SOLUTION MANUAL for Models of Network Reliability

SOLUTION MANUAL for Models of Network Reliability: Analysis, Combinatorics and Monte Carlo I. Gertsbakh and Y. Shpungin August 29, 2009

1

Contents 1 Solutions to Chapter 1

2

2 Solutions to Chapter 2

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1

1

Chapter 1.

1. First, let us explain how works the Mathematica program given in Section 1.5. The first line calls the Statistics package and the Continuous distribution generators. dist calls for a uniform distribution with the support [0,2], Q is the number of samples, sig counts the number of cases when the average lies inside the confidence interval. x1, x2, x3, x4 are the replicas of random variable uniformly distributed in [10,12], representing one sample of four items. xbar and stdev are the average and the standard deviation of a single sample, according to the formulas given in Problem 1. lef t, right are the limits of the confidence interval, according to (1.5.2). A bit tricky is operator alpha which produces either 2 if lef t < 11 < right or zero, otherwise. (Why ?). So, each time the true mean lies inside the confidence interval, the counter sig increases by 1. Operator For[....] repeats all the above Q times. Finally, conf gives the coverage proportion in decimals, and Print types the result. To repeat the experiment for exponential distribution with mean 11, change the operator dist = ExponentialDistribution[1]. The probability of coverage is 0.88 for the Uniform case and 0.83 for the Exponential case. We doubt that your boss will accept your proposal to use the normal confidence interval for the population mean. 2. Search algorithm for optimal schedule a. Set Max=100, Q=10,000. Put N=0. Define a matrix A with 10 rows and 20 columns; denote by a(i, j) the j-th element of the i-th row. Put initially all a(i, j) = 0. b. For i = 1 to i = 10, do the following: generate X(i) uniformly distributed in the interval [−δl (i)], δr (i)]; The actual beginning and completion of job i will be b(i) + X(i) = beg(i) and c(i) + X(i) = comp(i), respectively. Determine the actual schedule S = a(i, j) = 1 for j = [beg(i), ..., comp(i)], for i = 1, ...10. a(i, j) will be equal one for the time points which are occupied by job i, and zero, otherwise ∑ Calculate K = maxj=1,...,20 [ 20 i=1 a(i, j)]. 2

If K is less than Max, put Max:=K. Remember the respective schedule S for all jobs. In this way, the program remembers the schedule which has produced so far the best result. Set N:=N+1. If N =Q, Stop, else put all a(ij) = 0 and GOTO b. There is a little need to explain this algorithm. The main operator b locates randomly all jobs into the time matrix. K is the maximal number of jobs carried out simultaneously, which, by the Dilworth theorem, equals the minimal number of machines. The program remembers the number of machines and the corresponding schedule each time better result is achieved. 3. Below is a Mathematica program which counts the lucky numbers in R = 100, 000 experiments. Random[ ] produces a random variable uniformly distributed in [0,1]. When multiplied by 10, this becomes a random variable U (0, 10). Operator Floor takes its integer part. Therefore x is the sum of 6 digits in the lottery. A bit tricky is the expression f . It equals zero if x is a lucky number- 4, 9, 16, 25, 36 or 49. Otherwise f > 0. Operator Sign is zero if the argument is zero and 1 if it is positive. Thus the counter sig counts the number of cases when the sum is not a lucky number. r is the desired result in decimal form. Algorithm LuckyNumbers. N · (min fi ) · Then

∑

ai . ∑

(V ar[R])0.5 (max fi )0.5 · ( a2i )0.5 ∑ . < E[R] N 0.5 · (min fi ) · ai The result follows since

∑

∑

a2i < ( ∫

ai )2 .

