![LP Problems with Integral Coefficients
• Consider the LP problem modeled with integral coefficients:
min
𝒙
𝑧 = −𝒄𝑇𝒙 , Subject to 𝑨𝒙 ≤ 𝒃, 𝒙 ≥ 𝟎,
where 𝑨 ∈ ℤ𝑚×𝑛, 𝑟𝑎𝑛𝑘 𝑨 = 𝑚, 𝒃 ∈ ℤ𝑚, 𝒄 ∈ ℤ𝑛
Include slack variables to write the constraint as: 𝑨𝒙 + 𝑰𝒔 = 𝒃;
• If 𝑨 is totally unimodular, so is [𝑨 𝑰]; hence if 𝑨 ∈ ℤ𝑚×𝑛 is totally
unimodular and 𝒃 ∈ ℤ𝑚, all BFSs to the problem are integral.
• Thus total unimodularity of 𝑨 is a sufficient condition for integral
solution to the LP problem modeled with integral coefficients.](https://image.slidesharecdn.com/optimumengineeringdesign-day4-240221164756-7cf50022/85/Optimum-Engineering-Design-Day-4-Clasical-methods-of-optimization-6-320.jpg)
![Example: Project Management
%project management example
A=[-1 1 0 0 0; 0 -1 1 0 0; -1 0 1 0 0; 0 -1 0 1 0; 0 0
-1 1 0; 0 0 0 -1 1];
b=[3;0;1;4;2;5];
c=[-1;0;0;0;1];
Opt.f=c;
Opt.Aineq=-A;
Opt.bineq=-b;
Opt.lb=zeros(size(c));
Opt.solver='linprog';
Opt.options=optimset('Display',‘final');
[x,fval]=linprog(Opt)
x =
0 3 3 7 12
fval =
12](https://image.slidesharecdn.com/optimumengineeringdesign-day4-240221164756-7cf50022/85/Optimum-Engineering-Design-Day-4-Clasical-methods-of-optimization-10-320.jpg)
![Example: Project Management
%Transportation example
A=[1 1 1 0 0 0; 0 0 0 1 1 1; 1 0 0 1 0 0; 0 1 0 0 1 0;
0 0 1 0 0 1];
b=[10;10;5;9;6];
c=[3;4;6;5;3;5];
Opt.f=c;
Opt.Aeq=A;
Opt.beq=b;
Opt.lb=zeros(size(c));
Opt.solver='linprog';
Opt.options=optimset('Display',‘final');
[x,fval]=linprog(Opt)
x =
5 5 0 0 4 6
fval =
77](https://image.slidesharecdn.com/optimumengineeringdesign-day4-240221164756-7cf50022/85/Optimum-Engineering-Design-Day-4-Clasical-methods-of-optimization-12-320.jpg)
Optimum Engineering Design - Day 4 - Clasical methods of optimization
- 1. Optimization Methods in Engineering Design Day-4
- 2. Course Materials • Arora, Introduction to Optimum Design, 3e, Elsevier, (https://www.researchgate.net/publication/273120102_Introductio n_to_Optimum_design) • Parkinson, Optimization Methods for Engineering Design, Brigham Young University (http://apmonitor.com/me575/index.php/Main/BookChapters) • Iqbal, Fundamental Engineering Optimization Methods, BookBoon (https://bookboon.com/en/fundamental-engineering-optimization- methods-ebook)
