CCB4333=lectureSet8 (1)

Process Optimization

Mixed Integer Programming

Professor Dr. Saibal Ganguly

Universiti Teknologi PETRONAS

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Lecture Set 8:

September 2014

CCB 4333: Professor Saibal Ganguly

Mixed Integer Programming

At the end of the topic, students should be able to:

Understand, discuss and solve the concepts of mixed

integer programming

Formulate and solve simple mixed integer programming

based engineering optimization problems

Understand the overview of MILP, MINLP and MI

problems

Understand branch and bound methods

Understand depth and breadth based strategies.

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Lecture Set 8:


What is Mixed Integer Programming?

The objective function depends on two sets of

variables, x and y where

X represents vector of continuous variables

Y represents vector of integer variables

Minimize f(x,y)

Subject to g(x,y) 0

h(x,y) = 0

xi 0 for i = 1, 2, .., n and

yj = [0,1] for j = 1, 2, .., m

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Introduction to Integer Variables

In chemical plant design and operation, there are

many variables which have integer values

Examples:

Number of distillation columns or reactors

Type of separator or reactor

Decisions in time to keep or replace an equipment

Catalyst and promoters types

Sequence of heat exchangers or distillation

columns

etc.

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Introduction to Integer Variables

Some integer variables can only assume (0 or 1)

values, such as decision variables:

Whether to Install a new piece of equipment

Whether to Accept a new source of raw material

Whether to Mix the streams

Whether to Use the bypass connection

Such integer variables, y, are called binary

integer variables and y = [ 0 , 1 ]

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Types of Mixed Integer Programming

Problems

If the objective function or at least one

of the constraints are nonlinear MINLP

If the objective function and all the

constraints are linear MILP

If there are no continuous variables

involved IP

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Mixed Integer Problems: Classical

Formulations

(1) Blending problem: A given list of

possible inputs with known properties have

to be blended into products or outputs with

given properties and limits.

(2) Traveling salesman problem: A

salesman or a delivery system moves from

one delivery point to the next in an optimal

way and returns to the original point.

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Mixed Integer Problems: Classical Formulations

(Continued)

(3) Knapsack problem: A series of objects

with known weights and values need to be

optimally selected in subsets so that the

total weight does not exceed the limit but the

maximum profit is obtained.

(4) Plant Location problem: A production

demand versus time is available. The way to

take or generate the items from a known

common source with distributed properties.

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Continuous vs. Integer programming

Sometimes the integer variables can be treated as if they

were continuous, and round the optimal solution to the

nearest integer value.

Example 1:

Maximize f(x) = 3x1 + 4x

2

Subject to 3x1 - x

2 12

3x1 + 11x

2 66

where xi 0 for i = 1, 2

and x1 & x

2 are integers

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Yaxis: x2 & X axis: x

1

: Actual Optimal point (IP) : [5, 4] with f = 31

-20

-10

0

10

20

30

40

50

60

0 22

Constraint2 Constraint1

LP Optimal point (5.5, 4.5)

with f = 34.5


Truncation of fractional part of an LP problem

does not always give the solution of the integer

LP problem

Example 2: Replace 3x1 + 11x

2 66 by 7x

1 + 11x

2 88

Maximize f(x) = 3x1 + 4x

2

Subject to 3x1 - x

2 12

7x1 + 11x

2 88

where xi 0 for i = 1, 2

and x1 & x

2 are integers

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In this case: the optimal LP point remains the

same as before [ 5.5, 4.5] with f = 34.5

The truncated solution remains at

[5, 4] with f = 31

However: Actual IP solution is obtained at:

[0, 8] with f = 32 (different from truncated solution)

Therefore, specialized algorithms specific to

Mixed Integer problems have been developed.

