4. Duality Duality 4.1 Duality of LPs and the duality theorem. min c T x x R n, c R n. s.t. ai Tx = b i i M a i R n
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1 2 4. Duality of LPs and the duality theorem Complementary slackness The shortest path problem and its dual Farkas' Lemma Dual information in the tableau The dual Simplex algorithm Duality of LPs and the duality theorem 22- The dual of an LP in general form Derivation of the dual Consider an LP in general form: (4.) min c T x x R n, c R n s.t. ai Tx = b i i M a i R n ai Tx b i i M x j 0 j N x j unconstrained j N we transform it to standard form according to Lemma 3.2 with s surplus variables x i for the inequalities split variables + x j = x j - - xj with + xj, - xj! 0 This gives
2 4. Duality of LPs and the duality theorem 22-2 min ĉ T ˆx s.t.  ˆx = b, ˆx 0 with  = A j, j N (A j, A j ), j N I, 0, i M i M (4.2) ˆx = x j, j N (x + j, x j ), j N x s i, i M T ĉ = c j, j N (c j, c j ), j N 0, i M T where, w.o.l.g., matrix! has full row rank, and where A j denotes the column of x j in (4.) The previous results on the simplex algorithm give: If (4.2) has an optimal solution, then there is a basis!! of! with! "! "! "!#!#!# $!% " ' %&$ " i.e., reduced cost! 0 Let m be the number of constraints in (4.). Then 4. Duality of LPs and the duality theorem 22-3 π T := ĉ Ṱ B ˆB R m is a feasible solution for inequalities π T  ĉ T (4.3) Inequalities (4.3) have 3 groups w.r.t. their columns: Group Group 2 Group 3! " # $! % $ & $ '!"("#! " # $! % $ "! " # $! "% $!" # $! % $ & $ ' "#($%!! " "!! " #!# " $ "#%$% Definition of the dual of LP (4.4) - (4.6) define constraints for a new LP with variables π,..., π m. These constraints, together with the
3 4. Duality of LPs and the duality theorem 22-4 objective function max π T b constitute the dual LP of (4.). The initial problem (4.) is called the primal LP. Transformation rules primal -> dual (follow from (4.4) - (4.6)) primal dual min c T x max π T b a T i x = b i i M π i unconstrained a T i x b i i M π i 0 x j 0 j N π T A j c j x j unconstrained j N π T A j = c j Observe: The dual LP is obtained from the optimality criterion of the primal. The variables π,..., π m correspond to multipliers of the rows of! that fulfill the primal optimality criterion. 4. Theorem (dual dual = primal) The dual of the dual is the primal. We therefore speak of primal-dual pairs of LPs Proof 4. Duality of LPs and the duality theorem 22-5 Write the dual in primal form: min π T ( b) such that ( A T j )π c i j N ( A T j )π = c i j N π i 0 j M π i unconstrained j M The transformation rules yield the following dual LP max x T ( c) such that x j 0 j N x j unconstrained j N a T i x b i i M a T i x = b i i M which is the primal LP! The Duality Theorem 4.2 Theorem (Weak and Strong Duality Theorem)
4 4. Duality of LPs and the duality theorem 22-6 Let x be a primal feasible solution and π be a dual feasible solution. Then (Weak Duality Theorem)! " #! $ " %!"&#$ If an LP has an optimal solution, so has its dual, and the optimal objective values are the same (Strong Duality Theorem) Proof Let x be a primal feasible solution and π be a dual feasible solution. Then c T x π dual feasible (π T A)x = π T x primal feasible (Ax) π T b Assume w.o.l.g. that the LP is in primal form (4.2) and has an optimal solution => has an basic optimal feasible solution! with associated basis! and π T =ĉ Ṱ ˆB B is feasible for the dual by construction For this π we obtain! " #! "#$ " #%#%!$ %#! #$ " #% " #%!$ #%! #$ " #% #& %! #$ " #& So π and! have the same objective function value. Weak Duality (4.7) then implies that π is a dual optimal solution! 4. Duality of LPs and the duality theorem Theorem (Possible primal-dual pairs) Primal-dual pairs exist exactly in one of the following cases: () both LPs have a finite optimal solution and their objective values are equal (2) both LPs have no feasible solution (3) one LP has an unbounded objective function and the other has no feasible solution
