modelling of lp-problems (2wo09)jkeijspe/onderwijs/2wo09/slides.pdf · modelling of lp-problems...

Modelling of LP-problems (2WO09)

assignor: Judith Keijsperroom: HG 9.31

email: [email protected] info : http://www.win.tue.nl/∼jkeijspe

Technische Universiteit Eindhoven

meeting 1

J.Keijsper (TUE) Modelling of LP-problems (2WO09) meeting 1 1 / 70

Contents

1 Motivation MIP and LP models

2 Definition LP

3 LP modeling

4 LP solving

5 LP: Sensitivity analysis

6 Mixed Integer Programming (MIP)

7 Solving MIP’s


Motivation MIP and LP models

A discrete optimization problem can often be modeled as a (Mixed)Integer (linear) Program (MIP).

If integrality restrictions are removed from a MIP, a Linear Program (LP)is obtained.

LP’s can be solved fast. Solving an MIP usually involves solving (very)many LP’s.


Definition LPA Linear Program (LP) is the task of optimizing (i.e. minimizing ormaximizing) a linear function subject to linear inequalities.

Example

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0

Example

max x1 − 3x2 + 5x3 + 7x4

s.t. x1 − x3 ≤ 93x1 − 2x4 ≥ 3

2x2 − x3 − x4 = 5x1 ≥ 2

x3 ≤ 0J.Keijsper (TUE) Modelling of LP-problems (2WO09) meeting 1 4 / 70

Here is an example of an LP in compact notation (using indexedparameters ai ,j , bi , cj , indexed variables xj and indexed constraints):

Example

max∑n

j=1 cjxj

s.t.∑n

j=1 ai ,jxj ≤ bi for all i = 1 . . . m

xj ≥ 0 for all j = 1 . . . n

is compact notation for

max c1x1 + c2x2 + · · ·+ cnxn

s.t. a1,1x1 + a1,2x2 + · · ·+ a1,nxn ≤ b1

a2,1x1 + a2,2x2 + · · ·+ a2,nxn ≤ b2

· · · · · ·am,1x1 + am,2x2 + · · ·+ am,nxn ≤ bm

x1 ≥ 0· · · · · ·xn ≥ 0


LP modeling

1 Variables: decisions.

2 Objective function: quantity (linear function of the variables) to beoptimized.

3 Constraints: restrictions (linear inequalities in the variables) thatdecisions must satisfy.


Example

Chips production (From Aimms documentation: Optimization Modeling)How to maximize profit for the factory with the following data?

time required [min/kg] plain chips Mexican chips availability [min]

slicing 2 4 345frying 4 5 480packing 4 2 330

net profit [dollar/kg] 2 1.5

x = production of plain chips [kg]y = production of mexican chips [kg]

max 2x + 1.5y2x + 4y ≤ 3454x + 5y ≤ 4804x + 2y ≤ 330x , y ≥ 0


LP modeling in Aimms

0. Choose sets for indexing.

1. Declare parameters and variables (give meaningful names). Both canbe indexed by sets.

2. Declare a scalar (non-indexed) variable: the objective function. It hasa definition (as a linear function of the variables).

3. Declare constraints. A constraint can be indexed by a set. Everyconstraint has a definition (as a linear inequality in the variables).Simple bounds are specified in the range of a variable.

4. Declare a Mathematical Program (MP): specify the variable thatdefines the objective function, the direction of optimization(min/max), the set of constraints you want to use (default: all), theset of variables (default: all), the type of program (LP).

5. This abstract model is saved separately from the case (concretedataset) that you would like to compute an optimal solution for.Cases can be modified easily without changing the abstract model.


LP solving by hand: graphical method (2 variables)

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0



O x

y

x+2y=4

3x+2y=6

y=0

x=0



O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3



O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3

2x+3y=6



O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3



O x

y

x+2y=4

3x+2y=6

y=0

x=0



Conclusion: solving the system

x + 2y = 4

3x + 2y = 6

for x and y yields the optimal solution

x = 1

y = 1.5

where the objective function 2x + 3y attains its maximum value 6.5.



1 Draw lines (half-planes) in the x , y -plane corresponding to theconstraints.

2 Shade the feasible region in the x , y -plane, consisting of those pointsthat satisfy all the constraints.

3 Draw a contour of the objective function, and indicate the directionof optimization.

4 Shift this contour as far as possible in the direction of optimization sothat it still touches the feasible region.

5 Now derive from the picture which constraint-lines are intersecting inthe optimal feasible point, and solve the corresponding system ofequalities to obtain the cooordinates of this optimal point.


