1. Flow Feasibility Problems
Flow feasibility problem: Given (𝐺, 𝑢, 𝑟, 𝑠, 𝑘), determine whether there exists a feasible flow from 𝑟 to 𝑠 of value at least 𝑘.
algorithm and good characterization with max-flow min-cut thm.
Other types of flow feasibility problems?
Transportation Problem
Given 𝐺=(𝑉,𝐸) with partition {𝑃,𝑄} of nodes and vectors 𝑎∈ 𝑍 + 𝑃 , 𝑏∈ 𝑍 + 𝑄 , find 𝑥∈ 𝑅 𝐸 satisfying
( 𝑥 𝑝𝑞 :𝑞∈𝑄, 𝑝𝑞∈𝐸 )≤ 𝑎 𝑝 , for all 𝑝∈𝑃
( 𝑥 𝑝𝑞 :𝑝∈𝑃, 𝑝𝑞∈𝐸 )= 𝑏 𝑞 , for all 𝑞∈𝑄
𝑥𝑝𝑞≥0, for all 𝑝𝑞∈𝐸
𝑥𝑝𝑞 integral, for all 𝑝𝑞∈𝐸.  1 3 1 3 𝑟 𝑠 2 1 𝑃 2 1 𝑄 4 2 4 2
Combinatorial Optimization 2016

2. For any finite cut ′(𝐴∪𝐵∪{𝑟}), where 𝐴⊆𝑃, 𝐵⊆𝑄, the capacity is
There exists integral feasible flow in TP iff there exists an integral feasible flow in 𝐺′ from 𝑟 to 𝑠 of value (𝑏𝑞 :𝑞∈𝑄)
For any finite cut ′(𝐴∪𝐵∪{𝑟}), where 𝐴⊆𝑃, 𝐵⊆𝑄, the capacity is
( 𝑎 𝑖 :𝑖∈𝑃∖𝐴)+ ( 𝑏 𝑗 :𝑗∈𝐵)
𝑃∖𝐴
𝑄∖𝐵 𝑟 𝑠 𝐴 𝐵
(𝐴∪𝐵∪{𝑟})
Cut capacity is at least (𝑏𝑗 :𝑗∈𝑄) iff (𝑎𝑖 :𝑖∈𝑃\A)≥(𝑏𝑗 :𝑗∈𝑄\B)
Check this condition for the sets 𝐴 s.t. every node in 𝑃\A is adjacent to a node in 𝑄\B, since any node violating this could be added to 𝐴.
( Define Neighborset 𝑁(𝐶) of 𝐶⊆𝑉 : {𝑤: 𝑣𝑤∈𝐸 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑣∈𝐶}.
Then 𝑁(𝑄\B)=𝑃\A )
Necessary and sufficient conditions for existence of solution:
𝑎(𝑁(𝐶))≥𝑏(𝐶) , for all 𝐶⊆𝑄.
Combinatorial Optimization 2016

3. Does there exist 𝑥∈ 𝑅 𝐸 (or 𝑍 𝐸 ) such that 𝑎𝑣≤𝑓𝑥(𝑣)≤𝑏𝑣 , for all 𝑣∈𝑉
General problem:
Does there exist 𝑥∈ 𝑅 𝐸 (or 𝑍 𝐸 ) such that
𝑎𝑣≤𝑓𝑥(𝑣)≤𝑏𝑣 , for all 𝑣∈𝑉
𝑙𝑒≤𝑥𝑒≤𝑢𝑒, for all 𝑒∈𝐸
Special case: 𝑙=0, 𝑎=𝑏: want 𝑥∈ 𝑅 𝐸 such that
𝑓𝑥(𝑣)=𝑏𝑣 , for all 𝑣∈𝑉 (3.8)
0≤𝑥𝑒≤𝑢𝑒 , for all 𝑒∈𝐸
Note that 𝑏(𝑉)=0
Form 𝐺′ with 𝑉′=𝑉∪{𝑟, 𝑠}
𝑣𝑤∈𝐸 ⇒ capacity 𝑢𝑣𝑤
𝑣∈𝑉, 𝑏𝑣<0 ⇒ arc 𝑟𝑣 with 𝑢𝑟𝑣=− 𝑏𝑣 (𝑏𝑣<0: supply node)
𝑣∈𝑉, 𝑏𝑣>0 ⇒ arc 𝑣𝑠 with 𝑢𝑣𝑠=𝑏𝑣 (𝑏𝑣>0: demand node)
Then 𝐺 has a feasible flow ⇔ There is an (𝑟,𝑠)-flow in 𝐺′ of value ( 𝑏 𝑣 :𝑣∈𝑉, 𝑏 𝑣 >0)
Combinatorial Optimization 2016

