Difference between revisions of "Mixed-Integer linear programming"

From Optimization for Chemical Engineering: 4G3
Jump to navigation Jump to search
 
(31 intermediate revisions by the same user not shown)
Line 1: Line 1:
__NOTOC__
__NOTOC__
{{ClassSidebar
{{ClassSidebarYouTube
| date = 25 March 2015
| date = 25 March 2015
| dates_alt_text =  
| dates_alt_text =  
| vimeoID1 = 123236313
| vimeoID1 = 8sIC8WFyEQ0
| vimeoID2 =  
| vimeoID2 = NHiTIGi5I5o
| vimeoID3 =
| vimeoID3 = RGVepaP7AXk
| vimeoID4 =
| vimeoID4 = VJC-0rPN7aY
| vimeoID5 =
| vimeoID5 = IREdFiZXq2c
| vimeoID6 =
| vimeoID6 =
| vimeoID7 =
| vimeoID7 =
Line 19: Line 19:
| assignment_instructions =  
| assignment_instructions =  
| assignment_solutions =  
| assignment_solutions =  
| video_download_link_MP4 = https://www.dropbox.com/s/g4se3kw4zwmv3ri/2015-4G3-Class-11B.mp4?dl=1
| video_download_link_MP4 = http://learnche.mcmaster.ca/media/2015-4G3-Class-11B.mp4
| video_download_link_MP4_size = 787 M
| video_download_link_MP4_size = 787 M
| video_notes1 =
| video_notes1 =
| video_download_link2_MP4 =  
| video_download_link2_MP4 = http://learnche.mcmaster.ca/media/2015-4G3-Class-12A.mp4
| video_download_link2_MP4_size =  M
| video_download_link2_MP4_size =  905 M
| video_notes2 =
| video_notes2 =
| video_download_link3_MP4 =  
| video_download_link3_MP4 = http://learnche.mcmaster.ca/media/2015-4G3-Class-12B.mp4
| video_download_link3_MP4_size = M
| video_download_link3_MP4_size = 884 M
| video_notes3 =
| video_notes3 =
| video_download_link4_MP4 =  
| video_download_link4_MP4 = http://learnche.mcmaster.ca/media/2015-4G3-Class-13A.mp4
| video_download_link4_MP4_size = M
| video_download_link4_MP4_size = 952 M
| video_notes4 =
| video_notes4 =
| video_download_link5_MP4 =  
| video_download_link5_MP4 = http://learnche.mcmaster.ca/media/2015-4G3-Class-13B.mp4
| video_download_link5_MP4_size = M
| video_download_link5_MP4_size = 994 M
| video_notes5 =
| video_notes5 =
}}
}}
Line 45: Line 45:
! Topic
! Topic
! Slides/handouts for class
! Slides/handouts for class
! Video file
! References and Notes
! References and Notes
|-
|-
Line 55: Line 54:
| align="left" colspan="1"|
| align="left" colspan="1"|
[https://docs.google.com/document/d/12jmwaBfvSu0l8Wn3KC0JIJGBqjdloteNQJMgkW6aQyY  Handout from class]