5 4. Duality of LPs and the duality theorem 22-8 primal dual finite optimal solution feasible solution, unbounded objective no feasible solution finite optimal solution () feasible solution, unbounded objective (3) no feasible solution (3) (2) Proof Strong Duality Theorem => Case () occurs in row and column of the table, and this is the only table entry in which it occurs Consider now row 2 of the table, i.e., x is a primal feasible solution but c T x unbounded from below. If there is a dual feasible solution π, we obtain π T b " c T x with the Weak Duality Theorem 4. Duality of LPs and the duality theorem 22-9 => c T x is bounded from below, a contradiction. Therefore case (3) can only occur at positions (2,3) and (3,2) An example for (3) (P) min x s.t. x + x 2!, - x - x 2!, x, x 2! 0 => (P) has no feasible solution (D) max π + π 2 s.t. π - π 2 ", π - π 2 " 0, π, π 2! 0! 2! T b! => π T b is unbounded So only entry (3,3) remains. This case can occur
6 4. Duality of LPs and the duality theorem 22-0 An example for (2) (P) min x s.t. x + x 2!, - x - x 2!, x, x 2 unconstrained x 2 x => (P) has no feasible solutions (D) max π + π 2 s.t. π - π 2 =, π - π 2 = 0, π, π 2! 0 => (D) has no feasible solution! The transportation problem and its dual Hitchcock problem or transportation problem (Hitchcock 94) is a special minimum cost flow problem, see ADM 4. Duality of LPs and the duality theorem 22- I supply in A demand in B A u =! B G "bipartite" We want to transport a good (oil, grain, coal) at minimum cost from the supply locations to the demand locations Vertex i A (i =,..., m) supplies a i units Vertex j B (j =,..., n) demands b j units, total supply = total demand. Edges (i,j) A x B have cost c ij per transported unit and infinite capacity u ij An LP formulation for the transportation problem x ij = number of units transported from i to j min # i,j c ij x ij s.t. # j x ij = a i for all i (pick up supply a i from vertex i)
7 4. Duality of LPs and the duality theorem 22-2 # i x ij = b j for all j (deliver demand b j to vertex j ) x ij! 0 for all i, j The associated matrix A of coefficients has the form i =,...,m j =,...,n!!!!!" " " "!!#! "!! "" " " "! "# " " "! $!! $" " " "! $#!! " " "! # # " " " # " " " # # " " " # # # " " " #!! " " "! " " " # # " " " # $ $$ $ $$ $ $$ $ $$ # # " " " # # # " " " # " " "!! " " "!! # " " " #! # " " " # " " "! # " " " # #! " " " # #! " " " # " " " #! " " " # $ $$ $ $$ $ $$ $ $$ # # " " "! # # " " "! " " " # # " " "! The dual of the transportation problem Introduce dual variables u i, v j for the constraints as follows u i v j - # j x ij = - a i for all i # i x ij = b j for all j The dual LP reads 4. Duality of LPs and the duality theorem 22-3 max # i - a i u i + # j b j v j s.t. - u i + v j " c ij for all i, j u i, v j unconstrained Interpretation of the dual LP "Dual" entrepreneur offers to do the transportation for pairs (i,j) He can buy the supply a i at location i from the primal entrepreneur, transport it to j and sell it there u i v j v j - u i = price to buy a unit of the good at vertex i = returns per unit at vertex j = profit per unit bought in i and sold in j v j - u i " c ij dual entrepreneur must stay below primal transportation cost in order to get the transport (i,j) from the primal entrepreneur (otherwise primal entrepreneur will do it himself) Dual entrepreneur wants to maximize his total profit # j b j v j - # i a i u i under these conditions The dual of the diet problem The primal problem (see example 3.) min c T x s.t. Ax! r
