subject to the constraints y1 ≥ , y2 ≥ , y3 ≥ , and y1 + 4y2 − y3 ≥ (4) 2y1 + 2y2 + y3 ≥ If the standard minimum problem (2) is transformed into a standard maximum problem (by multiplying A, b , and c by −1 ), its dual by the definition above is a standard minimum problem which, when transformed to a standard maximum problem (again by changing the signs of all coefficients) becomes exactly (1) Therefore, the dual of the standard minimum problem (2) is the standard maximum problem (1) The problems (1) and (2) are said to be duals The general standard maximum problem and the dual standard minimum problem may be simultaneously exhibited in the display: y1 y2 ym x1 a11 a21 x2 a12 a22 ··· ··· ··· xn a1n a2n am1 ≥ c1 am2 ≥ c2 ··· ··· amn ≥ cn 10 ≤ b1 ≤ b2 ≤ bm (5) Solutions to Exercises of Section Find x1 , x2 , x3 and x4 to maximize 2x1 + 4x2 + 6x3 + 2x4 , subject to x1 − x2 − 2x3 + x4 ≤ −2x1 + x2 + x4 ≤ x1 + x2 + x3 + x4 ≤ and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 2 We check that y = ( 23 , 0, 14 ) and x = (0, , , 0) are feasible for their respective problems, and that they have the same value Clearly, y ≥ and x ≥ Substituting 16 14 y into the main constraints of Exercise 1, we find 23 + 14 = ≥ 2, −3 + = ≥ 4, + 14 = ≥ , and 23 + 14 = 16 ≥ , so y is feasible Similarly substituting x into 3 3 the main constraints of the solution of Exercise 1, we find − 13 + 43 = ≤ , 13 ≤ , and 2 14 16 + = ≤ , so x is feasible The value of y is + = , and the value of x is 12 16 + = Since these are equal, both are optimal (a) Find y1 , y2 , y3 and y4 to minimize 4y1 + 6y2 + 3y3 + 8y4 , subject to y1 + 2y2 − y3 + y4 ≥ −y1 + y2 + y4 ≥ y1 + 3y2 + 2y3 + y4 ≥ and y1 ≥ 0, y2 ≥ 0, y3 ≥ 0, y4 ≥ (b) If x = (0, 6, 0) is optimal for the maximum problem, then the strict inequality in the first, third and fourth main constraints implies that y1 = y3 = y4 = Also, x2 > implies that there is equality in the second constraint in the minimum problem Solving this gives y2 = After checking that y = (0, 2, 0, 0) is feasible, we conclude it is optimal (a) Find x = (x1 , , xI , xI+1 , , xI+J ) to maximize − subject to −xi + xI+j ≤ bij for all i and j and xi ≥ I i=1 si xi + J j=1 rj xI+j for all i (b) Since bij is measured in cost, say in dollars, per unit of the commodity, the main constraints show that the xi are also measured in dollars per unit of the commodity The objective function is therefore measured in dollars (This is the dimensional analysis.) We may imagine an independent entrepreneur who offers to buy the commodities at port Pi 52 for xi dollars per unit, and to sell them at market Mj for xI+j dollars per unit If the constraints are satisfied, we can essentially get all our goods transported from the ports to the markets at no increase in cost, so we would accept the offer Naturally he will choose the I J xi to maximize his profit, which is just the objective function, − i=1 si xi + j=1 rj