[https://docs.google.com/document/d/12jmwaBfvSu0l8Wn3KC0JIJGBqjdloteNQJMgkW6aQyY  Handout from class]
| [https://www.dropbox.com/s/g4se3kw4zwmv3ri/2015-4G3-Class-11B.mp4?dl=1 Video]
|align="left" colspan="1"|  
|align="left" colspan="1"|  
See the GAMS codes below
See the GAMS codes below
Line 69: Line 67:
| align="left" colspan="1"|
| align="left" colspan="1"|
[https://docs.google.com/document/d/1EhXsp-Whc-DfgxvSkAAqu3jRyOcBo-HubWvOqpQ1LQg  Handout from class]
[https://docs.google.com/document/d/1EhXsp-Whc-DfgxvSkAAqu3jRyOcBo-HubWvOqpQ1LQg  Handout from class]
| <!-- [https://www.dropbox.com/s/g4se3kw4zwmv3ri/2015-4G3-Class-11B.mp4?dl=1 Video] -->
|align="left" colspan="1"|  
|align="left" colspan="1"|  
See the GAMS codes below
See the GAMS codes below
|-
| 01 April
| 12B
| align="left" colspan="1"|
The branch and bound procedure to solve integer problems
| align="left" colspan="1"|
[https://docs.google.com/document/d/1MxIsIPG1uTNvAUFvyaHFUln1ClpUgTADi2HgbFPdkeU/edit?usp=sharing  Handout from class]
[[Media:Sequence-of-branch-and-bound-steps-4G3-2015.pdf |Sequence of the branch and bound steps]]
|align="left" colspan="1"|
See code below to solve the original problem, and then the relaxed problem.
See another example problem below (with solution).
|-
| 06 April
| 13A
| align="left" colspan="1"|
Working towards understanding schedule problems in engineering
* Gantt chart approach
* What the search variables mean
* What objective function choices exist
* Setting up the integer (disjunctive) variables constraints
| align="left" colspan="1"|
[https://docs.google.com/document/d/1lmBL1Bh251VYb7NmBD2ER-JPD2DSExGR87nu7AWs7oo/  Handout from class]
|align="left" colspan="1"|
See code below for the simple scheduling problem (3 products on 1 unit).
|-
| 08 April
| 13B
| align="left" colspan="1"|
Scheduling problems
Course wrap-up
| align="left" colspan="1"|
[https://docs.google.com/document/d/1VZRQ_aaQjMiK_brME9y1dhHYtkhTI_2NLt4PQ51OyhM  Handout from class]
|align="left" colspan="1"|
* The first 10 pages of [http://www.inf.ufpr.br/aurora/disciplinas/topicosia2/livros/search/integer.pdf this textbook chapter] give some great background to integer programs, and how they are solved.
* See code below for the multi-unit multi-product (3 products on 4 units) scheduling problem.
|}
|}




