1 Introduction. Linear Programming. Questions. A general optimization problem is of the form: choose x to. max f(x) subject to x S. where.


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1 Introduction Linear Programming Neil Laws TT 00 A general optimization problem is of the form: choose x to maximise f(x) subject to x S where x = (x,..., x n ) T, f : R n R is the objective function, S R n is the feasible set. We might write this problem: max f(x) subject to x S. x.. For example f(x) = c T x for some vector c R n, S = {x : Ax b} vector b R m. for some m n matrix A and some If f is linear and S R n can be described by linear equalities/inequalities then we have a linear programming (LP) problem. If x S then x is called a feasible solution. Questions In general: does a feasible solution x S exist? if so, does an optimal solution exist? if so, is it unique? If the maximum of f(x) over x S occurs at x = x then x is an optimal solution, and f(x ) is the optimal value..3.4
2 Example A company produces drugs A and B using machines M and M. ton of drug A requires hour of processing on M and hours on M ton of drug B requires 3 hours of processing on M and hour on M 9 hours of processing on M and 8 hours on M are available each day Each ton of drug produced (of either type) yields million profit Solution Let x = number of tons of A produced x = number of tons of B produced P : maximise x + x (profit in million) subject to x + 3x 9 (M processing) x + x 8 (M processing) x, x 0 To maximise its profit, how much of each drug should the company make per day?.5.6 x Diet problem A pigfarmer can choose between four different varieties of food, providing different quantities of various nutrients. x + x = 5 x + x = 8 x x + 3x = 9 food required 3 4 amount/wk A nutrient B C cost/kg The (i, j) entry is the amount of nutrient i per kg of food j. The shaded region is the feasible set for P. The maximum occurs at x = (3, ) with value
3 General form of the diet problem Foods j =,..., n, nutrients i =,..., m. Problem P : minimise 5x + 7x + 7x 3 + 9x 4 subject to.5x + x + x x 4 4 x + 3.x + x x +.5x + 5.6x 3 +.x x, x, x 3, x 4 0 Data: a ij = amount of nutrient i in one unit of food j b i = required amount of nutrient i c j = cost per unit of food j Let x j = number of units of food j in the diet. The diet problem is minimise c x + + c n x n subject to a i x + + a in x n b i for i =,..., m x,..., x n Real applications Programming = planning In matrix notation the diet problem is min x c T x subject to Ax b, x 0. Note that our vectors are always column vectors. We write x 0 to mean x i 0 for all i. (0 is a vector of zeros.) Similarly Ax b means (Ax) i b i for all i. Maybe many thousands of variables or constraints Production management: realistic versions of P, large manufacturing plants, farms, etc Scheduling, e.g. airline crews: need all flights covered restrictions on working hours and patterns minimise costs: wages, accommodation, use of seats by nonworking staff shift workers (call centres, factories, etc) Yield management (airline ticket pricing: multihops, business/economy mix, discounts, etc) Network problems: transportation capacity planning in telecoms networks Game theory: economics, evolution, animal behaviour..
