Part II
General Integer Programming
II.1
The Theory of Valid Inequalities
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 Let S = {xZ+n: Ax  b}
P = {xR+n: Ax  b}
S = P  Zn
 Have max{cx: xS} = max{cx: xconv(S)}. How can we construct
inequalities describing conv(S)? Use integrality and valid inequalities for P to
construct valid inequalities for S.
 Def: Valid inequalities x  0 and x  0 are said to be equivalent if (, 0) =
(, 0) for some  > 0.
x  0 dominates or is stronger than x  0 if they are not equivalent and
there exists  > 0 such that    and 0  0.
A maximal valid inequality is one that is not dominated by any other
inequality.
 A maximal inequality for S defines a nonempty face of conv(S), but not
conversely.
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 Valid inequality for P = {xR+n: Ax  b} ( not conv(S))
For all uR+m, v R+n, and  R+1,
(uA - v)x  ub +  is valid for P
uAx  ub is valid  uAx  ub + is valid
-x  0 is valid  -vx  0 is valid (weakening0
 (uA - v)x  ub +  is valid
 Prop 1.1: (, 0) valid for P = {xR+n: Ax  b}. Then x  0 equivalent to or
dominated by uAx  ub, uR+m, if any of the following conditions hold:
a. P   (In this case, no more than min(m, n) components of u need be positive)
b. {uR+m: uA  }  
c. A = 𝐴′
𝐼
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Pf) a) max{x: xP}  0
 Dual has optimal solution. Dual is min{ub: uA  , u  0}
Take basic dual optimal solution
 u0A  , u0  0, u0b  0
 min(m, n) of components of u0 is positive.
b) {uR+m: uA  }   . Dual feasible.
If P  , case a)
If P = , assume 𝑢A  .
If 𝑢b  0, done
Otherwise P =    𝑢  R+m such that 𝑢A  0, 𝑢b < 0.
 for some  > 0, (𝑢 + 𝑢)A  , (𝑢 + 𝑢)b  0.
c) A = 𝐴′ . Clear that  u  R+n such that uA  . Reduces to case b)
𝐼
 Frequently assume A = 𝐴′ to avoid trouble when P = , D = .
𝐼
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Integer Rounding
 Ex: Matching problem on G = (V, E)
Constraints:
𝑒∈𝛿(𝑖) 𝑥𝑒
(1.2)
(1.3)
≤ 1 for all 𝑖 ∈ 𝑉
𝑥 ∈ 𝑍+|E|,
Can add constraints
(1.4)
𝑒∈𝐸(𝑈) 𝑥𝑒
≤ 𝑘, if |U| = 2k+1, k  1.
Constraints (1.4) cannot be obtained by taking nonnegative linear combination
of (1.2)
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 How to generate (1.4)?
½  (

𝑒∈𝛿(𝑖) 𝑥𝑒
𝑒∈𝐸(𝑈) 𝑥𝑒
≤ 1 ), 𝑖 ∈ 𝑈, and add up
1
2
+
𝑒∈𝛿(𝑈) 𝑥𝑒
−12𝑥𝑒 ≤ 0
≤ 12 𝑈
for all 𝑒 ∈ 𝛿(𝑈)
Add up (weakening in this step)
1
𝑥
≤
𝑈
𝑒
𝑒∈𝐸(𝑈)
2
Since 𝑥𝑒 ∈ {0, 1}, left-hand side is integral.
Therefore, we can replace the right-hand side by  .
If 𝑈 is odd,
𝑒∈𝐸(𝑈) 𝑥𝑒 ≤
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2
𝑈
is a valid inequality for S. (strengthening)
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Chvatal-Gomory (C-G) Rounding Method
 For the set 𝑆 = {𝑥 ∈ Z+n: 𝐴𝑥 ≤ 𝑏},
𝐴 = (𝑎1 , … , 𝑎𝑛 )
1.
𝑗∈𝑁(𝑢𝑎𝑗 )𝑥𝑗
≤ 𝑢𝑏 for all 𝑢 ≥ 0;
2.
𝑗∈𝑁(
3.
𝑛
(
𝑢𝑎
)𝑥
≤
𝑢𝑏
,
since
𝑥
∈
𝑍
implies
𝑗
𝑗
𝑗∈𝑁
𝑢𝑎𝑗 )𝑥𝑗 ≤ 𝑢𝑏, since 𝑥 ≥ 0 implies −
𝑗∈𝑁
𝑗∈𝑁(
𝑢𝑎𝑗 − 𝑢𝑎𝑗 𝑥𝑗 ≤ 0
𝑢𝑎𝑗 )𝑥𝑗 is an integer.