4. (3.8) has a solution ⇔ there exists no 𝐴⊆𝑉 s.t.
𝑢(𝛿′(𝐴∪{𝑟}))<(𝑏𝑣: 𝑣∈𝑉, 𝑏𝑣>0)
(from max-flow min-cut theorem)
⇔ (−𝑏𝑣:𝑣∉𝐴, 𝑏𝑣<0)+(𝑏𝑣:𝑣∈𝐴, 𝑏𝑣>0)+𝑢((𝐴))<(𝑏𝑣:𝑣∈𝑉, 𝑏𝑣>0)
⇔ (−𝑏𝑣:𝑣∉𝐴, 𝑏𝑣<0)+(𝑏𝑣:𝑣∈𝐴, 𝑏𝑣>0)+𝑢((𝐴))<
(𝑏𝑣:𝑣∈𝐴, 𝑏𝑣>0)+(𝑏𝑣:𝑣∉𝐴, 𝑏𝑣>0)
⇔ no 𝐴 with 𝑢((𝐴))<(𝑏𝑣:𝑣∉𝐴)
⇔ for all 𝐴, have 𝑢((𝐴))≥(𝑏𝑣:𝑣∉𝐴) { or 𝑢(( 𝐴 ))≥(𝑏𝑣:𝑣∈𝐴) }
𝑉\A
𝑉\A 𝑟 𝑏𝑣 𝑠
−𝑏𝑣 𝐴 𝐴
{𝑣:𝑏𝑣<0}
{𝑣:𝑏𝑣>0}
There cannot be a subset of nodes whose total demand exceeds its “import capacity”.
Combinatorial Optimization 2016

5. A circulation is a vector 𝑥∈ 𝑅 𝐸 with 𝑓𝑥(𝑣)=0 for all 𝑣∈𝑉.
Thm 3.15(Gale, 1957) : There exists a solution to (3.8) ⇔ 𝑏(𝑉)=0 and, for every 𝐴⊆𝑉, 𝑏(𝐴)≤𝑢(( 𝐴 )). If 𝑏 and 𝑢 integral, then (3.8) has an integral solution iff the same conditions hold.
A circulation is a vector 𝑥∈ 𝑅 𝐸 with 𝑓𝑥(𝑣)=0 for all 𝑣∈𝑉.
Thm 3.17 (Hoffman’s Circulation Theorem, 1960):
Given a digraph 𝐺, 𝑙∈(𝑅∪{−})𝐸, and 𝑢∈(𝑅∪{})𝐸, with 𝑙≤𝑢,
There is a circulation 𝑥 with 𝑙≤𝑥≤𝑢 ⇔ every 𝐴⊆𝑉 satisfies 𝑢(( 𝐴 ))≥𝑙((𝐴)). Also for integral version.
(pf) 𝑙=− ⇒ 𝑙=−𝑀 (𝑀 large integer)
Let 𝑙≡0, 𝑢′≡𝑢−𝑙, 𝑥′≡𝑥−𝑙 Then 𝑓 𝑥 ′ 𝑣 =𝑓𝑥(𝑣)−𝑓𝑙(𝑣)
Hence 𝑥 is a circulation ⇔ 𝑓 𝑥 ′ 𝑣 =−𝑓𝑙(𝑣)
Apply Thm 3.15 with demands −𝑓𝑙(𝑣) and capacities 𝑢−𝑙
Note that (−𝑓𝑙(𝑣) :𝑣∈𝑉)=0 (nec condition for feasible flow)(why true?)
So there exists a feasible circulation ⇔ for every 𝐴⊆𝑉 we have
𝑢−𝑙  𝐴 ≥−(𝑓𝑙(𝑣) :𝑣∈𝐴)=𝑙((𝐴))−𝑙( 𝐴 )
⇒ 𝑢(( 𝐴 ))≥𝑙((𝐴)) 
Combinatorial Optimization 2016