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Branch and Bound Method

This is the most widely used and probably the most

successful method for solving both IP and MILP

problems

BASIC IDEA:

1) Solve the problem as if it contains only continuous

variables

2) If the optimal solution contains noninteger values

for some integer variables, use partitioning method

for these integer variables to divide up all the

possible solutions into subsets

3) Avoid exhaustive search using bounding methods


Branch and Bound Method The starting node (e.g. node 1) is called root node or

initial node. The ends node (e.g. nodes 2,4,5) are

called terminal nodes and nodes in between (e.g.

node 3) are regarded as intermediate nodes.

Each tree of the solutions have the root node,

intermediate nodes, and terminal nodes.

Every path from root node to terminal nodes

defines a complete solution.

The optimum solution should be chosen among all

terminal nodes.

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TREE REPRESENTATION Example: MILP with 3 binary variables


BRANCHING STRATEGIES

Depth-first (1) Perform branching on most recently

created node

(2) When no nodes can be expanded,

backtrack to node whose successor have

not been examined

Breadth-first (1) Select the node with the best value at

each level and expand all its successor

nodes


Example 3 Minimize the overall cost

Node 2: 10; Node 3: 17; Node 4: 6; Node 5: 2; Node 6: 2; Node 7: 2;

Node 8: 2; Node 9: 4; Node 10: 2; Node 11: 4;

Node 12: 4; Node 13: 4; Node 14: 2; Node 15: 4


Strategy: Depth first

Step 1: Select Initial Sequence: 1-2-4-9: 10+6+4 = 20;

Set expansion node = 4;

Set optimal sequence to 1-2-4-9 with bound = 20.

Step 2: Retrace to Sequence: 1-2-4-8: 10+6+2 = 18;

Since 18 < 20, change optimal sequence.

Set new optimal sequence to 1-2-4-8 with bound = 18.


Set expansion node = 5;

Since 14 < 18, change optimal sequence.

Set new optimal sequence to 1-2-5-10 with bound = 14.

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Since 16 > 14, donot change optimal sequence.

Step 5: Retrace to Sequence: 1-3: 17 > 14;

Prune and discard all branches of 1-3 and beyond.

The optimal sequence is 1-2-5-10 with bound = 14.

In breadth first strategy, compare

horizontally side nodes first and arrive at the

optimum.

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COMPARISON BETWEEN BRANCH AND

BOUND STRATEGIES

Depth-first

Requires less storage of nodes since maximum

nodes to be stored at any point is the number of

levels in the tree

Has the tendency of finding the optimal solution

early in the enumeration procedure.

Breadth-first

In general requires examination of fewer nodes

no backtracking

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BRANCH AND BOUND METHOD USING LP

RELAXATIONS In a mixed integer MILP formulation, often integer

variables can be binary that is either 0 or 1. Quite

often the approach to solving the MILP is obtained by

relaxation of the only 0 or 1 constraints to

anywhere between 0 and 1 constraints. This

procedure of relaxation is known as LP Relaxation

The subsequent problem formed is known as an LP

subproblem

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Example 4:

Maximize f = 86x1 + 4x

2 + 40x

3

Subject to 774x1 + 76x

2 + 42x

3 875

67x1 + 27x

2 + 53x

3 875

where xi 0 for i = 1, 2,3

and x1 x

2 & x

3 are integers [0,1]

Solution:

A decomposition is attempted using Branch and

Bound method.

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NODE 1: Continuous LP optimum is solved for

constraints 0 x1 1; 0 x

2 1 & 0 x

3 1 to yield

an optimal solution at [1, 0.776, 1] with f = 129.1

NODE 2 NODE 3 Assume x

2 = 0 Assume x

2 = 1

0 x1 1; 0 x

1 1;

x2 = 0 & x

2 = 1 &

0 x3 1 0 x

3 1

LP optimum: [1,0,1] LP optimum:[0.978,1,1]

f= 126.0 f= 128.11


NODE 3 Assume x

2 = 1

0 x1 1; x

2 = 1 & 0 x

3 1

LP optimum:[0.978,1,1] with f= 128.11

NODE 4 NODE 5 Assume x

1 = 0 Assume x

1 = 1

x1 = 0; x

1 = 1;

x2 = 0 & x

2 = 1 &

0 x3 1 0 x

3 1

LP optimum: [0,1,1] LP optimum:[1,1, 0.595]

f= 44.0 f= 113.81

Global optimum was obtained at Node 2.