xI+j Thus, xi for ≤ i ≤ I represents the shadow value of a unit of the commodity at port Pi , and xI+j for ≤ j ≤ J represents the shadow value of a unit of the commodity at market Mj 53 Solution to the Exercise of Section y1 y2 y3 y4 s1 s2 s3 s4 1 2 −1 −1 −2 −→ s1 s3 y3 s2 −→ y1 y4 y2 s4 −1 −2 −2 3 −2 −4 −1 Therefore, A−1 y1 s2 s3 s4 1 2 −1 1 −1 −3 −1 s1 y2 y3 y4 −→  s1 s3 s4 s2  −6 =  −9 −→ y1 y4 y2 y3 − 13 − 32 − 13 −3 83 13 −3 − 23 − 34 31 −2 53 10 −3 −2 −1 10 −1 13 −1 −4 54  −1   −2 s1 s3 y3 y4 s1 = s2 s3 s4 y1 s2 y2 s4 1 −2 −1 −1 −1 y1 y2 − 23 −2 10 −3 13 − 43 y3 y4 − 13 − 31 − 13 53 − 13 83 −3 Solutions to the Exercises of Section Minimize 4y1 + 2y2 + y3 subject to y1 −y1 −2y1 −y1 + 2y2 − 2y3 + + y3 + y2 − 4y2 + y3 ≥ ≥ −2 ≥ −3 ≥ −1 and y1 ≥ 0, y2 ≥ 0, y3 ≥ Pivoting according to the simplex rules gives y1 y2 y3 x1 −2 −1 x2 −1 x3 −2 x4 −1 −4 1 y2 x2 x3 − 12 −1 − 25 12 1 1 72 y1 x1 y3 −→ y2 x2 x3 y1 x4 x1 y3 −→ 1 x4 −2 −3 −1 3 12 From this we see that the value for both programs is 4, the solution to the minimum problem is y1 = , y2 = , y3 = , and the solution to the maximum problem is x1 = , x2 = , x3 = , x4 = Maximize −x1 + 2x2 + x3 subject to 2x1 + x2 − 7x3 ≤ −x1 + 4x3 ≤ −1 x1 + 2x2 − 6x3 ≤ y1 y2 y3 x1 −1 1 x2 −2 x3 −7 −1 −6 −1 −→ y1 x1 y3 y2 −1 1 and x2 −2 x1 ≥ 0, x2 ≥ 0, x3 ≥ x3 1 −4 −2 −1 y2 y3 x3 −→ y1 x1 x2 1 3/2 1/2 The value is zero, the solution to the minimum problem is y1 = , y2 = , y3 = , and the solution to the maximum problem is x1 = , x2 = 1/2 , x3 = 55 y1 y2 y3 x1 −3 −1 x2 −1 x3 −2 −2 1 −→ y1 y2 x3 x1 −2 −1 −1 x2 −1 x1 −3 x2 −1 −1 y3 −1 2 The first column shows that the maximum problem is unbounded feasible y1 y2 y3 x1 −2 x2 −1 −1 x3 −2 −1 −1 0 −→ y1 y2 x3 y3 2 −4 −1 0 The second row shows that the maximum problem is infeasible It follows that the minimum problem is unbounded feasible since the vector (y1 , y2 , y3 ) = (1/2, 0, 0) is feasible In general, to show whether or not the minimum problem is feasible, one may have to use the dual simplex method Using the Dual Simplex Method, y1 y2 y3 x1 −1 −2 x2 x3 −3 −2 1 y2 x2 x3 −→ y1 x1 y3 2 1 −3 The value is −3 , the solution to the minimum problem is y1 = , y2 = 3/2 , y3 = , and the solution to the maximum problem is x1 = , x2 = , x3 = Using the Dual Simplex Method , case 2, y1 y2 y3 x1 −3 −2 −3 x2 −1 −4 x3 1 −1 −1 −5 y1 x2 x3 −→ x1 y2 y3 1 The value is , the solution to the minimum problem is y1 = , y2 = , y3 = , and the solution to the maximum problem is x1 = , x2 = , x3 = 56 Solutions to the Exercises of Section (a) Maximize j=1 cj xj n j= +1 cj xj + n aij xj + for i = 1, , k (−aij )xj ≤ bi for i = k + 1, , m j= +1 n n aij xj + j= +1 j= +1 n (−aij )xj + j=1 (−aij )xj ≤ bi j= +1 aij xj + j=1 n (−aij )xj + j= +1 and subject to n aij xj + j=1 n j= +1 (−cj )xj + j= +1 for i = k + 1, , m aij xj ≤ −bi xj ≥ for j = 1, , xj ≥ and xj ≥ for j = + 1, , n (b) Minimize k i=1 bi yi k m yi aij + i=1 i=k+1 i=k+1 k m