4.4 Geometry and solution of Lagrangean duals
 Solving Lagrangean dual: 𝑍𝐷 = max 𝑍(𝜆)
𝜆≥0
𝑘
𝑍 𝜆 = min(𝑐′𝑥 𝑘 + 𝜆′(𝑏 − 𝐴𝑥 )) (unless 𝑍 𝜆 = −∞)
𝑘∈𝐾
𝑘
Let 𝑓𝑘 = 𝑏 − 𝐴𝑥
and ℎ𝑘 = 𝑐′𝑥 𝑘 . Then
𝑍 𝜆 = min(ℎ𝑘 + 𝑓𝑘′ 𝜆) (piecewise linear and concave)
𝑘∈𝐾
 𝑍(𝜆) is nondifferentiable, but mimic the idea for differentiable function
𝜆𝑡=1 = 𝜆𝑡 + 𝜃𝑡 𝛻𝑍 𝜆𝑡 ,
𝑡 = 1,2, …
 Prop 4.5: A function 𝑓: 𝑅𝑛 → 𝑅 is concave if and only if for any 𝑥 ∗ ∈ 𝑅𝑛 ,
there exists a vector 𝑠 ∈ 𝑅𝑛 such that
𝑓 𝑥 ≤ 𝑓 𝑥 ∗ + 𝑠′ 𝑥 − 𝑥 ∗ ,
for all 𝑥 ∈ 𝑅𝑛 .
 The vector 𝑠 is called the gradient of 𝑓 at 𝑥 ∗ (if differentiable). Generalize the
above property to nondifferentiable functions.
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 Def 4.3: Let 𝑓 be a concave function. A vector 𝑠 such that
𝑓 𝑥 ≤ 𝑓 𝑥 ∗ + 𝑠′ 𝑥 − 𝑥 ∗ ,
for all 𝑥 ∈ 𝑅𝑛 , is called a subgradient of 𝑓 at 𝑥 ∗ . The set of all subgradients of
𝑓 at 𝑥 ∗ is denoted as 𝜕𝑓(𝑥 ∗ ) and is called the subdifferential of 𝑓 at 𝑥 ∗ .
 Prop: The subdifferential 𝜕𝑓(𝑥 ∗ ) is closed and convex.
Pf) 𝜕𝑓(𝑥 ∗ ) is the intersection of closed half spaces.
 If 𝑓 is differentiable at 𝑥 ∗ , then 𝜕𝑓 𝑥 ∗ = 𝛻𝑓 𝑥 ∗ .
For convex functions, subgradient is defined as vector 𝑠 which satisfies
𝑓 𝑥 ≥ 𝑓 𝑥 ∗ + 𝑠′(𝑥 − 𝑥 ∗ ).
 Geometric intuition: Think in 𝑅𝑛+1 space
𝑓 𝑥 ≤ 𝑓 𝑥 ∗ + 𝑠′(𝑥 − 𝑥 ∗ )
⟺ 𝑠′𝑥 ∗ − 𝑓 𝑥 ∗ ≤ 𝑠 ′ 𝑥 − 𝑓(𝑥)
⟺ (𝑠, −1)′(𝑥 ∗ , 𝑓 𝑥 ∗ ) ≤ (𝑠, −1)′(𝑥, 𝑓 𝑥 )
⟺
𝑠, −1 ′ { 𝑥, 𝑓 𝑥
Integer Programming 2015
− 𝑥∗, 𝑓 𝑥∗ } ≥ 0
2
 Picture:
𝑠, −1 ′ { 𝑥, 𝑓 𝑥
− 𝑥∗, 𝑓 𝑥∗ } ≥ 0
𝑓 𝑥 = 𝑓 𝑥 ∗ + 𝑠′(𝑥 − 𝑥 ∗ )
𝑓(𝑥)
𝑥, 𝑓 𝑥
− (𝑥 ∗ , 𝑓 𝑥 ∗ )
(𝑥, 𝑓 𝑥 )
(𝑥 ∗ , 𝑓 𝑥 ∗ )
−1
𝑥∗
Integer Programming 2015
𝑠
(𝑠, −1)
𝑥
𝑥
3
𝑠, −1 ′ { 𝑥, 𝑓 𝑥
 For convex functions:
− 𝑥 ∗, 𝑓 𝑥∗ } ≤ 0
𝑓 𝑥 = 𝑓 𝑥 ∗ + 𝑠′(𝑥 − 𝑥 ∗ )
𝑓(𝑥)
𝑥, 𝑓 𝑥
− (𝑥 ∗ , 𝑓 𝑥 ∗ )
(𝑥 ∗ , 𝑓 𝑥 ∗ )
(𝑥, 𝑓 𝑥 )
−1
𝑥
Integer Programming 2015
𝑥∗
𝑠
(𝑠, −1)
𝑥
4
 Prop 4.6: Let 𝑓: 𝑅𝑛 → 𝑅 be a concave function. A vector 𝑥 ∗ maximizes 𝑓
over 𝑅𝑛 if and only if 0 ∈ 𝜕𝑓 𝑥 ∗ .