7. By the definition, E[X] = 01 t · fX (t) · dt = 1/2. E[X] is the center of mass with respect to the density function. By the definition, V ar[X] = ∫1 2 0 (t − 0.5) · fX (t)dt =[after elementary calculations]= 1/12. FX (t) = 0 for t < 0 and obviously equals∫1 for t > 1. For t ∈ [0, 1], P (X ≤ t) = 0t fX√ (v)dv = t. 0.5 c.v[X] = (1/12) /0.5 = 1/ 3. 8. We present an intuitive solution. Let X ∼ U (0, 1). The probability mass is uniformly spread in the interval [0, 1]. Suppose we multiply X by C. Then the probability mass of the random variable CX will be uniformly spread in the interval [0, C]. After adding constant D, the probability mass is shifted by D and will be uniformly spread in the interval [D, D + C]. To guarantee that the integral of this mass equals 1, the density of Z = CX + D must be 1/C and zero outside this interval. So, Z ∼ U (D, D + C). By using the rules of computing the mean and variance (see Section 1.4), E[Z] = C · E[X] + D = C/2 + D, V ar[Z] = C 2 · V ar[X] = C 2 /12. FZ (t) = 0 for t < D and FZ (t) = 1 for t > D + C. For t ∈ [D, D + C], FZ (t) = (t − D)/C. 9. Y = 2 · Random [ ] − 1, where Random[ ] produces U (0, 1). 10. The key to the solution is the following fact. If the maximum Y of X1 , X2 , ..., Xn is less or equal t, then all Xi , i = 1, ..., n must be less or ∏ equal t. Since Xi are i.r.v.’s, P (Y ≤ t) = ni=1 P (Xi ≤ t). For t ∈ [0, 1], it gives FY (t) = tn . For t < 0, FY (t) = 0 and for t > 1, FY (t) = 1. The density function of Y is fY (t) = ntn−1∫ for t ∈ [0, 1], and zero, otherwise. By integration, we find out that E[Y ] = 0t v · nv n−1 dv = n/(n + 1). 11. The probability to have k successes in the first k experiments and later on no successes is, obviously, pk (1 − p)n−k . But there are n!/(k!(n − k)!) different ways of positioning k successes in n experiments. This explains the result.

5

2

Chapter 2.

1. From Example 2.3.1-continued we obtain P (P1 ∪ P2 ∪ P3 ) = p1 p2 + p1 p3 + p2 p3 − 2p1 p2 p3 . Replacing in (2.3.3) qi = 1 − pi , we arrive at the expression p1 p2 (1 − p3 ) + p1 p3 (1 − p2 ) + p2 p3 (1 − p1 ) + p1 p2 p2 , which after simple algebra is identical to the previous expression. 2. There are two minimal cut sets of size 2, (1, 7), (7, 4) and (3, 6), (6, 5). They isolate nodes 6 and 7. The cut set (1, 2), (2, 4), (2, 3) isolates node 2. All other minimal cut sets contain 4 or more edges. 3. Associate with each edge e the weight w(e) = ln pe . Then use Kruskal algorithm to construct the maximal spanning tree. It will produce a tree with the maximal value of ln P (ST ), i.e. with maximal P (ST ). 4. The upper bound is RU = 1 − (1 − p2 )2 (1 − p3 )2 . The lower bound is RL = [1 − (1 − p)2 ]2 [1 − (1 − p)3 )2 ]. The numerical results are: p = 0.90, RL = 0.978141, RU = 0.997349. p = 0.95, RL = 0.994758, RU = 0.999807. p = 0.99, RL = 0.9999798, RU = 1.000000. 5. The explanation of (2.3.4) is given in the text. We give here the explanation of (2.3.5) in a similar manner. It is easy to see that each DOW N state contains some minimal cut in the sense that all elements of the cut are down. Suppose, for example, that all edges of the network in the Fig. 2.7 are down. Then this DOW N state contains 3 minimal cuts, each consisting of some pair of edges. On the other hand, each minimal cut defines some DOW N state. So, the sufficient and necessary condition for N being DOW N is that all elements of at least one min cut are down. 6. For a series system, ϕ(x) =

n ∏

xi =

i=1

min

{i=1,...,n}

xi .

The proof is simple: the left-hand side is 1 if and only if all xi = 1 For a parallel system, ϕ(x) = 1 −

n ∏

(1 − xi ) =

i=1

max

{i=1,...,n}

xi .

