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Integer Programming

- Introduction to Integer Programming (IP)
- Difficulties of LP relaxation
- IP Formulations
- Branch and Bound Algorithms

Integer Programming Model

- An Integer Programming model is a linear programming problem where some or all of the variables are required to be non-negative integers.
- These models are in general substantially harder than solving linear programming models.
- Network models are special cases of integer programming models and are very efficiently solvable.
- We will discuss several applications of integer programming models.
- We will study the branch and bound technique, one of the most popular algorithm to solve integer programming models.

Classifications of IP Models

Pure IP Model: Where all variables must take integer values.

Maximize z = 3x1 + 2x2

subject to x1 + x2£ 6 x1, x2³ 0, x1 and x2 integer

Mixed IP Model: Where some variables must be integer while others can take real values.

Maximize z = 3x1 + 2x2

subject to x1 + x2£ 6 x1, x2³ 0, x1 integer

0-1 IP Model: Where all variables must take values 0 or 1 .

Maximize z = x1 - x2

subject to x1 + 2x2£ 2 2x1 - x2£ 1, x1, x2 = 0 or 1

Classifications of IP Models (contd.)

LP Relaxation: The LP obtained by omitting all integer or 0-1 constraints on variables is called the LP relaxation of IP.

IP:

Maximize z = 21x1 + 11x2 subject to 7x1 + 4x2£ 13 x1, x2³ 0, x1 and x2 integer

LP Relaxation:

Maximize z = 21x1 + 11x2 subject to 7x1 + 4x2£ 13 x1, x2³ 0

Result:

Optimal objective function value of IP £ Optimal objective function value of LP relaxation

Simple Approaches for Solving IP

Approach 1:

- Enumerate all possible solutions
- Determine their objective function values
- Select the solution with the maximum (or, minimum) value.

Any potential difficulty with this approach?

-- may be time-consuming

Approach 2:

- Solve the LP relaxation
- Round-off the solution to the nearest feasible integer solution

Any potential difficulty with this approach?

-- may not be optimal solution to the original IP

The use of binary variables in constraints

- Any decision situation that can be modeled by “yes”/“no”, “good”/“bad” etc., falls into the binary category.
- To illustrate

The use of binary variables in constraints

- Example
- A decision is to be made whether each of three plants should be built (Yi = 1) or not built (Yi = 0)

RequirementBinary Representation

At least 2 plants must be built Y1 + Y2 +Y3 2

If plant 1 is built, plant 2 must not be built Y1 + Y2£ 1

If plant 1 is built, plant 2 must be built Y1– Y2£ 0

One, but not both plants must be built Y1+ Y2 = 1

Both or neither plants must be built Y1– Y2 =0

Plant construction cannot exceed $17 million

given the costs to build plants are $5, $8, $10 million 5Y1+8Y2+10Y3£ 17

Capital Budgeting Problem

- Stockco Co. is considering four investments
- It has $14,000 available for investment
- Formulate an IP model to maximize the NPV obtained from the investments

IP:

Maximize z = 16x1 + 22x2 + 12x3 + 8x4

subject to

5x1 + 7x2 + 4x3 + 3x4£ 14

x1, x2,,x3, x4= 0, 1

Fixed Charge Problem

- Gandhi cloth company manufactures three types of clothing: shirts, shorts, and pants
- Machinery must be rented on a weekly basis to make each type of clothing. Rental Cost:
- $200 per week for shirt machinery
- $150 per week for shorts machinery
- $100 per week for pants machinery
- There are 150 hours of labor available per week and 160 square yards of cloth
- Find a solution to maximize the weekly profit

Fixed Charge Problem (contd.)

