NETWORK FLOWS AND COMBINATORIAL OPTIMIZATION by dffhrtcv3

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									               Integer Programming




 Introduction to Integer Programming (IP)

 Difficulties of LP relaxation

 IP Formulations

 Branch and Bound Algorithms




                              0
              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.



                                1
               Classifications of IP Models
Pure IP Model: Where all variables must take integer values.

Maximize     z = 3x1 + 2x2
subject to   x1 + x 2  6
                x1, x2  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 + x 2  6
             x1, x2  0, x1 integer

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

Maximize     z = x 1 - x2
subject to   x1 + 2x2  2
             2x1 - x2  1, x1, x2 = 0 or 1

                                   2
          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, x2  0, x1 and x2 integer

LP Relaxation:
          Maximize z = 21x1 + 11x2
          subject to
                       7x1 + 4x2  13
                       x1 , x2  0

Result:
           Optimal objective function value of IP 
           Optimal objective function value of LP relaxation

                                  3
             IP and LP Relaxation




     3


     2


x2                      7x1 + 4x2= 13
     1

         x     x          x             x
               1          2             3

                   x1




                           4
            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


                                  5
     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



                      1 If a new health care plan is adopted
                  X  0 If it is not
                      

           1 If a new police station is built downtown
       X  0 If it is not
           

               1 If a particular constraint must hold
           X  0 If it is not
               

                                    6
            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)

   Requirement                                      Binary 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  
   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




                                           7
                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

               Investment 1        2        3        4
               choice
               Cash       $5000    $7000    $4000    $3000
               outflow
               NPV        $16000   $22000   $12000   $8000
IP:

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

             5x1 + 7x2 + 4x3 + 3x4  14
             x1, x2,,x3, x4  0, 1

                                       8
                     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
                      Labor hr Cloth yd   Price
            Shirts       3        4        $6
            Shorts       2        3        $4
            Pants        6        4        $8

                                     9
                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  0, and integer; y1, y2,,y3  0 or 1

                                      10
                      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.
                         Compact    Midsize    Large
              Steel Req. 1.5 tons   3 tons     5 tons
              Labor Req. 30 hours   25 hours   40 hours
              Profit     $2000      $3000      $4000


                                       11
              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, x3  0 and integer; y1, y2, y3 = 0 or 1

                                    12
                 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
                                         Cities within 1000 miles
                  Atlanta (AT)           AT, CH, HO, NO, NY, PI
                  Boston (BO)            BO, NY, PI
                  Chicago (CH)           AT, CH, NY, NO, PI
                  Denver (DE)            DE, SL
                  Houston (HO)           AT, HO, NO
                  Los Angeles (LA)       LA, SL, SF
                  New Orleans (NO)       AT, CH, HO, NO
                  New York (NY)          AT, BO, CH, NY, PI
                  Pittsburgh (PI)        AT, BO, CH, NY, PI
                  Salt Lake City (SL)    DE, LA, SL, SF, SE
                  San Francisco (SF)     LA, SL, SF, SE
                  Seattle (SE)           SL, SF, SE


                                        13
        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
                   AT BO CH DE HO LA NO NY PI SL SF SE   Required
subject to    AT    1 0 1 0 1 0 1 1 1 0 0 0 xAT          >=     1
              BO    0 1 0 0 0 0 0 1 1 0 0 0 xBO          >=     1
              CH    1 0 1 0 0 0 1 1 1 0 0 0 xCH          >=     1
              DE    0 0 0 1 0 0 0 0 0 1 0 0 xDE          >=     1
              HO    1 0 0 0 1 0 1 0 0 0 0 0 xHO          >=     1
              LA    0 0 0 0 0 1 0 0 0 1 1 0 xLA          >=     1
              NO    1 0 1 0 1 0 1 0 0 0 0 0 xNO          >=     1
              NY    1 1 1 0 0 0 0 1 1 0 0 0 xNY          >=     1
              PI    1 1 1 0 0 0 0 1 1 0 0 0 xPI          >=     1
              SL    0 0 0 1 0 1 0 0 0 1 1 1 xSL          >=     1
              SF    0 0 0 0 0 1 0 0 0 1 1 1 xSF          >=     1
              SE    0 0 0 0 0 0 0 0 0 1 1 1 xSE          >=     1




                                         14
                  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




                                15
             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.




