1 System Planning 2013 Lecture 7: Optimization Appendix A Contents: –General about optimization –Formulating optimization problems –Linear Programming

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System Planning 2013

• Lecture 7: Optimization• Appendix A• Contents:

– General about optimization– Formulating optimization problems– Linear Programming (LP)– Mixed Integer Linear Programming (MILP)

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Optimization – General

• Discipline in applied mathematics

• Used when:– Want to maximize or minimize something

(e.g. profit or cost) under various conditions (e.g. generation capacities, transmission limits)

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Optimization – GeneralTypical problems• In what sequence should parts be produced on a

machine in order to minimize the change-over time?• How can a dress manufacturer lay out its patterns on

rolls of cloth to minimize wasted material?• How many elevators should be installed in a new office building to achieve an acceptable expected waiting

time?• What is the lowest-cost formula for chicken food which

will provide required quantities of necessary minerals and other nutrients?

• What is the generation plan for the power plants during the next six hours?

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Optimization - General

The general optimization problem:

min ( )

s.t. ( )

( )

f x

g x b

h x c

x x x

x x

• f(x): objective function• x: optimization variables (vector)• g(x), h(x): constraints (vectors) • and : variabel limits (vector)

Defines the feasible set

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Optimization – Formulation

Formulation of optimization problems:

1. Describe the problem in words:• Think through the problem

2. Define symbols• Review the parameters and variables

involved3. Write the problem mathematically

• Translate the problem defined in words to mathematics

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The problem:

A manufacturer owns a number of factories. How should the manufacturer ship his goods to the customers?

??

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1. The problem in words:

How should the merchandise be shipped in order to minimize the transportation costs? The following must be fulfilled:

• The factories cannot produce more than their production capacities.

• The customers must at least receive their demand

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2. Define symbols:

• Index and parameters:m factories, factory i has capacity ai

n customers, customer j demand bj units of the commodityThe transportation cost from factory i to customer j is cij per unit

• VariablesLet xij, i = 1...m, j = 1...n denote the number of units of the commodity shipped from factory i to customer j


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3. Mathematical formulation:

• Objective function

Transportation cost:

• Constraints

Production capacity:

Demand:

Variable limits:

m

i

n

jijij xc

1 1

miax i

n

jij ...1

1

njbx j

m

iij ...1

1

njmixij ...1,...10


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The optimization problem:

njmix

njbx

miax

xc

ij

j

m

iij

i

n

jij

m

i

n

jijij

...1,...1,0

...1,

...1,s.t.

min

1

1

1 1


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LP – General

Linear Programming problems (LP problems)• Class of optimization problems with linear objective

function and constraints• All LP problems can be written as (standard form):

0

s.t.

min

x

bAx

xcT

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LP – General

• LP problems will here be explained through examples (similar as in Appendix A)

• Will have to look at the following:– Extreme points– Slack variables– Non existing solution– Not active constraints– No finite solution or unbounded problems– Degenerated solution– Flat optimum– Duality– Solution methods

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LP – Example• The problem:

0,0

12124

2s.t.

105min

21

21

21

21

xx

xx

xx

xxz

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LP – Example

Optimum in:

x1+x2 2

4x1+12x2 12

x1 0

x2

0

z = 20

z = 30

z = 40

Objective:

5.0

5.1

2

1

x

x

5.12z

x2

1

2

3

4

x11 2 3 4

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LP – Extreme points

• The “corners” of the feasible set are called extreme points

• Optimum is always reached in one or more extreme points!

x1+x2 2

4x1+12x2 12

x1 0

x2

0

z = 20

z = 30

z = 40

x2

1

2

3

4

x11 2 3 4

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LP – Slack variables

• LP problem on standard form:

• Introduce slack variables in order to write inequality constraints using equality

0

s.t.

min

x

bAx

xcT

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LP – Slack variables• Without slack variables:

(Ax b)

0,0

12124

2

21

21

21

xx

xx

xx

0,0,0,0

12124

2

4321

421

321

xxxx

xxx

xxx• With slack variables:

(Ax = b)

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• Can happen that the constraints makes no solution feasible! (Bad formulation of problem)

0,0

1

12124

2s.t.

105min

21

21

21

21

21

xx

xx

xx

xx

xxz

New constraint

LP – Non existing solution

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B: Feasible set defined by (c)

A: Feasible set defined by (a) & (b)

x1+x2 2

4x1+12x2 12

x1 0

x2

0

x2

1

2

3

4

x11 2 3 4

(a) (b)

x1+x2 1(c)

Feasible set is empty!

LP – Non existing solution

A B

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LP – Not active constraints

• Some constraints might not be active in optimum.

0,0

7

12124

2s.t.