Decision Variables:

x1 = number of shirts produced each week

x2 = number of shorts produced each week

x3 = number of pants produced each week

y1 = 1 if shirts are produced and 0 otherwise

y2 = 1 if shorts are produced and 0 otherwise

y3 = 1 if pants are produced and 0 otherwise

Formulation:

Max. z = 6x1 + 4x2 + 8x3 - 200y1 - 150 y2 - 100y3

subject to

3x1 + 2x2 + 6x3 £ 150

4x1 + 3x2 + 4x3 £ 160 x1 £ M y1, x2 £ M y2, x3 £ M y3

x1, x2,,x3 ³0, and integer; y1, y2,,y3 =0 or 1

Either-Or Constraints

- Dorian Auto is considering manufacturing three types of auto: compact, midsize, large.
- Resources required and profits obtained from these cars are given below.
- We have 6,000 tons of steel and 60,000 hours of labor available.
- If any car is produced, we must produce at least 1,000 units of that car.
- Find a production plan to maximize the profit.

Either-Or Constraints (contd.)

Decision Variables:

x1, x2, x3 = number of compact, midsize and large cars produced

y1, y2, y3 = 1 if compact , midsize and large cars are produced or not

Formulation:

Maximize z = 2x1 + 3x2 + 4x3

subject to

x1£ My1; x2£ My2; x3£ My3

1000 - x1£ M(1-y1)

1000 - x2£ M(1-y2)

1000 - x3£ M(1-y3)

1.5 x1 + 3x2 + 5x3£ 6000

30 x1 + 25x2 + 40 x3£ 60000

x1, x2, x3³ 0 and integer; y1, y2, y3 = 0 or 1

Set Covering Problems

- Western Airlines has decided to have hubs in USA.
- Western runs flights between the following cities: Atlanta, Boston, Chicago, Denver, Houston, Los Angeles, New Orleans, New York, Pittsburgh, Salt Lake City, San Francisco, and Seattle.
- Western needs to have a hub within 1000 miles of each of these cities.
- Determine the minimum number of hubs

Formulation of Set Covering Problems

Decision Variables:

xi = 1 if a hub is located in city i

xi = 0 if a hub is not located in city i

Minimize xAT + xBO + xCH + xDE + xHO + xLA + xNO + xNY + xPI + xSL + xSF + xSE

subject to

Additional Applications

- Location of fire stations needed to cover all cities
- Location of fire stations to cover all regions
- Truck dispatching problem
- Political redistricting
- Capital investments

Branch and Bound Algorithm

- Branch and bound algorithms are the most popular methods for solving integer programming problems
- They enumerate the entire solution space but only implicitly; hence they are called implicit enumeration algorithms.
- A general-purpose solution technique which must be specialized for individual IP's.
- Running time grows exponentially with the problem size, but small to moderate size problems can be solved in reasonable time.

s.t.

An Example

- Telfa Corporation makes tables and chairs
- A table requires one hour of labor and 9 square board feet of wood
- A chair requires one hour of labor and 5 square board feet of wood
- Each table contributes $8 to profit, and each

chair contributes $5 to profit.

- 6 hours of labor and 45 square board feet is

available

- Find a product mix to maximize the profit

Maximize z = 8x1 + 5x2

subject to x1 + x2£ 6; 9x1 + 5x2£ 45; x1, x2³ 0; x1, x2 integer

Feasible Region for Telfa’s Problem

Subproblem 1 : The LP relaxation of original

Optimal LP Solution: x1 = 3.75 and x2 = 2.25 and z = 41.25

Subproblem 2: Subproblem 1 + Constraint x1³ 4

Subproblem 3: Subproblem 1 + Constraint x1£ 3

Feasible Region for Subproblems

Branching : The process of decomposing a subproblem into two or more subproblems is called branching.

