dtu-management technical university of denmark · 2011. 3. 10. · thomas stidsen 28 dtu-management...
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1Thomas Stidsen
DTU-Management / Operations Research
Benders (once more ...)Thomas Stidsen
DTU-Management
Technical University of Denmark
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2Thomas Stidsen
DTU-Management / Operations Research
OutlineThe algorithm
The assignment from last week
GAMS Implementation
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3Thomas Stidsen
DTU-Management / Operations Research
DefinitionsRBMP: Relaxed Benders Master Problem
BPSP: Benders Primal Sub-Problem: Thedirect formulation for finding the u values.
BDSP: Benders Dual Sub-Problem: Theoriginal problem with fixed y values
RGP: Ray Generation Problem
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4Thomas Stidsen
DTU-Management / Operations Research
But ...When we solve the BPSP ( the u problem) what if:
Problem solved: No problem (we add optimalitycut)
Problem infeasible: We have a solution which isinfeasible because of the y values. Hence weneed a feasibility cut based on an extreme ray.
Problem un-bounded: Original problemun-bounded !
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5Thomas Stidsen
DTU-Management / Operations Research
Benders AlorithmInitialize RBMP, RGP, BDSP/BPSPZLO = −∞, ZUP =∞, y, ǫwhile ZUP − ZLO > ǫ
Solve BDSP(y) or BPSP(y)getting uif feasible
update ZUP
Add Optimality constraintelse
Solve RGP(y) getting uAdd Feasibility constraint
Solve RBMP getting yupdate ZLO
endwhile
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6Thomas Stidsen
DTU-Management / Operations Research
ExampleMin:
z0 = 5x− 3y
s.t.:
x + 2y ≥ 4
2x− y ≥ 0
x− 3y ≥ −13
x ≥ 0
y ∈ {0, . . . , 10}
Optimal value: (x=2.5, y=5), z=-2.5
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7Thomas Stidsen
DTU-Management / Operations Research
Example 10.6: The B matrix
A =
2
6
6
4
1
2
1
3
7
7
5
B =
2
6
6
4
2
-1
-3
3
7
7
5
bT = [4, 0,−13]
cT = [5]
fT = [−3]
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8Thomas Stidsen
DTU-Management / Operations Research
The Optimality cutsThe optimality cuts we are looking for are:
z0 ≥ fTy + (ui)T (b−By) i = 1, . . . , q
Which becomes:z0 ≥ −3y + (ui)T ([4, 0,−13]T − [2,−1,−3]T y)
Hence, every time we find new optimality cut values,i.e. u values, we only have to plug them in, in the
above equation.
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9Thomas Stidsen
DTU-Management / Operations Research
ExampleMin:
z0 = 5x− 3y
s.t.:
x ≥ 4− 2y
2x ≥ 0 + y
x ≥ −13 + 3y
x ≥ 0
y ∈ {0, . . . , 10}
Optimal value: (x=2.5, y=5), z=-2.5
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10Thomas Stidsen
DTU-Management / Operations Research
1. Iteration: SubproblemMax:
z1 = (4− 2y)u1 + y · u2 + (−13 + 3y)u3
s.t.:
u1 + 2u2 + u3 ≤ 5
ui ≥ 0
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11Thomas Stidsen
DTU-Management / Operations Research
InititialisationInitial solution
y0 = 0.
Upper bound:UP0 =∞.
Lower bound:LO0 = −∞.