- 3. Discrete Optimization Problems • An linear integer programming (IP) problem is formulated as: max 𝒙 𝑧 = 𝒄𝑇𝒙, subject to 𝑨𝒙 ≤ 𝒃, 𝒙 ∈ ℤ𝑛, 𝒙 ≥ 𝟎 • A binary integer programming (BIP) problem is formulated as: max 𝒙 𝑧 = 𝒄𝑇 𝒙, subject to 𝑨𝒙 ≤ 𝒃, 𝒙 ∈ 0,1 𝑛 • A Mixed integer programming (MIP) problem is formulated as: max 𝒙 𝑧 = 𝒄𝑇 𝒙, subject to 𝑨𝒙 ≤ 𝒃, 𝑥𝑖 ≥ 0, 𝑥𝑖 ∈ ℤ, 𝑖 = 1, … , 𝑛𝑑; 𝑥𝑖𝐿 ≤ 𝑥𝑖 ≤ 𝑥𝑖𝑈, 𝑖 = 𝑛𝑑 + 1, … , 𝑛 • A general mixed design optimization problem is formulated as: min 𝒙 𝑓 𝒙 , subject to ℎ𝑖 𝒙 = 0, 𝑖 = 1, … , 𝑙; 𝑔𝑗 𝑥 ≤ 0, 𝑗 = 𝑖, … , 𝑚; 𝑥𝑖 ∈ 𝐷𝑖, 𝑖 = 1, … , 𝑛𝑑; 𝑥𝑖𝐿 ≤ 𝑥𝑖 ≤ 𝑥𝑖𝑈, 𝑖 = 𝑛𝑑 + 1, … , 𝑛
- 4. Enumeration Method • Discrete optimization problems may be solved by enumeration, i.e., through an ordered listing of all solutions. • The number of solution combinations to be evaluated is: 𝑁𝑐 = 𝑞𝑖 𝑛𝑑 𝑖=1 , where 𝑛𝑑 is the number of design variables and 𝑞𝑖 is the number of discrete values for the design variable 𝑥𝑖. Note, 𝑁𝑐 increases rapidly with increase in 𝑛𝑑 and 𝑞𝑖.
- 5. LP Problems with Integral Coefficients • Consider the standard LPP modeled with integral coefficients: min 𝒙 𝑧 = 𝒄𝑇𝒙 , Subject to 𝑨𝒙 = 𝒃, 𝒙 ≥ 𝟎, where 𝑨 ∈ ℤ𝑚×𝑛, 𝑟𝑎𝑛𝑘 𝑨 = 𝑚, 𝒃 ∈ ℤ𝑚, 𝒄 ∈ ℤ𝑛 • Assume that 𝑨 is totally unimodular, i.e., every square submatrix 𝑪 of 𝑨 has det 𝑪 ∈ 0, ±1 . Then every basis matrix 𝑩 has det 𝑩 = ± 1; hence 𝑩−1 = ±𝐴𝑑𝑗 𝑩, and all basic solutions: 𝒙𝐵 = 𝑩−1𝒃 are integral. • Thus a necessary condition for integral solutions to the problem is that every 𝑚 × 𝑚 submatrix 𝑩 of 𝑨 has det 𝑩 = ±1.
- 6. LP Problems with Integral Coefficients • Consider the LP problem modeled with integral coefficients: min 𝒙 𝑧 = −𝒄𝑇𝒙 , Subject to 𝑨𝒙 ≤ 𝒃, 𝒙 ≥ 𝟎, where 𝑨 ∈ ℤ𝑚×𝑛, 𝑟𝑎𝑛𝑘 𝑨 = 𝑚, 𝒃 ∈ ℤ𝑚, 𝒄 ∈ ℤ𝑛 Include slack variables to write the constraint as: 𝑨𝒙 + 𝑰𝒔 = 𝒃; • If 𝑨 is totally unimodular, so is [𝑨 𝑰]; hence if 𝑨 ∈ ℤ𝑚×𝑛 is totally unimodular and 𝒃 ∈ ℤ𝑚, all BFSs to the problem are integral. • Thus total unimodularity of 𝑨 is a sufficient condition for integral solution to the LP problem modeled with integral coefficients.
- 7. Example: Integer BFS • Consider the LP problem with integral coefficients: max 𝒙 𝑧 = 2𝑥1 + 3𝑥2 Subject to: 𝑥1 ≤ 3, 𝑥2 ≤ 5, 𝑥1 + 𝑥2 ≤ 7, 𝒙 ∈ ℤ2, 𝒙 ≥ 𝟎 • Following the introduction of slack variables, the constraints are given as: 𝑨 = 1 0 1 0 1 1 1 0 0 0 1 0 0 0 1 , 𝒃 = 3 5 7 Note, 𝑨 is totally unimodular and 𝒃 ∈ ℤ3. • Using the simplex method, the optimal integral solution is given as: 𝒙𝑇 = 2,5,1,0,0 , with 𝑧∗ = 19.