Optimal solution of Subproblem 2:

z = 41, x1 = 4, x2 = 9/5 = 1.8

Subproblem 4: Subproblem 2 + Constraint x2³ 2

Subproblem 5: Subproblem 2 + Constraint x2 £ 1

x1 = 3.75

x2 = 2.25

1

x1£ 3

x1³ 4

Subproblem 3

2

Subproblem 2z = 41

x1 = 4

x2 = 1.8

x2£ 1

x2³ 2

Subproblem 5

4

Subproblem 4

Infeasible

The Branch and Bound Tree3

Optimal solution of Subproblem 5:

z = 40.05, x1 = 4.44, x2 = 1

Subproblem 6: Subproblem 5 + Constraint x1³ 5

Subproblem 7: Subproblem 5 + Constraint x1 £ 4

Feasible Region for Subproblems 6 & 7

Optimal solution of Subproblem 7:

z = 37, x1 = 4, x2 = 1

Optimal solution of Subproblem 6:

z = 40, x1 = 5, x2 = 0

z = 41.25

x1 = 3.75

x2 = 2.25

Subproblem 2

z = 41

x1 = 4

x2 = 1.8

Subproblem 5

z = 40.55

x1 = 4.44

x2 = 1

Subproblem 4Infeasible

Subproblem 7

z = 37

x1 = 4

x2 = 1

Subproblem 6

z = 40

x1 = 5

x2 = 0,

The Branch and Bound Tree1

x1³ 4

x1£ 3

Subproblem 3

z = 39

x1 = 3

x2 = 3,

7

2

x2£ 1

x2³ 2

3

4

5

6

Solving Knapsack Problems

Max z = 16x1+ 22x2 + 12x3 + 8x4

subject to

5x1+ 7x2 + 4x3 + 3x4 £ 14

xi = 0 or 1 for all i = 1, 2, 3, 4

LP Relaxation:

Max z = 16x1+ 22x2 + 12x3 + 8x4

subject to

5x1+ 7x2 + 4x3 + 3x4 £ 14

0 £ xi £ 1 for all i = 1, 2, 3, 4

Solving the LP Relaxation:

- Order xi’s in the decreasing order of ci/ai where ci are the cost coefficients and ai’s are the coefficients in the constraint

( Here: x1→x2 → x3 → x4)

- Select items in this order until the constraint is satisfied with equality

z = 44

x1 = x2 = 1

x3 =.5

1

x3 = 1

x3 = 0

Subproblem 3

z = 43.7

x1 =x3= 1,

x2 = .7, x4=0

Subproblem 2

z = 43.3, LB=42

x1 = x2=1

x3 = 0, x4 =.67

2

7

x2 = 1

x2 = 0

x4 = 1

x4 = 0

4

3

Subproblem 5

z = 43.6

x1 =.6, x2=x3=1

x4 = 0, LB = 36

Subproblem 4

z = 36

x1 = x3=1

x2 = 0, x4 =1

Subproblem 8

z = 38, LB=42

x1 = x2=1

x3 = x4 = 0

Subproblem 9

z= 42.85, LB=42

x1 = x4 =1

x3 = 0, x2 = .85

9

8

x1 = 0

x1 = 1

Subproblem 6

z = 42

x1 =0, x2=x3=1

x4 = 1, LB = 42

Subproblem 7

LB = 42

Infeasible

6

5

The Branch and Bound TreeStrategies of Branch and Bound

The branch and bound algorithm is a divide and conquer algorithm, where a problem is divided into smaller and smaller subproblems. Each subproblem is solved separately, and the best solution is taken.

Lower Bound (LB): Objective function value of the best solution found so far.

Branching Strategy : The process of decomposing a subproblem into two or more subproblems is called branching.

Strategies of Branch and Bound (contd.)

Upper Bounding Strategy: The process of obtaining an upper bound (UB) for each subproblem is called an upper bounding strategy.

Pruning Strategy: If for a subproblem, UB £ LB, then the subproblem need not be explored further.

(Illustrate how to fathom nodes in a search tree )

Searching Strategy: The order in which subproblems are examined. Popular search strategies: LIFO and FIFO.

分枝界限法的步驟

- 分枝
- 界限
- 洞悉

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