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12Thomas Stidsen
DTU-Management / Operations Research
Graphical BMP, initialisation101 Y
Z
Z
120
100
80
20
60
40
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13Thomas Stidsen
DTU-Management / Operations Research
1. Iteration: SubproblemMax:
z1 = 4u1 − 13u3
s.t.:
u1 + 2u2 + u3 ≤ 5
ui ≥ 0
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14Thomas Stidsen
DTU-Management / Operations Research
1. Iteration: SubproblemOptimal solution:
z1 = 20
u = (5, 0, 0)
Giving a new upper bound:UP1 = min{z1, UP0(=∞)} = 20. And arriving at thecut:
z0 ≥ −3y + (5, 0, 0)T ([4, 0,−13]T − [2,−1,−3]T y)
=
z0 ≥ −13y + 20
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15Thomas Stidsen
DTU-Management / Operations Research
1. Iteration: Benders Master ProblemMin:
z1
s.t.:
z1 ≥ 20− 13y
y ∈ {0, . . . , 10}
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16Thomas Stidsen
DTU-Management / Operations Research
1. Iteration: Benders Master ProblemOptimal solution
z1 = −110
y = 10
New lower bound:LO1 = max{z1, LO0(= −∞)} = −110.
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17Thomas Stidsen
DTU-Management / Operations Research
Graphical BMP, 1. iteration
Z >= 20 − 13y
Y
Z
Optimum
20
40
60
80
100
120
1
10
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18Thomas Stidsen
DTU-Management / Operations Research
2. iteration: Primal SubproblemMax:
z2 = −16u1 + 10u2 + 17u3 − 30
s.t.:
u1 + 2u2 + u3 ≤ 5
ui ≥ 0
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19Thomas Stidsen
DTU-Management / Operations Research
2. iteration: Primal SubproblemOptimal solution:
z2 = 55
u = (0, 0, 5)
New upper bound: UP2 = min{z2, UP1(= 20)} = 20.
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20Thomas Stidsen
DTU-Management / Operations Research
2. iteration: Benders Master ProblemMin:
z2
s.t.:
z2 ≥ 20− 13y
z2 ≥ −65 + 12y
y ∈ {0, . . . , 10}
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21Thomas Stidsen
DTU-Management / Operations Research
2. iteration: Benders Master ProblemOptimal solution:
z2 = −19
y = 3
New upper bound:LO2 = min{z2, LO1(−110)} = −19.
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22Thomas Stidsen
DTU-Management / Operations Research
Graphical BMP, 2. iteration
Optimum
Y
Z
20
40
60
80
100
120
1
10
Z >= 20 − 13y
Z >= −65 + 12y
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23Thomas Stidsen
DTU-Management / Operations Research
3. iteration: Primal SubproblemMax:
z3 = −2u1 + 3u2 − 4u3 − 9
s.t.:
u1 + 2u2 + u3 ≤ 5
ui ≥ 0
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24Thomas Stidsen
DTU-Management / Operations Research
3. iteration: Primal SubproblemOptimal solution:
z2 = −1.5
u = (0, 2.5, 0)
New upper bound:UP3 = min{z3, UP2(= 19)} = −2.5.
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25Thomas Stidsen
DTU-Management / Operations Research
3. iteration: Benders Master ProblemMin:
z3
s.t.:
z3 ≥ 20− 13y
z3 ≥ −65 + 12y
z3 ≥ −0.5y
y ∈ {0, . . . , 10}
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26Thomas Stidsen
DTU-Management / Operations Research
3. iteration: Benders Master ProblemOptimal solution:
z3 = −2.5
y = 5
Newlowerbound:LO3 = max{z1, LO3(= −19)} = −2.5.Now: LO3 = UP3
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27Thomas Stidsen
DTU-Management / Operations Research
Graphical BMP, 3. iteration
Z >= 20 − 13y
Y
Z
Optimum 20
40
60
80
100
120
1
10
Z >= −65 + 12y
Z >= −0.5y
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28Thomas Stidsen
DTU-Management / Operations Research
Exercise 10.1 (RKM)Several constraints are added without improvingbound (in case of minimization, non-increasing)why: Primal degeneracy: Possible to include newvariables into the simplex tableau without changingthe objective value (the new variables may stay atvalue zero). Dual degeneracy: If the current“solution” has several active constraints.
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29Thomas Stidsen
DTU-Management / Operations Research
Exercise 10.2 (RKM)If the Bender’s subproblem is infeasible what doesthat imply about the original problem ?
