Computational testing of exact separation for mixed-integer knapsack problems

Transcription

1 Computational testing of exact separation for mixed-integer knapsack problems Pasquale Avella (joint work with Maurizio Boccia and Igor Vasiliev ) DING - Università del Sannio Russian Academy of Sciences - Siberian Branch MIP Columbia University

2 MIP solvers & cutting planes MIP solvers include cut generation routines looking at single-row relaxations:

3 MIP solvers & cutting planes MIP solvers include cut generation routines looking at single-row relaxations: Knapsack Lifted Cover Inequalities

4 MIP solvers & cutting planes MIP solvers include cut generation routines looking at single-row relaxations: Knapsack Lifted Cover Inequalities Mixed knapsack Mixed-Integer Rounding (MIR) inequalities

5 MIP solvers & cutting planes MIP solvers include cut generation routines looking at single-row relaxations: Knapsack Lifted Cover Inequalities Mixed knapsack Mixed-Integer Rounding (MIR) inequalities Tableau rows Gomory Cuts

6 Cuts outside the template paradigm Can we do anything more to tighten MIP formulations by looking at single-row relaxations?

7 Cuts outside the template paradigm Can we do anything more to tighten MIP formulations by looking at single-row relaxations? We can try to generate cuts outside the template paradigm (local cuts: Applegate, Bixby, Chvátal and Cook, 2000)

8 Cuts outside the template paradigm Can we do anything more to tighten MIP formulations by looking at single-row relaxations? We can try to generate cuts outside the template paradigm (local cuts: Applegate, Bixby, Chvátal and Cook, 2000) Local cuts proved to be successful for the TSP

9 Cuts outside the template paradigm Can we do anything more to tighten MIP formulations by looking at single-row relaxations? We can try to generate cuts outside the template paradigm (local cuts: Applegate, Bixby, Chvátal and Cook, 2000) Local cuts proved to be successful for the TSP Based on exact separation.

10 Exact separation Given: a polyhedron P R n and a point x R n.

11 Exact separation Given: a polyhedron P R n and a point x R n. A separation algorithm is said exact if it either guarantees to provide a valid inequality for P cutting off x or concludes that x P.

12 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u}

13 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u} The exact separation LP SEPLP(X K ): max ȳπ π 0 wπ π 0, w X K (1) 1π = 1 (2) π, π 0 0

14 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u} The exact separation LP SEPLP(X K ): max ȳπ π 0 wπ π 0, w X K (1) 1π = 1 (2) π, π 0 0 ȳ R n is the fractional point to cut-off.

15 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u} The exact separation LP SEPLP(X K ): max ȳπ π 0 wπ π 0, w X K (1) 1π = 1 (2) π, π 0 0 Inequalities (1) ensure that the inequality is satisfied from every feasible solution in X K.

16 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u} The exact separation LP SEPLP(X K ): (2) is a normalization constraint. max ȳπ π 0 wπ π 0, w X K (1) 1π = 1 (2) π, π 0 0

17 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u} The exact separation LP SEPLP(X K ): max ȳπ π 0 wπ π 0, w X K (1) 1π = 1 (2) π, π 0 0 Let π, π 0 be the optimal solution of SEPLP(X K ).

18 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u} The exact separation LP SEPLP(X K ): max ȳπ π 0 wπ π 0, w X K (1) 1π = 1 (2) π, π 0 0 The inequality π y π 0 is valid for conv(x K ).

19 Exact separation of valid inequalities for the knaspack polytope The knapsack set (Boyd, 1988) X K = {y Z n + : ay b, y u} The exact separation LP SEPLP(X K ): max ȳπ π 0 wπ π 0, w X K (1) 1π = 1 (2) π, π 0 0 Extreme points of SEPLP(X K ) are in one-to-one correspondence with the facets of conv(x K ).

