integer programming (ip) - andrew david joseph · pdf file1 integer programming (ip) ......

January 8, 2002 FOMGT 353 Introduction to Management Science

1

Integer Programming (IP)

An Introduction


2

27.57689 Pancakes• In the Transport Problem from Lecture 11

we used Linear Programming (having assumed that the variables were continuous).

• As it happened the solution was in round numbers of pancakes, and it probably wouldn’t have mattered if we had a solution involving fractional pancakes. (As we would have rounded the numbers down with little practical effect on the answers!)

• Rounding to a solution is called the “LP Relaxation. Care should be taken as we will see in the next slides!


3

Take 2 Examples• Min 78 x1 + 1050 x2

s.t.x1 <= 8- x1 + 10 x2 <= 19.711.72 x1 + 1.23 x2 => 9x1 + 10 x2 => 9.9x1 – x2 <= 7.5

x1, x2 => 0 and Integer

• Max –78 x1 + 1050 x2

s.t.x1 <= 8- x1 + 10 x2 <= 19.711.72 x1 + 1.23 x2 => 9x1 + 10 x2 => 9.9x1 – x2 <= 7.5

x1, x2 => 0 and Integer


4

The Feasible Points

−1 1 2 3 4 5 6 7 8 9

11

23

4

x1

x2

x1 = 1 x2 = 2

x1 = 2 x2 = 2

x1 = 3 x2 = 2

x1 = 4 x2 = 2

x1 = 5 x2 = 2 x1 = 6

x2 = 2 x1 =7 x2 = 2 x1 = 8

x2 = 2

x1 = 8 x2 = 1

x1 = 7 x2 = 1

x1 = 6 x2 = 1

x1 = 5 x2 = 1

x1 = 4 x2 = 1

x1 = 3 x2 = 1

x1 = 2 x2 = 1

x1 = 1 x2 = 1


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Graphical Solutions

−1 1 2 3 4 5 6 7 8 9

11

23

4

x1

x2

The Solution to the IP

Maximization problem is x1 = 1 and

x2 = 2!!

The Solution to the LP

Maximization problem is x1 = 8 and x2 = 2.77!!

Objective Function: max -78 x1 + 1050 x2


Relaxation to the

Maximization problem is x1 = 8 and

x2 = 2!!


6

Graphical Solutions

−1 1 2 3 4 5 6 7 8 9

11

23

4

x1

x2


Minimization problem is x1 = 1 and

x2 = 1!!

The Solution to the LP

Minimization problem is

x1 = 7.718 & x2 = 0.218!!

Objective Function: min 78 x1 + 1050 x2


Relaxation to the Minimization

problem is x1 = 7 and

x2 = 1!!


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Issues With LP Relaxation

• Completely wrong answers, as in the examples.

• Which way do you round? Up or Down?• Sensitivity Analysis from LP is

probably more or less useless in the LP Relaxation.

• Would work well for the pancake problem!


8

LP Relaxation to put Bounds on the Problem

• An LP Relaxation of an IP will allow us to place Bounds on the values of the Decision Variables.

• In our maximization example, from the LP solution we knew that x1 <= 8 and that x2 <= 2.77 and z <= (-78) * 8 + 1050 * 2.77.

• In our minimization example, from the LP solution we knew that x1 => 1 and x2 => 1 z => 78 * 1 + 1050 * 1.


9

Binary Variables

• Binary or “0-1” variables are a special case of Integer variables in that not only must they take on integer values but the range of values is restricted to either zero or one.


10

Mixed LP/IP Problems

• There is no reason not to specify some continuous decision variables and some integer decision variables in the same problem, for example:

Max x1 + x2s.t.• x1 <= 3.2• x2 <= 0.9• x1 => 0 and integer, x2 => 0


11

Computational Aspects of Mixed LP/IP Problems

• Integer Programs are computationally harder to solve than Linear Programs!

• When we solve our example as an LP we only have to look at 5 corners and we have an algorithm which allows us to do so

systematically, improvingthe objective Functionvalue at each step!

−1 1 2 3 4 5 6 7 8 9

11

23

4

x1

x21 2

344


12


• As an IP, our example gives us 16 points that are feasible and which have to be evaluated against one another.

• If the top constraint were one further up the graph, we would have 24 feasible points!!

−1 1 2 3 4 5 6 7 8 9

11

23

4

x1

x2

−1 1 2 3 4 5 6 7 8 9

11

23

4

x1

x2


13


• IP Problems become hard to solve with relatively few Decision Variables.

• Mixed LP/IP problems are easier to solve than IP problems with the same number of Decision Variables.

• Mixed LP/IP (0-1) problems are very useful, as we will see!


14

Product Distribution

• UMass Ventures has employed you to help with their logistics.

• They are looking to expand into China and have identified 4 potential sites for warehouses.

• Management have decided to have at least two warehouses.

• How are you going to model this?


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Enumeration!!• One approach would be to formulate

(4Choose2 + 4Choose3 + 4Choose4) models that will allow us to look at all the possible combinations of two warehouses from four possible warehouses.