6. Generalization of Thm 3.15 and 3.17
Thm 3.18: Given a digraph 𝐺, 𝑏∈ 𝑅 𝑉 such that 𝑏(𝑉)=0, 𝑙∈(𝑅∪{−})𝐸, and 𝑢∈(𝑅∪{})𝐸, with 𝑙≤𝑢, there exists 𝑥∈ 𝑅 𝐸 such that
𝑓𝑥(𝑣)=𝑏𝑣 , for all 𝑣∈𝑉
𝑙𝑒 ≤𝑥𝑒 ≤𝑢𝑒 , for all 𝑒∈𝐸
⇔ every 𝐴⊆𝑉 satisfies 𝑢(( 𝐴 ))≥𝑏(𝐴)+𝑙((𝐴))
Min (𝑟, 𝑠) flow subject to lower bounds on the arcs
min 𝑓𝑥(𝑠) (3.9)
subject to 𝑓𝑥(𝑣)=0, for all 𝑣∈𝑉∖{𝑟, 𝑠}
𝑥𝑒≥𝑙𝑒, for all 𝑒∈𝐸.
( 𝑙≥0 )
Assume every arc 𝑒 having 𝑙𝑒>0 is in an (𝑟, 𝑠)-dipath ( o.w. there may be no feasible solution) and there is no (𝑠, 𝑟)-dipath ( o.w. may be unbounded)
⇒ There exists 𝑅, 𝑟∈𝑅, 𝑅⊆𝑉∖{𝑠} such that ( 𝑅 )=∅
⇒ for any solution 𝑥 of (3.9), have 𝑓𝑥(𝑠)=𝑥  𝑅 −{𝑥(( 𝑅 ))}≥𝑙((𝑅)) for any 𝑅 with ( 𝑅 )=∅. (The bound can be made tight)
Combinatorial Optimization 2016

7. Thm 3.19: (Min-Flow Max-Cut Theorem)
There exists a solution to (3.9) having 𝑓𝑥(𝑠)≤𝑘 ⇔
There does not exist 𝑅⊆𝑉 with 𝑟∈𝑅, 𝑠∉𝑅, ( 𝑅 )=∅, and 𝑙  𝑅 >𝑘.
Also for integral version.
(pf) Add arc 𝑠𝑟 to 𝐺 with 𝑙𝑠𝑟=0, 𝑢𝑠𝑟=𝑘 and put 𝑢𝑒=∞ for all 𝑒∈𝐸.
Then new 𝐺′ has a circulation ⇔ (3.9) has solution of value at most 𝑘.
⇔ There does not exist 𝐴⊆𝑉 with 𝑢(′(𝐴))<𝑙(′( 𝐴 )).
Since 𝑢𝑒=∞ for all 𝑒∈𝐸, such 𝐴 satisfies (𝐴)=∅.
Since 𝑙(( 𝐴 ))>0, there exists 𝑒∈( 𝐴 ) with 𝑙𝑒>0
⇒ since there must exist an (𝑟, 𝑠)-dipath using 𝑒, we have 𝑟∈ 𝐴 , 𝑠∈𝐴.
⇒ 𝑠𝑟∈′(𝐴)
⇒ 𝑘=𝑢(′(𝐴))<𝑙(′( 𝐴 ))=𝑙(( 𝐴 )). So take 𝑅= 𝐴 . 
𝑙=0, 𝑢=𝑘
𝐴( 𝑅 )
𝐴 (𝑅) 𝑟 𝑠
Combinatorial Optimization 2016