20 Recent results Extension of the local cuts technique to MIP problems Espinoza (2006)

21 Recent results Extension of the local cuts technique to MIP problems Espinoza (2006) MIPLIB instances Kaparis and Letchford (2007) yielded tighter lower bounds for several MIPLIB instances

22 Recent results Extension of the local cuts technique to MIP problems Espinoza (2006) MIPLIB instances Kaparis and Letchford (2007) yielded tighter lower bounds for several MIPLIB instances Generalized Assignment problem Medium-size Generalized Assignment instances d10200 and d20200 solved to optimality for the first time. Integrality gap reduced on many larger benchmark instances (up to 80x1600) (A., Boccia and Vasilyev, 2007).

23 Recent results (cont.) Single Source Capacitated Facility Location Problems Reformulation based on dicut inequalities + exact separation (Boccia, 2007). Many benchmark instances solved to optimality (MIP solvers failed).

24 Recent results (cont.) Single Source Capacitated Facility Location Problems Reformulation based on dicut inequalities + exact separation (Boccia, 2007). Many benchmark instances solved to optimality (MIP solvers failed). Set Covering Exact separation for subsets of formulation constraints (A., Boccia and Vasyliev, 2007). seymour solved to optimality on a single workstation.

25 A step further: the mixed-integer knapsack set X MI We consider single-row mixed-integer knapsack relaxations of MIP problems: X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v}

26 A step further: the mixed-integer knapsack set X MI We consider single-row mixed-integer knapsack relaxations of MIP problems: X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v} Atamturk (2002) studied the polyhedral structure of conv(x MI ).

27 A step further: the mixed-integer knapsack set X MI We consider single-row mixed-integer knapsack relaxations of MIP problems: X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v} Atamturk (2002) studied the polyhedral structure of conv(x MI ). Fukasawa and Goycoolea (2007) proposed an exact separation routine for X MI. The core of their separation procedure is a sophisticated Branch-and-Bound algorithm for the mixed-integer knapsack problem.

28 The knapsack set with a single continuous variable X MK If in X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v} we remove bounds v and aggregate the continuous variables we get the weaker knapsack set with a single continuous variable X MK : X MK = {(y, s) Z n + R + : ay s b, y u}

29 The knapsack set with a single continuous variable X MK If in X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v} we remove bounds v and aggregate the continuous variables we get the weaker knapsack set with a single continuous variable X MK : X MK = {(y, s) Z n + R + : ay s b, y u} Why we focus on X MK The set X MK is a better candidate for a lightweight exact separation routine.

30 A few remarks on conv(x MK ) The polyhedron conv(x MK ) was investigated by Marchand and Wolsey (1999)

31 A few remarks on conv(x MK ) The polyhedron conv(x MK ) was investigated by Marchand and Wolsey (1999) They showed that Mixed-Integer Rounding (MIR) inequalities n ( a j + (f a j f b ) + ) x j b + s 1 f b 1 f b j=1 (where f d = d d ) can be easily derived from X MK.

32 A few remarks on conv(x MK ) The polyhedron conv(x MK ) was investigated by Marchand and Wolsey (1999) They showed that Mixed-Integer Rounding (MIR) inequalities n ( a j + (f a j f b ) + ) x j b + s 1 f b 1 f b j=1 (where f d = d d ) can be easily derived from X MK. They characterized several other classes of valid inequalities for conv(x MK )

33 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: with π, σ and π 0 nonnegative. πy σs π 0,

34 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: πy σs π 0, with π, σ and π 0 nonnegative. Solve SEPLP(X MK ): max ȳπ sσ π 0 wπ tσ π 0, (w, t) X MK (3) 1π + σ = 1 (4) π, σ, π 0 0

35 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: πy σs π 0, with π, σ and π 0 nonnegative. Solve SEPLP(X MK ): max ȳπ sσ π 0 wπ tσ π 0, (w, t) X MK (3) 1π + σ = 1 (4) π, σ, π 0 0 (ȳ, s) R n is the fractional point to cut-off.