8 4. Duality of LPs and the duality theorem 22-4 x! 0 The associated dual problem max π T r s.t. π T A " c T π T! 0 Interpretation The dual entrepreneur makes nutrient pills for each of the m ingredients (magnesium, vitamin C,...) He asks the price π i per unit of nutrient i π T A j " c j <=> the total price of all pills substituting one unit of food j must not exceed the price c j of one unit of food j (pills will not be bought otherwise) max π T r <=> maximizing total profit of the dual entrepreneur Dual LPs often have a natural interpretation in practice 4.2 Complementary slackness 23- Complementary slackness provides simple necessary and sufficient conditions for optimality of a pair of primal feasible and dual feasible solutions. They have far reaching consequences for the design of algorithms (primaldual algorithms, primal-dual approximation algorithms) 4.4 Theorem (Complementary slackness) Let x be a primal feasible solution and π be a dual feasible solution. The following statements are equivalent: x, π are optimal (in the primal and the dual, respectively) u i := π i (a T i x - b i ) = 0 for all i =,..., m (4.8) v j := (c j - π T A j ) x j = 0 for all j =,..., n (4.9) i.e.,: (slack of primal or dual constraint) (value of associated dual or primal variable) = 0 Proof u i! 0 since a T i x - b i = 0 => u i = 0 a T i x - b i! 0 => π i! 0 => u i! 0 v j! 0
9 4.2 Complementary slackness 23-2 since x j unconstrained => π T A j = c j => v j = 0 x j! 0 => π T A j " c j => v j! 0 Set u := # i u i, v := # j v j => u, v! 0. Then u = 0 <=> (4.8) holds v = 0 <=> (4.9) holds Then u + v = # i π i (a i T x - b i ) + # j (c j - π T A j ) x j = - # i π i b i + # j c j x j + # i π i a T i x - # j π T A j x j = - π T b + c T x + (π T A)x - π T (Ax) = - π T b + c T x Hence: u + v = - π T b + c T x Suppose (4.8) and (4.9) hold => u + v = 0 => c T x = π T b Weak Duality Theorem => x, π are optimal Suppose that x and π are optimal Strong Duality Theorem => c T x = π T b => u + v = 0 => (4.8) and (4.9)! 4.2 Complementary slackness 23-3
10 4.3 The shortest path problem and its dual 24- The shortest path problem as primal LP Shortest Path Problem (SP) Instance Digraph G Rational edge weights c(e), e E(G) Vertices s, t V(G) Task Determine an elementary s,t-path P of minimum weight c(p) (shortest s,t-path)!!"" # # $!""!!#"c(w) = e E(P) c(e) (SP) is an instance of (LP) 4.3 The shortest path problem and its dual 24-2 The vertex-edge-incidence matrix A = (a ij ) of G is defined as #$ %&''( if i e j! "#!"!$ %&''( if e j i ) (*+(, otherwise where V(G) = {,..., n } and E(G) = { e,..., e m } Example G a A!!! "! #! $! % e e 4 "!! & & & s e 2 e 3 b e 5 t # & & &!!!! $!! &!! & % &!!!! &! The vertex-edge-incidence matrix of a digraph has per column exactly one, exactly one -, and 0 otherwise => sum of rows is 0 => rank(a) < n
11 4.3 The shortest path problem and its dual 24-3 Later: rank(a) = n- if G is connected (in the undirected sense) Let f j be a variable representing the amount of flow on edge e j, and let f := (f,..., f m ) T Flow conservation in node i is then expressed as a i T f = v 4 inflow in v = 5 = outflow from v An s,t-path is a flow of flow value from s to t (all f j = on the path and 0 otherwise) => every s,t-path is a solution of the linear system Af!" =! b # with "#$ #! %$!$ & ' & row s row t flow conservation with v = Of course, this linear system has also solutions that do not correspond to s,t-paths. But we have 4.3 The shortest path problem and its dual Lemma () If min c T f Af = b f! 0 has an optimal solution, then also one with f j { 0, }. Every such solution corresponds to an s,t-path (2) The simplex algorithm finds such a solution Proof: () follows from the algorithm for minimum cost s,t-flows in ADM I (2) can easily be shown directly, but follows also from the fact that matrix A is totally unimodular and b is integer. Then all basic feasible solutions of the LP are integer. We will show this more general result in Chapter 7.2.! Solving (SP) with the simplex algorithm We formulate (SP) as (LP)