 Step 2 is weakening and step 3 is strengthening
The procedure can be used recursively
Need at most n inequalities in the procedure (Prop 1.1)
Can generate all valid inequalities for S using C-G procedure
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Optimizing over the First Chvátal closure
 Ref: B&W, p187
 𝑃 = {𝑥 ∈ R+n: 𝐴𝑥 ≤ 𝑏}
𝑃1 = {𝑥 ∈ R+n: 𝐴𝑥 ≤ 𝑏,
𝑛
𝑗=1(
𝑢′𝐴𝑗 )𝑥𝑗 ≤ 𝑢′ 𝑏 , ∀ 𝑢 ∈ R+m}
 Separation problem for 𝑃1 : Given an 𝑥 ∗ ∈ 𝑃, we want to find a 𝑢 ∈ 𝑅+m such
that 𝛼′𝑥 ∗ > 𝛼0 , where 𝛼 = 𝑢′ 𝐴 and 𝛼0 = 𝑢′ 𝑏 , or prove that no such u exists.
(Formulation as a mixed integer program)
We can assume that 𝑢𝑖 < 1 when 𝑎𝑖 ∈ 𝑍 𝑛 , 𝑏𝑖 ∈ 𝑍.
Given 𝑥 ∗ , let 𝐽 𝑥 ∗ = {𝑗: 𝑥 j* > 0}.
maximize
subject to
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𝑗∈𝐽(𝑥∗) 𝛼𝑗 𝑥 j −𝛼0
0 ≤ 𝑢′ 𝐴𝑗 − 𝛼𝑗 < 1,
0 ≤ 𝑢′ 𝑏 − 𝛼0 < 1,
𝑗 ∈ 𝐽 𝑥∗ ,
0 ≤ 𝑢𝑖 < 1,
𝑖 = 1, … , 𝑚,
𝛼𝑗 ∈ 𝑍,
𝑗 ∈ 𝐽 𝑥 ∗ ∪ {0}
*
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 Practically, we replace the strict inequalities, e.g. 𝑢𝑖 < 1, with the inequality
𝑢𝑖 ≤ 1 − 𝛿 for a small 𝛿 > 0.
Also using small number of positive 𝑢𝑖 ’s usually finds much more effective
cuts. Hence we introduce the term −𝑤 𝑚
𝑖=1 𝑢𝑖 for a small 𝑤 > 0 in the
objective function.
maximize
*
𝑗∈𝐽(𝑥∗) 𝛼𝑗 𝑥 j −𝛼0 − 𝑤
subject to 0 ≤ 𝑢′ 𝐴𝑗 − 𝛼𝑗 ≤ 1 − 𝛿,
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𝑚
𝑖=1 𝑢𝑖
𝑗 ∈ 𝐽 𝑥∗ ,
0 ≤ 𝑢′ 𝑏 − 𝛼0 ≤ 1 − 𝛿,
0 ≤ 𝑢𝑖 ≤ 1 − 𝛿,
𝑖 = 1, … , 𝑚,
𝛼𝑗 ∈ 𝑍,
𝑗 ∈ 𝐽 𝑥 ∗ ∪ {0}
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 If we find good but not necessarily optimal solutions to the MIP, we find very
effective valid inequalities. Also heuristic methods to find good feasible
solutions to the MIP are helpful.
 MIP model may not be intended as computational tools to solve real problems.
But we can examine the strength of rank-1 C-G inequalities to describe the
convex hull of S for various problems.
 For some structured problems, e.g. knapsack problem, the separation problem
for the first Chvatal closure may have some structure which enables us to
handle the problem more effectively.
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Modular Arithmetic
 Valid inequality for the solutions of one linear equation.
S = {𝑥 ∈ Z+n:
𝑗∈𝑁 𝑎𝑗 𝑥𝑗
Sd = {𝑥 ∈ Z+n:
𝑗∈𝑁 𝑎𝑗 𝑥𝑗
= 𝑎0 }, 𝑎𝑗 ∈ 𝑅1 for all j.
= 𝑎0 + 𝑘𝑑 for some 𝑘 ∈ 𝑍1 }, 𝑑 is positive integer.