Line 160: Line 196:
model projects /all/;
model projects /all/;
solve projects using mip maximizing value;
solve projects using mip maximizing value;
</syntaxhighlight>
===Binary problem relaxation ===
This is the problem we looked at in class for the generators.
Notice that the only differences are:
* the <tt>binary</tt> variables become <tt>positive</tt> variables, thepr
* the problem is solved as an <tt>LP</tt> not an <tt>MIP</tt>
* the constraints for <tt>.LO</tt> and <tt>.UP</tt> are added, where required, depending on the partial solution solved
{| class="wikitable" style="vertical-align:top"
|-
! Original problem solved in class
! Relaxed problem solved in class
|-
| style="vertical-align:top" |
<syntaxhighlight lang="ocaml">
free variable        cost "total cost";
sets                j    "item j" /1*4/;
binary variables    x(j)  "whether to use generator";
parameter  c(j) "cost of using generator j"
/1  7,
2  12,
3  5,
4  14/;
parameter  p(j) "power of generator j"
/1  300,
2  600,
3  500,
4  1600/;
EQUATIONS
obj    "cost",
power_constraint;
obj..      cost =e= sum(j, c(j)*x(j));
power_constraint..  sum(j, p(j)*x(j)) =G= 700;
MODEL purchase /all/;
SOLVE purchase using MIP minimizing cost;
</syntaxhighlight>
|
<syntaxhighlight lang="ocaml">
free variable        cost "total cost";
sets                j    "item j" /1*4/;
positive variables  x(j)  "whether to use generator";
parameter  c(j) "cost of using generator j"
/1  7,
2  12,
3  5,
4  14/;
parameter  p(j) "power of generator j"
/1  300,
2  600,
3  500,
4  1600/;
EQUATIONS
obj    "cost",
power_constraint;
obj..      cost =e= sum(j, c(j)*x(j));
power_constraint..  sum(j, p(j)*x(j)) =G= 700;
* Solve the problem for the partial solution: (#, 0, #, 1)
x.LO('1')=0;
x.UP('1')=1;
x.LO('2')=0;
x.UP('2')=0;
x.LO('3')=0;
x.UP('3')=1;
x.LO('4')=1;
x.UP('4')=1;
MODEL purchase /all/;
SOLVE purchase using LP minimizing cost;
</syntaxhighlight>
|}
=== Example problem to practice ===
The objective is to '''''maximize''''' the objective function value, \(z\). The table or partial solutions is provided below. The only constraints are that all the 3 search variables must be binary (0 or 1). The "relaxed" refers to the relaxed problem solution.
[[Image:4G3-2015-MIP-Practice-problem.png|750px]]
Use the depth-first search method, and when branching, choose the branch with \(x_i = 0\) before the \(x_i=1\) branch.
If you do this, your nodes will have the following solution order:
Node 0 \((z=82.80)\)
Node 1 \((z=80.67)\)
Node 2 \((z=28.00)\) [incumbent]
Node 3 \((z=79.4)\)
Node 4 \((z=\text{Infeasible})\)
Node 5 \((z=77.00)\) [incumbent; turns out this is the eventual optimum]
Node 6 \((z=74.00)\)
[The end: can you prove why?]
<!--
\begin{center}\begin{tabular}{ccc|ccc}
Partial                            &  ${\bf x}_{\rm rel}^{(k)}$                  &  $z_{\rm rel}^{(k)}$              &  Partial                            &  ${\bf x}_{\rm rel}^{(k)}$                  &  $z_{\rm rel}^{(k)}$ \\ \hline
(\#,\#,\#)                            &  (0.2,1,0)                                &  82.80                            &  (0,0,1)                            &  \multicolumn{2}{c}{Infeasible} \\
(\#,\#,0)                            &  (0.2,1,0)                                &  82.80                            &  (0,1,\#)                            &  (0,1,0.67)                                                &  80.67 \\
(\#,\#,1)                            &  (0,0.8,1)                                &  79.40                            &  (0,1,0)                            &  (0,1,0)                                                &  28.00 \\
(\#,0,\#)                            &  (0.7,0,0)                                &  81.80                            &  (0,1,1)                            &  (0,1,1)                                                  &  77.00 \\
(\#,0,0)                            &  (0.7,0,0)                                &  81.80                            &  (1,\#,\#)                            &  (1,0,0)                                                &  74.00 \\
(\#,0,1)                            &  (0.4,0,1)                                &  78.60                            &  (1,\#,0)                            &  (1,0,0)                                                &  74.00 \\