- 8. Example: Project Management • Problem: Given a list of project tasks with task completion times, what is the minimum time to complete the project? • The LP problem with integral coefficients is defined as: Objective: min 𝑡𝑗 𝑡5 − 𝑡1 Subject to: 𝑡2 − 𝑡1 ≥ 3, 𝑡3 − 𝑡1 ≥ 1, 𝑡3 − 𝑡2 ≥ 0, 𝑡4 − 𝑡2 ≥ 4, 𝑡4 − 𝑡3 ≥ 2, 𝑡5 − 𝑡4 ≥ 5, 𝑡𝑗 ≥ 0, 𝑗 = 1 − 5 2 A,3 1 3 4 5 C,4 D,2 B,1 E,5 0
- 9. Example: Project Management • Then 𝑨 = −1 1 −1 −1 1 1 −1 1 −1 1 −1 1 ; 𝒃 = 3 0 1 4 2 5 ; 𝒄𝑇 = −1 0 0 0 1 Note, A is totally unimodular. • The optimal solution to the LPP is given as: 𝑡1 = 0, 𝑡2 = 3, 𝑡3 = 3, 𝑡4 = 7, 𝑡5 = 12; 𝑓∗ = 12
- 10. Example: Project Management %project management example A=[-1 1 0 0 0; 0 -1 1 0 0; -1 0 1 0 0; 0 -1 0 1 0; 0 0 -1 1 0; 0 0 0 -1 1]; b=[3;0;1;4;2;5]; c=[-1;0;0;0;1]; Opt.f=c; Opt.Aineq=-A; Opt.bineq=-b; Opt.lb=zeros(size(c)); Opt.solver='linprog'; Opt.options=optimset('Display',‘final'); [x,fval]=linprog(Opt) x = 0 3 3 7 12 fval = 12
- 11. Example: Transportation Problem • Consider a transportation problem with 𝑚 = 2, 𝑛 = 3 nodes Objective: min xij 𝑐𝑖𝑗𝑥𝑖𝑗 𝑛 𝑗=1 𝑚 𝑖=1 , 𝑖 = 1,2; 𝑗 = 1,2,3 Subject to: 𝑥𝑖𝑗 𝑚 𝑖=1 = 𝑎𝑖, 𝑥𝑖𝑗 𝑛 𝑗=1 = 𝑏𝑗 Use following representation to represent a solution with the given capacities and costs: Note, every BFS has 𝑚 + 𝑛 − 1 = 4 nonzero components • For example, possible BFS are: 𝑥11 𝑥12 𝑥13 10 𝑥21 𝑥22 𝑥23 10 5 9 6 0 4 6 10 5 5 0 10 5 9 6 1 9 0 10 4 0 6 10 5 9 6 5 0 5 10 0 9 1 10 5 9 6 3 4 6 5 3 5
- 12. Example: Project Management %Transportation example A=[1 1 1 0 0 0; 0 0 0 1 1 1; 1 0 0 1 0 0; 0 1 0 0 1 0; 0 0 1 0 0 1]; b=[10;10;5;9;6]; c=[3;4;6;5;3;5]; Opt.f=c; Opt.Aeq=A; Opt.beq=b; Opt.lb=zeros(size(c)); Opt.solver='linprog'; Opt.options=optimset('Display',‘final'); [x,fval]=linprog(Opt) x = 5 5 0 0 4 6 fval = 77
- 13. Binary Integer Programming Problems • Most IP problems can be reformulated in the BIP framework. – Assuming the number of variables is small, let 𝑥𝑚𝑖𝑛 < 𝑥𝑖 < 𝑥𝑚𝑎𝑥, then each 𝑥𝑖 can be represented as a binary number using 𝑘 bits, where 2𝑘+1 ≥ 𝑥𝑚𝑎𝑥 − 𝑥𝑚𝑖𝑛. – The resulting optimization problem involves selection of bits and in a BIP problem. • BIP problems can be solved via implicit enumeration. – The search starts from 𝑥𝑖 = 0, 𝑖 = 1, … , 𝑛, which is optimal. If this is not feasible, then we systematically adjust individual variable values till feasibility is attained. – In this process, obvious infeasible solutions are eliminated and the remaining ones are evaluated to find the optimum.