Original problem un-bounded ! Why ?
Bender’s Dual Sub-Problem is thenun-bounded.
Bender’s Dual Sub-Problem is a restrictedversion of the original problem meaning, hencethe original problem is also un-bounded.
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30Thomas Stidsen
DTU-Management / Operations Research
How did we get here ?
Original Problem
Full BendersMaster Problem
Project out x variables
Relaxed
Benders Problem
Benders DualSup−Problem
(Original Problem)
Ray GenerationProblem
Drop some (all ?) constraints
BendersSub−Problem
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31Thomas Stidsen
DTU-Management / Operations Research
Master Problem: LowerboundFarka’s lemma guarantees directcorrespondence between original problem andthe Full Benders Master Problem
The Relaxed Benders Master Problem willhence always give a lower bound ...
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32Thomas Stidsen
DTU-Management / Operations Research
Sub Problem: UpperboundBender’s dual problem is a restricted version ofthe original problem, hence a solution to thesub-problem is also a solution to the originalproblem
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33Thomas Stidsen
DTU-Management / Operations Research
The Benders Sub-problemMax:
fTy + (b−By)Tu
s.t.:
ATu ≤ c
u ≥ 0
Problem: If the objective is un-bounded, how to find
rays ?
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34Thomas Stidsen
DTU-Management / Operations Research
Extreme raysWe need to find the extreme rays in the cone{u|ATu ≤ 0, u ≥ 0}, such that a cut(ui)T (b−By) > 0 is found, see Proposition 10.3,hence:Max:
(b−By)Tu
s.t.:
ATu ≤ 0
0 ≤ u ≤ L
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35Thomas Stidsen
DTU-Management / Operations Research
Ray generation: An exampleMin:
2x1 + 6x2 + 2y1 + 3y2
s.t.:
−x1 + 2x2 + 3y1 − y2 ≥ 5
x1 − 3x2 + 2y1 + 2y2 ≥ 4
x1, x2 ≥ 0 y ∈ {0, 1, 2}
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36Thomas Stidsen
DTU-Management / Operations Research
Primal SubproblemMax:
u1 · [5− 3y1 + y2] + u2 · [4− 2y1 − 2y2]
s.t.:
−u1 + u2 ≤ 2
2u1 − 3u2 ≤ 6
u1, u2 ≥ 0
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37Thomas Stidsen
DTU-Management / Operations Research
Visually
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u2=(0,2)
u3=(3,0)
(u1,u2)=(1,1)
u1=(0,0)
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(u1,u2)=(3,2)
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38Thomas Stidsen
DTU-Management / Operations Research
Objectives for subproblem(5− 3y1 + y2, 4− 2y1 − 2y2), and y1, y2 ∈ {0, 1, 2}
(0, 0) : (5, 4)
(0, 1) : (6, 2)
(0, 2) : (7, 0)
(1, 0) : (2, 2)
(1, 1) : (3, 0)
(1, 2) : (4,−2)
(2, 0) : (−1, 0)
(2, 1) : (0,−2)
(2, 2) : (1,−4)
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39Thomas Stidsen
DTU-Management / Operations Research
Visually with possible objectives
(0,−2)
(3,0)
(7,0)
(4,−2)
(5,4)
(6,2)
(1,−4)
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(−1,0)
(2,2)
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40Thomas Stidsen
DTU-Management / Operations Research
First iterationInitial guess for (y1, y2) = (1, 1):Max:
u1 · [5− 3y1 + y2] + u2 · [4− 2y1 − 2y2] = 3u1
s.t.:
−u1 + u2 ≤ 2
2u1 − 3u2 ≤ 6
u1, u2 ≥ 0
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41Thomas Stidsen
DTU-Management / Operations Research
Tabular LP form, first iterationB.V. u1 u2 d1 d2 R.S. UBZ -3 0 0 0 0d1 -1 1 1 0 2 no limitd2 2 -3 0 1 6 u1 ≤
6
2= 3 ← min
B.V.: Basic VariablesR.S.: Right hand side
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42Thomas Stidsen
DTU-Management / Operations Research
Tabular LP form, first iterationB.V. u1 u2 d1 d2 R.S. UBZ 0 -9
20 3
29
d1 0 -1
21 1
25 no limit
u1 1 -3
20 1
23 no limit
We can increase u2 as much as we like !!! Ok, sowhat is the extreme ray ?
