Strong formulation for fixed charge network flow problem

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1 Strong formulation for fixed charge network flow problem Consider the network topology given by the graph G = (V, E) in the figure, where V = {a, b, c, d}. Each node represents a router of the network, and each arc (i, j) E represents a link. Next to each arc, the figure reports the bandwith (arc capacity) u ij > 0 in Mbps and the fixed cost c ij > 0 of using the arc. s a 4/1 b 5 6/3 3/3 6/2 3/1 c 3 3/1 4/2 2/1 d 1 There is a single source node s := a and a subset T V \ {s} of destinations. Next to each destination node we report its flow (bandwith) requirement b i. For the destination nodes i T, with i s, we assume b i 0. For the source node s, b s represents the total availability, that we assume equivalent to the sum of the requirements of all nodes i T, that is, b s = i T b i. The aim is to determine the minimum cost feasible flow that satisfies the requirements of all destinations i T. 1) Give a mixed-integer linear program for the problem. 2) Solve the linear relaxation. Do the relaxed integer variables take values that are integer or close to integer? 3) Propose a stronger alternative formulation. [Suggestion: consider an independent flow for each destination node.] How does the solution to the linear relaxation change? The optimal solution of the linear relaxation in 2) is still feasible? Document prepared by L. Taccari 1

2 AMPL model (file network-i.mod) # SETS set V set E within {V,V} # PARAMETERS param s symbolic in V param u{(i,j) in E} >= 0 param c{(i,j) in E} >= 0 param b{i in V} AMPL runfile (file network.run) reset model network.mod data network.dat option solver cplexamp solve display x,y option relax_integrality 1 solve display{(i,j) in E} (x[i,j],y[i,j]) Data (file network.dat), optimal value: 6 data set V := a b c d param s := a param: E: u := a b 4 a c 6 a d 6 b c 3 c b 3 c d 4 d b 3 d c 2 param: c := a b 1 a c 2 a d 3 b c 3 Document prepared by L. Taccari 2

3 c b 1 c d 2 d b 1 d c 1 param: b := b 5 c 3 d 1 Document prepared by L. Taccari 3

4 Solution Formulation Sets V : nodes E V V : arcs T V : destination nodes Parameters s V \ T : source node u ij : capacity of the arc (i, j), with (i, j) E c ij : cost of the arc (i, j), with (i, j) E > 0 i T b i : flow requirements in node i, with b i = < 0 i = s 0 otherwise Decision variables x ij : flow sent over the arc (i, j), with (i, j) E y ij : binary variable indicating if (i, j) E is used Model For each node i V, let δ + (i) = {j V : (i, j) E}, δ (i) = {j V : (j, i) E}. min c ij y ij (value) (i,j) E s.t. x ji x ij = b i i V (balance) j δ (i) j δ + (i) x ij u ij y ij (i, j) E (capacity) x ij 0 (i, j) E (nonnegative var.) y ij {0, 1} (i, j) E (binary var.) Document prepared by L. Taccari 4

5 AMPL model (file network.mod) # SETS set V set E within {V,V} # PARAMETERS param s symbolic in V param u{(i,j) in E}, >=0 param c{(i,j) in E}, >= 0 param b{i in V} default if i=s then (- sum{j in V: j<>s} b[j]) # VARIABLES var x{(i,j) in E}, >=0 var y{(i,j) in E}, binary # OBJECTIVE FUNCTION minimize costi_fissi: sum{(i,j) in E} c[i,j]*y[i,j] # CONSTRAINTS subject to bilancio{i in V}: sum{j in V : (j,i) in E} x[j,i] - sum{j in V : (i,j) in E} x[i,j] = b[i] subject to capacita{(i,j) in E}: x[i,j] <= u[i,j]*y[i,j] AMPL runfile (file network.run) reset model network.mod data network.dat option solver cplexamp solve display x,y option relax_integrality 1 solve display{(i,j) in E} (x[i,j],y[i,j]) Document prepared by L. Taccari 5

6 Data (file network.dat), optimal value: 6 data set V := a b c d param s := a param: E: u := a b 4 a c 6 a d 6 b c 3 c b 3 c d 4 d b 3 d c 2 param: c := a b 1 a c 2 a d 3 b c 3 c b 1 c d 2 d b 1 d c 1 param: b := b 5 c 3 d 1 Soluzione CPLEX : optimal integer solution objective 6 7 MIP simplex iterations 0 branch-and-bound nodes : x y := a b 3 1 a c 6 1 a d 0 0 b c 0 0 c b 2 1 c d 1 1 d b 0 0 d c 0 0 Document prepared by L. Taccari 6

7 CPLEX : optimal solution objective dual simplex iterations (0 in phase I) : x[i,j] y[i,j] := a b 4 1 a c a d b c 0 0 c b c d 0 0 d b 0 0 d c 0 0 Document prepared by L. Taccari 7