- 14. Implicit Enumeration Algorithm • The algorithm is given as: 1. Initialize: set 𝑥𝑖 = 0, 𝑖 = 1, … , 𝑛; quit, if it is feasible. 2. For some 𝑖, set 𝑥𝑖 = 1. If the resulting solution is feasible, then record it if it is the first feasible solution or if it improves upon a previous feasible solution. 3. Backtrack (set 𝑥𝑖 = 0) if a feasible solution was reached in the previous step or if feasibility appears impossible in this branch. 4. Choose another 𝑖 and return to 2. • The progress of the algorithm is recorded in a decision-tree using nodes and arcs, with node 0 representing the initial solution (𝑥𝑖 = 0, 𝑖 = 1, … , 𝑛), and node 𝑖 representing a change in the value of variable 𝑥𝑖. An arc from node 𝑘 to node 𝑖 indicates that 𝑥𝑖 is raised to one.
- 15. Implicit Enumeration Algorithm • At node 𝑖 the following possibilities exist: – The resulting solution is feasible, hence no further improvement in this branch is possible. – Feasibility is impossible from this branch. – The resulting solution is not feasible, but feasibility or further improvements are possible. • In the first two cases, the branch is said to have been fathomed. If that happens, we then backtrack to node 𝑘, where variable 𝑥𝑖 is returned to zero. • The algorithm continues till all branches have been fathomed, and returns an optimum 0-1 solution.
- 16. Example: Implicit Enumeration • Problem: A machine shaft is to be cut at two locations to given dimensions using one of the two available processes, A and B. The following information on process cost and three-sigma standard deviation is available. The problem requires that the combined maximum allowable tolerance be limited to 12mils: • Let 𝑥𝑖, 𝑖 = 1, … , 4 denote the available processes for both jobs; let 𝑡𝑖 denote their tolerances. Then, the BIP problem is formulated as: Objective: min 𝒙 𝑧 = 65𝑥1 + 57𝑥2 + 42𝑥3 + 20𝑥4 Subject to: 𝑥1 + 𝑥2 = 1, 𝑥3 + 𝑥4 = 1, 𝑥𝑖𝑡𝑖 𝑖 ≤ 12, 𝑥𝑖 ∈ 0,1 ProcessJob Job1 Job2 Cost SD Cost SD Process A $65 ±4mils $57 ±5mils Process B $42 ±6mils $20 ±10mils
- 17. Example: Implicit Enumeration Objective: min 𝒙 𝑧 = 65𝑥1 + 57𝑥2 + 42𝑥3 + 20𝑥4 Subject to: 𝑥1 + 𝑥2 = 1, 𝑥3 + 𝑥4 = 1, 𝑥𝑖𝑡𝑖 𝑖 ≤ 12, 𝑥𝑖 ∈ 0,1 • The problem is solved using Implicit Enumeration algorithm 1 4 3 𝑓 = 107 NFI 0 NFI 𝑥3 = 1 𝑥4 = 1 2 4 3 𝑓 = 99 Optimum 𝑥3 = 1 𝑥4 = 1 NFI 𝑥2 = 1 𝑥1 = 1
