Please click on one of the flags to reset Reading-Direction if you consider the current setting invalid

Mocus Algorithm for obtaining minimal cut setes (mcs) of a random computer network

Views  315
Rating  0

 زاهر عبد الهادي حسن الشنون
01/02/2020 19:33:22
تصفح هذه الورقة الالكترونية بتقنية Media To Flash Paper
Zahir Al Shanoon, [31.01.20 20:51]
Fault Tree and Success Tree Methods:
The operation of a system can be considered from two opposite viewpoints: the various ways that a system fails or the various ways that a system succeeds. Most of the construction and analysis methods used are, in principle, the same for both fault trees and success trees. First we will discuss the fault tree method and then describe the success tree method.
1- Fault Tree Analysis (FTA)
This method was developed at the Bell Laboratories in the early
1960s to evaluate the reliability and safety of the Minuteman Launch
Control System [3]. Fault tree analysis (FTA) starts by defining the
undesirable state (event) of the system or item under consideration
and then analyzes the system to determine all possible situations
that can result in the occurrence of the undesirable event. Thus, it
identifies all possible failure causes at all possible levels associated
2
with a system as well as the relationship between causes. FTA can
be used to analyze various types of maintainability-related problems. FTA uses various types of symbols [3]. Four commonly used symbols in fault tree construction are shown in Figure 2. The circle denotes a basic fault event or the failure of an elementary component. The event s occurrence probability and failure and repair rates are normally obtained from empirical data. The rectangle denotes a
fault event that results from the combination of fault events through
the input of a logic gate.
Figure 2: Four commonly used fault tree symbols: (a) AND gate,
(b) OR gate, (c) rectangle and (d) circle.
The OR gate denotes that an output fault occurs if one or more
of the input fault events occur. Finally, the AND gate denotes that
an output fault event occurs only if all the input fault events occur.
The probabilities of the occurrence of the output fault events of logic
gates (OR and AND) are given by
3
The OR gate
??(??0)=1??1???(????????=1) (1)
where ??(??0) is the probability of occurrence of the OR gate output
fault event, ??0 , n is the number of independent input fault events,
and ??(????) is the probability of occurrence of input fault event ????
for i = 1, 2, 3, . . . , n .
The AND gate
??(??0)=???(????????=1) (2)
where ??(??0) is the probability of occurrence of the AND gate
output fault event, ??0 and ??(????) is the probability of occurrence
of input fault event ???? for i = 1, 2, 3, . . . , n.
4
Figure 3: a fault tree of a random computer network system
By using Boolean algebra operations we get : T=E1.E2.E3
E13=X8+X12, E12=X9.E13=X8X9+X9X12 E3=X3+E12= X3+X8X9+X9X12 (3) E11=X8+X12, E10=X9.E11=X8X9+X9X12
E8=X6+E10=X6+X8X9+X9X12, E9=X7+X11
5
E7=E8E9=(X6+X8X9+X9X12).(X7+X11)
=X6X7+X7X8X9+X7X9X12+X6X11+X8X9X11+X9X11X12
E2=X2+E7=X2+X6X7+X7X8X9+X7X9X12+X6X11+X8X9X11+X9X11X12 (4)
E6=X5+X11, E5=X4+X10
E4=E5.E6=(X4+X10).(X5+X11)=X4X5+X4X11+X5X10+X10X11
E1=X1+E4=X1+X4X5+X4X11+X5X10+X10X11 (5)
From 3,4 and 5 we get :