36 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: πy σs π 0, with π, σ and π 0 nonnegative. Solve SEPLP(X MK ): max ȳπ sσ π 0 wπ tσ π 0, (w, t) X MK (3) 1π + σ = 1 (4) π, σ, π 0 0 Inequalities (3) ensure that the inequality is satisfied from every feasible solution in X MK.

37 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: πy σs π 0, with π, σ and π 0 nonnegative. Solve SEPLP(X MK ): max ȳπ sσ π 0 wπ tσ π 0, (w, t) X MK (3) 1π + σ = 1 (4) π, σ, π 0 0 (4) is a normalization constraint.

38 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: πy σs π 0, with π, σ and π 0 nonnegative. Solve SEPLP(X MK ): max ȳπ sσ π 0 wπ tσ π 0, (w, t) X MK (3) 1π + σ = 1 (4) π, σ, π 0 0 Let π, σ, π 0 be the optimal solution of SEPLP(X MK ).

39 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: πy σs π 0, with π, σ and π 0 nonnegative. Solve SEPLP(X MK ): max ȳπ sσ π 0 wπ tσ π 0, (w, t) X MK (3) 1π + σ = 1 (4) π, σ, π 0 0 The inequality π y σ s π 0 is valid for conv(x MK ).

40 Exact separation for conv(x MK ) Any valid inequality for conv(x MK ) has the form: πy σs π 0, with π, σ and π 0 nonnegative. Solve SEPLP(X MK ): max ȳπ sσ π 0 wπ tσ π 0, (w, t) X MK (3) 1π + σ = 1 (4) π, σ, π 0 0 Extreme points of SEPLP(X MK ) are in one-to-one correspondence with the facets of conv(x MK ).

41 Solving SEPLP(X MK ) by row generation Step 1 Let S X MK be a subset of the feasible solutions.

42 Solving SEPLP(X MK ) by row generation Step 1 Let S X MK be a subset of the feasible solutions. Step 2 Solve the partial separation problem SEPLP(S): max ȳπ sσ π 0 wπ tσ π 0, π + σ = 1 π, π 0 0 (w, t) S Let (π, σ, π 0 ) be the optimal solution of SEPLP(S).

43 Solving SEPLP(X MK ) by row generation Step 1 Let S X MK be a subset of the feasible solutions. Step 2 Solve the partial separation problem SEPLP(S): max ȳπ sσ π 0 wπ tσ π 0, π + σ = 1 π, π 0 0 (w, t) S Let (π, σ, π 0 ) be the optimal solution of SEPLP(S). Step 3 Solve the mixed-integer knapsack problem MKNAP max π w σ t (w, t) X MK to check whether the candidate inequality π y σ s π 0 is valid for conv(x MK ).

44 Solving SEPLP(X MK ) by row generation (cont.) Step 4 Let (ŵ,ˆt) be the optimal solution of MKNAP. If π ŵ σ ˆt > π 0 then set S = S {(ŵ,ˆt)} and goto Step 1.

45 Solving SEPLP(X MK ) by row generation (cont.) Step 4 Let (ŵ,ˆt) be the optimal solution of MKNAP. If π ŵ σ ˆt > π 0 then set S = S {(ŵ,ˆt)} and goto Step 1. Step 5 (π, σ, π 0 ) is the optimal solution of SEPLP(X MK ) and the inequality π y σ s π 0 is valid for conv(x MK ).

46 Solving MKNAP efficiently

47 Solving MKNAP efficiently The mixed-integer knapsack problem MKNAP: max π w σ t aw t b w Z n t 0 must be solved repeatedly.

48 Solving MKNAP efficiently The mixed-integer knapsack problem MKNAP: max π w σ t aw t b w Z n t 0 must be solved repeatedly. We need a very efficient algorithm.

49 Solving MKNAP efficiently The mixed-integer knapsack problem MKNAP: max π w σ t aw t b w Z n t 0 must be solved repeatedly. We need a very efficient algorithm. Proposition For any optimal solution (ŵ,ˆt) of MKNAP we have ˆt = max(0, aŵ b).