4 Free variables Slack variables In P we had Some variables may be positive or negative, e.g. omit the constraint x 0. Such a free variable can be replaced by We can rewrite by maximise x + x subject to x + 3x 9 x + x 8 x, x 0. where u, v 0. x = u v maximise x + x subject to x + 3x + x 3 = 9 x + x + x 4 = 8 x,..., x 4 0. x 3 = unused time on machine M x 4 = unused time on machine M x 3 and x 4 are called slack variables..3.4 Two standard forms With the slack variables included, the problem has the form max x ct x subject to Ax = b x 0. In fact any LP (with equality constraints, weak inequality constraints, or a mixture) can be converted to the form since: max x ct x subject to Ax = b x 0 minimising c T x is equivalent to maximising c T x, inequalities can be converted to equalities by adding slack variables, free variables can be replaced as above..5.6
5 Remark Similarly, any LP can be put into the form since e.g. max x ct x subject to Ax b Ax = b { Ax b x 0 Ax b (more efficient rewriting may be possible!). So it is OK for us to concentrate on LPs in these forms. We always assume that the underlying space is R n. In particular x,..., x n need not be integers. If we restrict to x Z n we have an integer linear program (ILP). ILPs are in some sense harder than LPs. Note that the optimal value of an LP gives a bound on the optimal value of the associated ILP
6 Geometry of linear programming Definition. A set S R n is called convex if for all u, v S and all λ (0, ), we have λu + ( λ)v S. For now we will consider LPs in the form u v v u u v max c T x subject to Ax = b x 0. convex convex not convex That is, a set is convex if all points on the line segment joining u and v are in S, for all possible line segments... Theorem. The feasible set is convex. S = {x R n : Ax = b, x 0} Extreme points Definition.3 A point x in a convex set S is called an extreme point of S if there are no two distinct points u, v S, and λ (0, ), such that x = λu + ( λ)v. Proof. Suppose u, v S, λ (0, ). Let w = λu + ( λ)v. Then That is, an extreme point x is not in the interior of any line segment lying in S. Aw = A[λu + ( λ)v] = λau + ( λ)av = [λ + ( λ)]b = b and w λ0 + ( λ)0 = 0. So w S..3.4
7 Theorem.4 If an LP has an optimal solution, then it has an optimal solution at an extreme point of the feasible set. Proof. Idea: If the optimum is not extremal, it s on some line within S all of which is optimal: go to the end of that line, repeat if necessary. Since there exists an optimal solution, there exists an optimal solution x with a minimal number of nonzero components. Suppose x is not extremal, so that x = λu + ( λ)v Since x is optimal, c T u c T x and c T v c T x. But also c T x = λc T u + ( λ)c T v so in fact c T u = c T v = c T x. Now consider the line defined by x(ε) = x + ε(u v), ε R. Then (a) Ax = Au = Av = b so Ax(ε) = b for all ε, (b) c T x(ε) = c T x for all ε, (c) if x i = 0 then u i = v i = 0, which implies x(ε) i = 0 for all ε, (d) if x i > 0 then x(0) i > 0, and x(ε) i is continuous in ε. for some u v S, λ (0, )..5.6 So we can increase ε from zero, in a positive or a negative direction as appropriate, until at least one extra component of x(ε) becomes zero. This gives an optimal solution with fewer nonzero components than x. So x must be extreme. Basic solutions Let a i be the ith column of A, so that n Ax = b x i a i = b. Definition.5 () A solution x of Ax = b is called a basic solution if the vectors {a i : x i 0} are linearly independent. (That is, columns of A corresponding to nonzero variables x i are linearly independent.) () A basic solution satisfying x 0 is called a basic feasible solution (BFS). i= Note: If A has m rows, then at most m columns can be linearly independent. So any basic solution x has at least n m zero components. More later..7.8
8 Theorem.6 x is an extreme point of if and only if x is a BFS. Proof. S = {x : Ax = b, x 0} () Let x be a BFS. Suppose x = λu + ( λ)v for u, v S, λ (0, ). To show x is extreme we need to show u = v. Let I = {i : x i > 0}. Then (a) if i I then x i = 0, which implies u i = v i = 0. (b) Au = Av = b, so A(u v) = 0 = n (u i v i )a i = 0 i= = i I (u i v i )a i = 0 since u i = v i = 0 for i I which implies u i = v i for i I since {a i : i I} are linearly independent. Hence u = v, so x is an extreme point..9.0 () Suppose x is not a BFS, i.e. {a i : i I} are linearly dependent. Then there exists u 0 with u i = 0 for i I such that Au = 0. For small enough ε, x ± εu are feasible, and x = (x + εu) + (x εu) Corollary.7 If there is an optimal solution, then there is an optimal BFS. Proof. This is immediate from Theorems.4 and.6. so x is not extreme...
