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APPLIED INTEGER PROGRAMMING,
MODELING AND SOLUTION
CHAPTERS 14.4-14.15
Sara Gestrelius
Der-San Chen, Robert G. Batson, Yu Dang
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CHAPTER 14.5: LAGRANGIAN
DUAL
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LAGRANGIAN RELAXATION
(IP)
Complicated constraints.
Easy constraints.
nZ
ts
x
bxA
bxA
cx
22
11..
min
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LAGRANGIAN RELAXATION
(IP) (LRIP)
nZ
ts
x
bxA
bxA
cx
22
11..
min
nZ
ts
x
bxA
xAbucx
22
11
..
)(min
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LAGRANGIAN RELAXATIONNOTE: Less than or equal
to zero if original IP
feasible!
(IP) (LRIP)
nZ
ts
x
bxA
bxA
cx
22
11..
min
nZ
ts
x
bxA
xAbucx
22
11
..
)(min
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LAGRANGIAN RELAXATION
nZ
ts
x
bxA
bxA
cx
22
11..
min
nZ
ts
x
bxA
xAbucx
22
11
..
)(min
NOTE: Less than or equal
to zero if original IP
feasible!
So the Lagrangian relaxation optimal value can be used
as a lower bound to the original IP. But we want a tight
bound…
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LAGRANGIAN DUAL PROBLEM
22
11Zxu
bxA
xAbucx
..
)(minmax0
ts
n
…and to find the largest lower bound we solve the
Lagrangian dual problem.
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LAGRANGIAN DUAL PROBLEM
22
11Zxu
bxA
xAbucx
..
)(minmax0
ts
n
…and to find the largest lower bound we solve the
Lagrangian dual problem.
Dual variable aka
Lagrange multiplier
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LAGRANGIAN DUAL PROBLEM
22
11Zxu
bxA
xAbucx
..
)(minmax0
ts
n
…and to find the largest lower bound we solve the
Lagrangian dual problem.
Dual variable aka
Lagrange multiplier
22
1Z
1
bxA
xuAcub
..
)(minmax0
ts
nxu
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LAGRANGIAN DUAL PROBLEM
Proposition 14.8: Given a specific , if the
following two conditions are satisfied, then the
optimal vector of LRIP(u) is optimal for IP:
1.
2. whenever
11 )(* bxA u
iiu 111 )(* bbxA 0iu
(u)*x
1mRu
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EQUALITY INSTEAD OF
INEQUALITY
22
1Z
1
bxA
xuAcub
..
)(minmax
ts
nxu
nZ
ts
x
bxA
bxA
cx
22
11..
minNo sign restriction on u!
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LP DUAL OF IP
Proposition 14.5: The integer program (IP) and the dual of its
linear programming relaxation (Relaxed IP) form a weak dual
pair.
IP
Relaxed IP
Primal (minimization)
Dual (maximization)
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CHAPTER 14.6: PRIMAL –DUAL
SOLUTION VIA BENDERS’
PARTITIONING
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PRIMAL-DUAL SOLUTION VIA
BENDERS’ PARTITIONING
integerand0,0
..
min
2
21
yx
bDy
byAxA
ycxc
2
11ts
MIPs are hard. Let’s try to partition it.
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PRIMAL-DUAL SOLUTION VIA
BENDERS’ PARTITIONING
integerand0,0
..
min
2
21
yx
bDy
byAxA
ycxc
2
11ts
integerand0
..
)(min 2
y
bDy
yyc
2ts
q
0
..
min)(
21
1
x
yAbxA
xc
1ts
yq
MASTER SUB-PROBLEM
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PRIMAL-DUAL SOLUTION VIA
BENDERS’ PARTITIONING
integerand0,0
..
min
2
21
yx
bDy
byAxA
ycxc
2
11ts
integerand0
..
)(min 2
y
bDy
yyc
2ts
q
0
..
min)(
21
1
x
yAbxA
xc
1ts
yq
It’d be great if the y’s weren’t in the constraints of q(y).
DUAL!
MASTER SUB-PROBLEM
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DUAL OF SUB-PROBLEM
0
..