50 Solving MKNAP efficiently The mixed-integer knapsack problem MKNAP: max π w σ t aw t b w Z n t 0 must be solved repeatedly. We need a very efficient algorithm. Proposition For any optimal solution (ŵ,ˆt) of MKNAP we have ˆt = max(0, aŵ b). It follows that: (ˆt = 0) (ˆt = aŵ b > 0)

51 Solving MKNAP efficiently (cont.) Proposition The optimal solution of MKNAP is the best between the optimal solutions of the two following knapsack problems:

52 Solving MKNAP efficiently (cont.) Proposition The optimal solution of MKNAP is the best between the optimal solutions of the two following knapsack problems: KNAP1 (t = 0): max π w aw b w Z n

53 Solving MKNAP efficiently (cont.) Proposition The optimal solution of MKNAP is the best between the optimal solutions of the two following knapsack problems: KNAP1 (t = 0): max π w aw b w Z n KNAP2 (t = aw b): min ( σ a π )w aw b + 1 w Z n

54 Solving MKNAP efficiently (cont.) Proposition The optimal solution of MKNAP is the best between the optimal solutions of the two following knapsack problems: KNAP1 (t = 0): max π w aw b w Z n KNAP2 (t = aw b): min ( σ a π )w aw b + 1 w Z n!! Both the knapsack problems can be solved very fast by dynamic programming (Pisinger, 2004).

55 Implementation details When embedded into a cutting plane algorithm, SEPLP(X MK ) is applied to each row defining a mixed-integer knapsack set: X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v}. Some operations are required to put the row in the right form:

56 Implementation details When embedded into a cutting plane algorithm, SEPLP(X MK ) is applied to each row defining a mixed-integer knapsack set: X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v}. Some operations are required to put the row in the right form: Bound substitution: replace a subset of continuous variable by their simple or variable bounds.

57 Implementation details When embedded into a cutting plane algorithm, SEPLP(X MK ) is applied to each row defining a mixed-integer knapsack set: X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v}. Some operations are required to put the row in the right form: Bound substitution: replace a subset of continuous variable by their simple or variable bounds. Preprocessing: transform the mixed integer set X MI into the mixed-integer knapsack set X MK.

58 Implementation details When embedded into a cutting plane algorithm, SEPLP(X MK ) is applied to each row defining a mixed-integer knapsack set: X MI = {(y, x) Z n + R p + : ay + gx b, y u, x v}. Some operations are required to put the row in the right form: Bound substitution: replace a subset of continuous variable by their simple or variable bounds. Preprocessing: transform the mixed integer set X MI into the mixed-integer knapsack set X MK. Convert coefficients into integers (required to use dynamic programming)

59 Bound substitution Consider the mixed-integer set X MI = {(y, x) Z n + R p + : ay + gx b, y u, lx v}.

60 Bound substitution Consider the mixed-integer set X MI = {(y, x) Z n + R p + : ay + gx b, y u, lx v}. The MIP formulation can also include some additional variable bounds on the continuous variables.

61 Bound substitution Consider the mixed-integer set X MI = {(y, x) Z n + R p + : ay + gx b, y u, lx v}. The MIP formulation can also include some additional variable bounds on the continuous variables. Bound substitution consists of replacing some continuous variables by their respective simple/variable bounds. It is done heuristically by performing one of the following substitutions: x j = l j + x j ; x j = v j x j ; x j = l j y i + x j ; w j = ṽ j y k x j

62 Bound substitution Consider the mixed-integer set X MI = {(y, x) Z n + R p + : ay + gx b, y u, lx v}. The MIP formulation can also include some additional variable bounds on the continuous variables. Bound substitution consists of replacing some continuous variables by their respective simple/variable bounds. It is done heuristically by performing one of the following substitutions: x j = l j + x j ; x j = v j x j ; x j = l j y i + x j ; w j = ṽ j y k x j Let (ȳ, x) be the current fractional solution. The bound with smallest slack is selected for substitution. That is, let µ = min{ x j l j, v j x j, x j l j ȳ i, ṽ j ȳ k x j }.