(\#,1,\#)                            &  (0.2,1,0)                                &  82.80                            &  (1,\#,1)                            &  (1,0,1)                                                &  63.00 \\
(\#,1,0)                            &  (0.2,1,0)                                &  82.80                            &  (1,0,\#)                            &  (1,0,0)                                                &  74.00 \\
(\#,1,1)                            &  (0,1,1)                                  &  77.00                            &  (1,0,0)                            &  (1,0,0)                                                &  74.00 \\
(0,\#,\#)                            &  (0,1,0.67)                                &  80.67                            &  (1,0,1)                            &  (1,0,1)                                                &  63.00 \\
(0,\#,0)                            &  (0,1,0)                                &  28.00                            &  (1,1,\#)                            &  (1,1,0)                                                &  62.00 \\
(0,\#,1)                            &  (0,0.8,1)                                &  79.40                            &  (1,1,0)                            &  (1,1,0)                                                &  62.00 \\
(0,0,\#)                            &  \multicolumn{2}{c|}{Infeasible}                            & (1,1,1)                            &  (1,1,1)                                                &  51.00 \\
(0,0,0)                            &  \multicolumn{2}{c|}{Infeasible}                            &  \\
\end{tabular}\end{center}
-->
=== Simple schedule: one unit, 3 products ===
This example was covered in [https://docs.google.com/document/d/1lmBL1Bh251VYb7NmBD2ER-JPD2DSExGR87nu7AWs7oo class 13A]:
<syntaxhighlight lang="python">
set j "jobs" /1*3/;
alias (j,jp);
parameter p(j) "process time of j"
/ 1  15,    2  6,      3  9/;
scalar M "bigM";
M = sum(j, p(j));
free variable MeanComp  "Mean completion time";
positive variables x(j)  "Start time of job j";
binary variables y(j,jp) "Disjunctive variable for j and jp";
EQUATIONS
MeanCeqn    "Mean completion time",
Disj1(j,jp) "Disjunctive part 1",
Disj2(j,jp) "Disjunctive part 2";
MeanCeqn..  MeanComp =E= sum(j, x(j)+p(j) ) / card(j);
Disj1(j,jp)$(ord(j) lt ord(jp)).. x(j) + p(j)  =L= x(jp) + M*(1-y(j,jp));
Disj2(j,jp)$(ord(j) lt ord(jp)).. x(jp)+ p(jp) =L= x(j)  + M*y(j,jp);
MODEL simplesched /all/;
SOLVE simplesched using MIP minimizing MeanComp;
</syntaxhighlight>
=== Scheduling 3 products on 4 units ===
This example was covered in [https://docs.google.com/document/d/1VZRQ_aaQjMiK_brME9y1dhHYtkhTI_2NLt4PQ51OyhM/edit class 13B]:
<syntaxhighlight lang="python">
* Based on: https://comp.uark.edu/~rrardin/oorbook/software/gams/custommw.gms
set j "Products" /A*C/,
k    "Units"    /1*4/;
alias (j,jp);
alias (k,kp);
set succ(j,k,kp) "Product j successor pairs"
/A.1.2, A.2.3, A.3.4,
B.3.4,
C.3.2, C.2.4/;
TABLE p(j,k) "Process time of product j on unit k"
  1  2  3      4
A  1  5  4.0  1.5
B  0  0  4.5  1.0
C  0  3  5.0  1.5; 
scalar M "bigM";
M = sum((j,k), p(j,k));
FREE variable
AverageC    "Average completion time";
POSITIVE variables x(j,k)  "Start time of job (product) j on unit k:  j.k";
BINARY variables y(j,jp,k) "Disjunctive variable for j and jp on k";
EQUATIONS
avgComplete  "Average completeness",
pred(j,k,kp)  "Precedence within jobs",
disj1(j,jp,k) "Disjunctive part 1 for j and jp on k",
disj2(j,jp,k) "Disjunctive part 2 for j and jp on k";
avgComplete..                AverageC =E= sum((j,k), x(j,k)+p(j,k) );
* These are the precedence constraints
pred(j,k,kp)$succ(j,k,kp)..  x(j,k) + p(j,k) =L= x(j,kp);
disj1(j,jp,k)$(ord(j) lt ord(jp) and p(j,k) gt 0 and p(jp,k) gt 0)..
                                  x(j,k) + p(j,k) =L= x(jp,k) + M*(1-y(j,jp,k));
disj2(j,jp,k)$(ord(j) lt ord(jp) and p(j,k) gt 0 and p(jp,k) gt 0)..
                                  x(jp,k) + p(jp,k) =L= x(j,k) + M*y(j,jp,k);
MODEL Jobshop /all/;
option mip = mosek;
SOLVE Jobshop USING MIP minimizing AverageC;
</syntaxhighlight>
</syntaxhighlight>