12 4.3 The shortest path problem and its dual 24-5 min c T f Af = b (A = vertex-edge-incidence matrix ) f! 0 and solve it with the simplex algorithm. Since rank(a) < n, we may delete a row => delete the row for vertex t, this yields b! 0 In the example we obtain the following tableau for cost vector c = (, 2, 2, 3, ) Initial tableau, not yet transformed w.r.t. a basis, and graph G with edge costs!!! "! #! $! %! " " #! "!!! & & & # &!! &!! & $ & &!!!! &! G s 2 a 2 b 3 t Choose {, 4, 5 } as basis and transform the tableau w.r.t. that basis. Interpret the associated basic feasible solution in the graph. 4.3 The shortest path problem and its dual 24-6!!! "! #! $! %!$ &!! & & &!!!!! & & &! $! &!!! &! % & &!!!! &! s a e e 4 3 t e 5 b f5 = 0 The basic solution has n- = V - variables, but not every s,t-path has so many edges => many basic feasible solutions are degenerate (a common phenomenon in combinatorial optimization problem) Next tableau and basic feasible solution in the graph!!! "! #! $! % a!# & &!! &!! &! &!!!! &! "! &!!! &! %! & & &!! s e e 2 2 e 5 b t => optimal solution found, shortest path has length 3 The dual of the shortest path problem
13 4.3 The shortest path problem and its dual 24-7 We formulate it w.r.t. the full tableau containing also the row for vertex t => dual variables π i correspond to a node potential in graph G Tableau in the example:!!! "! #! $! % " #! " " #! $ %!! & & & '! $ & & & &!!!!!! $ '!! &!! & & $ " &!!!! &! & Dual LP: max π s - π t π i - π j " c ij for all edges (i, j) E(G) π i unconstrained Interpretation of the dual LP Along any path 4.3 The shortest path problem and its dual 24-8 i k p... q t from i to t we have (π i - π k ) + (π k - π p ) (π q - π t ) = π i - π t " c ik " c kp " c qt => π i - π t " c ik + c kp c qt = length of the path from i to t Since this holds for every such path, π i - π t " length of a shortest path from i to t => max π s - π t is equivalent to finding the greatest lower bound for the length of a shortest path from s to t Complementary slackness conditions Path f and node potential π are primal-dual optimal <=> () f ij > 0 => π i - π j = c ij i.e., edge (i,j) lies on a shortest path => potential difference = cost (2) π i - π j < c ij => f ij = 0
14 4.3 The shortest path problem and its dual 24-9 i.e., potential difference < cost => edge (i,j) does not lie on a shortest path Interpretation: the lower bounds π i - π t are tight along any shortest path The cord model (for c ij! 0) edge (i, j) <-> cord with length c ij π i - π j π i - π j " c ij <-> pulling vertices i and j apart <-> pulling is bounded from above by length c ij max π s - π t <-> pull s and t apart as far as possible complementary slackness: the cords on shortest paths are the tight ones 4.3 The shortest path problem and its dual 24-0 Remarks Deleting the row for vertex t => we have no variable π t => the dual objective function is max π s But: edges (i,t) yield the dual constraint π i " c it, so that π s cannot get arbitrarily large. We obtains the same dual constraint π i " c it if we set π t = 0 (which we may do w.o.l.g. since we only have potential differences in the dual). Dijkstra's algorithm (ADM I) applied to the dual graph (in which the direction of all edges of G is reversed) iteratively computes the π i, where π t is set to 0.