63 Bound substitution Consider the mixed-integer set X MI = {(y, x) Z n + R p + : ay + gx b, y u, lx v}. The MIP formulation can also include some additional variable bounds on the continuous variables. Bound substitution consists of replacing some continuous variables by their respective simple/variable bounds. It is done heuristically by performing one of the following substitutions: x j = l j + x j ; x j = v j x j ; x j = l j y i + x j ; w j = ṽ j y k x j Let (ȳ, x) be the current fractional solution. The bound with smallest slack is selected for substitution. That is, let Let: µ = min{ x j l j, v j x j, x j l j ȳ i, ṽ j ȳ k x j }. x j = l j + x j v j x lj y i + x j ṽ j y k x j j if µ = x j l j if µ = v j x j if µ = x j l j if µ = ṽ j ȳ k x j

64 Preprocessing Let a i y i + i I j P g j x j b, with 0 y i u i j I and x j 0 j P, be the mixed-integer inequality after bound substitution.

65 Preprocessing Let a i y i + i I j P g j x j b, with 0 y i u i j I and x j 0 j P, be the mixed-integer inequality after bound substitution. All the continuous variables with positive coefficients can be discarded (Atamturk, 2000).

66 Preprocessing Let a i y i + i I j P g j x j b, with 0 y i u i j I and x j 0 j P, be the mixed-integer inequality after bound substitution. All the continuous variables with positive coefficients can be discarded (Atamturk, 2000). All the continuous variables with negative coefficients are aggregated into the same variable s: s = g j x j, j P where P = {j P : g j < 0}.

67 Preprocessing Let a i y i + i I j P g j x j b, with 0 y i u i j I and x j 0 j P, be the mixed-integer inequality after bound substitution. All the continuous variables with positive coefficients can be discarded (Atamturk, 2000). All the continuous variables with negative coefficients are aggregated into the same variable s: s = g j x j, j P where P = {j P : g j < 0}. All the integer variables with negative coefficients are complemented: { uj y j if a j < 0 y j = otherwise y j

68 Convert all the coefficients into integers The integer knapsack problems of MKNAP are solved by the dynamic programming algorithm of Pisinger (2001).

69 Convert all the coefficients into integers The integer knapsack problems of MKNAP are solved by the dynamic programming algorithm of Pisinger (2001). Dynamic programming is fast, but there is a price to pay: it requires that all the knapsack coefficients are integers.

70 Convert all the coefficients into integers The integer knapsack problems of MKNAP are solved by the dynamic programming algorithm of Pisinger (2001). Dynamic programming is fast, but there is a price to pay: it requires that all the knapsack coefficients are integers. The coefficients of the integer variables must be converted into suitably small integers before running exact separation.

71 Convert all the coefficients into integers The integer knapsack problems of MKNAP are solved by the dynamic programming algorithm of Pisinger (2001). Dynamic programming is fast, but there is a price to pay: it requires that all the knapsack coefficients are integers. The coefficients of the integer variables must be converted into suitably small integers before running exact separation. We adopt a brute-force approach: enumerate all the q N in the interval [1, 10 4 ], stopping when qb qb ε and qa j qa j ε for each j I. In our experiments we set ε = 10 5.

72 Convert all the coefficients into integers The integer knapsack problems of MKNAP are solved by the dynamic programming algorithm of Pisinger (2001). Dynamic programming is fast, but there is a price to pay: it requires that all the knapsack coefficients are integers. The coefficients of the integer variables must be converted into suitably small integers before running exact separation. We adopt a brute-force approach: enumerate all the q N in the interval [1, 10 4 ], stopping when qb qb ε and qa j qa j ε for each j I. In our experiments we set ε = If the procedure fails, we discard the inequality since too large coefficients may cause numerical problems.

73 Lifting Exact separation runs over the fractional support.

74 Lifting Exact separation runs over the fractional support. Then standard sequential lifting is used to get globally valid inequalities.