Latest revision as of 11:55, 12 August 2018

Class date(s): 25 March 2015





Resources

Scroll down, if necessary, to see the resources.

Date Class number Topic Slides/handouts for class References and Notes
25 March 11B
  • Representation of integer variables
  • Types of problems that can be solved with integer variables

Handout from class

See the GAMS codes below

30 March 12A

More problems that can be solved with integer variables

  • Knapsack problem
  • Mutual exclusivity constraints
  • Dependence constraints
  • Allocation problems

Handout from class

See the GAMS codes below

01 April 12B

The branch and bound procedure to solve integer problems

Handout from class

Sequence of the branch and bound steps

See code below to solve the original problem, and then the relaxed problem.

See another example problem below (with solution).

06 April 13A

Working towards understanding schedule problems in engineering

  • Gantt chart approach
  • What the search variables mean
  • What objective function choices exist
  • Setting up the integer (disjunctive) variables constraints

Handout from class

See code below for the simple scheduling problem (3 products on 1 unit).

08 April 13B

Scheduling problems

Course wrap-up

Handout from class

  • The first 10 pages of this textbook chapter give some great background to integer programs, and how they are solved.
  • See code below for the multi-unit multi-product (3 products on 4 units) scheduling problem.


Solving a basic ILP

free variable        income    "total income";
positive variables   x1, x2;
binary variable      delta     "use ingredient x3 or not at all";

EQUATIONS
obj    "maximize income",
blend  "blending constraint";
obj..   income =E= 18*x1 - 3*x2 - 9*(20*delta);
blend.. 2*x1 + x2 + 7*(20*delta) =L= 150;
x1.up = 25;
x2.up = 30;

model recipe /all/;
SOLVE recipe using MIP maximizing income;

Solving the knapsack problem

free variable        value "total value";
sets                 j     "item j" /1*5/;
binary variables     x(j)  "whether to include item in the knapsack";

parameter   v(j) "value of object j"
/1   4,
 2   2,
 3   10,
 4   1,
 5   2/;

parameter   w(j) "weight of object j"
/1  12,
 2   1,
 3   4,
 4   1,
 5   2/;

EQUATIONS
obj     "maximize value",
weight  "weight constraint";

obj..    value =e= sum(j, v(j)*x(j));
weight.. sum(j, w(j)*x(j)) =L= 15;

model knapsack /all/;
solve knapsack using mip maximizing value;

Solving the knapsack problem for selecting amount projects, with constraints

free variable        value "total value";
sets                 j     "item j" /1*5/;
binary variables     x(j)  "whether to include item in the project";

parameter   v(j) "value of object j"
/1   50,
 2   72,
 3   25,
 4   41,
 5   17/;

parameter   w(j) "cost of object j"
/1   8,
 2   21,
 3   15,
 4   10,
 5   7/;

EQUATIONS
obj     "maximize value",
weight  "weight constraint",
me1v2   "1 and 2 are mutually exclusive",
me1v3v5 "1, 3 and 5 are mutually exclusive",
d4v3    "4 depends on 3";

obj..    value =e= sum(j, v(j)*x(j));
weight.. sum(j, w(j)*x(j)) =L= 35;

me1v2..    x('1') + x('2') =L= 1;
me1v3v5..  x('1') + x('3') + x('5') =L= 1;
d4v3..     x('4') =L= x('3');

model projects /all/;
solve projects using mip maximizing value;

Binary problem relaxation

This is the problem we looked at in class for the generators.

Notice that the only differences are:

  • the binary variables become positive variables, thepr
  • the problem is solved as an LP not an MIP
  • the constraints for .LO and .UP are added, where required, depending on the partial solution solved
Original problem solved in class Relaxed problem solved in class
free variable        cost "total cost";
sets                 j     "item j" /1*4/;
binary variables     x(j)  "whether to use generator";

parameter   c(j) "cost of using generator j"
/1   7,
 2   12,
 3   5,
 4   14/;

parameter   p(j) "power of generator j"
/1   300,
 2   600,
 3   500,
 4   1600/;

EQUATIONS
obj    "cost",
power_constraint;

obj..      cost =e= sum(j, c(j)*x(j));
power_constraint..  sum(j, p(j)*x(j)) =G= 700;

MODEL purchase /all/;
SOLVE purchase using MIP minimizing cost;
free variable        cost "total cost";
sets                 j     "item j" /1*4/;
positive variables   x(j)  "whether to use generator";

parameter   c(j) "cost of using generator j"
/1   7,
 2   12,
 3   5,
 4   14/;

parameter   p(j) "power of generator j"
/1   300,
 2   600,
 3   500,
 4   1600/;

EQUATIONS
obj    "cost",
power_constraint;

obj..      cost =e= sum(j, c(j)*x(j));
power_constraint..  sum(j, p(j)*x(j)) =G= 700;

* Solve the problem for the partial solution: (#, 0, #, 1)
x.LO('1')=0;
x.UP('1')=1;

x.LO('2')=0;
x.UP('2')=0;

x.LO('3')=0;
x.UP('3')=1;

x.LO('4')=1;
x.UP('4')=1;

MODEL purchase /all/;
SOLVE purchase using LP minimizing cost;

Example problem to practice

The objective is to maximize the objective function value, \(z\). The table or partial solutions is provided below. The only constraints are that all the 3 search variables must be binary (0 or 1). The "relaxed" refers to the relaxed problem solution.