T=x1x2x3 + x2x3x4x5 + x1x3x6x7 + x1x2x8x9 + x2x3x4x11 + x2x3x5x10 + x1x3x6x11 + x1x2x9x12 + x3x4x6x11 + x1x7x8x9 + x2x3x10x11 + x1x7x9x12 + x1x8x9x11 + x3x6x10x11 + x4x8x9x11 + x1x9x11x12 + x4x9x11x12 + x8x9x10x11 + x9x10x11x12 + x3x4x5x6x7 + x2x4x5x8x9 + x3x5x6x7x10 + x2x4x5x9x12 + x4x5x7x8x9 + x2x5x8x9x10 + x4x5x7x9x12 + x4x5x8x9x11 + x2x5x9x10x12 + x5x7x8x9x10
the all minimal cut sets of the system are
{x1,x2,x3}, {x2,x3,x4,x5}, {x1,x3,x6,x7},{x1,x2,x8,x9}, {x2,x3,x4,x11},
{x2,x3,x5,x10 },{x1,x3,x6,x11},{ x1,x2,x9,x12}, {x3,x4,x6,x11}, {x1,x7,x8,x9},
{x2,x3,x10,x11},{x1,x7,x9,x12},{x1,x8,x9,x11},{x3,x6,x10,x11},{x4,x8,x9,x11},
{x1,x9,x11,x12},{x4,x9,x11,x12},{x8,x9,x10,x11},{x9,x10,x11,x12},{x3,x4,x5,x6,x7},
{x2,x4,x5,x8,x9},{x3,,x5,x6,x7,x10},{x2,x4,x5,x9,x12},{x4,x5,x7,x8,x9},{x2,x5,x8,x9,x10},r
{x4,x5,x7,x9,x12},{x4,x5,x8,x9,x11},{x2,x5,x9,x10,x12},{x5,x7,x8,x9,x10}
6
Figure 4: Parallel- series system structure
7
The Reliability of a parallel-series system is
????= ?(1??(1?????)) (6)????=1????=1

Zahir Al Shanoon, [31.01.20 20:51]
????= R3R9 + R1R4R10 + R1R5R11 + R2R6R9 + R2R7R11 + R3R8R12 - R2R3R6R9 + R2R6R8R12 - R3R8R9R12 - R1R2R5R7R11 - R1R3R4R9R10 - R1R3R5R9R11 - R1R4R5R10R11 - R2R3R6R8R12 - R2R3R7R9R11 - R2R6R7R9R11 - R2R6R8R9R12 - R1R2R4R6R9R10 - R1R2R5R6R9R11 - R1R2R4R7R10R11 - R1R3R4R8R10R12 + R2R3R6R7R9R11 - R1R3R5R8R11R12 + R2R3R6R8R9R12 - R2R3R7R8R11R12 - R2R6R7R8R11R12 + R1R2R3R4R6R9R10 + R1R2R3R5R6R9R11 + R1R2R3R5R7R9R11 + R1R2R4R5R7R10R11 + R1R2R5R6R7R9R11 - R1R2R4R6R8R10R12 + R1R3R4R5R9R10R11 - R1R2R5R6R8R11R12 + R1R3R4R8R9R10R12 + R1R3R5R8R9R11R12 + R2R3R6R7R8R11R12 + R2R3R7R8R9R11R12 + R2R6R7R8R9R11R12 - R1R2R3R5R6R7R9R11 + R1R2R3R4R6R8R10R12 + R1R2R3R4R7R9R10R11 + R1R2R3R5R6R8R11R12 + R1R2R4R5R6R9R10R11 + R1R2R3R5R7R8R11R12 + R1R2R4R6R7R9R10R11 + R1R2R4R6R8R9R10R12 + R1R2R5R6R7R8R11R12 + R1R2R5R6R8R9R11R12 + R1R3R4R5R8R10R11R12 - R2R3R6R7R8R9R11R12 - R1R2R3R4R5R6R9R10R11 - R1R2R3R4R5R7R9R10R11 - R1R2R3R4R6R7R9R10R11 - R1R2R3R4R6R8R9R10R12 - R1R2R3R5R6R7R8R11R12 - R1R2R4R5R6R7R9R10R11 - R1R2R3R5R6R8R9R11R12 + R1R2R3R4R7R8R10R11R12 - R1R2R3R5R7R8R9R11R12 + R1R2R4R5R6R8R10R11R12 + R1R2R4R6R7R8R10R11R12 - R1R2R5R6R7R8R9R11R12 - R1R3R4R5R8R9R10R11R12 + R1R2R3R4R5R6R7R9R10R11 - R1R2R3R4R5R6R8R10R11R12 - R1R2R3R4R5R7R8R10R11R12 - R1R2R3R4R6R7R8R10R11R12 + R1R2R3R5R6R7R8R9R11R12 - R1R2R4R5R6R7R8R10R11R12 - R1R2R3R4R7R8R9R10R11R12 - R1R2R4R5R6R8R9R10R11R12 - R1R2R4R6R7R8R9R10R11R12 + R1R2R3R4R5R6R7R8R10R11R12 + R1R2R3R4R5R6R8R9R10R11R12 + R1R2R3R4R5R7R8R9R10R11R12 + R1R2R3R4R6R7R8R9R10R11R12 + R1R2R4R5R6R7R8R9R10R11R12 - R1R2R3R4R5R6R7R8R9R10R11R12
8
2-Success Tree Analysis (STA)
The success tree method is conceptually the same as the fault tree method. By defining the desirable top event, all intermediate and primary events that guarantee the occurrence of this desirable event are deductively postulated. Therefore, if the logical complement of the top event of a fault tree is used as the top event of a success tree, the Boolean structure represented by the fault tree is the Boolean complement of the success tree. Thus, the success tree, which shows the various combinations of success events that guarantee the occurrence of the top event, can be logically represented by path sets instead of cut sets. to better understand this problem, consider a random computer network shown in Figure 1. The success tree for this system is shown in Figure 5.