The proof is similar. The left-hand side side is 0 if and only if all xi = 0. 6

7. If any pair of edges is up, then the network is UP. and the corresponding term in the parentheses will be zero. This makes ϕ(x) = 1. 8. (i) means that the system is DOWN if all its components are down and is UP if all its components are up. (ii) means that if at least one component changes its state from down to up, the system state can not get worse. 9. The bridge has four minimal path sets: (1, 4), (3, 5), (1, 2, 5) and (3, 2, 4). Each of these sets is a path without redundant components. The set (1, 2, 3, 4) is a path set but not a minimal path set since components 2,3 are redundant. 10. Assume that there is at least one minimal path set P1 , all elements of ∏ which are up. Then i∈P1 xi = 1 which leads to ϕ(x) = 1. Suppose now that the system is UP. Then there must be at least one minimal path set having all its elements in the up state. Then the right-hand side of the formula is 1. Therefore, ϕ(x) = 1 if and only if there is one minimal path set having all its elements in the up state. This completes the proof. 11. The collection of all min cut cuts is (1, 3), (4, 5), (2, 3, 4), (1, 2, 5). (1, 2, 3) is a cut set, but not a minimal cut set. 12. Any monotone system can be represented in two equivalent ways: as a series connection of parallel subsystems each being a minimal cut set, or as a parallel connection of series subsystems each being a minimal path set. (2.5.3) is the structure function corresponding to the first way of representation. Indeed, ϕ(x) = 0 if and only if one of the multiples in the product is zero. This will happen if all xi in one of the min cuts are zeroes, which means that all components of one min cut are down. 13. Represent the system as a series connection of four parallel subsystems corresponding to the given minimal cut sets. Imagine a source s before the first min cut and the terminal t after the last min cut. Let us find out all minimal paths leading from s to t. Each such path must go via element 1 or element 2. If the path goes via 1, then there are four minimal paths: (1, 2, 3), (1, 2, 4), (1, 5, 3), (1, 5, 4). If the element 1 is down, each path must contain element 2, and there are two minimal path sets (2, 4, 5), (2, 5, 3). In total, there are 6 minimal paths. 14. For a binary random variable, the mean equals the probability that ∏ this variable equals 1. For series system, ϕ(X) = ni=1 Xi . Since the binary variables are independent, the mean of their product equals to the product of their means. Therefore, ∏ R = E[ϕ(X)] = ni=1 pi .

7

This gives the reliability for series system: Rser = ∏ a parallel system, Rpar = 1 − ni=1 (1 − pi ).

∏n

i=1 pi .

Similarly, for

15. By the result of Problem 10, ϕ(X) = 1 − (1 − X1 X2 X3 )(1 − X2 X4 ). Since the minimal path sets are not disjoint (component 2 appears in both of them), X1 X2 X3 and X2 X4 are not independent, and we must first open the brackets, simplify the expression for ϕ(X) and only then take the expectation. Note that for binary variables Xi2 = Xi . This gives the result ϕ(X) = X1 X2 X3 + X2 X4 − X1 X2 X3 X4 and R = p1 p2 p3 + p2 p4 − p1 p2 p3 p4 16. LB is the reliability of a parallel connection of series subsystems corresponding to non overlapping minimal paths, i.e the probability that at least one of these paths is operational. This implies that the system is operational, and LB ≤ R. If the system is UP then all its minimal cuts are UP, and thus all non overlapping minimal cuts are UP. Thus R ≤ U B. (By non overlapping paths we mean paths that have no common elements; similar non overlapping cuts have no common elements). Remark. Since there might be several sets of non overlapping paths and non overlapping cuts, it is possible to construct several lower bounds and use the largest of them. Similarly, one can construct several upper bounds and use the smallest of them. This method is in fact the Ushakov-Litvak original suggestion.# 17. It is easy to check that zi ≥ xi , zi ≥ yi ⇒ zi ≥ max(xi , yi ) ⇒ ϕ(z) ≥ max(ϕ(x), ϕ(y)) = 1 − (1 − ϕ(x))(1 − ϕ(y)), Q.E.D. 18. Option (i) corresponds to the structure function ϕ(z). Option (ii) corresponds to the structure function 1 − (1 − ϕ(x))(1 − ϕ(y)). The result of the previous problem implies that the first option is preferable. In words: it is always better to reinforce each element by a parallel one than to reinforce the whole system by a similar one in parallel. 19. The new structure function will be h(ϕ(y (1) ), ϕ(y (2) ), ..., ϕ(y (n) )). A particular case of the above described situation is replacement of each component of the system by a subsystem identical to the original system. Such systems are called recurrent. 20. The following Mathematica program BLOCK-1 generates variable A which is positive if the graph is connected and zero, otherwise. BLOCK-1