8. Minimum Cuts and Linear Programming
LP interpretation of the min cut problem:
max 𝑓𝑥(𝑠)
s.t. (𝑥𝑤𝑣:𝑤∈𝑉, 𝑤𝑣∈𝐸)−(𝑥𝑣𝑤:𝑤∈𝑉, 𝑣𝑤∈𝐸)=0, for all 𝑣∈𝑉∖{𝑟, 𝑠}
0≤𝑥𝑣𝑤≤𝑢𝑣𝑤, for all 𝑣𝑤∈𝐸
Dual Problem:
min (𝑢𝑒𝑧𝑒 :𝑒∈𝐸) (3.10)
s.t. −𝑦𝑣+𝑦𝑤+𝑧𝑣𝑤 ≥0, for all 𝑣𝑤∈𝐸, 𝑣, 𝑤∈𝑉∖{𝑟, 𝑠}
𝑦𝑤+𝑧𝑣𝑤≥0, for all 𝑟𝑤∈𝐸
− 𝑦𝑣+𝑧𝑣𝑟 ≥0, for all 𝑣𝑟∈𝐸
− 𝑦𝑣+𝑧𝑣𝑠≥1, for all 𝑣𝑠∈𝐸
𝑦𝑤+𝑧𝑠𝑤≥−1, for all 𝑠𝑤∈𝐸
𝑧𝑒≥0, for all 𝑒∈𝐸
Define 𝑦𝑟=0, 𝑦𝑠=−1, we can unify the constraints as
−𝑦𝑣+𝑦𝑤+𝑧𝑣𝑤≥0, for all 𝑣𝑤∈𝐸
Then add 1 to each 𝑦𝑣 (dual feasibility not changed)
Combinatorial Optimization 2016

9. −𝑦𝑣+𝑦𝑤+𝑧𝑣𝑤≥0, for all 𝑣𝑤∈𝐸 𝑧𝑒 ≥0, for all 𝑒∈𝐸
min (𝑢𝑒𝑧𝑒 :𝑒∈𝐸) (3.11)
s.t. 𝑦𝑟=1, 𝑦𝑠=0
−𝑦𝑣+𝑦𝑤+𝑧𝑣𝑤≥0, for all 𝑣𝑤∈𝐸
𝑧𝑒 ≥0, for all 𝑒∈𝐸
Thm 3.20: If (3.11) has an optimal solution, it has one of the form:
For some (𝑟, 𝑠)-cut (𝑅), 𝑦 is the characteristic vector of 𝑅 and 𝑧 is the characteristic vector of (𝑅).
(pf) Choose (𝑅) to be a min cut. Then (𝑦, 𝑧) feasible to (3.11) and objective value is  𝑢𝑒𝑧𝑒 :𝑒∈𝐸 =𝑢  𝑅 = max flow value
By LP duality, this is optimal to (3.11) and (3.10) 
(See a different proof in the text.)
We may identify the cut using the bounded variable simplex method applied to a converted problem.
Note that a spanning tree corresponds to a maximum linearly independent columns of the network matrix.
Combinatorial Optimization 2016