Figure 5: a success tree of a random computer network system
9
By Using Boolean algebra operations, we get:
T=E1+E2+E3
E13=X8X12, E12=X9+ E13=X9+X8X12
E3=X3E12=X3X9+X3X8X12 (7)
E11=X8X9, E10=X9+X8X12
E8=X6.E10=X6X9+X6X8X12, E9=X7X11
E7=E8+E9=X7X11+X6X9+X6X8X12
E2=X2E7=X2X7X11+X2X6X9+X2X6X8X12 (8)
E6=X5X11, E5=X4X10
E4=E5+E6=X5X11+X4X10
E1=X1E4=X1X4X10+X1X5X11 (9)
from 7,8 and 9 we get:
T= X1X4X10+X1X5X11+X2X7X11+X2X6X9+X2X6X8X12+X3X9+X3X8X12
The all minimal path sets of the system are {x1,x4,x10}, {x1,x5,x11}, {x2,x7,x11}, {x2,x6,x9},{x2,x6,x8,x12},{x3,x9},{x3,x8,x12}
Figure 6: series - Parallel system structure
10
The reliability of a series-parallel system is
????= 1??(1??????) (10)????=1????=1

Zahir Al Shanoon, [31.01.20 20:51]
????= R3R9 + R1R4R10 + R1R5R11 + R2R6R9 + R2R7R11 + R3R8R12 - R2R3R6R9 + R2R6R8R12 - R3R8R9R12 - R1R2R5R7R11 - R1R3R4R9R10 - R1R3R5R9R11 - R1R4R5R10R11 - R2R3R6R8R12 - R2R3R7R9R11 - R2R6R7R9R11 - R2R6R8R9R12 - R1R2R4R6R9R10 - R1R2R5R6R9R11 - R1R2R4R7R10R11 - R1R3R4R8R10R12 + R2R3R6R7R9R11 - R1R3R5R8R11R12 + R2R3R6R8R9R12 - R2R3R7R8R11R12 - R2R6R7R8R11R12 + R1R2R3R4R6R9R10 + R1R2R3R5R6R9R11 + R1R2R3R5R7R9R11 + R1R2R4R5R7R10R11 + R1R2R5R6R7R9R11 - R1R2R4R6R8R10R12 + R1R3R4R5R9R10R11 - R1R2R5R6R8R11R12 + R1R3R4R8R9R10R12 + R1R3R5R8R9R11R12 + R2R3R6R7R8R11R12 + R2R3R7R8R9R11R12 + R2R6R7R8R9R11R12 - R1R2R3R5R6R7R9R11 + R1R2R3R4R6R8R10R12 + R1R2R3R4R7R9R10R11 + R1R2R3R5R6R8R11R12 + R1R2R4R5R6R9R10R11 + R1R2R3R5R7R8R11R12 + R1R2R4R6R7R9R10R11 + R1R2R4R6R8R9R10R12 + R1R2R5R6R7R8R11R12 + R1R2R5R6R8R9R11R12 + R1R3R4R5R8R10R11R12 - R2R3R6R7R8R9R11R12 - R1R2R3R4R5R6R9R10R11 - R1R2R3R4R5R7R9R10R11 - R1R2R3R4R6R7R9R10R11 - R1R2R3R4R6R8R9R10R12 - R1R2R3R5R6R7R8R11R12 - R1R2R4R5R6R7R9R10R11 - R1R2R3R5R6R8R9R11R12 + R1R2R3R4R7R8R10R11R12 - R1R2R3R5R7R8R9R11R12 + R1R2R4R5R6R8R10R11R12 + R1R2R4R6R7R8R10R11R12 - R1R2R5R6R7R8R9R11R12 - R1R3R4R5R8R9R10R11R12 + R1R2R3R4R5R6R7R9R10R11 - R1R2R3R4R5R6R8R10R11R12 - R1R2R3R4R5R7R8R10R11R12 - R1R2R3R4R6R7R8R10R11R12 + R1R2R3R5R6R7R8R9R11R12 - R1R2R4R5R6R7R8R10R11R12 - R1R2R3R4R7R8R9R10R11R12 - R1R2R4R5R6R8R9R10R11R12 - R1R2R4R6R7R8R9R10R11R12 + R1R2R3R4R5R6R7R8R10R11R12 + R1R2R3R4R5R6R8R9R10R11R12 + R1R2R3R4R5R7R8R9R10R11R12 + R1R2R3R4R6R7R8R9R10R11R12 + R1R2R4R5R6R7R8R9R10R11R12 - R1R2R3R4R5R6R7R8R9R10R11R12