- 18. Integer Programming Problems • Consider the IP problem: max 𝒙 𝑧 = 𝒄𝑇𝒙, Subject to 𝑨𝒙 ≤ 𝒃, 𝒙 ∈ ℤ𝑛, 𝒙 ≥ 𝟎 • The LP relaxation of the IP problem is defined as: max 𝒙 𝑧 = 𝒄𝑇 𝒙, Subject to 𝑨𝒙 ≤ 𝒃, 𝒙 ≥ 𝟎 The LP relaxation solution serves as upper bound on the IP solution • Two popular approaches to solve IP problems are: – Branch and bound method – Cutting plane method
- 19. Branch and Bound Method • Start from the LP relaxation solution; successively introduce the integrality constraints on the design variables – Each integrality constraint on variable 𝑥𝑖 introduces two new subprograms (branching) that can be solved via LP methods – Each subprogram results in one of the following: • There is no feasible solution. • The solution does not improve upon available IP solution. • An improved IP solution is returned and is recorded as current optimal. • The progress of the algorithm is recorded in a decision-tree • The algorithm ends when all branches have been fathomed
- 20. Example: Branch and Bound Method • Consider the following IP problem: Max 𝑧 = 60𝑥1 + 90𝑥2 + 120𝑥3, Subject to: 35𝑥1 + 60𝑥2 + 140𝑥3 ≤ 1000, 15𝑥1 + 30𝑥2 + 60𝑥3 ≥ 2000, 𝑥1 + 𝑥2 − 2𝑥3 ≤ 0; 𝒙 ∈ ℤ3, 𝒙 ≥ 𝟎 • 𝑆0: LP relaxation (𝐹0), 𝑥1 ∗ = 0, 𝑥2 ∗ = 7.69, 𝑥3 ∗ = 3.85, 𝑓∗ = 1153.8; – Record as upper bound • 𝑆1: 𝐹0 ∪ 𝑥3 ≤ 3 , integer solution: 𝑥1 ∗ = 0, 𝑥2 ∗ = 6, 𝑥3 ∗ = 3, 𝑓∗ = 900; – Record as current optimum • 𝑆2: 𝐹0 ∪ 𝑥3 ≥ 4 , non-integer solution: 𝑥1 ∗ = 1.6, 𝑥2 ∗ = 6.4, 𝑥3 ∗ = 4, 𝑓∗ = 1152; further improvement possible • 𝑆3: 𝐹2 ∪ 𝑥2 ≤ 6 , non-integer solution: 𝑥1 ∗ = 2.1, 𝑥2 ∗ = 6, 𝑥3 ∗ = 4.05, 𝑓∗ = 1151.4; further improvement possible
- 21. Example: Branch and Bound Method • 𝑆4: 𝐹3 ∪ 𝑥3 ≤ 4 , integer solution: 𝑥1 ∗ = 2, 𝑥2 ∗ = 6, 𝑥3 ∗ = 4, 𝑓∗ = 1140; – Record as the new optimum • 𝑆5: 𝐹3 ∪ 𝑥3 ≥ 5 , non-integer solution: 𝑥1 ∗ = 8.57, 𝑥2 ∗ = 0, 𝑥3 ∗ = 5, 𝑓∗ = 1114.3, no improvement over current optimum. The branch is fathomed. • 𝑆6: 𝐹2 ∪ 𝑥2 ≥ 7 , non-integer solution: 𝑥1 ∗ = 0.57, 𝑥2 ∗ = 7, 𝑥3 ∗ = 4, 𝑓∗ = 1144.3; further improvement possible • 𝑆7: 𝐹6 ∪ 𝑥1 ≤ 0 , non-integer solution: 𝑥1 ∗ = 0, 𝑥2 ∗ = 7.33, 𝑥3 ∗ = 4, 𝑓∗ = 1140; no improvement over current optimum. The branch is fathomed. • 𝑆8: 𝐹6 ∪ 𝑥1 ≥ 1 has no feasible solution. The branch is fathomed. • All branches having been fathomed, the optimal solution is: 𝑥∗ = (2,6,4), 𝑓∗ = 1140.