15 4.4 Farkas' Lemma 25- This is a central and very useful lemma in duality theory. It has several variants also known as Theorems of the Alternative. Cones and projections The cone C(a,..., a m ) generated by a,..., a m Let a,..., a m R n (e.g. the rows of matrix A). The cone C(a,..., a m ) generated by a,..., a m is defined as!!" " # $ $ $ # " % # $%! & R ' " & % % ) ( " ( # ) ( # & $ (%" = set of non-negative linear combinations of a,..., a n a 2 C(a,a 2 ) a vectors in the green angle have a non-negative projection onto a und a Farkas' Lemma 25-2 The projection of y onto a α y cos α = a! " #!! # projection of y onto a =! cos α => the projection of y onto a is non-negative <=> y T a is non-negative 4.5 Theorem (Farkas' Lemma) Let a,..., a m R n and c R n. The following are equivalent () for all y R n : y T a i! 0 for all i =,...,m => y T c! 0 i.e., for all y : y has a non-negative projection onto each a i => y has a non-negative projection onto c (2) c C(a,..., a m ) i.e., c lies in the cone generated by a,..., a m
16 4.4 Farkas' Lemma 25-3 Proof () => (2) Consider the LP min c T y a i T y! 0 i =,...,m y unconstrained => y = 0 is a feasible solution of the LP The objective function is bounded from below since the constraints of the LP imply c T y! 0 because of (),. => LP has a finite optimal solution => the dual LP max 0 π T A j = c j π! 0 has a feasible solution => there are numbers π,..., π m! 0 with c = π T A = # i π i a i => c C(a,..., a m ) 4.4 Farkas' Lemma 25-4 (2) => () c C(a,..., a m ) => there are numbers π i! 0 with c = # i π i a i consider y with y T a i! 0 for all i =,...m => y T c = # i π i y T a i! # i π i 0 = 0! There are many equivalent formulations of Farkas' Lemma. Examples are (A) y (y T a i! 0 i => y T b! 0) <=> x! 0 with A T x = b (original version by Farkas 894) (B) y! 0 (y T a i! 0 i => y T b! 0) <=> x! 0 with A T x " b More in Chapter 7.5 An application of Farkas' Lemma: necessary conditions for the disjoint path problem Disjoint Path Problem Instance Undirected graph G Pairs of vertices { s, t },..., { s k, t k } Task Determine pairwise edge disjoint paths from s i to t i (i =,..., k)
17 4.4 Farkas' Lemma 25-5 An example: minimum cost embeddings of VPNs into the base net of Telekom 4.4 Farkas' Lemma 25-6 The decision version of the disjoint path problem is NP-complete. We therefore look for strong necessary and hopefully also sufficient criteria for the existence of a solution. Cut criterion Let H be the graph with V(H) := V(G) and E(H) := { { s, t },..., { s k, t k } }. A necessary condition for the existence of a solution is the cut criterion δ G (X) δ H (X) for all = X V(G) i.e., there are at least as many edges leaving X in G as there are pairs in H to be connected G H V-X X V-X X V-X The cut criterion is not sufficient 4.6 Example
18 4.4 Farkas' Lemma 25-7 G 2 4 H Cut criterion holds, but there is no solution Distance criterion Let dist G,z (s,t) be the length of a shortest path from s to t in G w.r.t. edge weights z(e)! 0, e E(G). An instance of the disjoint path problem fulfills the distance criterion :<=> for any choice of edge weights z(e)! 0, e E(G), &'!# (")!!" #" # )!*"!!"#" $!%" * $!