Valid inequality for Sd  valid for S.
Let 𝑎𝑗 = 𝑏𝑗 + 𝛼𝑗 𝑑, 0 ≤ 𝑏𝑗 < 𝑑, 𝛼𝑗 is integer (𝑏𝑗 = remainder of 𝑎𝑗 /𝑑)
Sd = {𝑥 ∈ Z+n:
𝑗∈𝑁 𝑏𝑗 𝑥𝑗
𝑗∈𝑁 𝑏𝑗 𝑥𝑗
(

= 𝑏0 + 𝑘𝑑 for some 𝑘 ∈ 𝑍1 },
 0, 𝑏0 < 𝑑, 𝑘 ≥ 0)
𝑗∈𝑁 𝑏𝑗 𝑥𝑗
 𝑏0
 Ex: 37𝑥1 − 68𝑥2 + 78𝑥3 + 𝑥4 = 141, xZ+4
d = 12  𝑥1 + 4𝑥2 + 6𝑥3 + 𝑥4  9,
𝑑=1 
𝑗∈𝑁
𝑎𝑗 − 𝑎𝑗 𝑥𝑗 ≥ 𝑎0 − 𝑎0 : Gomory cutting plane
ex) 𝑥0 = 334 − 12𝑥1 + 74𝑥2 − 11
𝑥 , 𝑥𝑖 ∈ Z+1 for all i.
4 3
1
2
𝑥1 + 14𝑥2 + 34𝑥3 ≥ 34
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Disjunctive Constraints
 Prop 1.3: If
then,
𝑗∈𝑁 𝜋j
1𝑥
≤ 𝜋01 valid for 𝑆1 ⊂ 𝑅+n,
𝑗∈𝑁 𝜋j
2𝑥
𝑗
≤ 𝜋02 valid for 𝑆2 ⊂ 𝑅+n,
𝑗
𝑗∈𝑁 min(𝜋j
1,
j2)xj  max(01, 02) is valid for 𝑆1 ⋃𝑆2 .
 Disjunctive procedure: S = {xZ+n: Ax  b}
(1)
𝑗∈𝑁(𝑢𝑎𝑗 )𝑥𝑗
≤ 𝑢𝑏, 𝑢 ≥ 0
(2) Given 𝛿 ∈ 𝑍+1, if
a)
𝑗∈𝑁 𝜋 j𝑥𝑗
− (𝑥k  ) ≤ 𝜋0 is valid for S for some   0, and
b)
𝑗∈𝑁 𝜋j𝑥𝑗
+ (𝑥 k  1) ≤ 𝜋0 is valid for S for some   0, then
c)
𝑗∈𝑁 𝜋 j𝑥𝑗
≤ 𝜋0 is valid for S.
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 a)
− (𝑥 k  ) ≤ 𝜋0 is valid for S
𝑗∈𝑁 𝜋j𝑥𝑗
𝛼 × ( 𝑥𝑘 ≤ 𝛿 )

b)
𝑗∈𝑁 𝜋j𝑥𝑗
𝑗∈𝑁 𝜋 j𝑥𝑗
≤ 𝜋0 is valid for 𝑆1 = 𝑆⋂{𝑥 ∈ 𝑍 +n : 𝑥𝑘 ≤ 𝛿}
+ (𝑥k  1) ≤ 𝜋0 is valid for S
𝛽 × ( −𝑥𝑘 ≤ − 𝛿 + 1 )

𝑗∈𝑁 𝜋j𝑥𝑗
≤ 𝜋0 is valid for 𝑆2 = 𝑆⋂{𝑥 ∈ 𝑍+n : 𝑥𝑘 ≥ 𝛿 + 1}
From Proposition 1.3,
𝑗∈𝑁 𝜋j𝑥𝑗
≤ 𝜋0 is valid for 𝑆1 ⋃𝑆2 .
 D-inequalities
 Ex: Figure 1.5
P = {xR+2: −𝑥1 + 𝑥2 ≤ 12,
1
2
𝑥1 + 𝑥2 ≤ 54, 𝑥1 ≤ 2}
1st ineq  −14𝑥1 + 𝑥2 − 34𝑥1
≤ 12
2nd ineq  −14𝑥1 + 𝑥2 − 34(1 − 𝑥1 ) ≤ 12
Using the disjunction 𝑥1 ≤ 0 or 𝑥1 ≥ 1 leads to valid inequality
−14𝑥1 + 𝑥2 ≤ 12 for 𝑆 = 𝑃⋂𝑍 2 .