9 Discussion Typically we may assume: n > m (more variables than constraints), A has rank m (its rows are linearly independent; if not, either we have a contradiction, or redundancy). Then: x is a basic solution there is a set B {,..., n} of size m such that x i = 0 if i B, {a i : i B} are linearly independent. Proof. Simple exercise. Take I = {i : x i 0} and augment it to a larger linearly independent set B if necessary. Then to look for basic solutions: choose n m of the n variables to be 0 (x i = 0 for i B), look at remaining m columns {a i : i B}. Are they linearly independent? If so we have an invertible m m matrix. Solve for {x i : i B} to give i B x ia i = b. So also Ax = n x i a i = i= n 0a i + i B n x i a i i B = b as required. This way we obtain all basic solutions (at most ( n m) of them)..3.4 Bad algorithm: look through all basic solutions, which are feasible? what is the value of the objective function? We can do much better! Simplex algorithm: will move from one BFS to another, improving the value of the objective function at each step..5.6
10 3 The simplex algorithm () The simplex algorithm works as follows.. Start with an initial BFS.. Is the current BFS optimal? 3. If YES, stop. If NO, move to a new and improved BFS, then return to. Recall P, expressed without slack variables: maximise x + x subject to x + 3x 9 x + x 8 x, x 0 From Corollary.7, it is sufficient to consider only BFSs when searching for an optimal solution x Rewrite: E x +3x +x 3 = 9 () x + x +x 4 = 8 () x + x = f(x) (3) Put x, x = 0, giving x 3 = 9, x 4 = 8, f = 0 (we re at the BFS x = (0, 0, 9, 8)). B A C D x + x = 8 F x x + 3x = 9 Note: In writing the three equations as () (3) we are effectively expressing x 3, x 4, f in terms of x, x
11 . Start at the initial BFS x = (0, 0, 9, 8), vertex A, where f = 0.. From (3), increasing x or x will increase f(x). Let s increase x. (a) From (): we can increase x to 9, if we decrease x 3 to 0. (b) From (): we can increase x to 4, if we decrease x 4 to 0. The stricter restriction on x is (b). 3. So (keeping x = 0), (a) increase x to 4, decrease x 4 to 0 using (), this maintains equality in (), (b) and, using (), decreasing x 3 to 5 maintains equality in (). With these changes we move to a new and improved BFS x = (4, 0, 5, 0), f(x) = 4, vertex D. To see if this new BFS is optimal, rewrite () (3) so that each nonzero variable appears in exactly one constraint, f is in terms of variables which are zero at vertex D. Alternatively, we want to express x, x 3, f in terms of x, x 4. How? Add multiples of () to the other equations We are at vertex D, x = (4, 0, 5, 0) and f = 4. () () : 5 x +x 3 x 4 = 5 (4) () : x + x + x 4 = 4 (5) (3) () : x x 4 = f 4 (6) Now f = 4 + x x 4. So we should increase x to increase f.. From (6), increasing x will increase f (increasing x 4 would decrease f). (a) From (4): we can increase x to, if we decrease x 3 to 0. (b) From (5): we can increase x to 8, if we decrease x to 0. The stricter restriction on x is (a). 3. So increase x to, decrease x 3 to 0 (x 4 stays at 0, and from (5) x decreases to 3). With these changes we move to the BFS x = (3,, 0, 0), vertex C
12 Rewrite (4) (6) so that they correspond to vertex C: 5 (4) : x + 5 x 3 5 x 4 = (7) (5) 5 (4) : x 5 x x 4 = 3 (8) (6) 5 (4) : 5 x 3 5 x 4 = f 5 (9). We are at vertex C, x = (3,, 0, 0) and f = 5.. We have deduced f = 5 5 x 3 5 x 4 5. So x 3 = x 4 = 0 is the best we can do! In that case we can read off x = 3 and x =. So x = (3,, 0, 0), which has f = 5, is optimal. Summary At each stage: B = basic variables, we express x i, i B and f in terms of x i, i B, setting x i = 0, i B, we can read off f and x i, i B (gives a BFS!). At each update: look at f in terms of x i, i B, which x i, i B, would we like to increase? if none, STOP! otherwise, choose one and increase it as much as possible, i.e. until one variable x i, i B, becomes Summary continued So one new variable enters B (becomes nonzero, becomes basic), another one leaves B (becomes 0, becomes nonbasic). This gives a new BFS. We update our expressions to correspond to the new B. Simplex algorithm We can write equations as a tableau x +3x +x 3 = 9 () x + x +x 4 = 8 () x + x = f 0 (3) x x x 3 x ρ 0 8 ρ ρ 3 This initial tableau represents the BFS x = (0, 0, 9, 8) at which f = 0. Note the identity matrix in the x 3, x 4 columns (first two rows), and the zeros in the bottom row below it