(" The cut criterion reduces to the distance criterion for edge weights if e δ(x) z(e) := 0 otherwise 4.4 Farkas' Lemma Theorem (The distance criterion is necessary) The distance criterion is necessary and sufficient for the existence of a fractional solution of the disjoint path problem. In particular, it is necessary for the existence of a solution of a disjoint path problem Proof Consider the disjoint path problem as a cycle packing problem cycles = all elementary cycles in G + H that contain exactly one edge of H k := number of these cycles integer cycle packing = union of pairwise edge disjoint cycles that contain every edge of H in exactly one cycle (existence <=> feasibility of the disjoint path problem) fractional cycle packing = non-negative linear combination (of incidence vectors) of all these cycles such that the resulting vector has the value at the entries corresponding to the edges of H, and is at most at every entry corresponding to an edge of G. (they contain integer cycle packings as special case) Example 4.6 has the following cycles in the cycle packing problem
19 4.4 Farkas' Lemma 25-9 A formulation of the fractional cycle packing 4.4 Farkas' Lemma 25-0 Let M be the E(G)-cycle-incidence matrix, i.e., rows of M <-> edges of G columns of M <-> incidence vectors of all cycles of G + H M e,c = <=> e lies on cycle C Let N be the E(H)-cycle-incidence matrix, i.e., rows of N columns of N <-> edges of H <-> incidence vectors of all cycles of G + H N e,c = <=> e lies on cycle C Observe: every column of N contains exactly one => fractional cycle packing = π' R k with π'! 0, Mπ' ", Nπ' = Add slack variables to obtain a linear system and denote the enlarged vector again by π => fractional cycle packing = π R k+m (m = E(G) ) with π! 0, Mπ =, Nπ = Write it as M I Aπ =, π 0 with A = N 0 i.e., the all ones vector lies in the cone C(A,..., A k+m ) generated by the columns A j = of A
20 4.4 Farkas' Lemma 25- Applying Farkas' Lemma gives condition (3) Farkas' Lemma yields: there is such a vector π <=> for all y R E(G) + E(H) : y T A j! 0 for all j =,...,k+m => y T! 0 Partition y into (z,v) T, such that z corresponds to the rows of M (edges of G) and v to the rows of N (edges of H). We then get: y T A j! 0 => z i! 0 y T A j! 0 => z T M j + v T N j! 0 for columns A j of slack variables for the other columns A j Let C j be the cycle of column A j => C j decomposes into a path P j in G and an edge f from H Then z T M j = length z(p j ) of the path P j w.r.t. edge weights z(e) v T N j = edge weight v(f), where f is the edge of H lying on cycle C j Hence y T A j! 0 => z(p j ) + v(f)! 0 for all cycles C j containing edge f So z(p j ) + v(f)! 0 is equivalent to 4.4 Farkas' Lemma 25-2 dist G,z (s,t) + v(f)! 0 with f = { s, t } () The constraint y T! 0 becomes # e! E(G) z(e) + # e! v(e)! 0 (2) E(H) Farkas' Lemma then yields for arbitrary (z,v) (3) z(e)! 0, dist G,z (s,t) + v(f)! 0 for all edges f = { s, t } in H => # e! E(G) z(e) + # f! E(H) v(f)! 0 Condition (3) is equivalent to the distance criterion is (by proving that their negations are equivalent) (3) violated => distance criterion violated (3) violated => there are z, v with z(e)! 0, dist G,z (s,t) + v(f)! 0 for all edges f = { s, t } in H and # e! E(G) z(e) + # f! E(H) v(f) < 0 => 0 " # f! E(H) dist G,z (s,t) + # f! E(H) v(f) < # f! E(H) dist G,z (s,t) - # e! E(G) z(e) => # e! E(G) z(e) < # f! E(H) dist G,z (s,t) => distance criterion violated distance criterion violated => (3) violated