4G3-2015-MIP-Practice-problem.png

Use the depth-first search method, and when branching, choose the branch with \(x_i = 0\) before the \(x_i=1\) branch.

If you do this, your nodes will have the following solution order:

Node 0 \((z=82.80)\)

Node 1 \((z=80.67)\)

Node 2 \((z=28.00)\) [incumbent]

Node 3 \((z=79.4)\)

Node 4 \((z=\text{Infeasible})\)

Node 5 \((z=77.00)\) [incumbent; turns out this is the eventual optimum]

Node 6 \((z=74.00)\)

[The end: can you prove why?]


Simple schedule: one unit, 3 products

This example was covered in class 13A:

set j "jobs" /1*3/;
alias (j,jp);

parameter p(j) "process time of j"
/ 1  15,    2  6,       3  9/;

scalar M "bigM";
M = sum(j, p(j));
free variable MeanComp   "Mean completion time";
positive variables x(j)  "Start time of job j";
binary variables y(j,jp) "Disjunctive variable for j and jp";

EQUATIONS
MeanCeqn    "Mean completion time",
Disj1(j,jp) "Disjunctive part 1",
Disj2(j,jp) "Disjunctive part 2";

MeanCeqn..   MeanComp =E= sum(j, x(j)+p(j) ) / card(j);
Disj1(j,jp)$(ord(j) lt ord(jp)).. x(j) + p(j)  =L= x(jp) + M*(1-y(j,jp));
Disj2(j,jp)$(ord(j) lt ord(jp)).. x(jp)+ p(jp) =L= x(j)  + M*y(j,jp);

MODEL simplesched /all/;
SOLVE simplesched using MIP minimizing MeanComp;

Scheduling 3 products on 4 units

This example was covered in class 13B:

* Based on: https://comp.uark.edu/~rrardin/oorbook/software/gams/custommw.gms
set j "Products" /A*C/,
k     "Units"    /1*4/;
alias (j,jp);
alias (k,kp);

set succ(j,k,kp) "Product j successor pairs"
/A.1.2, A.2.3, A.3.4,
 B.3.4,
 C.3.2, C.2.4/;

TABLE p(j,k) "Process time of product j on unit k"
   1   2   3      4
A  1   5   4.0  1.5
B  0   0   4.5  1.0
C  0   3   5.0  1.5;   

scalar M "bigM";
M = sum((j,k), p(j,k));
FREE variable
AverageC     "Average completion time";

POSITIVE variables x(j,k)  "Start time of job (product) j on unit k:  j.k";
BINARY variables y(j,jp,k) "Disjunctive variable for j and jp on k";

EQUATIONS
avgComplete   "Average completeness",
pred(j,k,kp)  "Precedence within jobs",
disj1(j,jp,k) "Disjunctive part 1 for j and jp on k",
disj2(j,jp,k) "Disjunctive part 2 for j and jp on k";

avgComplete..                AverageC =E= sum((j,k), x(j,k)+p(j,k) ); 

* These are the precedence constraints
pred(j,k,kp)$succ(j,k,kp)..  x(j,k) + p(j,k) =L= x(j,kp);

disj1(j,jp,k)$(ord(j) lt ord(jp) and p(j,k) gt 0 and p(jp,k) gt 0)..
                                   x(j,k) + p(j,k) =L= x(jp,k) + M*(1-y(j,jp,k));

disj2(j,jp,k)$(ord(j) lt ord(jp) and p(j,k) gt 0 and p(jp,k) gt 0)..
                                   x(jp,k) + p(jp,k) =L= x(j,k) + M*y(j,jp,k);
MODEL Jobshop /all/;
option mip = mosek;
SOLVE Jobshop USING MIP minimizing AverageC;