)'(max
11
2
u
cuA
yAbu 1
ts
0
..
min)(
21
1
x
yAbxA
xc
1ts
yq
SUB-PROBLEM
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DUAL OF SUB-PROBLEM
0
..
)'(max
11
2
u
MuE
cuA
yAbu 1
ts0
..
min)(
21
1
x
yAbxA
xc
1ts
yq
SUB-PROBLEM
Avoid unboundedness.
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PRIMAL-DUAL SOLUTION VIA
BENDERS’ PARTITIONING
0
..
)'(max
11
2
u
MuE
cuA
yAbu 1
ts
1. For every y’ we will pick out a vertex ui of the dual
feasible solution space. From weak duality:
)()( 2yAbuy 1 iq
Dual of sub-problem
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PRIMAL-DUAL SOLUTION VIA
BENDERS’ PARTITIONING
0
..
)'(max
11
2
u
MuE
cuA
yAbu 1
ts)()( 2yAbuy 1 iq
)()( 2yAbuycyyc 122 iqz
2. So, we have that the master objective
Dual of sub-problem1. For every y’ we will pick out a vertex ui of the dual
feasible solution space. From weak duality:
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PRIMAL-DUAL SOLUTION VIA
BENDERS’ PARTITIONING
0
..
)'(max
11
2
u
MuE
cuA
yAbu 1
ts)()( 2yAbuy 1 iq
)()( 2yAbuycyyc 122 iqz
2. So, we have that the master objective
3. ADD CONSTRAINT TO MASTER!
integerand0
..
)(min 2
y
bDy
yyc
2ts
q
Dual of sub-problem
Master
1. For every y’ we will pick out a vertex ui of the dual
feasible solution space. From weak duality:
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PRIMAL-DUAL SOLUTION VIA
BENDERS’ PARTITIONING
0
..
)'(max
11
2
u
MuE
cuA
yAbu 1
ts)()( 2yAbuy 1 iq
)()( 2yAbuycyyc 122 iqz
2. So, we have that the master objective
3. ADD CONSTRAINT TO MASTER!
integerand0
...1)(..
min
22
y
bDy
yAbuyc
2
1 Tizts
z
i integerand0
..
)(min 2
y
bDy
yyc
2ts
q
Dual of sub-problem
Master
For every y’ we will pick out a vertex ui of the dual feasible
solution space. From weak duality:
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THE MASTER
integerand0
...1)(..
min
22
y
bDy
yAbuyc
2
1 Tizts
z
i
The number of vertices (T) can be very large,
so instead of solving the Master with all T, start
with one and then generate more constraints
as needed.
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BENDERS’ ALGORITHM FOR MIP
Step 0: Initialization
Set 𝑡 = 1,𝐵𝑢 = +∞ and select some 𝜀 (convergence criterion). Select some u1
that is feasible for the dual sub-problem.
Step1: Solve master in iteration t
Solve the relaxed pure integer program:
Let zt and yt be the solution. If z is unbounded from below, take yt to be some
value that gives zt some arbitrarily large negative value.
integerand0
...1)(..
min
22
y
bDy
yAbuyc
2
1 tizts
z
i
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BENDERS’ ALGORITHM FOR MIP
Step 2: Solve sub-problem in iteration t
Generate the most violated constraint of IP by solving the linear program:
Let the solution of this LP be optimal value 𝑢0𝑡+1 at 𝒖𝑡+1.
Step 3: Optimality check
Check convergence criterion. Set 𝐵𝑢 = 𝑚𝑖𝑛 𝐵𝑢, 𝑢0𝑡+1 + 𝒄1𝒚
𝑡 . If 𝑧𝑡 > 𝐵𝑢 − 𝜀, stop; the optimal solution has been reached. Otherwise, add the constraint 𝑧 ≥𝒄2𝒚 + 𝒖𝑡+1(𝒃1 − 𝑨2𝒚) to the master problem. Set 𝑡 = 𝑡 + 1. Return to step 1.
0
..
)'(max
11
2
u
MuE
cuA
yAbu 1
ts