8

m = {{1, 1, 0, 0, 0}, {1, 0, 1, 0, 0}, {0, 0, 0, 1, 1}, {0, 0, 1, 1, 0}, {0, 1, 0, 0, 1}, {0, 1, 0, 1, 0}} For[j = 1, j < 7, j + +, For[i = 1, i < 7, i + +, If [m[ [j] ].m[ [i] ]≤ 0, m[ [j] ] = m[ [j] ], m[ [j] ] = m[ [j] ] + m[ [i] ] ] ] A=Sum [ Product [m [ [ i,j ] ],{j, 1, 5}], {i, 1, 6}] Comments. m is a matrix, with 5 columns and six rows. The edge vectors are the rows of this matrix. m[ [ i ] ] calls for the i-th edge vector. The If operator checks the scalar product of the j-th and i-th vectors. If it is zero, the j-vector remains unchanged; otherwise, the j-th vector is replaced by the sum of j-th and i-th vectors. Internal For does this operation for all i running from 1 to 6. The external For does the If operation for all j running from 1 to 6. Finally, Product multiplies all coordinates in a row, and Sum gives the sum of all these products. If there is a row vector with all nonzero elements, at least one of the products is nonzero and A is positive. If, on the contrary, all row vectors contain at least one zero element, all products will be zero and A=0. From the programming point of view, the above program is far from being efficient. It takes n2 If comparisons to analyze a network with n edges. The students with sufficient programming background are strongly advised to implement the DDS described in Section 2.2. 21. Mathematica Monte Carlo program for computing probability of network connectivity. Define the vector p = {p1 , p2 , p3 , p4 , p5 , p6 } of edge up probabilities. Define vector b = {b1 , b2 , b3 , b4 , b5 , b6 } of six constants. Define the matrix m as follows: m = {b[[1]] ∗ {1, 1, 0, 0, 0}, b[[2]] ∗ {1, 0, 1, 0, 0}, b[[3]]∗{0, 0, 0, 1, 1}, b[[4]]∗{0, 0, 1, 1, 0}, b[[5]]∗{0, 1, 0, 0, 1}, b[[6]]∗{0, 1, 0, 1, 0}}. Now denote the whole program given above in solution of problem 20 as BLOCK-1, with m modified as above. The whole program is as follows: p = {p1 , p2 , p3 , p4 , p5 , p6 } b = {b1 , b2 , b3 , b4 , b5 , b6 } For[Q = 0; r = 1, r < 1001, r + +, For[ i=1, i < 7, i++, If [Random[ ] > p[[i]], b[[i]]=0, b[[i]]=1] ]; 9

BLOCK-1; If [A > 0, Q=Q+1,Q=Q ] ] Rel=Q/1000; Print[”Rel=”Rel] Comments. The external For repeats the connectivity evaluation 1000 times. Q is the counter. The first internal For checks the presence of each of the 6 edges. With probability pi edge ei exists, and the b[[i]] multiple at the corresponding edge vector is 1. If edge is erased, the multiple becomes zero. Then BLOCK-1 is carried out and it produces the parameter A. The last If adds 1 to the counter, iff A is positive. Rel is the reliability estimate. 22. In addition to the arrays defined for the Kruskal’s algorithm, let us define one more. EdgeState[1, ..., m] is the array of edge states, so that EdgeState[i] = 1, if the edge i is up, and EdgeState[i] = 0, otherwise. In fact, there is no need for us to construct the minimal spanning tree, it is sufficient to check whether all terminals are connected by the edges in up. Below we give the appropriate pseudocode. It uses slightly modified Kruskal’s algorithm. Procedure CMC // The main cycle for j ← 1 to M // DSS initialization for i ← 1 to n H[i] ← 0 Comp[i] ← i T comp[i] ← 0 for i ← 1 to k j ← T [i] T comp[j] ← 1 // iT comp - the current maximal number // of terminals in one component. iT comp ← 1 i←0 for i ← 1 to m simulate state othe edge i if i is up then EdgeState[i] = 1 else 10