• 4Choose2 = 6, 4Choose3 = 4 and 4Choose4 = 1 so in this case we would have to set up and solve 11 models and then select the “best” solution.

• 5 sites 2 or more !!! 6, 7 !!!


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A Little Math - Shreek !• k ! = k * (k-1) * (k-2) * … * (1)• 0 ! = 1 by definition• Examples

3 ! = 3 * (3-1) * (3-2)= 3 * 2 * 1 = 6

4 ! = 4 * (4-1) * (4-2) * (4-3)= 4 * 3 * 2 * 1 = 24


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Combinations (Order Unimportant!!)

• The number of subsets is called the number of Combinations of r symbols taken k at a time, which is denoted:-

rCk or said r choose k

• Example… r = 4; k = 2

kr

)!(!!rCkkrk

rkr

−=

=

613*2

1*23*4

)!2)(1*2(1*2*3*4

)!24(!2!4

2424C ====

−=

=


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A Single Modelk out of n Alternatives Constraint

• Another alternative is to set up a single model with Binary Decision Variables “To Open (1) or Not (0) Warehouse #”,

Say W1, W2, W3 and W4.• Then we need a constraint to force at least 2 of

the 4 to equal one.W1 + W2 + W3 + W4 >= 2 satisfies the requirement!

• If we wanted to choose exactly 2!W1 + W2 + W3 + W4 = 2

– Less than or equal to 2! W1 + W2 + W3 + W4 <= 2


19

A Single ModelConditional or Co-requisite Constraints

• We have set up a constraint to force two or more warehouses into the solution, but this has no effect on the flows into and out of the warehouses!!

• We need to make flows into and out of each warehouse “Conditional” on whether or not the warehouse is opened!

F1 <= 20 W1 (F1 – 20 W1 <= 0)!!!• Makes F1 conditional on W1, if W1 = 0 then F1 <= 0

else F1 <= 20.


20

A Single ModelConditional or Co-requisite Constraints

• Political considerations may require us to open the Warehouse in Beijing if we open one in Shanghai, but not vice versa.

Wb >= Ws will model this co-requisite.

• Wb can equal 1 when Ws is 0, but Wb cannot equal 0 when Ws is 1.

Wb = Ws would force them to be open at the same time or closed at the same time.

• As before we need to express the constraints in the proper form:

Wb – Ws >= 0. AndWb – Ws = 0.


21

Simple Example

• A factory, in Holyoke, capable of making 45,000 widgets a day, has the choice between two sites for warehouses, one in Chicopee and the other in Palmer.

• The CFO says that the daily capital cost of the Chicopee warehouse will be $124 and the Palmer warehouse will be $98.

• There are 3 outlets with demands of 12,000 each.


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ChicopeeWarehouse 1

Cost $124

HolyokeFactory

Capacity 45,000 per day

PalmerWarehouse 2

Cost $98

Outlet 1Demand12,000



F1 (0.35)

F2 (0.45)

W12 (0.12)

W11 (0.25)

W13 (0.21)

W21 (0.14)

W22 (0.27)

W23 (0.18)

W1

W2

Key to FlowsName (Cost in $ per piece)

Simple Example cont…


23

ChicopeeWarehouse 1

Cost $124

HolyokeFactory

Capacity 45,000 per day

PalmerWarehouse 2

Cost $98




F1 (0.35)

F2 (0.45)

W12 (0.12)

W11 (0.25)

W13 (0.21)

W21 (0.14)

W22 (0.27)

W23 (0.18)

W1

W2

Key to FlowsName (Cost in $ per piece)

Simple Example cont…Objective : Min 0.25W11 + 0.12W12 + 0.21W13 + 0.14W21 + 0.27W22 + 0.18W23 + 0.35F1 + 0.45F2 + 124W1 + 98W2

Only 1 Warehouse:

W1 + W2 = 1

W1, W2 Binary Variables

Factory Capacity:

45,000 W1 – F1 => 0

45,000 W2 – F2 => 0

Conservation of Flow:

F1 - W11 - W12 - W13 = 0

F2 - W21 - W22 - W23 = 0

Demand:

W11+ W21 =

12,000

W11+ W21 =

12,000

W11+ W21 =

12,000


24

Write Out The Model!


25

Simple Example cont…• Min 0.25W11 + 0.12W12 + 0.21W13 + 0.14W21 + 0.27W22 + 0.18W23 +

0.35F1 + 0.45F2 + 124W1 + 98W2

s.t.

• Factory Capacity:45,000 W1 – F1 => 045,000 W2 – F2 => 0

• Conservation of Flow:F1 - W11 - W12 - W13 = 0F2 - W21 - W22 - W23 = 0

• Demand:W11+ W21 = 12,000W11+ W21 = 12,000W11+ W21 = 12,000

• Only 1 Warehouse:W1 + W2 = 1

W1, W2 Binary Variables and W11, W12, W13, W21, W22, W33, F1, F2 =>0


26

Reading and Homework.

• Read Chapter 2, Section 2.9.

integer programming (ip) - andrew david joseph · pdf file1 integer programming (ip) ......

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