Pf) A vector 𝑥 ∗ maximizes 𝑓 over 𝑅𝑛 if and only if
𝑓 𝑥 ≤ 𝑓 𝑥 ∗ = 𝑓 𝑥 ∗ + 0′ 𝑥 − 𝑥 ∗ , for all 𝑥 ∈ 𝑅𝑛 ,
which is equivalent to 0 ∈ 𝜕𝑓 𝑥 ∗ .
 Prop 4.7: Let
𝑍 𝜆 = min (ℎ𝑘 + 𝑓𝑘′ 𝜆)
𝑘∈𝐾
𝐸 𝜆 = 𝑘 ∈ 𝐾: 𝑍 𝜆 = ℎ𝑘 + 𝑓𝑘′ 𝜆 .
Then, for every 𝜆∗ ≥ 0 the following relations hold:
(a) For every 𝑘 ∈ 𝐸 𝜆∗ , 𝑓𝑘′ is a subgradient of the function 𝑍() at 𝜆∗ .
(b) 𝜕𝑍 𝜆∗ = conv 𝑓𝑘 : 𝑘 ∈ 𝐸 𝜆∗ , i.e., a vector 𝑠 is a subgradient of the
function 𝑍() at 𝜆∗ if and only if 𝑠 is a convex combination of the vectors
𝑓𝑘 , 𝑘 ∈ 𝐸 𝜆∗ .
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5
Pf) (a) By definition of 𝑍(𝜆) we have for every 𝑘 ∈ 𝐸(𝜆∗ )
𝑍 𝜆 ≤ ℎ𝑘 + 𝑓𝑘′ 𝜆 = ℎ𝑘 + 𝑓𝑘′ 𝜆∗ + 𝑓𝑘′ 𝜆 − 𝜆∗ = 𝑍 𝜆∗ + 𝑓𝑘′ 𝜆 − 𝜆∗ .
(b) We have shown that {𝑓𝑘 : 𝑘 ∈ 𝐸 𝜆∗ } ⊆ 𝜕𝑍(𝜆∗ ) and since a convex
combination of two subgradients is also a subgradient , it follows that
conv 𝑓𝑘 : 𝑘 ∈ 𝐸 𝜆∗
⊆ 𝜕𝑍 𝜆∗ .
Assume that there is 𝑠 ∈ 𝜕𝑍 𝜆∗ , such that 𝑠 ∉ conv( 𝑓𝑘 : 𝑘 ∈ 𝐸 𝜆∗ ) and
derive a contradiction. By the separating hyperplane theorem, there exists a
vector 𝑑 and a scalar , such that
𝑑′𝑓𝑘 ≥ 𝛾,
for all 𝑘 ∈ 𝐸(𝜆∗ )
(4.29)
𝑑′ 𝑠 < 𝛾
Since 𝑍 𝜆∗ < ℎ𝑖 + 𝑓𝑖′ 𝜆∗ , for all 𝑖 ∉ 𝐸 𝜆∗ , then for sufficiently small 𝜀 > 0,
we have, 𝑍 𝜆∗ + 𝜀𝑑 = ℎ𝑘 + 𝑓𝑘′ 𝜆∗ + 𝜀𝑑 , for some 𝑘 ∈ 𝐸(𝜆∗ )
Since 𝑍 𝜆∗ = ℎ𝑘 + 𝑓𝑘′ 𝜆∗ , we obtain that
𝑍 𝜆∗ + 𝜀𝑑 − 𝑍 𝜆∗ = 𝜀𝑓𝑘′ 𝑑, for some 𝑘 ∈ 𝐸(𝜆∗ ).