EdgeState[i] = 0 repeat i←i+1 u ← F node[i] v ← Snode[i] ucomp ← f ind2(u) vcomp ← f ind2(v) // check if the nodes of the edge i // belong to different components. if (ucomp ̸= vcomp) then r ← merge2(ucomp, vcomp) // the number of terminals in the // resulting component after merging). j ← T comp[ucomp] + T comp[vcomp] if (j > iT comp) then iT comp ← j until(iT comp = k || i = m) if (iT comp = k) then ˆ=R ˆ+1 R ˆ = R/M ˆ R

11

3

Chapter 3.

1. Denote by τi the random lifetime of component i. We have to find P (3, 2, 1) = P (τ3 < τ2 < τ1 ). Suppose we know that τ2 =x. Then, given this fact, the result is P (τ3 < x < τ1 ) =[ since τ3 and τ1 are independent]= (1 − e−λ3 x )e−λ1 x = e−λ1 x − e−(λ1 x+λ3 x) . Now, by the total probability formula, the result is obtained by integrating this expression with respect to the density function λ2 e−λ2 x : ∫

∞

P (3, 2, 1) = 0

[e−λ1 x − e−λ1 x+λ3 x ]λ2 e−λ2 x dx =

(3.1)

= λ2 /(λ1 + λ2 ) − λ2 /(λ1 + λ2 + λ3 ). 2. The failure rate is defined only for t ∈ [0, 1). In this interval the density is 1 and the CDF of X is t. So, the result is h(t) = 3.

∫

1 , t ∈ [0, 1). 1−t ∫

t

t

h(x)dx = 0

0

f (x) dx = − 1 − F (x)

∫

t

0

d(1 − F (x)) = − ln(1−F (t)).(3.2) 1 − F (x)

The result follows since exp(ln a) = a. 4. By the definition, X represents a sum of n exponentially distributed i.r.v.’s with parameter λ. Similarly, Y is a sum of m such variables. Therefore Z is a sum of n + m exponentials, Z ∼ Gamma(n + m, λ). 5. For t > 0, P (Y ≤ t) = P (−λ−1 log X ≤ t) = P (log X ≥ −λt) = P (X ≥ e

−λt

)=1−e

−λt

(3.3)

.

Thus to generate Y ∼ Exp(λ), we have to generate X ∼ U (0, 1) and take Y = −λ−1 log X. 6. Let τi ∼ Fi (t) for i = 1, 2. τ1 and τ2 are positive i.r.v.’s. The event τ1 + τ2 ≤ t is a union of the events Ax = (τ1 ≤ t − x) and Bx = (τ2 ∈ [x, x + dx]), for all x ∈ [0, t]. P (Bx ) ≈ f2 (x)dx. Therefore, P (τ1 + τ2 ≤ t) = F (2) (t) =

∫

t

0

12

F1 (t − x) · f2 (x)dx.

7. ∫

∞ 0

A = P (τ1 ≤ (τ2 , ..., τn )) =

(

P (τ1 ∈ [x, x + dx])

∩

(3.4)

P (τ2 > x, ..., τn > x)

Since τi are i.r.v.’s, P (τ2 > x, ..., τn > x) = P (τ2 > x)·...·P (τn > x) = exp[−(λ2 +...+λn )x]. Now the integral becomes ∫

A=

∞

λ1 exp[− 0

n ∑

λ1 λi x] exp(−λ1 x)dx = ∑n

i=1 λi

i=2

.

(3.5)

8. Suppose we perform a series of independent identical binary experiments and stop at the first success. The probability of success is p. X is defined as the number of the experiment at which the success took place. Obviously, P (X = k) = q k−1 p. ∑ k−1 p = pq M (1 + q + q 2 + ...) = q M P (X > M ) = ∞ k=M +1 q The memoryless property: P (X > M + N |X > M ) = N )/P (X > M ) = q N = P (X > N ).