- 22. Example: Branch and Bound Method • The decision tree for the problem 𝑆0: 𝐹0 = 𝐹𝑅 𝒙∗ = (0, 7.69, 3.85), 𝑓∗ = 1153.8 𝑆1: 𝐹0 ∪ 𝑥3 ≤ 3 𝒙∗ = (0, 6,3), 𝑓∗ = 900 𝑆2: 𝐹0 ∪ 𝑥3 ≥ 4 𝒙∗ = (1.6, 6.4, 4), 𝑓∗ = 1152 𝑆3: 𝐹2 ∪ 𝑥2 ≤ 6 𝒙∗ = (2.1, 6, 4.05), 𝑓∗ = 1151.4 𝑆6: 𝐹2 ∪ 𝑥2 ≥ 7 𝒙∗ = (0.57, 7, 4), 𝑓∗ = 1144.3 𝑆4: 𝐹3 ∪ 𝑥3 ≤ 4 𝒙∗ = (2, 6, 4), 𝑓∗ = 1140 𝑆5: 𝐹3 ∪ 𝑥3 ≥ 5 𝒙∗ = (8.57, 0, 5), 𝑓∗ = 1114.3 𝑆7: 𝐹6 ∪ 𝑥1 ≤ 0 𝒙∗ = (0, 7.33, 4), 𝑓∗ = 1140 𝑆8: 𝐹6 ∪ 𝑥1 ≥ 1 𝑁𝐹𝑆 Optimum solution
- 23. Cutting Plane Method • Let the feasible region for LP relaxation problem be given as: Ω = 𝑥 ∈ ℝ𝑛: 𝑨𝒙 ≤ 𝒃 . • The cutting plane algorithm iteratively trims the feasible region by applying additional constraints (Gomory cuts) so as to exclude non- integer solutions. • It thereby generates a family of polyhedra that satisfy: Ω ⊃ Ω1 ⊃ Ω2 ⊃ ⋯ ⊃ Ω ∩ ℤ𝑛. • The algorithm terminates in finite steps generating an optimal integral BFS.
- 24. Cutting Plane Method • Consider the IP problem: max 𝒙 𝑧 = 𝒄𝑇𝒙, Subject to 𝑨𝒙 ≤ 𝒃, 𝒙 ∈ ℤ𝑛, 𝒙 ≥ 𝟎, • Let the optimal BFS for the LP relaxation be represented as: 𝑰𝒙𝐵 + 𝑨𝑁𝒙𝑁 = 𝒃 where the ith constraint is given as: 𝑥𝑖 + 𝑎𝑖𝑗𝑥𝑗 𝑛 𝑗=𝑚+1 = 𝑏𝑖 • Let 𝑎𝑖𝑗 represent the integer part of 𝑎𝑖𝑗; then for any BFS 𝑥𝑖 + 𝑎𝑖𝑗 𝑥𝑗 𝑛 𝑗=𝑚+1 ≤ 𝑏𝑖 • However, only an integral solution would satisfy 𝑥𝑖 + 𝑎𝑖𝑗 𝑥𝑗 𝑛 𝑗=𝑚+1 ≤ 𝑏𝑖
- 25. Cutting Plane Method • Subtract the second inequality from first to get: 𝑎𝑖𝑗 − 𝑎𝑖𝑗 𝑥𝑗 𝑛 𝑗=𝑚+1 ≥ 𝑏𝑖 − 𝑏𝑖 Or 𝛼𝑖𝑗𝑥𝑗 𝑛 𝑗=𝑚+1 ≥ 𝛽𝑖 where 𝛼𝑖𝑗, 𝛽𝑖 > 0 denote the fractional part • Imposing the above constraint (the cut) excludes the vertex associated with current optimal BFS from the feasible region • The cut involves non-basic variables; however, original constraint equations may be used to replace them with basic variables • Repeated application of additional cuts reduces the feasible region to where the optimal BFS is integral.