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(continued)
From (4.29), we have
𝑍 𝜆∗ + 𝜀𝑑 − 𝑍 𝜆∗ ≥ 𝜀𝛾.
Since 𝑠 ∈ 𝜕𝑍 𝜆∗ , we have
𝑍 𝜆∗ + 𝜀𝑑 ≤ 𝑍 𝜆∗ + 𝜀𝑠 ′ 𝑑.
From (4.29), we have
𝑍 𝜆∗ + 𝜀𝑑 − 𝑍 𝜆∗ < 𝜀𝛾,
contradicting (4.30).

Integer Programming 2015
(4.30)
(def’n of subgradient)
( 𝑑′ 𝑠 < 𝛾 )
7
 Subgradient algorithm
 For differentiable functions, the direction 𝛻𝑓(𝑥) is the direction of steepest
ascent of 𝑓 at 𝑥. But, for nondifferentiable functions, the direction of
subgradient may not be an ascent direction. However, moving along the
direction of a subgradient, we can move closer to the maximum point of a
concave function.
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 Ex: 𝑓 𝑥1 , 𝑥2 = min {−𝑥1 , 𝑥1 − 2𝑥2 , 𝑥1 + 2𝑥2 } : concave function
direction of subgradient may not be an ascent direction for 𝑓, but we can
move closer to the maximum point.
𝑥2
𝑓(𝑥) = −2
𝑠 2 = (1,2)
𝑠1 𝑥 − 𝑥 ∗ = 0
𝑓(𝑥) = −1
𝑥 ∗ = (−2,0)
𝑠1 = (1, −2)
Integer Programming 2015
𝑓(𝑥) = 0
𝑥1
𝑠2 𝑥 − 𝑥 ∗ = 0
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 The subgradient optimization algorithm
Input: A nondifferentiable concave function 𝑍(𝜆).
Output: A maximizer of 𝑍(𝜆) subject to 𝜆 ≥ 0.
Algorithm:
1. Choose a starting point 𝜆1 ≥ 0; let 𝑡 = 1.
2. Given 𝜆𝑡 , check whether 0 ∈ 𝜕𝑍(𝜆𝑡 ). If so, then 𝜆𝑡 is optimal and the algorithm
terminates. Else, choose a subgradient 𝑠 𝑡 of the function 𝑍(𝜆𝑡 ). (𝑓𝑘 → 𝑏 − 𝐴𝑥 𝑘 )
3. Let 𝜆𝑗𝑡+1 = max{𝜆𝑗𝑡 + 𝜃𝑡 𝑠𝑗𝑡 , 0} (projection), where 𝜃𝑡 is a positive step size
parameter. Increment 𝑡 and go to Step 2.
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 Choosing the step lengths:
 Thm :
(a) If
∞
𝑡=1 𝜃𝑡
→ ∞, and lim 𝜃𝑡 = 0, then 𝑍(𝜆𝑡 ) → 𝑍𝐷 the optimal value of
𝑡→∞
𝑍().
(b) If 𝜃𝑡 = 𝜃0 𝛼 𝑡 , 𝑡 = 1,2, … , for some parameter 0 < 𝛼 < 1, then 𝑍(𝜆𝑡 ) → 𝑍𝐷
if 𝜃0 and  are sufficiently large.
(c) If 𝜃𝑡 = 𝑓 𝑍𝐷∗ − 𝑍 𝜆𝑡 / 𝑠 𝑡 2 , where 0 < 𝑓 < 2, and 𝑍𝐷∗ ≤ 𝑍𝐷 , then
𝑍 𝜆𝑡 → 𝑍𝐷∗ , or the algorithm finds 𝜆𝑡 with 𝑍𝐷∗ ≤ 𝑍 𝜆𝑡 ≤ 𝑍𝐷 , for some finite 𝑡.
 (a) guarantees convergence, but slow (e.g., 𝜃𝑡 = 1/𝑡 )
(b) In practice, may use halving the value of 𝜃𝑡 after 𝑣 iterations
(c) 𝑍𝐷∗ typically unknown, may use a good primal upper bound in place of 𝑍𝐷∗ .
If not converge, decrease 𝑍𝐷∗ .