13 The tableau To update ( pivot ) At a given stage, the tableau has the form (a ij ) b which means Ax = b f(x) = c T x f We start from A = A, b = b, c = c and f = 0. c T f. Choose a pivot column Choose a j such that c j > 0 (corresponds to variable x j = 0 that we want to increase). Here we can take j =.. Choose a pivot row Among the i s with a ij > 0, choose i to minimize b i /a ij (strictest limit on how much we can increase x j ). Here take i = since 8/ < 9/. Updating the tableau is called pivoting In our example we pivot on j =, i =. The updated tableau is 3. Do row operations so that column j gets a in row i and 0s elsewhere: multiply row i by a ij, for i i, add a i j a ij (row i) to row i, add c j a ij (row i) to objective function row. which means x x x 3 x ρ = ρ ρ 0 4 ρ = ρ ρ 3 = ρ 3 ρ 5 x +x 3 x 4 = 5 x + x + x 4 = 4 x x 4 = f 4 nonbasic variables x, x 4 are 0, x = (4, 0, 5, 0). Note the identity matrix inside (a ij ) telling us this
14 Geometric picture for P Next pivot: column, row since 5 5/ < 4 /. x x x x 3 x ρ = 5 ρ ρ = ρ ρ ρ 3 = ρ 3 ρ Now we have only nonpositive entries in the bottom row: STOP. x = (3,, 0, 0), f(x) = 5 optimal. E B A C D x + x = 8 F x x + 3x = Comments on simplex tableaux x x x 3 x 4 A B C D A, B, C, D are BFSs. E, F are basic solutions but not feasible. Simplex algorithm: A D C (or A B C is we choose a different column for the first pivot). Higherdimensional problems less trivial! We always find an m m identity matrix embedded in (a ij ), in the columns corresponding to basic variables x i, i B, in the objective function row (bottom row) we find zeros in these columns. Hence f and x i, i B, are all written in terms of x i, i B. Since we set x i = 0 for i B, it s then trivial to read off f and x i, i B
15 Comments on simplex algorithm Choosing a pivot column We may choose any j such that c j > 0. In general, there is no way to tell which such j result in fewest pivot steps. Choosing a pivot row Having chosen pivot column j (which variable x j to increase), we look for rows with a ij > 0. If a ij 0, constraint i places no restriction on the increase of x j. If a ij 0 for all i, x j can be increased without limit: the objective function is unbounded. Otherwise, the most stringent limit comes from the i that minimises b i /a ij
16 4 The simplex algorithm () Initialisation Two issues to consider: can we always find a BFS from which to start the simplex algorithm? does the simplex algorithm always terminate, i.e. find an optimal BFS or prove the problem is unbounded? To start the simplex algorithm, we need to start from a BFS, with basic variables x i, i B, written in terms of nonbasic variables x i, i B. If A already contains I m as an m m submatrix, this is usually easy! This always happens if b 0 and A is created by adding slack variables to make inequalities into equalities Suppose the constraints are Ax b, x 0 where b 0. Then an initial BFS is immediate: introducing slack variables z = (z,..., z m ), so the initial tableau is Ax + z = b, x, z 0 x x n z z m A I m b c T 0 T 0 Example Add slack variables: max 6x + x + x 3 s.t. 9x + x + x 3 8 4x + x + 4x 3 4 x + 3x + 4x 3 96 x, x, x x + x + x 3 + w = 8 4x + x + 4x 3 + w = 4 x + 3x + 4x 3 + w 3 = 96 and an initial BFS is ( x z ) = ( 0b )