P ((X>M +N )∩(X>M )) P (X>M )

= P (X > M +

9. 9.1. Obviously, system lifetime τ = min(τ (e1 ), τ (e2 )). Then P (τ > t) = P (τ (e1 ) > t, τ (e2 ) > t). Because of independence of edge lifetimes we obtain 1 − F (t) = (1 − F1 (t))(1 − F2 (t)). 9.2. f (t) = dFdt(t) = f1 (t) + f2 (t) − f1 (t)F2 (t) − f2 (t)F1 (t) = f1 (t)(1 − F2 (t)) + f2 (t)(1 − F1 (t)). Dividing f (t) by (1 − F (t)) gives that the system failure rate h(t) = h1 (t) + h2 (t), where hi (t) = fi (t)/(1 − Fi (t)), i = 1, 2. 10. Since all lifetimes have the same distribution and are independent, all different orders of failure appearance have the same probability. Since there are 3!=6 different orders, each one of them has probability 1/6. 11. Note that τ , the lifetime, equals τ = min(τ1 , τ2 , ..., τk ), where τi is component i lifetime. Then, similarly to the solution of Problem 9.1, P (τ > t) = 1 − Fτ (t) =

k ∏

(1 − Fi (t)).

i=1

13

We arrive at the desired result since Fi (t) = 1 − e−λi t . To prove that component j fails first, repeat the reasoning given in the solution of Problem 7 and replace λ1 by λj . 12. Let us note first that by (3.4.2), the mean value of τ ∼ Exp(λ) equals ∫ ∞ 1 µ= e−λx dx = . λ 0 The time to the first failure of the system has exponential distribution with parameter k · λ. The mean time to the first failure is, therefore, 1/(kλ). After the first failure, the system will consist of (k − 1) independent exponential distributions (the memoryless property !) and therefore, the mean time between the first and the second failure will be 1/(k − 1)λ. Continuing this reasoning, we arrive at the result: (1/k + 1/(k − 1) + 1/(k − 2) + ... + 1)/λ. 13. The result follows from the fact that the mean of the sum of r.v.’s is the sum of the means of these r.v.’s. 14. ∫Let us use (3.4.1): t β−1 dx = (αtβ )/β. Thus 0 αx FX (t) = 1 − e−αβ

−1 tβ

, fX (t) = αtβ−1 exp[−αβ −1 tβ ].

(3.6)

15. By the result of Problem 9.2, the series system will have failure rate h(t) = (α1 + α2 )t(β−1) , which again is the failure rate of a Weibull distribution with parameters (α1 + α2 , β). β is called the shape parameter. So, the Weibull family with fixed β is closed under minimum operation. 16. Here F (t) = 1 − e−αβ respect to Y produces

−1 tβ

(see Problem 14). Solving F (Y ) = X with

Y = [−(β/α) log(1 − X)]1/β .

(3.7)

17. The event Ak = [τ0 + ... + τk ≤ t0 ] is equivalent to the event Bk = [ξ(t0 ) ≥ k]. Similarly, the event Ak+1 = [τ0 + ... + τk + τk+1 ≤ t0 ] is equivalent to the event Bk+1 = [ξ(t0 ) ≥ k + 1]. Therefore, (

)

P (Ak ) − P (Ak+1 ) = P ξ(t0 ) = k , Q.E.D. 14

4

Chapter 4.

1. Solve Problem 2 and set λ = 1 2. Let t ∈ (0, 1). P (−λ−1 log(1 − ξ) ≤ t) = P (log(1 − ξ) ≥ −λt) = P (1 − ξ ≥ exp(−λt)) = P (ξ ≤ 1 − e

−λt

)=1−e

−λt

(4.1) .