75 Lifting Exact separation runs over the fractional support. Then standard sequential lifting is used to get globally valid inequalities. Computing a lifting coefficient amounts to solve a knapsack problem with a single continuous variable. The problem can be solved by splitting into two integer knapsack problems.

76 Computational results Computational experiments were carried out on a 64bit Pentium Quad-core 2.6 GHz processor with 4 Gb RAM. The LP solver was Xpress 2007B.

77 Computational results Computational experiments were carried out on a 64bit Pentium Quad-core 2.6 GHz processor with 4 Gb RAM. The LP solver was Xpress 2007B. The test bed consists of all the MIPLIB 2003 mixed-integer instances and of the Mittleman instances bc1, bienst1, bienst2, binkar10 1, dano3-4, dano3-5. We set a limit of 300 CPU secs for the time spent in separation.

78 Computational results (cont.) We compare the lower bounds returned by exact separation with those provided by Mixed-Integer Rounding (MIR) inequalities

79 Computational results (cont.) We compare the lower bounds returned by exact separation with those provided by Mixed-Integer Rounding (MIR) inequalities We compare with the MIR separation procedure of K. Wolter (2007), embedded into SCIP version 1.00, linked to SoPlex and ZIMPL 2.07 (Achterberg, 2007).

80 Computational results (cont.) We compare the lower bounds returned by exact separation with those provided by Mixed-Integer Rounding (MIR) inequalities We compare with the MIR separation procedure of K. Wolter (2007), embedded into SCIP version 1.00, linked to SoPlex and ZIMPL 2.07 (Achterberg, 2007). We set SCIP parameters to perfom Wolter s procedure on single rows, i.e. to forbid constraint aggregation. Separation of Lifted Cover inequalities is enabled too.

81 Computational results (cont.) We compare the lower bounds returned by exact separation with those provided by Mixed-Integer Rounding (MIR) inequalities We compare with the MIR separation procedure of K. Wolter (2007), embedded into SCIP version 1.00, linked to SoPlex and ZIMPL 2.07 (Achterberg, 2007). We set SCIP parameters to perfom Wolter s procedure on single rows, i.e. to forbid constraint aggregation. Separation of Lifted Cover inequalities is enabled too. For simplicity of comparison, separation routines run on the original (i.e. not preprocessed) instances.

82 Computational results Name SCIP SCIP SCIP MK -SEP MK -SEP MK -SEP LB %Gap Time LB % Gap Time 10teams a1c1s aflow30a aflow40b arki atlanta-ip dano3mip danoint fiber fixnet gesa gesa2-o glass liu markshare markshare mas mas misc mkc modglob msc98-ip net nsrand-ipx roll swath timtab timtab tr vpm binkar bienst bienst dano dano rgn

83 The overall effect Some preliminary tests on non-trivial instances (Cplex 11.1) timtab1 After B&B nodes: the original formulation returned a relative gap of 12.30%. Using exact separation the relative gap is 10.23%.

84 The overall effect Some preliminary tests on non-trivial instances (Cplex 11.1) timtab1 After B&B nodes: the original formulation returned a relative gap of 12.30%. Using exact separation the relative gap is 10.23%. timtab2 After B&B nodes: with the original formulation the gap is 37.6%. Using exact separation the relative gap is 34.5%.

85 The overall effect Some preliminary tests on non-trivial instances (Cplex 11.1) timtab1 After B&B nodes: the original formulation returned a relative gap of 12.30%. Using exact separation the relative gap is 10.23%. timtab2 After B&B nodes: with the original formulation the gap is 37.6%. Using exact separation the relative gap is 34.5%. aflow40 Solved in nodes by Cplex 11.1 after exact separation. Solved in nodes on the original formulation.

86 The overall effect Some preliminary tests on non-trivial instances (Cplex 11.1) timtab1 After B&B nodes: the original formulation returned a relative gap of 12.30%. Using exact separation the relative gap is 10.23%. timtab2 After B&B nodes: with the original formulation the gap is 37.6%. Using exact separation the relative gap is 34.5%. aflow40 Solved in nodes by Cplex 11.1 after exact separation. Solved in nodes on the original formulation. tr12-30 After B&B nodes: with the original formulation the relative gap is 2.76%. Using exact separation the relative gap is 0.29%.