 In practice, 0 ∈ 𝜕𝑍(𝜆𝑡 ) is rarely met. Usually, only find approximate optimal
solution fast and resort to B-and-B. Convergence is not monotone.
If 𝑍𝐷 = 𝑍𝐿𝑃 , solving LP relaxation may give better results (monotone
convergence).
11
Integer Programming 2015
 Ex: Traveling salesman problem:
For TSP, we dualize degree constraints for nodes except node 1.
𝑍 𝜆 = min
( = min
𝑒∈𝐸 (𝑐𝑒
− 𝜆𝑖 − 𝜆𝑗 )𝑥𝑒 + 2
𝑒∈𝐸 𝑐𝑒 𝑥𝑒
𝑒∈𝛿( 1 ) 𝑥𝑒
𝑒∈𝐸(𝑆) 𝑥𝑒
+
𝑖∈𝑉∖{1} 𝜆𝑖 (2 −
𝑒∈𝛿 𝑖
𝑥𝑒 ) )
= 2,
≤ 𝑆 − 1,
𝑒∈𝐸(𝑉∖ 1 ) 𝑥𝑒
𝑖∈𝑉∖{1} 𝜆𝑖
𝑆 ⊂ 𝑉 ∖ 1 , 𝑆 ≠ ∅, 𝑉 ∖ 1 ,
= 𝑉 − 2,
𝑥𝑒 ∈ 0,1 .
step direction is
𝜆𝑡+1
= 𝜆𝑡𝑖 + 𝜃𝑡 (2 −
𝑖
𝑒∈𝛿 𝑖
𝑥𝑒 𝜆𝑡 ) ( 𝑠 𝑡 = 𝑏 − 𝐴𝑥 𝑘 )
step size using rule (c) is
𝜃𝑡 = 𝑓 𝑍𝐷∗ − 𝑍 𝜆𝑡 /
𝑡 2
𝑒∈𝛿( 𝑖 ) 𝑥𝑒 (𝜆 ))
𝐴𝑥(𝜆𝑡 ) 2 )
𝑖∈𝑉(2 −
(𝜃𝑡 = 𝑓 𝑍𝐷∗ − 𝑍 𝜆𝑡 / 𝑏 −
 Note that 𝑖 −th coordinate of subgradient direction is two minus the number of
edges incident to node 𝑖 in the optimal one-tree. We do not have 𝜆 ≥ 0 here
since the dualized constraints are equalities.
Integer Programming 2015
12
 Lagrangean heuristic and variable fixing:
 The solution obtained from solving the Lagrangian relaxation may not be
feasible to IP, but it can be close to a feasible solution to IP.
Obtain a feasible solution to IP using heuristic procedures. Then we may
obtain a good upper bound on optimal value.
Also called ‘primal heuristic’
 May fix values of some variables using information from Lagrangian
relaxation (refer to W, pp177-178).
 Branching variable selection? (solutions to Lagrangian relaxation are integer
valued)
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 Choosing a Lagrangean dual:
How to determine the constraints to relax?
Strength of the Lagrangean dual bound
Ease of solution of 𝑍(𝜆)
Ease of solution of Lagrangean dual problem 𝑍𝐷 = max 𝑍(𝜆)
𝜆≥0
 Ex: Generalized Assignment Problem (max problem)(refer to W, pp. 179):