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 Every valid inequality for disjunction  D-inequality
 Prop 1.4: 𝑃 = {𝑥 ∈ R+n: 𝐴𝑥 ≤ 𝑏}, suppose 𝐴 = 𝐴′ .
𝐼
𝜋𝑥 ≤ 𝜋0 valid for P’ = (𝑃 ∩ 𝑥: 𝑥𝑘 ≤ 𝛿 ) ∪ (𝑃 ∩ 𝑥: 𝑥𝑘 ≥ 𝛿 + 1 )
 ∃𝛼, 𝛽 ≥ 0 such that
𝑗∈𝑁 𝜋j𝑥𝑗
− (𝑥k  ) ≤ 𝜋0,
𝑗∈𝑁 𝜋 j𝑥𝑗
+ (𝑥k  1) ≤ 𝜋0 valid for P.
Pf) 𝜋𝑥 ≤ 𝜋0 valid for P’  𝜋𝑥 ≤ 𝜋0 valid for 𝑃 ∩ {𝑥: 𝑥𝑘 ≤ 𝛿} and 𝑃 ∩
{𝑥: 𝑥𝑘 ≥ 𝛿 + 1}.
 By Prop 1.1, ∃ 𝑢1 , 𝛼 , (𝑢2 , 𝛽) ≥ 0 such that
𝑢1 𝐴 + 𝛼𝑒𝑘 ≥ 𝜋, 𝑢1 𝑏 + 𝛼𝛿 ≤ 𝜋0 (𝑢1 𝐴 ≥ 𝜋 − 𝛼𝑒𝑘 , 𝑢1 𝑏 ≤ 𝜋0 − 𝛼𝛿)
𝑢2 𝐴 − 𝛽𝑒𝑘 ≥ 𝜋, 𝑢2 𝑏 − 𝛽(𝛿 + 1) ≤ 𝜋0 (𝑢2 𝐴 ≥ 𝜋 + 𝛽𝑒𝑘 , 𝑢2 𝑏 ≤ 𝜋0 +
𝛽(𝛿 + 1))
𝑢𝑖 𝐴𝑥 ≤ 𝑢𝑖 𝑏 are valid for P and are equal to or dominate
𝜋𝑥 − 𝛼(𝑥𝑘 − 𝛿) ≤ 𝜋0 and 𝜋𝑥 − 𝛽(𝛿 + 1 − 𝑥𝑘 ) ≤ 𝜋0 . Hence these are valid
for P.

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Boolean Implications
 𝑆 = {𝑥 ∈ 𝐵𝑛 :
𝑗∈𝑁 𝑎𝑗 𝑥𝑗
≤ 𝑏}, 𝑎𝑗′ 𝑠, 𝑏 are positive integers.
S is the feasible set for a 0-1 knapsack problem.
Let 𝐶 ⊂ 𝑁 be such that

𝑗∈𝐶 𝑥𝑗
𝑗∈𝐶 𝑎𝑗
>𝑏
≤ 𝐶 − 1 is valid for S. (important result)
 𝑇 = {𝑥 ∈ 𝐵1 , 𝑦 ∈R+n:
𝑗∈𝑁 𝑦𝑗
≤ 𝑛𝑥, 𝑦𝑗 ≤ 1, ∀𝑗 ∈ 𝑁}
 𝑦𝑗 ≤ 𝑥, 𝑗 ∈ 𝑁 is valid for T.
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Geometric or Combinatorial Implication
 Node packing: 𝑥𝑖 + 𝑥𝑗 ≤ 1, ∀ 𝑖, 𝑗 ∈ 𝐸, 𝑥 ∈ 𝐵|𝑉|
1
2
5
6
3
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7
 𝑥4 + 𝑥5 + 𝑥6 + 𝑥7 ≤ 1
(clique constraint)
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 ≤2
(odd hole constraint)
Using C-G to obtain odd hole constraint:
𝑥1 + 𝑥2 ≤ 1
𝑥2 + 𝑥3 ≤ 1
𝑥3 + 𝑥4 ≤ 1
𝑥4 + 𝑥5 ≤ 1
𝑥1
+ 𝑥5 ≤ 1
Multiply ½ on both sides and add
 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 ≤ 52 = 2
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