1. Estimation of the Number of Min-Cut Sets in a Network
Ilya Gertsbakh
Department of Mathematics, BGU
Yoseph Shpungin
Sami Shamoon Technological College 1

2. 1. What is a min size cut set. 2. Why is it important to know its size
and the number of such sets: useful B-P
approximation to system failure probability!
3. To learn how good the approximation is,
it is necessary to estimate system failure probability- [spectrum ].
4. Some methods to make the simulation/computations more efficient –Lomonosov’s lemma 2

3. Estimating the number of min cuts is a NP-problem.
What we will do, in fact, is an implementation of a special enumeration scheme which will enable to count the number of min cuts in a reasonable time
for medium-size networks (up to 80 edges)
2-A

4. The Network and Min Cut-Set: definitions
Min cut set is the “weakest link” of the system . The “easiest”
way to destroy the system is to invalidate its min-size cutset
its size by r 3

5. Example The network: W(r)=2 m=5 edges 4 1 s t 3 2 5
T={s,t} ; {1,2,3} is a CUTSET, but not a
Min CUTSET. {1,2} is a CUTSET of Min
size r=2. {1,3,5} is a CUTSET , but not
of minimal size.
Here W(r)=2,
r=2
Denote by W(r) the number of min-size CUTSETS. 4

6. Why is it important to know W(r) :
probability
independently
Will be proved later 5

7. Important:
Not only to know the number of min-size cuts of size , W(r),
but to know the DIMENSION
r of the min-size cut set.
How accurate is the apprioximation 6

8. A general remark: All what we say below is applicable to any
monotone system consisting of identical and independent components.
Network –as a system – and edges –as components (or nodes as components )
is easy to visualize
6-A

9. Network Destruction Spectrum
permutation
of m edges n n
the
goes 7

10. the
W(r)=
Will be proved
W(r) m 6 8

11. Example 4 1 3 t s 5 2
{1,2} is a min size CUTSET
W(2)= 9 7

12. Proof of the formula for W(r)10

13. Probability of missing a minimal size cut set
in M=10^7
100, 11

14. P[ to miss S in a single experiment]=
Suppose we did M experiments and composed a list of all different min cut sets (of size r=5). What is the probability that we will miss one particular set S of size 5 ?
The probability
P[ to miss S in a single experiment]=
1-5!75!/80!= ( /10,000,000);
[P to miss S in M= 300,000,000 (!!) experiments]=3.8/1,000,000 – practically negligible 12

15. Specially designed algorithm
Spectrum Monte Carlo
The simulation algorithm
Most important !!
Specially designed algorithm
Is used 13

16. Proof of the Burtin-Pittel Formula
For independent edges, 14

17. This term is instead of “2” because in case of equal
edge failure probabilities
all b(i) =1
Important: here we need more information: not only the number
of min cuts but the lists of edges in these cuts 15

18. Numerical Experiments
Hypercube H-5: 32 nodes, 80 edges
Nodes labelled from (0,0,0,0,0) to
(1,1,1,1,1).
Edges connect nodes differing by one digit:
(0,0,0,0,0) (1,0,0,0,0) (1,1,0,0,0)
(0,1,0,0,0) (1,0,1,0,0)
(0,0,1,0,0) (1,0,0,1,0)
(0,0,0,1,0) (1,0,0,0,1)
(0,0,0,0,1)
H-k has
k x 2^(k-1) edges 16

19. The Spectrum for H-5, 7 terminals:
f(5)=98x10^(-6); f(13)=49,540x10^(-6)
f(6)=465x10^(-6); f(14)=72,354x10^(-6)
f(7)=1427x10^(-6); f(15)=102,525x10^(-6)
f(8)=3291x10(-6); f(16)=137,073x10^(-6)
f(9)= 6,529x10^(-6); f(17)=162,745x10^(-6)
f(10)=12,076 x10^(-6); f(18)=161,968x10^(-6)
f(11)=20,354x10^(-6); f(19)=126,600x10^(-6)
f(12)=32,627x10^(-6); f(20)=73,243x10^(-6)
f(21)=29,151x10^(-6)
f(22)=7,024x10^(-6)
f(23)=854x10^(-6) 17

20. 18 runs, each 1,000,000 replications,
H-5 with 7 randomly chosen terminals, 25 nodes subject to failures, edges do not fail:
18 runs, each 1,000,000 replications,
The number of min size cuts has 95% Student small-sample confidence interval
on W(5) :[4.96, 5.38] -> 5 min cuts
The B-P approximation is
B-P=W(5)x a^5= 160x10^(-5), for a=0.2
= 38x10^(-5), for a=0.15
= 5 x 10^(-5), for a=0.1 18

21. How good is B-P approximation?
B-P=W(5)x a^5= , for a=0.3
= , for a=0.2
= , for a=0.1
Exact result
F(0.3)=0.0124, for a=0.3;
F(0.2)= , for a=0.2;
F(0.1)= , for a=0.1;
F(0.01)=5.14x10^(-10) ;B-P=>5.00x10^(-10)
Fantastic accuracy, probably we are just lucky! 19