17 Solve by simplex What if A doesn t have this form? This initial tableau is already in the form we need ρ = ρ ρ = ρ ρ ρ 3 = ρ 3 3ρ ρ 4 = ρ 4 ρ This solution is optimal: x = 8, x = 0, x 3 = 0, f = 8. In general we can write the constraints as Ax = b, x 0 (4.) where b 0 (if necessary, multiply rows by to get b 0). If there is no obvious initial BFS and we need to find one, we can introduce artificial variables w,..., w m and solve the LP problem min x,w subject to m i= w i Ax + w = b x, w 0. (4.) Two cases arise: () if (4.) has a feasible solution, then (4.) has optimal value 0 with w = 0. () if (4.) has no feasible solution, then the optimal value of (4.) is > 0. We can apply the simplex algorithm to determine whether it s case () or (). In case (), give up. In case (), the optimal BFS for (4.) with w i 0 gives a BFS for (4.)! This leads us to the twophase simplex algorithm. Twophase simplex algorithm Example: With artificial variables: maximize 3x + x 3 subject to x + x + x 3 = 30 x x + x 3 = 8 x 0 minimize w + w subject to x + x + x 3 + w = 30 x x + x 3 + w = 8 x, w
18 So start by adding row + row to objective row: Note: To minimise w + w, we can maximise w w. So start from the simple tableau x x x 3 w w The objective function row should be expressed in terms of nonbasic variables (the entries under the identity matrix should be 0). x x x 3 w w Now start with simplex pivot on a 3 : x x x 3 w w Deleting the w columns and replacing the objective row by the original objective function 3x + x 3 : Pivot on a : x x x 3 w w So we have found a point with w w = 0, i.e. w = 0. Phase I is finished. BFS of the original problem is x = (0, 7, 6). x x x Again we want zeros below the identity matrix subtract row from row 3: Now do simplex. x x x
19 Shadow prices Pivot on a : x x x Done! Maximum at x = 4, x = 3, x 3 = 0, f = 7. Recall P : max x + x s.t. x + 3x 9 x + x 8 x, x We had initial tableau and final tableau ρ 0 8 ρ ρ ρ ρ ρ 3 Directly from the tableaux (look at columns 3 and 4): ρ = 5 ρ 5 ρ ρ = 5 ρ ρ ρ 3 = ρ 3 5 ρ 5 ρ From the mechanics of the simplex algorithm: ρ, ρ are created by taking linear combinations of ρ, ρ ρ 3 is ρ 3 (a linear combination of ρ, ρ )
20 Suppose we change the constraints to x + 3x 9 + ε x + x 8 + ε. Then the final tableau will change to ε 5 ε ε ε ε 5 ε This is still a valid tableau as long as + 5 ε 5 ε ε ε 0. In that case we still get an optimal BFS from it, with optimal value ε + 5 ε. The objective function increases by 5 per extra hour on M and by 5 per extra hour on M (if the changes are small enough ). These shadow prices can always be read off from the initial and final tableaux Digression: Termination of simplex algorithm Typical situation : each BFS has exactly m nonzero and n m zero variables. Then each pivoting operation (moving from one BFS to another) strictly increases the new variable entering the basis and strictly increases the objective function. Since there are only finitely many BFSs, we have the following theorem. Theorem 4. If each BFS has exactly m nonzero variables, then the simplex algorithm terminates (i.e. finds an optimal solution or proves that the objective function is unbounded). What is some BFSs have extra zero variables? Then the problem is degenerate. Almost always: this is no problem. In rare cases, some choices of pivot columns/rows may cause the algorithm to cycle (repeat itself). There are various ways to avoid this (e.g. always choosing the leftmost column, and then the highest row, can be proved to work always.) See e.g. Chvátal book for a nice discussion
21 5 Duality: Introduction Recall P : Obvious bounds on f(x) = x + x : maximise x + x subject to x + 3x 9 (5.) x + x 8 (5.) x, x 0. x + x x + 3x 9 from (5.) x + x x + x 8 from (5.). By combining the constraints we can improve the bound, e.g. 3 [(5.) + (5.)]: x + x x x 7 3. More systematically? For y, y 0, consider y (5.) + y (5.). We obtain (y + y )x + (3y + y )x 9y + 8y Since we want an upper bound for x + x, we need coefficients : y + y 3y + y. How to get the best bound by this method? D : minimise 9y + 8y subject to y + y 3y + y y, y 0. P = primal problem, D = dual of P Duality: General Weak duality In general, given a primal problem P : maximise c T x subject to Ax b, x 0 the dual of P is defined by D : minimise b T y subject to A T y c, y 0. Exercise The dual of the dual is the primal. Theorem 5. (Weak duality theorem) If x is feasible for P, and y is feasible for D, then c T x b T y. Proof. Since x 0 and A T y c: c T x (A T y) T x = y T Ax. Since y 0 and Ax b: y T Ax y T b = b T y. Hence c T x y T Ax b T y