21 4.4 Farkas' Lemma 25-3 distance criterion violated => there is z! 0 with # e! E(G) z(e) < # f! E(H) dist G,z (s,t) choose v(f) := - dist G,z (s,t) for edge f = {s, t} in H => dist G,z (s,t) + v(f)! 0 for all edges f = { s, t } in H and # e! E(G) z(e) + # f! E(H) v(f) = # e! E(G) z(e) - # f! E(H) dist G,z (s,t) < 0 => (3) is violated! The distance criterion is stronger than the cut criterion Example 4.6 does not fulfill the distance criterion G 2 4 H Set z(e) = for all e in G => # f = {s,t}! E(H) dist G,z (s,t) = 8, # e! E(G) z(e) = Farkas' Lemma 25-4 The distance criterion is not sufficient for the existence of a solution of the disjoint path problem The instance of the disjoint path problem G 2 H 2 A fractional cycle packing! +! +! +! So the distance criterion holds because of Theorem 4.7 There is no solution for the disjoint path problem
22 4.5 Dual information in the tableau 26- How to get dual information from the optimal primal tableau? Suppose w.o.l.g. that the initial tableau (possibly with artificial variables from Phase I) has columns,...,m as basic columns and that the tableau is transformed w.r.t. to this basis... Then the following properties hold in the optimal tableau with basis B rows,...,m are obtained from the initial tableau by multiplying it with B - from the left the reduced cost are obtained as! " "! "! # $ % " " # $%&&#' where π is an optimal solution of the dual problem (Proof of the Strong Duality Theorem) In columns,...,m (which are unit vectors in the initial tableau) we get! " "! "! # $ % " "! "! # " #$&%%& Hence an optimal dual solution is obtained from the optimal tableau of the primal as 4.5 Dual information in the tableau 26-2! "! # "! "# " #"! $$ % % % $ &% #&%$'% Observe: this holds for the dual problem of the initial tableau (and not for dual versions of other, equivalent primal formulations). Moreover, the first m columns contain B - = B - I (4.3) c j -! j B Example (Example for the Two-Phase-Method continued) Initial tableau
23 4.5 Dual information in the tableau 26-3! "!! " "! " #!!! "! #! $! %!# & & & &!!!!!!$ &!!! & & & & &! "!!! & & # "! & &! " " # &! & %!!! &! " # $ & &! " "! &! Optimal tableau! "!! " "! " #!!! "! #! $! %!#!&$" %$"!!!! #$" ' #$" ' '!% '!!! ' ' ' ' '! "!$"!$" ' ' #$"!!$" ' '! $ %$"!!$"! ' ($" '!$"! '! % #$"!%$" '!!!!$" '!!$" '! (4.2) gives π = 0-5/2 = - 5/2 π 2 = 0 - (- ) = π 3 = 0 - (- ) = 4.5 Dual information in the tableau 26-4 for the values of the dual variables w.r.t. the dual problem obtained from the primal formulation with artificial a variables x i 4.9 Example (Example for the shortest path problem continued) Solving the primal problem Initial tableau has no identity matrix, but 2 unit vectors => add one artificial variable in Phase I! " #! # " # # # $ # %!$! & & & & &!% &! " " #! &! "!!!! & & & " # $ & &!! &!! & ' # % & & &!!!! &! Transform cost coefficients of ξ and z to reduced form (must become 0 for basic variables)
24 4.5 Dual information in the tableau 26-5! " #! # " # # # $ # %!$!! &!!!! & & &!% & & $ # & & & &! "!!!! & & & " # $ & &!! &!! & ' # % & & &!!!! &! Pivot step! " #! # " # # # $ # %!$ &! & & & & &!%!#!#! & & & & & # "!!!! & & & " # $ & &!! &!! & ' # %!!! &!! &! => ξ = 0 and x a is a non-basic variable => optimal w.r.t. z basic columns of the initial tableau Primal information (visualized in the graph) 4.5 Dual information in the tableau 26-6 s a e 4 e 2 2 e 5 b t the primal optimal solution displays the edges on the shortest path Dual information (obtained from the primal optimal tableau and displayed in the graph)! "! # $ %! "# $ %! #! $!%&! %! %! # '! "# '! %! #! %! &! # (! "# (! )! #! )! '! # π t = 0 since row t is not in the primal LP