10. 𝑙=0, 𝑢=∞ Basic arcs 𝑟 𝑠 𝑦𝑣=1 𝑦𝑣=0 𝑅
Add arc 𝑠𝑟 with 𝑙𝑠𝑟=0, 𝑢𝑠𝑟=∞, and max 𝑥𝑠𝑟 subject to 𝑓𝑥(𝑣)=0 for all 𝑣∈𝑉.
Add artificial variable 𝑥𝑎 with 𝑙=𝑢=0 and objective coefficient 0 to the constraint corresponding to node 𝑠.
As explained earlier, the artificial variable is always in the basis, hence dual variable 𝑦𝑠=0 in a b.f.s. (from 𝑦′𝐴𝑗= 𝑐 𝑗 for basic variables)
If flow value is positive 𝑥𝑠𝑟 is always in the basis, hence the spanning tree basis can be divided into two parts. Assigning 𝑦𝑣=1 for 𝑣∈𝑅, 0 for 𝑣∈ 𝑅 satisfies 𝑦′𝐴𝑗=𝑐𝑗 for basic arcs (We have −𝑦𝑣+𝑦𝑤=0 )
If current LP solution is optimal, the arcs in (𝑅) are at their upper bounds and the arcs in ( 𝑅 ) at 0 (and nonbasic). (at opt sol, have 𝑥 𝑒 = 𝑢 𝑒 when 𝑐 𝑒 −𝑦′ 𝐴 𝑒 >0, and 𝑥 𝑒 =0 when 𝑐 𝑒 −𝑦′ 𝐴 𝑒 <0 for nonbasic 𝑥 𝑒 )
Combinatorial Optimization 2016

11. Different interpretation: min (𝑢𝑒𝑧𝑒 :𝑒∈𝐸) (3.12) subject to
(𝑧𝑒 :𝑒∈𝑃)≥1, for all arc-sets P of simple (𝑟, 𝑠)-dipaths
𝑧𝑒≥0, for all 𝑒∈𝐸
Thm 3.21: If (3.12) has an optimal solution, then it has one that is the characteristic vector of an (𝑟, 𝑠)-cut
(pf) (3.12) has a feasible solution and it is unbounded if some 𝑢𝑒<0, so assume that 𝑢≥0.
Choose 𝑧′ to be the characteristic vector of min cut (𝑅).
Dual of (3.12) is
max (𝑤𝑃 :𝑃 a simple (𝑟, 𝑠)-dipath) (3.13)
s.t. (𝑤𝑃 :𝑒 an arc of 𝑃)≤𝑢𝑒, for all 𝑒∈𝐸
𝑤𝑃≥0 , for all 𝑃 a simple (𝑟, 𝑠)-dipath
Let 𝑥 be a max flow. Find simple (𝑟, 𝑠)-dipath 𝑃 such that 𝑥𝑒>0 for each arc of 𝑃, put 𝑤𝑃= min value of 𝑥𝑒 on 𝑃, subtract 𝑤𝑃 from 𝑥𝑒 for each arc 𝑒 of 𝑃.
Repeat until 𝑓𝑥(𝑠)=0, at which point 𝑤𝑃= max flow value.
Then, 𝑤 is feasible to (3.13) and since 𝑤𝑃=(𝑢𝑒𝑧𝑒′), 𝑧′ is optimal to (3.12). 
Combinatorial Optimization 2016

12. Thm 3.21 implies that the cut polyhedron defined by (3.12) is integral
Similar result also holds for path polyhedron. For the polyhedron
(𝑥𝑒 :𝑒∈𝐶)≥1, for all arc-sets 𝐶 of (𝑟, 𝑠)-cuts
𝑥𝑒≥0, for all 𝑒∈𝐸
Extreme points are incidence vectors of simple (𝑟, 𝑠)-dipaths
(may be seen from results for blocking polyhedron)
Let 𝑃={𝑥∈ 𝑅 + 𝑛 :𝐴𝑥≥1}, where 𝐴 is a nonnegative matrix.
Then the blocking polyhedron of 𝑃 is defined as
𝑃 𝐵 ={𝜋∈ 𝑅 + 𝑛 : 𝜋𝑥≥1 ∀𝑥∈𝑃}
Then 𝑃 𝐵 ={𝜋∈ 𝑅 + 𝑛 :𝐵𝜋≥1}, where rows of 𝐵 are extreme points of 𝑃.
Also 𝑃 𝐵 𝐵 =𝑃.
Here, each row of 𝐴 is the incidence vector of a simple 𝑟, 𝑠 −dipath. Each row of 𝐵 is the incidence vector of a 𝑟,𝑠 −cut.
(no more details here.)
Combinatorial Optimization 2016