m i=1 yj aij + i=k+1 k i=k+1 m yi (−aij ) + subject to yi (−aij ) ≥ cj for j = 1, , yi (−aij ) ≥ cj for j = + 1, , n m yj (−aij ) + i=k+1 and m i=k+1 (−bi )yi + m yj aij + yi aij + i=1 m i=k+1 bi yi + i=k+1 yi ≥ yi aij ≥ −cj for j = + 1, , n for i = 1, , k yi ≥ and yi ≥ for i = k + 1, , m (c) It is easy to check that these standard programs are dual One can mimic the proof of the Equilibrium Theorem for Standard Problems Let x and y ∗ be feasible for the General Maximum Problem and its dual If the condition (*) is satisfied, then ∗ m m yi∗ bi i=1 = n yi∗ ( i=1 m aij x∗j ) j=1 n = yi∗ aij x∗j i=1 j=1 57 n ∗ j=1 aij xj n m since yi∗ = whenever bi = n cj x∗j j=1 = ( Similarly, m yi∗ aij )x∗j j=1 i=1 ∗ n = yi∗ aij x∗j i=1 j=1 Since the values are equal, both x and y are optimal ∗ Now suppose x∗ and y ∗ are optimal By the Duality Theorem, their values are equal, that is, j cj x∗j = i yi∗ bi But j cj xj ≤ j i ∗ yi aij xj ≤ yi bi i for all feasible x and y So for x∗ and y , we get equality throughout Therefore, j (cj − aij yi )x∗j = i and since cj ≤ for all j , we must have that x∗j = whenever cj < Similarly, yi∗ = whenever j aij x∗j < bi , as was to be shown ∗ i aij yi i aij yi∗ (a) Minimize 5y1 + 2y2 + y3 subject to − y3 ≥ −y1 ≥5 5y1 + 3y2 and y1 ≥ , y2 and y3 unrestricted 2y1 + y3 = 5y1 + y2 + 2y3 ≥ (b) From the display in Example 1, we see that the optimal values are y1 = , and y2 = Then solving the equality constraint for y3 , we find y3 = − 2y1 = −1 We first pivot to interchange y1 x1 x2 x3 ← λ y1 −1 y2 −2 y3 −3 −4 y4 −1 ↑µ 1 1 0 −1 Then we interchange x2 and µ, delete of the simplex method: x1 x3 y1 y2 4 −1 y3 −2 −1 −→ y4 −1 x2 1 −2 1 and λ and delete the λ row −→ x1 x2 x3 y1 −3 −1 −4 −6 −1 −1 −1 1 −1 y2 y3 y4 ↑µ 0 the µ row, and pivot once according to the rules y3 1 −1 −→ y2 x1 y4 x2 x3 y1 5/7 4/7 3/7 3/7 2/7 3/7 5/7 1/7 The value of the game is 1/7 The optimal strategy for Player I is (4/7, 3/7, ), and the optimal strategy for Player II is (5/7, , 2/7, ) 58 Solutions to the Exercises of Section Pivoting according to the instructions gives x1 −3 y1 y2 y3 x2 −2 −3 x3 −1 −1 −1 x4 2 −→ −→ y1 −2 0 −→ x3 x2 y3 y1 −3 −8 y2 −1 −2 x1 −1 −1 x4 −4 −7 y1 x4 y3 x3 −1 1 y2 −2 −2 x1 −3 −7 x2 −3 11 x1 y2 y3 x2 −2 −1 x3 −1 1 −4 0 x4 −3 −→ 0 −→ y1 −3 −2 y2 x3 x4 −4 1 −3 0 1 −3 0 −→ x3 x4 y3 y1 1 −1 −1 y2 −2 −1 x1 −3 −1 −4 x2 −4 0 y1 y2 y3 x3 −1 −1 −1 x4 2 x1 −3 x2 −2 −3 0 x1 x2 y3 and we are back where we started with just some columns interchanged y1 y2 y3 x1 −3 x2 −2 −3 x3 −1 −1 −1 x4 2 −→ 0 1 0 y1 x1 x3 0 0 y1 x1 y3 −→ y2 x2 −.5 −.5 −1.5 0 1.5 y3 5 2.5 x4 1.5 3.5 y2 x2 x3 x4 −.5 −.5 −.5 1.5 −1.5 −.5 0 1.5 −2.5 3.5 5 2.5 0 1 −.5 0 0 1.5 −.5 5 0 1.5 2.5 The value is 2.5, the solution to the maximum problem is x = (.5, 0, 1, 0), and the solution to the minimum problem is y = (0, 1.5, 2.5) 59 The first four pivots are the same as for Problem Then x3 x4 y3 y1 1 −1 −1 y2 −2 −1 x1 −3 −1 −4 