3. Let t ∈ (0, 1). P (Y ≤ t) = P (F (X) ≤ t) = P (X ≤ F −1 (t)) = F F −1 (t) = t.

(4.2)

The inverse function F −1 (·) does exist since F (·) is continuous and strictly increasing. 4. Suppose Random [ ] generates X ∼ U (0, 1) Set X = Random[ ]; Set Y = Random[ ] +1; Set I = 1 if Random[ ] ≤ p , or I = 0, otherwise; Set Z = I · X + (1 − I) · Y . 5. By the description of the network, it is UP if all four edges are up, the probability of which is R = 0.952 0.972 . If an edge having up probability p is reinforced by another one in parallel with up probability p1 , the reinforced edge will have up probability r = 1 − (1 − p)(1 − p1 ). If we reinforce the ”weak” edge, it will have up probability r = 1−0.05·0.1 = 0.995. Similarly, if we reinforce the ”strong” edge, its reliability will be r = 1−0.03·0.1 = 0.997. So, we have to compare R1 = 0.95 · 0.995 · 0.972 with R2 = 0.952 · 0.97 · 0.997. Obviously R1 > R2 . So, reinforced should be the ”weak” edge, which is intuitively obvious. ∏

6. System UP probability is R = 41 pi . A reinforced edge i will have up probability poi = 1 − (1 − pi )0.1 = 0.9 + 0.1pi . The ratio a = R/Rio , where Rio is the reliability of system with i-th edge being reinforced, equals pi /(0.9 + 0.1pi ). This expression is an increasing function of pi . Thus, a is minimal, i.e. the reinforcement is optimal, if pi = min(p1 , p2 , p3 , p4 ). So, we always must reinforce the less reliable edge. We note that the less reliable edge in the case of series system is the most important. ∏ BIM for edge i equals k̸=i pk = R/pi . Thus the largest BIM has the edge with the smallest up probability.

15

7. By (4.3.1), each of the edges has stationary availability p = 100/(100 + 5) = 0.952 . It equals the edge stationary up probability. Network stationary UP probability is then Av = 0.9524 = 0.823. 8. The network has five 3-edge minimal size min cuts. Each one of them cuts off one of the nodes a, b, c, d, e. If we position the more reliable edges on the periphery, and the less reliable will be the radial ones, each min cut will have failure probability g = 0.012 · 0.05 = 5 · 10−6 and the failure probability of the network will be, by the BP approximation, F ≈ 25 · 10−6 . If one of the less reliable edges will be put on periphery, there will be at least one min cut of size three with two edges having failure probability 0.05 and one having failure probability 0.01. Thus there will be a cut of size 3 resulting in failure probability F . The presence of other cuts will only increase this probability. Thus the initial location is optimal. 9. Let us differentiate both sides of (4.6.2) with respect to p. It follows from (4.6.1) that the left-hand side will have a sign change for p ∈ [0, 1], see Fig. 4.5. The right-hand side is a sum of derivatives of the importance functions. If the left-hand side is negative, all derivatives in the right-hand side can not be positive. Therefore, there must be such p value for which at least one one of the importance function derivatives is negative. Therefore, it is impossible that all importance functions are increasing functions of p. In practical terms, it means that the BIM of a component is not always an increasing function of pin the interval p ∈ [0, 1]. For example, let us check a system which is a series connection of two two-component parallel subsystems. The reliability of first subsystem components are p1 , p2 , and of the second subsystem - p3 , p4 . It is easy to check that system reliability equals R = (1 − (1 − p1 )(1 − p2 ))(1 − (1 − p3 )(1 − p4 )). Taking derivative with respect to p1 and setting pi ≡ p, we obtain Imp1 = 2 − 6p + 3p2 . This is not an increasing function of p in [0, 1]. Indeed, Imp1 = 2 for p = 0 and Imp1 = −1 for p = 1. 10. For a parallel system, R = 1 −

∏n

∂R/∂pj = R/(1 − pj ).

i=1 (1 − pi ).

It is easy to establish that (4.3)

Thus the maximal importance has the component with maximal pj , i.e. the most reliable one. ∏ For a series system, R = ni=1 pi and ∂R/∂pj = R/pj .