11
3-Reduction Method (RM):
This is a simple and useful method for determining the reliability of systems consisting of independent series and parallel subsystems. The method sequentially
reduces the series and parallel configurations to equivalent units until the whole
system becomes a single hypothetical unit [2]. The primary advantage of this method is that it is easy to understand the use. The following example demonstrates the method.
Example1. Consider a system whose reliability block diagram is given
in Figure 1, it represent a graph of mixed system whose edges are the system
Components.
we will simplify by using series and parallel reduction step by step
? step1 X13=X4X10
? step2 X14=X7X11
? step3 X15 =1-(1-X8X12)(1-X9)
Figure7: after reduction1 of the system
? step4 X16=X5X6
? step5 X17= 1-(1-X1)(1-X2)
? step6 X18= 1-(1-X13)(1-X14)
Rewrite the Components by replacing
R1=X17, R2=X3,R3=X16, R4= X18, R5= X15
12
Figure8: Reduction of complex system
The reliability of system is
Rre=R1R4 + R2R5 + R1R3R5 - R1R2R3R5 - R1R2R4R5 - R1R3R4R5 + R1R2R3R4R5
13
Rules of Boolean Algebra
Mathematical Notation
Engineering Notation
Designation
X ?Y=Y?X
X.Y=Y.X
Commutative Law
X U Y=Y U X
X+Y=Y+X
X? (Y? Z) = (X?Y) ?Z
X. (Y. Z) = (X. Y) . Z
X(YZ)=(XY)Z
Associative Law
X U (Y U Z) = (X U Y) U Z
X + (Y + Z) = (X + Y) + Z
X ? (Y U Z) = (X ? Y) U (X? Z)
X .(Y + Z) =(X .Y) + (X? Z)
X(Y+Z)= XY + XZ
Distributive Law
X U (Y ? Z)= (X U Y) ? (X U Z)
X+ (Y. Z)= (X+ Y) . (X + Z)
X?X=X
X.X=X
Idempotent Law
XUX=X
X+X=X
X? (X U Y)=X
X . (X + Y)=X
Law of Absorption
X U(X ?Y)=X
X +(X .Y)=X
X???? = ?
X . ??? = ?
Complementation
XU??? = ? = I
X + ??? = ? =I
?? ????? = ??? U ??? ?? .???? = ??? + ??? De Morga??,s Rule
?? ?? ???? = ???? ??? ??+???? = ??? . ???
? ???=?
? . X=?
Operations with ? and ?
?UX=X
?+X=X
? ?X=X
? . X=X
? ?X=?
?+X=?

  • وصف الــ Tags لهذا الموضوع
  • Methods: Reliability, Mocus Algorithem, computer