22 Comments Suppose y is a feasible solution to D. Then any feasible solution x to P has value bounded above by b T y. So D feasible = P bounded. Similarly P feasible = D bounded. Corollary 5. If x is feasible for P, y is feasible for D, and c T x = b T y, then x is optimal for P and y is optimal for D. Proof. For all x feasible for P, c T x b T y by Theorem 5. = c T x and so x is optimal for P. Similarly, for all y feasible for D, and so y is optimal for D. b T y c T x = b T y Strong duality As an example of applying this result, look at x = (3, ), y = ( 5, 5 ) for P and D above. Both are feasible, both have value 5. So both are optimal. Does this nice situation always occur? Theorem 5.3 (Strong duality theorem) If P has an optimal solution x, then D has an optimal solution y such that c T x = b T y. Proof. Write the constraints of P as Ax + z = b, x, z 0. Consider the bottom row in the final tableau of the simplex algorithm applied to P
23 { xcolumns }} { { zcolumns }} { c c n y ym f Here f is the optimal value of c T x, c j 0 for all j (5.3) yi 0 for all i (5.4) and the bottom row tells us that c T x f = c T x y T z. Thus c T x = f + c T x y T z = f + c T x y T (b Ax) = f b T y + (c T + y T A)x. This is true for all x, so f = b T y (5.5) c = A T y + c. From (5.3) c 0, hence A T y c. (5.6) Comments So (5.4) and (5.6) show that y is feasible for D. And (5.5) shows that the objective function of D at y is b T y = f = optimal value of P. So from the weak duality theorem (Theorem 5.), y is optimal for D. Note: the coefficients y from the bottom row in the columns corresponding to slack variables give us the optimal solution to D, comparing with the shadow prices discussion: these optimal dual variables are the shadow prices! 5. 5.
24 Example Example It is possible that neither P nor D has a feasible solution: consider the problem maximise x x subject to x x x + x x, x 0. Consider the problem maximise x + 4x + x 3 + x 4 subject to x + 3x + x 4 4 x + x 3 x + 4x 3 + x 4 3 x,..., x The final tableau is x x x 3 x 4 x 5 x 6 x () By the proof of Theorem 5.3, y = ( 5 4, 4, 4 ) is optimal for the dual. () Suppose the RHSs of the original constraints become 4 + ε, 3 + ε, 3 + ε 3. Then the objective function becomes ε + 4 ε + 4 ε 3. If the original RHSs of 4, 3, 3 correspond to the amount of raw material i available, then the most you d be prepared to pay per additional unit of raw material i is y i (with y as in ())
25 (3) Suppose raw material is available at a price < 5 4 per unit. How much should you buy? With ε > 0, ε = ε 3 = 0, the final tableau would be + 5 ε 5 ε + 0 ε For this tableau to represent a BFS, the three entries in the final column must be 0, giving ε 5. So we should buy 5 additional units of raw material. (4) The optimal solution x = (,,, 0) is unique as the entries in the bottom row corresponding to nonbasic variables (i.e. the, 5 4, 4, 4 ) are < 0. (5) If say the 5 4 was zero, we could pivot in that column (observe that there would somewhere to pivot) to get a second optimal BFS x. Then λx + ( λ)x would be optimal for all λ [0, ]
26 6 Duality: Complementary slackness Recall P : maximise c T x subject to Ax b, x 0, D : minimise b T y subject to A T y c, y 0. The optimal solutions to P and D satisfy complementary slackness conditions that we can use to solve one problem when we know a solution of the other. Theorem 6. (Complementary slackness theorem) Suppose x is feasible for P and y is feasible for D. Then x and y are optimal (for P and D) if and only if and (A T y c) j x j = 0 for all j (6.) (Ax b) i y i = 0 for all i. (6.) Conditions (6.) and (6.) are called the complementary slackness conditions for P and D Interpretation Proof. As in the proof of the weak duality theorem, c T x (A T y) T x = y T Ax y T b. (6.3) () If dual constraint j is slack, then primal variable j is zero. If primal variable j is > 0, then dual constraint j is tight. () The same with primal dual. From the strong duality theorem, x, y both optimal c T x = b T y c T x = y T Ax = b T y from (6.3) (y T A c T )x = 0 and y T (Ax b) = 0 n (A T y c) j x j = 0 and j= m (Ax b) i y i = 0. i=