                                16
Example:




s.t.




           17
Solved as LP by Simplex method




                             18
                           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, x2  0; x1, x2 integer

                                     19
          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 x1  4

Subproblem 3: Subproblem 1 +
Constraint x1  3




                                20
            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 x2  2

Subproblem 5: Subproblem 2 +
Constraint x2  1



                                     21
Feasible Region for Subproblems 4 & 5




                  22
                            The Branch and Bound Tree
                                          Subproblem 1
                                            z = 41.25
                                 1           x1 = 3.75
                                            x2 = 2.25
                        x1  4                                   x1  3

                     Subproblem 2                               Subproblem 3
                        z = 41
              2
                         x1 = 4
                        x2 = 1.8
     x2  2                                   x2  1

        Subproblem 4                     Subproblem 5
3         Infeasible                                        4




    Optimal solution of Subproblem 5:

                  z = 40.05,         x1 = 4.44,    x2 = 1

    Subproblem 6: Subproblem 5 + Constraint x1  5

    Subproblem 7: Subproblem 5 + Constraint x1  4

                                                       23
          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




                             24
                         The Branch and Bound Tree
                                    Subproblem 1
                                1     z = 41.25
                                       x1 = 3.75
                                      x2 = 2.25
                      x1  4                                x1  3

                  Subproblem 2                             Subproblem 3
                     z = 41                                   z = 39      7
             2        x1 = 4                                   x1 = 3
                     x2 = 1.8                                 x2 = 3,
    x2  2                               x2  1
                                    Subproblem 5
       Subproblem 4
                                      z = 40.55
3        Infeasible                                    4
                                       x1 = 4.44
                                        x2 = 1



                 Subproblem 6                          Subproblem 7
                    z = 40                                z = 37
        6                                                  x1 = 4         5
                     x1 = 5
                    x2 = 0,                               x2 = 1




                                                  25
                    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


                                        26
                    The Branch and Bound Tree
                                Subproblem 1
                                    z = 44
                          1       x1 = x 2 = 1
                                    x3 =.5

                                                              x3 = 1
                 x3 = 0


                                                                  Subproblem 3
              Subproblem 2
                                                                     z = 43.7
             z = 43.3, LB=42                                                       2
                                                                    x1 =x3= 1,
       7         x1 = x2=1
              x3 = 0, x4 =.67                                      x2 = .7, x4=0

                                                                                       x2 = 1
    x4 = 0                          x4 = 1               x2 = 0
                                                                    3                        4
     Subproblem 8                Subproblem 9               Subproblem 4       Subproblem 5
     z = 38, LB=42              z= 42.85, LB=42                 z = 36             z = 43.6
8       x1 = x2=1             9     x1 = x4 =1                 x1 = x3=1       x1 =.6, x2=x3=1
       x3 = x 4 = 0              x3 = 0, x2 = .85            x2 = 0, x4 =1     x4 = 0, LB = 36

                                                             x1 = 0                    x1 = 1

                                                 Subproblem 6
                                                                             Subproblem 7
                                                     z = 42
                                                                                LB = 42
                                             5   x1 =0, x2=x3=1            6
                                                                               Infeasible
                                                 x4 = 1, LB = 42




                                                    27
              Strategies 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.




                                28
         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.




                                29
11.6 分枝界限法及其在二位元整數規劃的應用




        30
       分枝界限法的步驟

 分枝
 界限
 洞悉




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分枝




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界限




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洞悉




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洞悉測試摘要




   35
BIP 分枝界限演算法摘要




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完成例題解題過程




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應用分枝界限法的其他選擇




      41
11.7 混合整數規劃之分枝界限演算法




       42
MIP分枝界限演算法摘要




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