𝑧 = max
Integer Programming 2015
𝑛
𝑚
𝑗=1 𝑖=1 𝑐𝑖𝑗 𝑥𝑖𝑗
𝑛
𝑗=1 𝑥𝑖𝑗 ≤ 1
𝑚
𝑖=1 𝑎𝑖𝑗 𝑥𝑖𝑗 ≤ 𝑏𝑗
𝑥 ∈ 𝐵𝑚𝑛
for 𝑖 = 1, … , 𝑚
for 𝑗 = 1, … , 𝑛
14
 Dualize both sets of constraints:
𝑍𝐷1 = min 𝑍1 (𝑢, 𝑣) , where
𝑢≥0,𝑣≥0
𝑍1 𝑢, 𝑣 = max
𝑛
𝑗=1
𝑚
𝑖=1
𝑐𝑖𝑗 − 𝑢𝑖 − 𝑎𝑖𝑗 𝑣𝑗 𝑥𝑖𝑗 +
𝑚
𝑖=1 𝑢𝑖
+
𝑛
𝑗=1 𝑣𝑗 𝑏𝑗
𝑥 ∈ 𝐵𝑚𝑛
 Dualize first set of assignment constraints:
𝑍𝐷2 = min 𝑍 2 (𝑢) , where
𝑢≥0
𝑍 2 𝑢 = max
𝑛
𝑚
𝑗=1 𝑖=1(𝑐𝑖𝑗 − 𝑢𝑖 )𝑥𝑖𝑗
𝑚
𝑖=1 𝑎𝑖𝑗 𝑥𝑖𝑗 ≤ 𝑏𝑗
𝑚𝑛
+
𝑚
𝑖=1 𝑢𝑖
for 𝑗 = 1, … , 𝑛
𝑥∈𝐵
 Dualize the knapsack constraints:
𝑍𝐷3 = min 𝑍 3 𝑣 , where
𝑣≥0
𝑍 3 𝑣 = max
Integer Programming 2015
𝑛
𝑚
𝑗=1 𝑖=1(𝑐𝑖𝑗 − 𝑎𝑖𝑗 𝑣𝑗 )𝑥𝑖𝑗
𝑛
𝑗=1 𝑥𝑖𝑗 ≤ 1
𝑥 ∈ 𝐵𝑚𝑛
+
𝑛
𝑗=1 𝑣𝑗 𝑏𝑗
for 𝑖 = 1, … , 𝑚
15
 𝑍𝐷1 = 𝑍𝐷3 = 𝑍𝐿𝑃
For each 𝑖,
conv{𝑥:
= conv{𝑥:
𝑛
𝑗=1 𝑥𝑖𝑗 ≤ 1, 𝑥𝑖𝑗 ∈ {0,1} for 𝑗
𝑛
𝑗=1 𝑥𝑖𝑗 ≤ 1, 0 ≤ 𝑥𝑖𝑗 ≤ 1 for
= 1, … , 𝑛}
𝑗 = 1, … , 𝑛}
𝑍1 𝑢, 𝑣 , 𝑍 3 (𝑣) can be solved by inspection. Calculating 𝑍𝐷3 look easier than
calculating 𝑍𝐷1 since there are 𝑚 dual variables compared to 𝑚 + 𝑛 for 𝑍𝐷1 .
 𝑍𝐷2 ≤ 𝑍𝐿𝑃
 To find 𝑍 2 𝑢 , we need to solve 𝑛 0-1 knapsack problems. Also note that the
information that we may obtain while solving the knapsack cannot be stored
and used for subsequent optimization.
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 May solve the Lagrangean dual by constraint generation (NW p411-412):
 Recall that
𝑍𝐷 = maximize 𝑦
subject to 𝑦 + 𝜆′ 𝐴𝑥 𝑘 − 𝑏 ≤ 𝑐′𝑥 𝑘 ,
𝜆′ 𝐴𝑤 𝑗 ≤ 𝑐′𝑤 𝑗 ,
𝑘∈𝐾
𝑗∈𝐽
𝜆≥0
 Given 𝑦 ∗ , 𝜆∗ , with 𝜆∗ ≥ 0, calculate
𝑍 𝜆∗ =
min (𝑐 ′ 𝑥 + 𝜆∗ ′ 𝑏 − 𝐴𝑥 )
𝑥∈conv(𝐹)
(min 𝑐 ′ − 𝜆∗ ′𝐴 𝑥 + 𝜆′ 𝑏)
If 𝑦 ∗ ≤ 𝑍 𝜆∗ , stop 𝑦 ∗ = 𝑍 𝜆∗ , 𝜆 = 𝜆∗ is an optimal solution
If 𝑦 ∗ > 𝑍 𝜆∗ , an inequality is violated
𝑍 𝜆∗ → −∞, ∃ ray 𝑤 𝑗 s.t. 𝑐 ′ − 𝜆∗ ′𝐴 𝑤 𝑗 < 0, hence 𝜆′ 𝐴𝑤 𝑗 ≤ 𝑐′𝑤 𝑗 is violated
 𝑍 𝜆∗ < ∞, ∃ extreme point 𝑥 𝑘 s.t. 𝑍 𝜆∗ = 𝑐′𝑥 𝑘 + 𝜆∗ ′ 𝑏 − 𝐴𝑥 𝑘 . Since 𝑦 ∗ >
𝑍 𝜆∗ , 𝑦 + 𝜆′(𝐴𝑥 𝑘 − 𝑏) ≤ 𝑐′𝑥 𝑘 is violated.
