cutset lecture

8/18/2019 Cutset Lecture

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Cut set method: concept

• Minimal cut sets in the following systems

5Electrical & Computer Engineering Department

Minimal cut set # Components of the cut set

1 AB

2 CD

3 AED

4 BEC


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Application of cut sets

• Most, but not all, of the methods for deducing cuts are based on a

knowledge of the minimal paths between input and output.

• A minimal path can be defined as: – A path between the input and output is minimal if, in that path, no node or

intersection between branches is traversed more than once.

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Electrical & Computer Engineering Department

AC, BD, AED, BECMinimal paths


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• All components of each cut must fail in order for the system to fail.•

Consequently, the components of the cut set are effectively connected in parallel – The failure probabilities of the components in the cut set may be combined using the

principle of

parallel systems

• In addition, the system fails if anyone of the cut sets occurs and consequently eachcut is effectively in series with all the other cuts.


Minimal cut set # Components ofthe cut set

1 AB

2 CD

3 AED

4 BEC


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• Although these cut sets are in series, the concept of series systems cannot

be used because the same component can appear in two or more of the cutsets, e.g., component A appears in cuts C1 and C3.

• The concept of union does apply however and if the i th cut is designatedas and its probability of occurrence is designated as P().

• The unreliability of the system is given by

( ∪ ∪3…∪…∪)

8

Electrical & Computer Engineering Department


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Example 1

• Consider the following power distribution system. Components A, B, C, D,

and E indicate distribution lines. The system success requires that at leastone of the paths, AC, BD, AED, BEC is good. Determine the powerdistribution system unreliability if the reliability of A=B=C=D=E=0.99?



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Example 1

∪ ∪3 ∪ + + 3 +

∩ ∩3 ∩C P C ∩3 ∩ 3 ∩+ ∩ ∩3 + ∩ ∩ + ∩3 ∩ + ∩3 ∩ ∩ ∩3 ∩

∩

∩ ∩ . . ∩ ∩3 ∩ . . 3 . 10Electrical & Computer Engineering Department


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Example 1

+ + + +2

If , then

2 +23

5 +2Given that 0.99 → 1 0.01

0.00020195 1 0.99979805(similar to Example 1)


l


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Approximate evaluation

• The precise evaluation is always theoretically possible, but it can be an

exhaustive and time-consuming exercise which can become prohibitivewith large systems.

• The first approximation assumes that can be reduced to a summation ofunreliabilities:

+ +⋯ +⋯+ =

()

Note: the reliability of a component, i.e., , is itself an approximation not anexact value.


Cl i i 1


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Classroom activity 1• Consider the following power distribution system. Components A, B, C, D, and E

indicate distribution lines. The system success requires that at least one of the paths,AC, BD, AED, BEC is good. Use the firs approximation, and then determine the

power distribution system unreliability?


Cl i i 1


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Classroom activity 1• Solution:

+ + +If

2 +23 0.000202 0.999798

Compared with the precise value: – Error of

+0.02% – Error of 5×10− %When this approximation is made, the value of is always greater than the exactvalue and exhibits the greatest positive error.

– the upper bound of system unreliability

If the second order terms are also included in the analysis, the value of will be lessthan the exact value and will exhibit the greatest negative error.

– the lower bound of system unreliability


A i l i


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Approximate evaluation

• The second approximation is to neglect cut sets of an order greater than a

certain value – the order of a cut set being equal to the number of components comprising thatcut set, i.e., a cut set created by two components is known as a cut set of order2 or a second order cut set

• This approximation assumes that high order cut sets are much less probablethan low order cut sets.

• This is valid if all components have reliabilities of similar value but can beinvalid if a low order cut set involves components having very highreliabilities while a high order cut set involves components having very low

reliabilities.


E l 2


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Example 2

• Reconsider the following system which was discussed in the previousclassroom activity. If we neglect the third order (and up) cut sets (i.e., thefirst and the second approximations):

+If

2 0.000200 0.9998000

Error o f 1%Error of +2×10− %These are small and would decrease as component reliabilities increased.


Cl ti it 2


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Classroom activity 2

• Consider the following power system where components A, B, C, D, E, andF indicate the distribution feeders. Evaluate the reliability of the systemusing

a) conditional probability and b) cut set methodsif each component has a reliability of 0.99. Assume that source and load are

100% reliable.



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Classroom acti it 2


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Cut set method:The minimal cut sets are AB, AD, CE, AE, BFC, DFC

1) Consider first the result that will be obtained if only second order events are usedand the evaluation is reduced to one of summating the cut probabilities

+ + + 0.0004000 0.9996002) Consider now the result that will be obtained if all of the cuts are used and theevaluation is again limited to one of summating the cut probabilities

+ + + + + 0.000402 0.999598


cutset lecture

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