87 The overall effect Some preliminary tests on non-trivial instances (Cplex 11.1) timtab1 After B&B nodes: the original formulation returned a relative gap of 12.30%. Using exact separation the relative gap is 10.23%. timtab2 After B&B nodes: with the original formulation the gap is 37.6%. Using exact separation the relative gap is 34.5%. aflow40 Solved in nodes by Cplex 11.1 after exact separation. Solved in nodes on the original formulation. tr12-30 After B&B nodes: with the original formulation the relative gap is 2.76%. Using exact separation the relative gap is 0.29%. nsrand-ip After 5000 B&B nodes: with the original formulation the gap is 1.5%. Using exact separation the relative gap is 0.37%.

88 Some considerations Computational experience shows that exact separation for conv(x MK ) is effective in tightening MIP formulations.

89 Some considerations Computational experience shows that exact separation for conv(x MK ) is effective in tightening MIP formulations. Computation time is much larger than for MIR separation, but still reasonable when dealing with hard instances.

90 Some considerations Computational experience shows that exact separation for conv(x MK ) is effective in tightening MIP formulations. Computation time is much larger than for MIR separation, but still reasonable when dealing with hard instances. Exact separation not applicable to large and dense rows.

91 Back to the template paradigm: Mixed Knapsack Inequalities We focus on mixed knapsack inequalities (Marchand and Wolsey, 2002), which can described by the following procedure. Given: X BMK = {(y, s) B n + R + : ay s b, y u}

92 Back to the template paradigm: Mixed Knapsack Inequalities We focus on mixed knapsack inequalities (Marchand and Wolsey, 2002), which can described by the following procedure. Given: X BMK = {(y, s) B n + R + : ay s b, y u} i) Set the s = s;

93 Back to the template paradigm: Mixed Knapsack Inequalities We focus on mixed knapsack inequalities (Marchand and Wolsey, 2002), which can described by the following procedure. Given: X BMK = {(y, s) B n + R + : ay s b, y u} i) Set the s = s; ii) Find a valid inequality αy β for the resulting binary knapsack polytope; X BMK s = {y B n + : ay b + s, y u}

94 Back to the template paradigm: Mixed Knapsack Inequalities We focus on mixed knapsack inequalities (Marchand and Wolsey, 2002), which can described by the following procedure. Given: X BMK = {(y, s) B n + R + : ay s b, y u} i) Set the s = s; ii) Find a valid inequality αy β for the resulting binary knapsack polytope; X BMK s = {y B n + : ay b + s, y u} iii) lift the s to get a valid inequality for X BMK of the form αy γs β.

95 Mixed Knapsack Inequalities: lifting the s Let η(s) = max αy (5) ay b + s (6) y {0, 1} n (7)

96 Mixed Knapsack Inequalities: lifting the s Let η(s) = max αy (5) ay b + s (6) y {0, 1} n (7) Proposition The inequality αy β + γs is valid for conv(x BMK ) if η(s) β + γs for each s R +.

97 Mixed Knapsack Inequalities: lifting the s (cont.) A geometrical interpretation

98 Mixed Knapsack Inequalities: lifting the s (cont.) A geometrical interpretation η(s) is a step function

99 Mixed Knapsack Inequalities: lifting the s (cont.) A geometrical interpretation η(s) is a step function the line β + γs is a valid rhs if it defines an upper bound on the η(s), for each s R +.