8 Multi-commodity formulation Sets V : nodes E V V : arcs T V : destination nodes Parameters s V \ T : source node u ij : capacity of the arc (i, j), with (i, j) E c ij : cost of the arc (i, j), with (i, j) E > 0 i T b i : flow requirement in node i, with b i = < 0 i = s 0 otherwise b i i = k d k i : flow requirement directed to k in node i, that is dk i = b k i = s 0 otherwise Decision variables x k ij : flow sent on the arc (i, j) for the destination node k T y ij : binary variable indicating if arc (i, j) E is used Model min s.t. c ij y ij (value) (i,j) E x k ji x k ij = d k i i V, k T (balance) j δ (i) j δ + (i) x k ij min{u ij, b k }y ij (i, j) E, k T (capacity restricted to k) x k ij u ij (i, j) E (capacity) k T x k ij 0 (i, j) E, k T (nonnegative var.) y ij {0, 1} (i, j) E (binary var.) Document prepared by L. Taccari 8

9 Since for each arc (i, j) E the total flow quantity through it is k T xk ij, we clearly have that x ij = k T xk ij i, j E, where x ij are the flow variables of the previous, single-flow formulation. The multi-commodity formulation is obtained extending the space of the decision variables of the single-flow formulation. Document prepared by L. Taccari 9

10 AMPL model (file network-2.mod) # SETS set V set E within {V,V} set T within V # PARAMETERS param s symbolic in V param u{(i,j) in E}, >=0 param c{(i,j) in E}, >= 0 param b{i in V} default if i=s then (- sum{j in V: j<>s} b[j]) param d{i in V, k in T} default if i<>s and i=k then b[i] else if i=s then -b[k] else 0 # VARIABLES var y{(i,j) in E}, binary var z{(i,j) in E, k in T}, >=0, <=u[i,j] # OBJECTIVE FUNCTION minimize costi_fissi: sum{(i,j) in E} c[i,j]*y[i,j] # CONSTRAINTS subject to bilancio{i in V, k in T}: sum{(j,i) in E} z[j,i,k] - sum{(i,j) in E} z[i,j,k] = d[i,k] subject to capacita{(i,j) in E, k in T}: z[i,j,k] <= min(u[i,j],b[k])*y[i,j] subject to flusso_totale{(i,j) in E}: sum{k in T} z[i,j,k] <= u[i,j] AMPL runfile (file network-2.run) reset model network-2.mod data network-2.dat option solver cplexamp solve Document prepared by L. Taccari 10

11 display{k in T}:{(i,j) in E} z[i,j,k] display{(i,j) in E} (sum{k in T} z[i,j,k],y[i,j]) option relax_integrality 1 solve display{k in T}:{(i,j) in E} z[i,j,k] display{(i,j) in E} (sum{k in T} z[i,j,k],y[i,j]) data Data (file network-2.dat), optimal value: 6 set V := a b c d set T := b c d param s := a param: E: u := a b 4 a c 6 a d 6 b c 3 c b 3 c d 4 d b 3 d c 2 param: c := a b 1 a c 2 a d 3 b c 3 c b 1 c d 2 d b 1 d c 1 param: b := b 5 c 3 d 1 Soluzione CPLEX : optimal integer solution objective 6 Document prepared by L. Taccari 11

12 11 MIP simplex iterations 0 branch-and-bound nodes z[i,j, b ] := a b 3 a c 2 a d 0 b c 0 c b 2 c d 0 d b 0 d c 0 z[i,j, c ] := a b 0 a c 3 a d 0 b c 0 c b 0 c d 0 d b 0 d c 0 z[i,j, d ] := a b 0 a c 1 a d 0 b c 0 c b 0 c d 1 d b 0 d c 0 : sum{k in T} z[i,j,k] y[i,j] := a b 3 1 a c 6 1 a d 0 0 b c 0 0 c b 2 1 c d 1 1 d b 0 0 d c 0 0 CPLEX : optimal solution objective dual simplex iterations (0 in phase I) z[i,j, b ] := a b 4 a c 1 a d 0 b c 0 c b 1 Document prepared by L. Taccari 12

13 c d 0 d b 0 d c 0 z[i,j, c ] := a b 0 a c 3 a d 0 b c 0 c b 0 c d 0 d b 0 d c 0 z[i,j, d ] := a b 0 a c 1 a d 0 b c 0 c b 0 c d 1 d b 0 d c 0 : sum{k in T} z[i,j,k] y[i,j] := a b 4 1 a c 5 1 a d 0 0 b c 0 0 c b c d 1 1 d b 0 0 d c 0 0 The multi-commodity formulation is stronger than the initial one, as it is possible to notice from the optimal value of the linear relaxation, which is significantly closer to the optimal value of the original problem. The solution obtained with the relaxation of point 2) is not feasible for the new model, since the objective function value in that case is strictly smaller than the one of the multi-commodity formulation. Considering the multi-commodity formulation, we can observe that the constraints z k ij min{u ij, b k }y ij (i, j) E, k T force the variable y ij to take value 1 corresponding to partial flows zij k which are equivalent to the entire requirement b k of the destination k. As an example, the flow directed to d on the arc (a, c) is equal to 1, that is zac d = b d, therefore the capacity constraint implies y ac = 1. In Document prepared by L. Taccari 13

14 the solution to question 2), on the other hand, we have y ac = 2/3, not feasible for the new formulation. Document prepared by L. Taccari 14

Modeling Network Optimization Problems

Modeling Network Optimization Problems John E. Mitchell http://www.rpi.edu/~mitchj Mathematical Models of Operations Research MATP4700 / ISYE4770 October 19, 01 Mitchell (MATP4700) Network Optimization