27 Comments But A T y c and x 0, so n j= (AT y c) j x j is a sum of nonnegative terms. Also, Ax b and y 0, so m i= (Ax b) iy i is a sum of nonpositive terms. Hence n j= (AT y c) j x j = 0 and m i= (Ax b) iy i = 0 is equivalent to (6.) and (6.). What s the use of complementary slackness? Among other things, given an optimal solution of P (or D), it makes finding an optimal solution of D (or P ) very easy, because we know which the nonzero variables can be and which constraints must be tight. Sometimes one of P and D is much easier to solve than the other, e.g. with variables, 5 constraints, we can solve graphically, but 5 variables and constraints is not so easy Example Consider P and D with A = ( ) 4 0, b = 3 ( ), c = Is x = (0, 4, 3 4 ) optimal? It is feasible. If it is optimal, then x > 0 = (A T y) = c, that is 4y y = x 3 > 0 = (A T y) 3 = c 3, that is 0y + y = 3 So x = (0, 4, 3 4 ) and y = (, 3) are feasible and satisfy complementary slackness, therefore they are optimal by Theorem 6.. Alternatively, we could note that this x and y are feasible and c T x = 0 = b T y, so they are optimal by Corollary 5.. which gives y = (y, y ) = (, 3). The remaining dual constraint y + 3y 4 is also satisfied, so y = (, 3) is feasible for D
28 Example continued If we don t know the solution to P, we can first solve D graphically. y 4y y = feasible set y = 3 The optimal solution is at y = (, 3), and we can use this to solve P : y > 0 = (Ax) = b, that is x + 4x = y > 0 = (Ax) = b, that is 3x x + x 3 = 3 y + 3y > 4, that is (A T y) > c = x = 0 and so x = (0, 4, 3 4 ). y y + 3y = Example Consider the primal problem y maximise 0x + 0x + 0x 3 + 0x 4 subject to x + 8x + 6x 3 + 4x 4 0 3x + 6x + x 3 + 4x 4 0 x,..., x 4 0 dual feasible set with dual minimise 0y + 0y subject to 3y + 3y 0 () 8y + 6y 0 () 6y + y 0 (3) 4y + 4y 0 (4) y, y 0. (4) () () (3) The dual optimum is where () and (3) intersect. y 6. 6.
29 Example continued Since the second and fourth dual constraints are slack at the optimum, the optimal x has x = x 4 = 0. Also, since y, y > 0 at the optimum, } x + 6x 3 = 0 = x = 0, x 3 = 5. 3x + x 3 = 0 Hence the optimal x is (0, 0, 5, 0). Suppose the second 0 is replaced by 4. The new dual optimum is where (3) and (4) intersect, at which point the first two constraints are slack, so x = x = 0. Also, since y, y > 0 at the new optimum, } 6x 3 + 4x 4 = 0 = x 3 = 35 4 x 3 + 4x 4 = 4, x 4 = 6. So the new optimum is at x = (0, 0, 35 4, 6 )
30 7 Twoplayer zerosum games () Payoff matrix There is a payoff matrix A = (a ij ): We consider games that are zerosum in the sense that one player wins what the other loses. Each player has a choice of actions (the choices may be different for each player). Players move simultaneously. Player I plays i 3 Player II plays j If Player I plays i and Player II plays j, then Player I wins a ij from Player II. The game is defined by the payoff matrix. Note that our convention is that I wins a ij from II, so a ij > 0 is good for Player I = row player What s the worst that can happen to Player I if Player I chooses row? row? row 3? (We look at the smallest entry in the appropriate row.) What s the worst that can happen to Player II if Player II chooses a particular column? (We look at the largest entry in that column.) The matrix above has a special property. Entry a 3 = 4 is both the smallest entry in row the largest entry in column 3 (, 3) is a saddle point of A. Thus: Player I can guarantee to win at least 4 by choosing row. Player II can guarantee to lose at most 4 by choosing column 3. The above is still true if either player announces their strategy in advance. Hence the game is solved and it has value