LP solving with the simplex method (software)

O x

y

x+2y=4

3x+2y=6

y=0

x=0



O x

y

x+2y=4

3x+2y=6

y=0

x=0


Special situations: infeasible

O x

y

y=0

x=0


Special situations: unbounded

O x

y

y=0

x=0


Special situations: multiple optimal solutions (no Aimmswarning)

O x

y

y=0

x=0


LP: Sensitivity analysisA parametric curve depicts the sensitivity of an LP model to parametricchange by simply recomputing the optimal solution (or the optimal valueof the objective) for a range of values of the chosen parameter.

parameter

opt.

valueLP

10 90


Definition Shadowprice

The shadowprice of a constraint in an LP model is the rate of change ofthe optimal value of the objective function from a one unit increase in therighthand side of the constraint (for a small enough increase).

Example

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0


Shadowprice

O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3


Shadowprice


x + 2y = 4

3x + 2y = 6


x = 1

y = 1.5



Shadowprice

To compute the shadowprice of the constraint 3x + 2y ≤ 6,change

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0

tomax 2x + 3ys.t. 3x + 2y ≤ 7

x + 2y ≤ 4x ≥ 0

y ≥ 0


Shadowprice

O x

y

x+2y=4

y=0

x=0

2x+3y=3

3x+2y=7


O x

y

x+2y=4

y=0

x=0

2x+3y=3

3x+2y=7


Shadowprice


x + 2y = 4

3x + 2y = 7


x = 1.5

y = 1.25


The shadowprice of the constraint 3x + 2y ≤ 6 is therefore 0.25.


ShadowpriceSmall enough increase: optimum is intersection of the sameconstraint-lines (no other constraint becomes binding by the increase).

O x

y

x+2y=4

y=0

x=0

2x+3y=3

3x+2y=15


Shadowprice

Information on the shadowprices of constraints is implicitly available whensolving an LP by the simplex method. In Aimms, simply indicate that youwant this information explicitly.


Definition Reduced Cost

The reduced cost (or reduced profit) of a variable in an LP model is therate of change of the optimal value of the objective function from a oneunit increase in the bound of the variable that is binding the optimum.(for a small enough increase).

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0


Reduced cost

To compute the reduced cost of the variable x ,change

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0

tomax 2x + 3ys.t. 3x + 2y ≤ 7

x + 2y ≤ 4x ≥ 0.5

y ≥ 0


O x

y

x+2y=4

x=0

2x+3y=3

y=0

x=0.5

3x+2y=6


Reduced cost

Conclusion: Nothing changes.Solving the system

x + 2y = 4

3x + 2y = 6


x = 1

y = 1.5


The reduced cost of the variable x is therefore 0.


Definition MIP

A Mixed Integer Program (MIP) is the task of optimizing (i.e. minimizingor maximizing) a linear function subject to linear inequalities andintegrality constraints on (some of) the variables.

Example

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0x integraly integral

Integral variables can model decisions regarding integral numbers andyes/no decisions.


Example

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0x ≤ 1

y ≥ 0y ≤ 1

x integraly integral

is equivalent tomax 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x binaryy binary


LP-relaxationBy deleting all integrality constraints from a MIP, an LP is obtained: itsLP-relaxation.

Example

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0


has LP-relaxationmax 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0


MIP solving by hand: graphical method (2 variables)

O x

y

x+2y=4

3x+2y=6

y=0

x=0

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0



O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3



O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3

2x+3y=6



Conclusion from the picture: optimal solution of this MIP is

x = 0

y = 2

where the objective function 2x + 3y attains its maximum value 6.


Note: in general, the optimal solution of an MIP is not a corner point ofthe feasible region of the LP-relaxation (and hence can not be obtained asthe solution of a system of constraint-equalities).

O x

y

y=0

x=0


MIP solving using an LP solver: Branch and Bound

Idea: solve the LP-relaxation. If the optimal solution is not integral,branch: split the feasible region into disjoint parts that both do notcontain this LP-optimum. and solve an LP for each of these parts.


Branch and Bound

Example

MIP:max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0



Branch and Bound

Start by solving the LP-relaxation LP0:

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0

LP0 has optimal solution (x , y) = (1, 1.5) and optimal objective value 6.5.


Branch and Bound Tree

The LP-relaxation can be viewed as the root node of a tree, the branchand bound tree. It corresponds to the entire feasible region.The LP’s for the smaller parts of the feasible region are the other nodes ofthe tree. Every branch from a node corresponds to adding an inequality tothe LP of that node, thereby restricting the feasible region further.



