Integer Programming Computationally speaking, we can partition problems into two categories.
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Presentation on theme: "Integer Programming Computationally speaking, we can partition problems into two categories."— Presentation transcript:
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Integer Programming
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Computationally speaking, we can partition problems into two categories.
Easy Problems and Hard ProblemsWe can say that easy problem ( or in some languages polynomial problems) are those problems with their solution time proportional to nk. Where n is the number of variables in the problem, and k is a constant, say 2, 3, 4 … Let’s assume that k = 2. Therefore, easy problems are those problems with their solution time proportional to n2. Where n is the number of variables in the problem.Difficult problems are those problems with their solution time proportional to kn, or in our case 2n.
Computational Complexity
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Now
suppose we have an easy problem with n variables. Therefore, the solution time is proportional to n2 . We have solved this problem in 1 hour using a computer with 5 GHz CPU.Suppose we have a new computer with its processing capabilities 10000 times of a 5 GHz computer, and we have one century time. What is the size of the largest problem that we can solve with this revolutionary computer in one century (100 years and about 10000 hours per year).
) 2 = (10)10 n+x / n = (10)5 = 100000n+x = 100,000nThe number of variables in the new problem is 100,000 times greater that the number of variables in the old problem
.
Computational Complexity
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Now supposed we have a hard problem with
n variables. Therefore, the solution time is proportional to 2n . We have solved this problem in one hour using a computer with 900 MHz CPU.Suppose we have a new computer with 900,000 MHz CPU, and we have one century time. What is the size of the largest problem that we can solve in one century using a 900, 000 MHz CPU.1 2n(10)10 2(n+x)Computational Complexity
x = log (10)10x log 2 = 10 log (10)x ( .301) = 10(1)x = 33The number of variables in the new problem is 33 variables greater that the number of variables in the old problem.
Computational Complexity
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Working with Binary Variables
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1-
Exactly one of the two projects is selectedX1+X2 =1 2- At least one of the two projects is selectedX1+X2 ≥1 3-At most one of the two projects is selectedX1+X2 ≤1 4- None of the projects should be selectedX1+X2 =05- Both projects must be selectedX1+X2 =2
6-
none, or one or both of projects are selected7- If project 1 is selected then project 2 must be selectedX2 ≥ X18- If project 1 is selected then project 2 could not be selectedX1+X2 ≤1 !!!
Binary Variables
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More Practice
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Either project 1&2 or projects 3&4&5 are
selectedX1=X2X3=X4=X5X1+X3=1Suppose Y1 is our production of product 1 . Due to technical considerations, we want one of the following two constraints to be satisfied. If the other one is also satisfied it does not create any problem.If the other one is violated it does not create any problem. More Practice
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We can produce 3 products but due to managerial difficulties, we want to produce only two of them.
We can produce these products either in plant 1 or plant 2 but not in both. In other words, we should also decide whether we produce them in plant one or plant two.Other information are given below Required hrs Available hrs Product 1 Product 2 Product 3Plant 1 3 4 2 30Plant 2 4 6
2
40Unit profit 5 7 3 Sales potential 7 5 9
One constraint out of two
Xij
volume of production of product i in plant j, i=1,2,3, j=1,2..
3 ≥ 1600 Suppose that meeting 3 out of 4 of these constraints is “good enough”.
Meeting a Subset of Constraints
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We want two of the following 4 constraints to be satisfied.
The other 2 are free, they may be automatically satisfied, but if they are not satisfied there is no problemy1 + y2 100y1 +2 y3 160y2 + y3 50y1 + y2 + y3 170
In general we want to satisfy k constraints out of n constraints
Still Fun
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A Constraint With k Possible Values
y1 + y2 = 10 or 20 or 100Mastering Formulation of Binary Variables
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We have 5 demand centers, referred to as 1, 2, 3, 4,5.
We plan to open one or more Distribution Centers (DC) to serve these markets. There are 5 candidate locations for these DCs, referred to as A, B, C, D, and E. Annual cost of meeting all demand of each market from a DC located in each candidate location is given below DC1 DC2 DC3 DC4 DC5 Market 1 1 5 18 13 17 Market 2 15 2 26 14 14Market 3 28 18 7 8 8Market 4 100 120 8 8 9Market 5 30 20 20 30 40Each DC can satisfy the demand of one, two, three, four, or five marketA Location Allocation Problem
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Depreciated initial investment and operating cost of a DC in location A, B, C, D, and E is 199, 177, 96, 148, 111.