31 Mixed strategies Consider the game of ScissorsPaperStone: Scissors beats Paper, Paper beats Stone, Stone beats Scissors. No saddle point. Scissors Paper Stone Scissors Paper Stone If either player announces a fixed action in advance (e.g. play Paper ) the other player can take advantage. So we consider a mixed strategy: each action is played with a certain probability. (This is in contrast with a pure strategy which is to select a single action with probability.) Suppose Player I plays i with probability p i, i =,..., m. Then Player I s expected payoff if Player II plays j is m a ij p i. i= Suppose Player I wishes to maximise (over p) his minimal expected payoff m min a ij p i. j i= LP formulation Similarly Player II plays j with probability q j, j =,..., n, and looks to minimise (over q) max i n a ij q j. j= This aim for Player II may seem like only one of several sensible aims (and similarly for the earlier aim for Player I). Soon we will see that they lead to a solution in a very appropriate way, corresponding to the solution for the case of the saddle point above. Consider Player II s problem minimise maximal expected payout : n n min q max a ij q j i subject to q j =, q 0. j= j= This is not exactly an LP look at the objective function
32 Equivalent formulation An equivalent formulation is: min v subject to m a ij q j v q,v j= n q j = j= q 0. for i =,..., m since v, on being minimised, will decrease until it takes the value of max i n j= a ijq j. This is an LP but not exactly in a useful form for our methods. We will transform it! First add a constant k to each a ij so that a ij > 0 for all i, j. This doesn t change the nature of the game, but guarantees v > 0. So WLOG, assume a ij > 0 for all i, j. Now change variables to x j = q j /v. The problem becomes: min v subject to m a ij x j x,v j= n x j = /v j= x 0. for i =,..., m This transformed problem for Player II is equivalent to P : max x n x j subject to Ax, x 0 j= which is now in our standard form. ( denotes a vector of s.) Doing the same transformations for Player I s problem { } m m max min a ij p i subject to p i =, p 0 p j turns into D : min y i= i= m y i subject to A T y, y 0. i= (Check: on problem sheet.) P and D are dual and hence have the same optimal value
33 Conclusion Let x, y be optimal for P, D. Then: Player I can guarantee an expected gain of at least v = / m i= y i, by following strategy p = vy. Player II can guarantee an expected loss of at most v = / n j= x j, by following strategy q = vx. The above is still true if a player announces his strategy in advance. So the game is solved as in the saddle point case (this was just a special case where the strategies were pure). v is the value of the game (the amount that Player I should fairly pay to Player II for the chance to play the game)
34 8 Twoplayer zerosum games () x Some games are easy to solve without the LP formulation, e.g. ( ) A = 4 3 Suppose Player I chooses row with probability p, row with probability p. The he should maximise min( p + 4( p), p 3( p)) = min(4 6p, 5p 3) 5p 3 x 4 6p So the min is maximised when which occurs when p = 7. And then v = 35 3 =. 4 6p = 5p 3 (We could go on to find Player II s optimal strategy too.) A useful trick: dominated actions Consider Player II should never play column 3, since column is always at least as good as column 3 (column dominates column 3.) So we reduce to Now Player I will never play row 3 since row is always better, so ( 4 ) 3 has the same value (and optimal strategies) as A
35 Final example Consider add to each entry A = Ã = , Solve the LP for Player II s optimal strategy: 0 max x + x + x 3 subject to x,x,x x 0. x x x 3 This has value > 0 (e.g. strategy ( 3, 3, 3 ) for Player I) Initial simplex tableau: final tableau: Optimum: x = 5 6, x = 4, x 3 = 3 8, and x + x + x 3 = 5 6. So value v = /(x + x + x 3 ) = 6 5. Player II s optimal strategy: q = vx = 6 5 ( 5 6, 4, 3 8 ) = ( 3, 4 5, 5 ). Dual problem for Player I s strategy has solution y = 4, y =, y 3 = 3 6 (from bottom row of final tableau). So Player I s optmal strategy: p = vy = 6 5 ( 4,, 3 6 ) = ( 4 5, 8 5, 3 5 ). Ã has value 6 6 5, so the original game A has value 5 =
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