x2 −4 0 −→ x3 x4 x1 y1 y2 y3 x2 0 1 − 31 31 − 13 13 − 3 31 − 43 13 − 73 83 34 53 43 −→ x3 y1 x1 x4 y2 y3 x2 0 1 1 1 −2 −2 1 2 −2 2 This is the solution found in Problem 2, but with some columns interchanged and some rows interchanged 60 Solutions to the Exercises of Section There are n = activities, m = resource, and p = parts to be made We wish to maximize min{(2x1 + 4x2 + 6x3 )/2, (2x1 + x2 + 3x3 )/1, (x1 + 3x2 + 2x3 )/1} subject to 2x1 + 3x2 + 4x3 ≤ 12 and x1 ≥ 0, x2 ≥ 0, x3 ≥ We add an extra variable, λ unrestricted (though it might as well be restricted to be non-negative), less than or equal to each of the three linear forms in the minimum We wish to maximize λ subject to x1 2x1 x1 2x1 + + + + 2x2 x2 x2 3x2 + + + + 3x3 3x3 x3 4x3 ≥ ≥ ≥ ≤ λ λ λ 12 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ The simplex tableau is solved by the simplex method: y1 y2 y3 y4 x1 x2 x3 ← λ −1 −2 −3 −2 −1 −3 −1 −3 −2 0 0 −1 ← −→ λ x2 y3 y4 x1 y2 x3 −3 −3 −1 −1 1 −3 −3 −3 y1 −1 −1 −2 −1 ← −→ ← 0 12 λ x2 x3 x1 λ y2 y3 y4 −→ ← 0 12 y4 x1 x2 x3 y1 −1 −2 −3 −1 −1 −1 −1 −1 −2 −3 −→ y2 y3 λ x2 x3 y4 0 12 x1 y2 y3 y1 −6 −7 −1 −1 −1 1 −2 −7 −4 11 −6 −7 0 12 y1 4/3 4/3 4/3 2/3 1/3 1/3 1/3 The value is 8, and the solution to the maximum problem is x1 = x2 = x3 = 4/3 61 The problem is to minimize y3 +y4 +y5 subject to |y1 +y2 −1| ≤ y3 , |2y1 −y2 +1| ≤ y4 , and |y1 − y2 − 2| ≤ y5 and y1 ≥ and y2 ≥ Each inequality involving an absolute value becomes two inequalities: Minimize y3 + y4 + y5 subject to y1 −y1 2y1 −2y1 y1 −y1 + − − + − + y2 y2 y2 y2 y2 y2 + + + + + + y3 y3 y4 y4 y5 y5 ≥ ≥ −1 ≥ −1 ≥ ≥ ≥ −2 and y1 ≥ 0, y2 ≥ We pivot y3 , y4 and y5 to the top and delete Then we have feasible solution to the minimum problem, so we finish according to dual simplex method y1 y2 ↑y3 ↑y4 ↑y5 x1 x2 x3 x4 x5 x6 −1 −2 −1 −1 −1 −1 1 0 0 0 1 0 0 0 1 −1 1 −1 −2 y1 y2 −→ x1 x4 ↑y5 x2 −2 −2 0 x3 −2 x5 −1 0 −2 0 1 −→ x2 −2 −2 0 x6 y1 −1 y2 −1 −→ x1 x4 1 x5 y1 y2 x1 ↑y4 ↑y5 x3 −2 x2 x3 x4 x5 x6 −2 −2 −1 −2 −1 −1 1 0 0 1 0 0 1 −1 −2 x6 −2 y1 −1 x2 −→ x1 x4 x5 1 4 y2 x3 x6 −1 −1 1 1 5 1 The value is 3, and the optimal point is (y1 , y2 ) = (0, 1) We are to minimize µ subject to |y1 + y2 − 1| ≤ µ, |2y1 − y2 + 1| ≤ µ, and |y1 − y2 − 2| ≤ µ, all variables unrestricted Each inequality involving an absolute value becomes two inequalities: Minimize µ subject to y1 −y1 2y1 −2y1 y1 −y1 + − − + − + y2 y2 y2 y2 y2 y2 + + + + + + 62 µ µ µ µ µ µ ≥ ≥ −1 ≥ −1 ≥ ≥ ≥ −2 x1 x2 x3 x4 x5 x6 −1 −2 −1 −1 −1 −1 1 1 1 −1 1 −1 −2 ↑y1 ↑y2 ↑µ ↑y1 −→ x4 x5 x1 x2 −3 0 x3 ↑y2 ↑µ 1.5 5 −.5 −.5 1.5 0 x1 0 ↑y1 −→ x4 ↑µ x6 1 x1 −→ x4 x5 1.5 ↑y1 x2 x3 ↑y2 x5 −3 −1 −1 −1 −1 2 −1 0 −3 x2 x3 ↑y2 ↑µ x6 −.5 1.5 x6 1 0 1/6 1/3 1/2 1.5 The value is 1.5 and the Chebyshev point is (y1 , y2 ) = (0, −.5) Let xi be the intensity at which activity Ai is operated, i = 1, We wish to maximize the rate of return, (2x1 + 3x2 )/(x1 + 2x2 + 1), subject to x1 + 3x2 ≤ 3x1 + 2x2 ≤ and x1 ≥ , x2 ≥ Choose t so that t(x1 + 2x2 + 1) = , and let zi = txi for i = 1, The problem becomes: choose z1 , z2 and t to maximize 2z1 + 3z2 , subject to z1 + z2 + t = z1 + 3z2 − 2t ≤ 3z1 + 2z2 − 3t ≤ ↑y1 y2 y3 z1 1 −2 z2 −3 ← t −2 −3 0 z1 −2 ← −→ t y2 y3 ← −→ t z2 z1 and z2 −3 ↑y1 y3 z1 ≥ , z2 ≥ 3 y2 ↑y1 ← −→ t y2 z1 y3 z2 ↑y1 −1/6 2/3 1/2 1/2 −1/2 1/2 1/2 1/6 4/3 1/2 1/2 1/3 −1/3 1 7/18 1/6 5/18 5/18 1/9 19/18 19/18 The value is 19/18, and the solution is (z1 , z2 , t) = (5/18, 3/18, 7, 18) The optimal intensities are (x1 , x2 ) = (z1 /t, z2 /t) = (5/7, 3/7), and the optimal rate is 19/18 63 Solutions to the Exercises of Section The Northwest Corner Rule and the Least-Cost Rule lead to the same feasible shipping schedule Solve for the dual variables and find that the constraint vj − ui ≤ bij is not satisfied for i = , j = Add θ to that square, and add and subtract θ from other squares keeping the constraints satisfied The transporation cost can be improved by taking θ = 2 1 −→ −θ +θ +θ −θ 1 1 −→ −1 4 2 This leads to a transportation array that satisfies the optimality condition The value of the optimal shipping schedule is + + = As a check, vj rj − ui si = + = The Least-Cost Rule, which gives the same result no matter how the rows and columns are rearranged, leads to success on the first try The value is + + 15 + 35 = 57 1 3 −1 5 6 5 After finding the Least-Cost shipping schedule and the dual varaibles, we see that the constraint vj − ui ≤ bij is not satisfied for i = , j = This leads to an improvement by adding θ = to that square The new schedule is optimal, with value 535 −1 4 10 8 15 −θ 20 +θ 4 20 30 25 +θ 25 25 −θ −2 15 30 −→ 35 15 15 20 30 30 25 25 64 15 50 10 10 25 35 50 After finding an initial shipping schedule using the Least-Cost Rule, it takes two more applications of the improvment routine with θ = in each to arrive at the optimal schedule 4 +θ −θ 4 −θ +θ −θ +θ +θ −θ +θ −θ 1 7 7 6 2 −θ +θ 3 7 65 4 6 3 4 6 Related Texts [1] Vaˇsek Chv´atal, Linear Programming, (1983) W H Freeman & Co [2] G B Dantzig, Linear Programming and Extensions, (1963) Princeton University Press [3] David Gale, The Theory of Linear Economic Models, (1960) McGraw-Hill [4] Samuel Karlin, Mathematical Methods and Theory in Games, Programming and Economics, vol 1, (1959) Addison-Wesley [5] James K Strayer, Linear Programming and Applications, (1989) Springer-Verlag 66 ... 66 LINEAR PROGRAMMING Introduction A linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to linear constraints The... called optimal All Linear Programming Problems Can be Converted to Standard Form A linear programming problem was defined as maximizing or minimizing a linear function subject to linear constraints... solving linear programming problems.) We shall also see later that this theorem contains the Minimax Theorem for finite games of Game Theory The Duality Theorem If a standard linear programming