The objective function is to minimize total cost (Investment &Operating and Distribution) of the system.Suppose we want to open only one DC. Where is the optimal location.Suppose we don't impose any constraint on the number of DCs. What is the optimal number of DCs.A Location Allocation Problem
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Mercer Development is considering the potential of four different development projects. Each project would be completed in at most three years. The required cash outflow for each project is given in the table below, along with the net present value of each project to Mercer, and the cash that is available (from previous projects) each year.
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Example #1 (Capital Budgeting)
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Maximum Flow Problem : D.V. and OF
51222
4
1432
5
3
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Maximum Flow Problem : Arc Capacity Constraints
51222
4
1432
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Maximum Flow Problem : Flow Balance Constraint
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t
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=
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Maximum Flow Problem : Flow Balance Constraint
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Maximum Flow Problem with Restricted Number of Arcs
xij : The decision variable for the directed arc from node i to nod j.xij = 1 if arc ij is on the flow pathxij = 0 if arc ij is not on the flow path x
ij
4
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Maximum Flow Problem with Restricted Number of Arcs
The shortest route between two pointsl ij : The length of the directed arc ij. l ij is a parameter, not a decision variable. It could be the length in term of distance or in terms of time or cost ( the same as c ij )
For those nodes which we are sure that we go from i to j we only have one directed arc from i to j. For those node which we are not sure that we go from i to j or from j to i,
we have
two directed arcs
, one from
i to j,
the other from
j to i.
We may have symmetric or asymmetric network
.
In a
symmetric
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l
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ij
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this condition
does not hold
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Example
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Decision Variables and Formulation
xij : The decision variable for the directed arc from node i to nod j.xij = 1 if arc ij is on the shortest routexij = 0 if arc ij is not on the shortest route
x
ij - xji = 0 i N \ O and D xoj =1 xiD = 1
Min Z =
l
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ij
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Example
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= 0
+ x
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- x
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= 0
….
…..
Min Z = + 5x
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+ 4x
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+ 3x
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+ 2x
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+ 6x
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+ 2x
34
+ 3x
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+ 2x
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+ 2x
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+ 5x
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+ 3x
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+ 2x
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+ 3x
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+ 2x
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The ShR Problem : Binary Decision Variables
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Find the shortest route of these two networks.
But for red bi-directional edges you are not allowed to define two decision variables.
Only one.
Solve the small problem.
Only using 5 variables
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Do not worry about the length of the arcs, we do not need to write the objective functions.
Note that we do not know whether we may go from node 2 to 3 or from 3 to 2. Now we want to formulate this problem as a shortest route. The ShR Problem : Binary Decision Variables 1
4
32
Slide35
To formulate the problem as a shortest route, you probably want to define a pair of decision variables for arcs 2-3 and 3-2, and then for example write the constraint on node 2 as follows
X23+X24 = X12 + X32 But you are not allowed to define two variables for 2-3 and 3-2. You should formulated the problem only using the following variablesX23 is a binary variable corresponding to the NON-DIRECTED edge between nodes 2 and 3. It is equal to 1 if arc 2-3 or 3-2 is on the shortest route and it is 0 otherwise.
The ShR Problem : Binary Decision Variables
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As usual we can have the following variables
X12 X13 X24 X34 each corresponding to a directed arcThe solution of (X12 = 1 , X23 = 1, X34 = 1 all other variables are 0) means that the shortest route is 1-2-3-4.The solution of (X13
= 1
, X32 = 1, X24 = 1 all other variables are 0) ) means that the shortest route is 1-3-2-4.The solution of (X13 = 1 , X34= 1 all other variables are 0) ) means that the shortest route is 1-3-4.
The ShR Problem : Smaller Number of Binary Variables
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Now you should formulate the shortest route problem using defining only one variable for each edge.
Your formulation should be general, not only for this example. How many variables do you need to formulate this problem? The ShR Problem : Smaller Number of Binary Variables
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How Many Binary Variables for this ShR problem
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When are “non-integer” solutions okay?
Solution is naturally divisible Solution represents a rate Solution only for planning purposesWhen is rounding okay?Integer Programming
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Rounded solution may not be feasible
Rounded solution may not be close to optimal There can be many rounded solutionsExample:
Consider a problem with 30 variables that are non-integer in the LP solution. How many possible rounded solutions are there?
The Challenge of Rounding
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How Integer Programs are solved
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Spreadsheet Solution to Example #1
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Suppose the Washington State legislature is trying to decide on locations at which to base search-and-rescue teams. The teams are expensive, and hence they would like as few as possible. However, since response time is critical, they would like every county to either have a team located in that county, or in an adjacent county. Where should the teams be located?