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43Thomas Stidsen
DTU-Management / Operations Research
The simplex tableau as equations
−1
2u2 + d1 +
1
2d2 = 5
u1 −3
2u2 +
1
2d2 = 3
But when we use the simplex algorithm, we only
want to increase one variable at the time, and we
hence freeze the slack variable d2 = 0
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44Thomas Stidsen
DTU-Management / Operations Research
After the freezeu1 = 3 +
3
2u2
d1 = 5 +1
2u2
⇓
u2 =2
3u1 − 2
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45Thomas Stidsen
DTU-Management / Operations Research
To find extreme rays by problem modificationsOk, that’s all very nice, but what about usingcomputer codes for finding the rays ?
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46Thomas Stidsen
DTU-Management / Operations Research
Extreme RayThe extreme rays arises when our dual subproblemis unbounded:Max:
dummyz
s.t.:
dummyz = 0
fTy + (b−By)Tu = 1
ATu ≤ 0
u ≥ 0
What about the objective ???
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47Thomas Stidsen
DTU-Management / Operations Research
ModificationsZero right hand sides
Dummy objective (the objective does not matter!)
Fix value of objective function to 1
What solution will the simplex or dual-simplex solvefind ?
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48Thomas Stidsen
DTU-Management / Operations Research
ExampleA good example with the fixed chargeTransportation Problem ... see the ErwinKalvelagen note.
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49Thomas Stidsen
DTU-Management / Operations Research
The Fixed Charge Transportation problemMin: ∑
i,j
(fi,jyi,j + ci,jxi,j)
s.t.:∑
j
xi,j = si ∀i
∑
i
xi,j = dj ∀j
xi,j ≤ Miyi,j ∀i, j
xi,j ≥ 0 yi,j ∈ {0, 1}
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50Thomas Stidsen
DTU-Management / Operations Research
Benders (primal) subproblemMax:
∑
i
(−si)ui +∑
j
djvj +∑
i,j
(−Mi,jyi,j)wi,j
s.t.:
ui + vj − wi,j ≤ ci,j ∀i, j
ui, vj, wi,j ≥ 0
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51Thomas Stidsen
DTU-Management / Operations Research
Benders Master ProblemMin:
z
s.t.:
z ≥∑
i,j
fi,jyi,j +∑
i
(−si)uki +
∑
j
djvkj
+∑
i,j
(−Mi,jwki,j)yi,j ∀k
0 ≥∑
i
(−si)uk′
i +∑
j
djvk′
j
+∑
i,j
(−Mi,jwk′
i,j)yi,j ∀k′
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52Thomas Stidsen
DTU-Management / Operations Research
Extreme raysWhat about the extreme rays ? How doesKalvelagen handle these ?
How does he detect an extreme ray ? He addsan artificial upper bound on the objective !
If the optimal objective value is significantly lessthan the upper bound, he assumes it to becorrect ...
If there is an extreme ray, he adds the dummyobjective adds the objective constraint and fixthe original objective value to 1 and solve theproblem ...
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53Thomas Stidsen
DTU-Management / Operations Research
Changed Benders SubproblemMax:
dummy
s.t.:
dummy = 0∑
i
(−si)ui +∑
j
djvj +∑
i,j
(−Mi,jyi,j)wi,j = 1
ui + vj − wi,j ≤ ci,j ∀i, j
ui, vj, wi,j ≥ 0