25 4.5 Dual information in the tableau s 2 3 a 2 b 3 t 0 the dual solution displays the shortest distance from a vertex to t 4.6 The dual Simplex algorithm 27- Goal: use the primal tableau to solve the dual LP Characteristics of the dual LP The primal optimality condition!! " becomes a dual constraint => the primal simplex algorithm has a primal feasible solution and fulfills the dual constraint!! " only at termination when the optimum is reached This suggests the following characteristics for the dual simplex algorithm generate a sequence of dual feasible solutions establish primal feasibility only at termination when the optimum is reached Deriving the operations in the tableau Tableau X with basic solution
26 4.6 The dual Simplex algorithm 27-2!!! "! #! $! %!#! & & & &! "!!! & & &! $! & & &!!! #!!!! &! &!! => dual feasible, i.e.,!! " primal infeasible, i.e., x B! 0 Choose a pivot row r (instead of a pivot column) with x r0 < 0 (i.e., an infeasible entry x r0 < 0 in the primal basic solution) Choose a pivot column in row r the by considering entries x rj < 0 (as to obtain x r0! 0 after the pivot) Pivoting with x rs < 0 changes the cost row to!!" "!!"!! #"! #$!!$ " " #% & & & % ' 4.6 The dual Simplex algorithm 27-3 j s 0 x 0j x 0s must become 0 r x rj x rs must become To stay feasible in the dual, x 0j! 0 for all j!!"!!!$ for "#$%! #" %!! #"! #$ => choose column s is such a way that!!"! #$ " #$%!!!$! #$ "! #$ %!& $ " && ' ' ' & ( # Observe the symmetry with the primal simplex algorithm in particular: all x rj > 0 => dual LP has an unbounded objective function 4.0 Theorem (Interpretation of the dual simplex algorithm)
27 4.6 The dual Simplex algorithm 27-4 The dual simplex algorithm is the primal simplex algorithm applied to the primal formulation of the dual LP Proof: Check! 4. Example (Example for the shortest path problem continued) initial tableau, not yet transformed w.r.t. a basis and graph with costs!!! "! #! $! %! " " #! "!!! & & & # &!! &!! & $ & &!!!! &! G s 2 a 2 b 3 t choose B = { 2, 4, 3 } as basis and transform the tableau w.r.t. B, display the basic solution in the graph 4.6 The dual Simplex algorithm 27-5!!! "! #! $! %!#! & & & & => dual feasible! "!!! & & &! $! & & &!!! #!!!! &! &!! primal infeasible s a e e 3 e 2 b e 4 e 5 t the basic solution corresponds to the s,t-cut X = { s, a } ADM I: s,t-cuts are "dual structures" of s,t-flows. This is confirmed here by LP duality Choosing the pivot element r = 3 is the pivot row choosing the pivot column:!!"! #$ " #$%!!!$! #$ "! #$ %!& $ " && ' ' ' & ( # " #$%! & $& &! $& # "! j = j = 5 => s = 5 is the pivot column
28 4.6 The dual Simplex algorithm 27-6!!! "! #! $! %!#! & & & &! "!!! & & &! $! & & &!!! #!!!! &! &!! pivot operation!!! "! #! $! %!#! & & & &! "!!! & & &! $ &!! &!! &! %!! &!! &! s a e 4 e 2 2 e 5 b t => primal and dual feasible => optimal The dual optimal solution can be obtained from the inverse of the optimal basis as! "! # " $ $!" (Duality Theorem). The optimal basis is B = { 2, 4, 5 } with inverse 4.6 The dual Simplex algorithm!!! "! # # #! #! #! 27-7 So! "! # " $ $!"! #$% %% "& " ' ' ' " ' " ' "! #%% %% "& 3 s 2 3 a 2 b 3 t 0 Observe: in this case we could not obtain π and B - directly from the optimal tableau, since the dual LP is not the one constructed from the initial tableau with basis { 2, 4, 3 }, but the dual LP of Example 4.9.
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