22. 10 runs, each 10,000,000 replications each
H-5 with 12 randomly chosen terminals, 80 edges subject to failures, nodes do not fail:
10 runs, each 10,000,000 replications each
The number of min size cuts has 95% Student small-sample confidence interval
on W(5) :[12, 14] - min cuts
Assuming W(5)=12: a=
F x10^(-6)
BP x10^(-6) 20

23. Numerical results:H-5, 12 terminals, 80 edges, 30 edges randomly displaced
Minimal r=3. By our formula, W(3)=1.04
Exact check of all min cuts in 10^8 runs
Discovers only ONE min cut of size 3
Edge Failure Pr Pr(Down). Exact BP Appr
x10^(-6) x10^(-6) 21

24. How to compute or estimate the system DOWN probability ?
Natural question:
How to compute or estimate the system DOWN probability ?
For the case of equal edge failure probabilities,
we will use the spectrum approach 22

25. 3. I.I.D. edge lifetimes 2 a b Terminals are (s,t) s=1 t=4 3 c d
a b C d, a b D c -> k=3; a C b d, a D b c -> k= Total = 4!=24 23
Spectrum={0,2/3,1/3,0}

26. Example: graph K5: n=5, T=V24
K-th order statistic

27. failure occurs at t(1), the second at t(2), t(1) <t(2) < t(3),…
Think of edge failures as a sequence of events developing in time. The first
failure occurs at t(1), the second at t(2),
t(1) <t(2) < t(3),…
No matter which edge fails at t(1), the value of t(1) is a replica of the first order statistic, t(2) is a replica of the second order statistic, etc. If k is the critical number, then the k-th order statistic is the instant of system failure 25

28. For our purposes, we fix t=1 and set F(1)=alpha
Important:
For our purposes, we fix t=1 and set F(1)=alpha 26

29. This formula is important to establish the accuracy of the Spectrum
approach. For the case of H-5, 7 terminals, 80 edges,alpha=0.2,
F(exact) ~ and the standard deviation Sigma~ ;
For alpha =0.1, F(exact)~ , Sigma ~ 27

30. How to treat nonreliable nodes ?v
Node v fails at the instant t e1 e2 t
System has reliable nodes,
all edges incident v
have lifetime t t
time 28 t
e1,e2 fail at t

31. Key issue: fast estimation of the spectrum:
First what comes to mind:
When you move along a permutation.
check the terminal connectivity after each
edge removal Very time consuming.
By Lomonosov’s Lemma, it is enough
to construct only one maximal spanning tree !!
How it works: 29

32. Give weights to edges acc. to the permutation:30
The algorithm:
Take a permutation of edge numbers:
Edge (1,2,3,4,5,6,7,8) Weight=Lifetime
Permutation : (8,4,7,3,6,2,5,1)-edge No
Edge No 7 gets weight 3 4 6 8 5 1 2 2 7 6 3 3 5 4 7 8 1
All nodes are terminals !! Failure =loss connectivity
Give weights to edges acc. to the permutation:

33. Construct the maximal spanning tree-Kruskal algorithm31
The smallest link in the maximal
tree is 5 5 8 6 7
Claim: for the above permutation,
the critical number is 5 5
Let us delete edges acc. to the permutation: N8,N4, N7,N3,N6 2 3 4 1
The network breaks down on the
fifth step, i.e. after removal the edge N6
q(Perm)=5

34. We have the following edge numbers: (1,2,3,4,5,6,7,8) -Lifetimes
And their permutation:
(8,4,7,3,6,2,5,1) –No of edges:
edge number 8 has lifetime 1,
edge number 4 has lifetime 2, etc.
edge number 1 has lifetime 8.
At time t=1, fails edge 8, at t=2, fails edge
4,at time 3 fails edge 7,…, at t=5 fails edge
6. And this is the network failure. So, the network fails at the elimination of the fifth edge !-> Critical No r=5 32

35. =26
Construct the maximal spanning tree-Kruskal algorithm: the weight of it is maximal b
The smallest link in the maximal
tree is 5 5 8 d e a 6 7
Claim: for the above permutation,
the critical number is 5 c
Proof: Obviously the network is alive (connected) at time=5. Suppose
It can survive time T>5. Let us delete the edge (b,e). All nodes fall apart into
two connected groups S=(b,a,d,c) and Q=(e). There must be a spanning tree
with an edge, say, (a,e) connecting S and Q. This edge
therefore lives longer than 5 - contradiction because it means the existence
of a spanning tree with larger weight than the original one. 6
Let us delete edges acc. to the permutation: N8,N4, N7,N3,N6 4 7 3 8
The network breaks down on the
fifth step, i.e. after removal the edge N6
q(Perm)=5 33

36. Statistics is the science of producing unreliable facts from reliable figures.
Quips & Quotes, p.765

37. Thank you

38. Expressing C(r+k), k>0 via the Spectrum

39. How large the number of replications M should be ?
59,000,000 34

41. Another representation of P(S)C(
C(r+1),
C(r+2), C(
C(r+1) –is the number of system Down states with exactly (r+1)
components being down
How to find C(r+k) – see slide 14, without proof

42. How large the number of replications M should be ?
59,000,000 35