100 Mixed Knapsack Inequalities: lifting the s (cont.)

101 Mixed Knapsack Inequalities: lifting the s (cont.) The lifting algorithm

102 Mixed Knapsack Inequalities: lifting the s (cont.) The lifting algorithm Step 0 Inizialize γ.

103 Mixed Knapsack Inequalities: lifting the s (cont.) The lifting algorithm Step 0 Inizialize γ. Step 1 Solve the problem: ζ = max αy γs αy b + s y {0, 1} N s 0

104 Mixed Knapsack Inequalities: lifting the s (cont.) The lifting algorithm Step 0 Inizialize γ. Step 1 Solve the problem: ζ = max αy γs αy b + s y {0, 1} N s 0 Step 2 If ζ β then the inequality αy β + γs is valid for conv(x BMK ). STOP.

105 Mixed Knapsack Inequalities: lifting the s (cont.) The lifting algorithm Step 0 Inizialize γ. Step 1 Solve the problem: ζ = max αy γs αy b + s y {0, 1} N s 0 Step 2 If ζ β then the inequality αy β + γs is valid for conv(x BMK ). STOP. Step 3 Increase γ and Go to Step 1.

106 Mixed Knapsack Inequalities: lifting the s (cont.) A numerical example

107 Mixed Knapsack Inequalities: lifting the s (cont.) A numerical example Consider the set X BMK = {7y 1 + 6y 2 + 5y 3 + 3y 4 + 2y 5 s 6}.

108 Mixed Knapsack Inequalities: lifting the s (cont.) A numerical example Consider the set X BMK = {7y 1 + 6y 2 + 5y 3 + 3y 4 + 2y 5 s 6}. Let X BMK 1 {x X BMK : s = 1}.

109 Mixed Knapsack Inequalities: lifting the s (cont.) A numerical example Consider the set X BMK = {7y 1 + 6y 2 + 5y 3 + 3y 4 + 2y 5 s 6}. Let X BMK 1 {x X BMK : s = 1}. The inequality y 1 + y 2 + y 3 + y 4 1 is valid for conv(x BMK 1 ).

110 Mixed Knapsack Inequalities: lifting the s (cont.) A numerical example Consider the set X BMK = {7y 1 + 6y 2 + 5y 3 + 3y 4 + 2y 5 s 6}. Let X BMK 1 {x X BMK : s = 1}. The inequality y 1 + y 2 + y 3 + y 4 1 is valid for conv(x BMK 1 ). η(s) step function.

111 Mixed Knapsack Inequalities: lifting the s (cont.) A numerical example Consider the set X BMK = {7y 1 + 6y 2 + 5y 3 + 3y 4 + 2y 5 s 6}. Let X BMK 1 {x X BMK : s = 1}. The inequality y 1 + y 2 + y 3 + y 4 1 is valid for conv(x BMK 1 ). Initialization: γ = 3/15.

112 Mixed Knapsack Inequalities: lifting the s (cont.) A numerical example Consider the set X BMK = {7y 1 + 6y 2 + 5y 3 + 3y 4 + 2y 5 s 6}. Let X BMK 1 {x X BMK : s = 1}. The inequality y 1 + y 2 + y 3 + y 4 1 is valid for conv(x BMK 1 ). Iteration 1: update γ = 1; y 1 + y 2 + y 3 + y 4 s 1 is valid.

113 Computational results for Mixed Knapsack Inequalities Name SCIP SCIP SCIP LCI LCI LCI LB %Gap Time LB % Gap Time 10teams a1c1s aflow30a aflow40b arki atlanta-ip dano3mip danoint fiber fixnet gesa gesa2-o glass liu markshare markshare mas mas misc mkc modglob msc98-ip net nsrand-ipx , roll swath timtab timtab tr vpm , binkar bienst bienst dano dano rgn

114 Research directions More efficient ways of solving the exact separation LP

115 Research directions More efficient ways of solving the exact separation LP How to generate new rows by constraint aggregation?

116 Research directions More efficient ways of solving the exact separation LP How to generate new rows by constraint aggregation? Looking at more complex MIP substructures than the single row.

117 Research directions More efficient ways of solving the exact separation LP How to generate new rows by constraint aggregation? Looking at more complex MIP substructures than the single row. How to select MIP substructures to ensure that exact separation leads to violated cuts?