LP0

(1, 1.5)

6.5


Branch and BoundExpand node LP0: branch on variable y :LP1 adds the inequality y ≤ 1 to LP0. LP2 adds y ≥ 2 to LP0.LP1:

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0y ≤ 1

LP2:max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0y ≥ 2


Branch and Bound

Explore node LP1:max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0

y ≥ 0y ≤ 1

optimal solution (x , y) = (113 , 1), objective value 52

3 .


Branch and Bound

O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3

y=1

y=2



LP0

(1, 1.5)

6.5

LP1

LP2

5.67

y>=2y<=1

(1.33, 1)


Expand node LP1: branch on variable x :LP3 adds the inequality x ≤ 1 to LP1. LP4 adds x ≥ 2 to LP1.LP3:

max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0x ≤ 1

y ≥ 0y ≤ 1

LP4:max 2x + 3ys.t. 3x + 2y ≤ 6

x + 2y ≤ 4x ≥ 0x ≥ 2

y ≥ 0y ≤ 1



LP0

(1, 1.5)

6.5

LP1

LP2

5.67

y>=2y<=1

x<=1 x>=2

LP4

LP3

(1.33, 1)


Branch and Bound

Whenever an integral optimal solution for an LP in one of the nodes isfound, a bound for the best possible MIP objective value is updated. Thetree is pruned in this node.


x + 2y ≤ 4x ≥ 0x ≤ 1

y ≥ 0y ≤ 1

optimal solution (x , y) = (1, 1) (integer!), objective value 5.New bound: 5. Current best MIP solution: (x , y) = (1, 1).Prune by integrality.


Branch and Bound

O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3

y=1

x=1


Branch and Bound

The tree can also be pruned at a node if the LP has an optimal solutionwith objective value not better than the current bound.


x + 2y ≤ 4x ≥ 0x ≥ 2

y ≥ 0y ≤ 1

optimal solution (x , y) = (2, 0) (integer!), objective value 4.Current bound = 5 > 4. No update.Prune by integrality or by bound.


Branch and Bound

O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3

y=1

x=2



LP0

(1, 1.5)

6.5

LP1

LP2

5.67

y>=2y<=1

x<=1 x>=2

LP4

LP3(1,1)

5

(2,0)

4

INTINT

(1.33, 1)



x + 2y ≤ 4x ≥ 0

y ≥ 0y ≥ 2

optimal solution (x , y) = (0, 2) (integer!), objective value 6.New bound: 6. Current best MIP solution: (x , y) = (0, 2)Prune by integrality.All nodes are pruned: done.Conclusion: optimal MIP solution: (x , y) = (0, 2), optimal MIP objectivevalue 6.


Branch and Bound

O x

y

x+2y=4

3x+2y=6

y=0

x=0

2x+3y=3

y=2


Branch and Bound

LP0

(1, 1.5)

6.5

LP1

LP2

5.67

y>=2y<=1

x<=1 x>=2

(0,2)

6

LP4

LP3(1,1)

5

(2,0)

4

INT

INTINT

(1.33, 1)



The LP-relaxation can be viewed as the root node of a tree, the branchand bound tree. It corresponds to the entire feasible region. The LP’s forthe smaller parts of the feasible region are the other nodes of the tree.Every branch corresponds to adding an inequality to the LP, therebyrestricting the feasible region further. Whenever an integral optimalsolution for an LP in one of the nodes is found, a bound for the bestpossible MIP objective value is updated.

The branching stops (tree is pruned) at a node if the LP of this node

is infeasible (infeasibility),

has an integral optimal solution (integrality), or

has an optimal solution with objective value not better than thecurrent bound (bound)


Size of the branch and bound tree

For an MIP on n integer variables, each taking a value between 0 and 9,the number of nodes (LPs to solve) can become as large as 10n, soexponential in n.There are no general theorems about how many of those nodes actuallyhave to be explored.


Binary variables

In a MIP, an integer variable x which is also required to be ≥ 0 and ≤ 1 iscalled a binary variable.

0 ≤ x ≤ 1, x integral

A binary variable models yes/no decisions.


Sensitivity analysis for MIP

Parametric curves can be constructed for a MIP (but requires solvingmany MIPs so might take too much time for large MIPs).

Shadow prices and reduced costs have no meaning in a MIP. Ignore thevalues Aimms computes for these when solving a MIP!


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