- 26. Example: Cutting Plane Method • Consider simplified manufacturing problem given as: max 𝑥1,𝑥2 𝑓 = 7.5𝑥1 + 5𝑥2 Subject to: 10𝑥1 + 5𝑥2 ≤ 60, 5𝑥1 + 12𝑥2 ≤ 80; 𝑥1, 𝑥2 ∈ ℤ • Using Simplex method, the initial and final tableaus are: • The non-integer optimum solution is: 𝑥1 ∗ = 3.368, 𝑥2 ∗ = 5.263, 𝑓∗ = 52.58. • The first cut is given as: 𝑥1 − 𝑠2 ≤ 3 or 6𝑥1 + 12𝑥2 ≤ 83 Basic 𝒙𝟏 𝒙𝟐 𝒔𝟏 𝒔𝟐 Rhs 𝒙𝟏 1 0 0.126 -0.053 3.368 𝒙𝟐 0 1 -0.053 0.105 5.263 −𝐳 0 0 0.684 0.131 52.58 Basic 𝒙𝟏 𝒙𝟐 𝒔𝟏 𝒔𝟐 Rhs 𝒔𝟏 10 5 1 0 60 𝒔𝟐 5 12 0 1 80 −𝐳 -7.5 -5 0 0 0
- 27. Example: Cutting Plane Method • The series of Gomory cuts that result in the optimum solution to the IP problem are given as: No . Cut Optimal solution 1. 6𝑥1 + 12𝑥2 ≤ 83 𝑥1 ∗ = 3.389, 𝑥2 ∗ = 5.222, 𝑓∗ = 51.528 1. 10𝑥1 + 6𝑥2 ≤ 65 𝑥1 ∗ = 3.357, 𝑥2 ∗ = 5.238, 𝑓∗ = 51.369 1. 7𝑥1 + 10𝑥2 ≤ 70 𝑥1 ∗ = 3.846, 𝑥2 ∗ = 4.308, 𝑓∗ = 50.385 1. 7𝑥1 + 9𝑥2 ≤ 65 𝑥1 ∗ = 3.909, 𝑥2 ∗ = 4.182, 𝑓∗ = 50.227 1. 7𝑥1 + 8𝑥2 ≤ 60 𝑥1 ∗ = 4, 𝑥2 ∗ = 4, 𝑓∗ = 50
- 28. Example: Cutting Plane Method • Consider following IP problem: Max 𝑧 = 60𝑥1 + 90𝑥2 + 120𝑥3, Subject to: 35𝑥1 + 60𝑥2 + 140𝑥3 ≤ 1000, 15𝑥1 + 30𝑥2 + 60𝑥3 ≥ 2000, 𝑥1 + 𝑥2 − 2𝑥3 ≤ 0; 𝒙 ∈ ℤ3, 𝒙 ≥ 𝟎 LP relaxation solution: 𝑥1 ∗ = 0, 𝑥2 ∗ = 7.69, 𝑥3 ∗ = 3.85, 𝑓∗ = 1153.8 Final Simplex tableau is: Basic 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒔𝟏 𝒔𝟐 𝒔𝟑 Rhs 𝒙𝟐 0.808 1 0 0.539 0.039 0 7.69 𝒙𝟑 -0.096 0 1 -0.231 0.019 0 3.85 𝐬𝟑 0.173 0 0 0.115 0.115 1 123.1 −𝐳 0.115 0 0 2.077 0.577 0 1153.8
- 29. Gomory Cuts for the Problem • Consider the first constraint equation: 0.808𝑥1 + x2 + 0.539𝑠1 + 0.039𝑠2 = 7.692 • The first cut is developed as: 0.808𝑥1 + 0.539𝑠1 + 0.039𝑠2 ≥ 0.692 • The following series of cuts progressively squeezes the feasible region and produces an integer solution: 𝒙∗ = (2, 6, 4), 𝑓∗ = 1140 No. Cut Optimal solution 1. 0.808𝑥1 + 0.539𝑠1 + 0.039𝑠2 − 𝑠4 = 0.692 𝑥1 ∗ = 0.857, 𝑥2 ∗ = 7, 𝑥3 ∗ = 3.929, 𝑓∗ = 1152.9 1. 0.833𝑠1 + 0.024𝑠2 + 0.881𝑠4 − 𝑠5 = 0.929 𝑥1 ∗ = 2.162, 𝑥2 ∗ = 5.946, 𝑥3 ∗ = 4.054, 𝑓∗ = 1151.3 1. 0.054𝑠1 + 0.973𝑠2 + 0.135𝑠5 − 𝑠6 = 0.946 𝑥1 ∗ = 2.083, 𝑥2 ∗ = 5.972, 𝑥3 ∗ = 4.028, 𝑓∗ = 1145.8 1. 0.056𝑠1 + 0.139𝑠5 + 0.972𝑠6 − 𝑠7 = 0.972 𝑥1 ∗ = 2, 𝑥2 ∗ = 6, 𝑥3 ∗ = 4, 𝑓∗ = 1140 View publication stats View publication stats