105min

21

21

21

21

21

xx

xx

xx

xx

xxz

New constraint

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x2

x1

1

2

3

4

1 2 3 4

x1+x2 2

x2

0

4x1+12x2 12

x1 0

z = 20

z = 30

z = 40

x1+x2 7

Not active constraints

Active constraints

LP – Not active constraints

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LP – No finite solution

• If the constraints don’t limit the objective: | z|

0,0

12124

2s.t.

min

21

21

21

21

xx

xx

xx

xxzNew objective function

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x1+x2 2

x2

0

4x1+12x2 12x1 0

x2

x1

1

2

3

4

1 2 3 4

z = -5

z = -4

z = -3

min z -

LP – No finite solution

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LP – Degenerated solution

• More than one extreme point can be optimal. All linear combinations of these points are then also optimal

0,0

12124

2s.t.

1010min

21

21

21

21

xx

xx

xx

xxzNew objective

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x1+x2 2

x2

0

4x1+12x2 12x1 0

x2

1

2

3

4

1 2 3 4

z = 30

z = 40

z = 50The line between the extreme points are also optimal solutions

x1

LP – Degenerated solution

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LP – Flat optimum

• A number of extreme points can have almost the same objective function value.

0,0

12124

2s.t.

99.910min

21

21

21

21

xx

xx

xx

xxzNew objective

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LP – Flat optimum

x1+x2 2

x2

0

4x1+12x2 12x1 0

x2

1

2

3

4

1 2 3 4

z = 30

z = 40

z = 50

x1

z = 19.980

z = 19.995

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LP – Duality• All LP problems (primal problem) have a corresponding dual

problem

are called dual variables

0

s.t.

min

x

bAx

xcT

0

s.t.

max

cA

bT

T

Primal problem: Dual problem:

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LP – Duality

Theorem (strong duality):

If the primal problem has an optimal solution, then also the dual problem has an optimal solution and the objective values of these solutions are the same.

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LP – Duality

• Q: What does the dual variables describe?• A: Dual variables = Marginal value of the

constraint that the dual variable represents, i.e. how the objective function value changes when the right-hand side of the constraint changes

0

s.t.

min

x

bAx

xcT

One dual variable for each constraint!

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LP – Dualityx2

x1

1

2

3

4

1 2 3 4

x1+x2 2

x2

0

4x1+12x2 12

x1 0

z = 20

z = 30

z = 40

12

x1+x2 73

In optimum:1 > 0 (active)2 > 0 (active)3 = 0 (not active)

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• Dual variables • Small changes in right-hand side b

• Change in objective function value z:

LP – Duality

z = Tb

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MILP – General

Mixed Integer Linear Programming problems (MILP problems)

• Class of optimization problems with linear objective function and constraints

• Some variables can only take integer values

1

2

1

2

1

2

min

s.t.

0

0,1,2,...

T xc

x

xA b

x

x

x

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MILP – Example

Optimum in:

x1+x2 2

4x1+12x2 12

x1 0

x2

0

Objective:

)5.0(1

)5.1(1

2

1

x

x

)5.12(15z

x2

1

2

3

4

x11 2 3 4

z = 20

z = 30

z = 40

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MILP – Solving

• Easy to implement integer variables.• Hard to solve the problems.

– Execution times can increase exponentially with the number of integer variables

• Avoid integer variables if possible!

• Special case of integer variables: Binary variables

x {0,1}

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MILP – Example

• Minimize the cost for buying ”something”• Variable, x: Quantity of ”something”, x 0.

• Not constant cost per unit - cost function.• For example: Discount when buying more than a

specified quantity:Cost [SEK]

Number of ”units”

xb

x1

x2 Split x into two different variables, x1 and x2. Observe that x1≤ xb. Also note that both x1 and x2 are 0.

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MILP – Example

• Cheaper to buy in segment 2 Must force the problem to fill the first segment before entering segment 2.

• Can be performed by introducing a binary variable.

Cost [SEK]


xb

x1

x2

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MILP – Example

• Introduce the following constraints:

0

0

2

1

Msx

sxx bs=0 s=1

Cost [SEK]


xb

x1

x2

Check if s=0: 000,0 2221 xxxx

Check if s=1:MxMx

xxxxxx bbb

22

111

0

,0

where M is an arbitrarely large number.

• Let s be an integer variable indicating whether being in segment 1 or 2.

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LP – Solution methods

• Simplex:Returns optimum in extreme point (also whendegenerated solution)

• Interior point methods:If degenerated solution: Returns solution between twoextreme points

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LP – Solution methods

x1+x2 2

x2

0

4x1+12x2 12x1 0

x2

1

2

3

4

1 2 3 4

z = 30

z = 40

z = 50

Simplex

Interior point method

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PROBLEM 11 - Solutions to LP problems

Assume that readymade software is used tosolve an LP problem. In which of the followingcases do you get an optimal solution to the LPproblem and in which cases do you have toreformulate the problem (or correct an error inthe code).

a) The problem has no feasible solution.b) The problem is degenerated.c) The problem does not have a finite solution.

Documents

1 System Planning 2013 Lecture 7: Optimization Appendix A Contents: –General about optimization –Formulating optimization problems –Linear Programming