 Note that max/min is interchanged in NW.
Integer Programming 2015
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Nonlinear Optimization Problems
 Geometric understanding of the strength of the bounds provided by the
Lagrangean dual
 Let 𝑓: 𝑅𝑛 ↦ 𝑅 and 𝑔: 𝑅𝑛 ↦ 𝑅𝑚 be continuous functions. Let ℱ ⊂ 𝑅𝑛 .
minimize 𝑓(𝑥)
subject to 𝑔(𝑥) ≤ 0,
(4.31)
𝑥 ∈ ℱ.
Let 𝑍𝑃 be the optimal cost. Consider Lagrangean function
𝑍 𝜆 = 𝑚𝑖𝑛𝑥∈ℱ {𝑓 𝑥 + 𝜆′ 𝑔 𝑥 }.
 For all 𝜆 ≥ 0, 𝑍(𝜆) ≤ 𝑍𝑃 and 𝑍(𝜆) is a concave function.
Lagrangean dual: 𝑍𝐷 = 𝑚𝑎𝑥𝜆≥0 𝑍(𝜆).
 Let 𝑌 = { 𝑦, 𝑧 ′ ∈ 𝑅𝑚+1 : 𝑦 ≥ (=)𝑓 𝑥 , 𝑧 ≥ (=)𝑔 𝑥 , for all 𝑥 ∈ ℱ}.
Problem (4.31) can be restated as
minimize 𝑦
subject to (𝑦, 0)′ ∈ 𝑌. ( 𝑦, 𝑧 ′ ∈ 𝑌, 𝑧 ≤ 0 )
Integer Programming 2015
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
𝑓, (𝑦)
𝑐𝑜𝑛𝑣(𝑌)
𝑍𝑃
𝑍 𝜆 = 𝑓 + 𝜆𝑔
𝑍𝐷
𝑔, (𝑧)
Figure 4.7: The geometric interpretation of the Lagrangean dual
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 Given that 𝑍 𝜆 = 𝑚𝑖𝑛𝑥∈ℱ 𝑓 𝑥 + 𝜆′ 𝑔(𝑥), we have for a fixed  and for all
𝑥 ∈ ℱ that
𝑍 𝜆 ≤ 𝑦 + 𝜆′ 𝑧, ∀(𝑦, 𝑧)′ ∈ 𝑌.
Geometrically, this means that the hyperplane 𝑍 𝜆 = 𝑦 + 𝜆′ 𝑧 lies below the
set Y.
For 𝑧 = 0, we obtain 𝑦 = 𝑍(𝜆), that is 𝑍(𝜆) is the intercept of the hyperplane
𝑍 𝜆 = 𝑦 + 𝜆′ 𝑧 with the vertical axis.
To maximize 𝑍(𝜆), this corresponds to finding a hyperplane 𝑍 𝜆 = 𝑦 + 𝜆′ 𝑧,
which lies below the set Y, such that the intercept of the hyperplane 𝑍 𝜆 =
𝑦 + 𝜆′ 𝑧 with the vertical axis is the largest.
 Thm 4.10: The value 𝑍𝐷 of the Lagrangean dual is equal to the value of the
following optimization problem:
minimize 𝑦
subject to (𝑦, 0)′ ∈ 𝑐𝑜𝑛𝑣(𝑌). ( 𝑦, 𝑧 ∈ 𝑐𝑜𝑛𝑣 𝑌 , 𝑧 ≤ 0 )
Pf) not given here.
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 Ex 4.7: Ex 4.3 revisited
𝑓 𝑥1 , 𝑥2 = 3𝑥1 − 𝑥2 and 𝑔 𝑥1 , 𝑥2 = −1 − 𝑥1 + 𝑥2 ,
ℱ = {(1, 0), (2,0), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2), (1, 3), (2, 3)}
𝑓, 𝑔 = {(3, -2), (6, -3), (2, -1), (5, -2), (-2, 1), (1, 0), (4, -1), (0, 1), (3, 0)}
𝑓
𝑐𝑜𝑛𝑣(𝑌)
𝑍𝑃 = 1
𝑍𝐷 = −13
Integer Programming 2015
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𝑔
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