Mixed Integer Nonlinear Programming (MINLP)

Example 5:

Reaction: A B

Reactor I: conversion 0.8; Cost = 5.5 (Feed)0.6

/hr

Reactor II: conversion 0.667; Cost = 4 (Feed)0.6

/hr

x0: feed flowrate (kmol/hr) of A at 5 /kmol

Desired product, B, at 10 kmol/hr

Select the cheapest combination of the above reactors

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Solution:

Objective function:

Minimise Cost = 5.5 (x1)

0.6

+ 4 (x1)

0.6

+ 5 x0

Operation cost Material cost

Constraints:

Mass balance at initial split: x0 = x

1 + x

2

Mass balance at reactor I: z1 = 0.8 x

1

Mass balance at reactor II: z2 = 0.667 x

2

Mass balance at mixer: z1 + z

2 = 10

Non-negativity constraints: x0 , x

1, x

2 , z

1, z

2 0

DOF = NV NE =5 - 4 = 1 (optimisation problem)


Eliminate z1, z

2 and x

0 from the constraints

Minimise 5.5 (x1)0.6

+ 4 (x1)

0.6

+ 5 x0

subject to 0.8 x1 + 0.667 x

2 = 10

If we eliminate x1, the problem can easily be solved

Let us plot Cost as a function of x2


Discussion

Note that cost function exhibits two minima

at the extreme values 0 and 15 of x2

At x2 = 0 the global optimum (87.5/hr) occurs

at which it corresponds to selecting reactor I

only

At x2 = 15 we have a local optimum which

corresponds to selecting reactor II only

Clearly, this is an undesirable feature as it

means that when using standard NLP

algorithms our solution will be dependent on

the starting point


MINLP Formulation

Introduce binary variables

How to incorporate these binary

variables into the formulation?

Define new constraint:

With upper limits of x1 and x

2

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Define Binary variables

If y1 = 0 (reactor I is not selected); x

1 = 0

If y1 = 1 (reactor I is selected); x

1 = m1

If y2 = 0 (reactor II is not selected); x

2 = 0

If y2 = 1 (reactor II is selected); x

2 = m2

x1 (m1) y

1 0

x2 (m2) y

2 0

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Upper Limit Calculation

Upper limit of x1 is when 10 kmol/hr of B is to be

produced from x1

Similarly upper limit of x2 is when 10 kmol/hr of B

is to be produced from x2

Thus, new constraints are calculated :

m1 = 10/0.8 = 12.5 kmol/h

m2 = 10/0.667 = 15 kmol/h

x1 (12.5) y

1 0

x2 (15) y

2 0


The full MINLP formulation

Minimise 5.5 (x1)0.6

+ 4 (x1)

0.6

+ 5 x0

subject to 0.8 x1 + 0.667 x

2 = 10

x1 (12.5) y

1 0

x2 (15) y

2 0

Non-negativity constraints: x0 , x

1, x

2 , z

1, z

2 0

Integer constraints: y1 and

y

1 = [0,1]

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N COMPUTER SOFTWARE

1. For LP and MILP:

LINDO by Linus Schrage. Interactive easy to use.

ZOOM by Roy Marsten.

LSSOL, LPSOL from Stanford Software Centre

2. For NLP:

GINO by leon Lasdon. Interactive program.

MINOS by Murtagh and Saunders (Stanford Software).

CONOPT by Drud in Denmark

3. For MINLP:

DICOPT++ by Viswanathan and Grossmann.

The program GAMS provides a computer interface that

facilitates the formulation and solution of LP, MILP, NLP,

an MINLP problems. GAMS interfaces with other

commercial packages also.


Thank you

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End of Lecture Set 8


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