(4.4) 16

For this system, the most important component is the less reliable one. 11. a). The structure function of the system is ϕ(x) = x1 [1 − (1 − x2 x3 )(1 − x4 x5 )]. b). Opening the brackets we obtain ϕ(x) = x1 x2 x3 + x1 x4 x5 − x1 x2 x3 x4 x5 . Replacing each xi by a binary variable Xi with P (Xi = 1) = pi and taking the expectation, we find out that system reliability r(p) = E[ϕ(X)] = p1 p2 p3 + p1 p4 p5 − p1 p2 p3 p4 p5 c). Imp1 = p2 p3 + p4 p5 − p2 p3 p4 p5 = [pi = p] = 2p2 − p4 . Imp2 = p1 p3 − p1 p3 p4 p5 = [pi = p] = p2 − p4 It is easy to see that components 2, 3, 4, 5 have equal importance measures. Obviously, the most important is component 1. 12. It follows from (4.5.1) that system reliability function can be obtained by replacing each pi in the expression of R by component i reliability Ri (t) = 1 − Fi (t). In our case, pi must be replaced by exp[−λi t]. Thus we arrive at the expression Rs (t) = exp[−(λ1 + λ2 + λ3 )t] + exp[−(λ1 + λ4 + λ5 )t] −

(4.5)

exp[−(λ1 + λ2 + λ3 + λ4 + λ5 )t]. 13. The radar system is a so-called two-out-of-three system. It is UP if two stations are up and one is down, or all three stations are up. Denote the station lifetime by F (t) = 1 − exp[−λt]. Rs (t) = 3(1 − F (t))2 F (t) + (1 − F (t))3 = 3e−2λt − 2e−3λt .

(4.6)

Using the formula (4.2.4) for system mean UP time we obtain E[UP] = 5/(6λ). The stationary availability is, therefore, 5/(6λ)/(5/(6λ) + 0.1/λ) = 0.893. 14. The system is not monotone since ϕ(0, 0, 0) = 0, ϕ(1, 0, 0) = 1, ϕ(0, 1, 0) = 1, ϕ(0, 0, 1) = 1, ϕ(1, 1, 0) = 1, ϕ(0, 1, 1) = 1, ϕ(1, 0, 1) = 1 but ϕ(1, 1, 1) = 0. 15. This network has 6 minimal path sets: (1, 3, 7), (1, 4, 8), (2, 5, 7), (2, 6, 8) and (1, 3, 5, 6, 8), (2, 6, 4, 3, 7). (To check it, sketch the network !) The network is UP if at least one of the minimal paths is UP. If all min paths are DOWN the network is DOWN. Below is the algorithm. We denote by Xi the U (0, 1) random variables generated via operator Random[ ], i = 1, 2, ..., 8. Define I(1) = X1 X3 X7 , I(2) = X1 X4 X8 I(3) = X2 X5 X7 , I(4) = X2 X6 X8 , 17

I(5) = X1 X3 X7 X7 X8 , I(6) = X2 X6 X4 X3 X7 . Algorithm: small channel network reliability. 1. Set Q := 0; N = 10, 000;. 2. For k = 1 to k = 8 set I(k) = 1 if Xk < pi ; 3. For k = 1 to k = 6 calculate I(k); 4. Calculate I = I(1) + ... + I(6); 5. If I ≥ 1, set Q := Q + 1;; Comment: If I ≥ 1, there is at leat one operational path. 6. If Q ≤ N GOTO 2; 7. R = Q/N ; Print R. ∏

16. For a series system, the structure function is ϕ(X) = ni=1 Xi . Taking expectation, we obtain ∏ Rser = E[ϕ(X)] = ni=1 pi . Replacing pi by e−λi t we obtain that Rser = ∑ exp[− n1 λi t]. ∏

For a parallel system, the structure function is ϕ(X) = 1 − ni=1 (1 − Xi ). Taking expectation, we obtain ∏ Rpar = E[ϕ(X)] = 1 − ni=1 (1 − pi ). ∏ Replacing pi by e−λi t we obtain that Rpar = 1 − ni=1 (1 − exp[−λi t]). 17. Rs (t0 ) is the probability that the system is UP at the instant t0 . If the system is not monotone, it could happen that the system went DOWN for the first time at some instant t1 , returned to UP at some instant t2 and remained in UP till t0 , t1 < t2 < t0 . Therefore, it is not true that always the system lifetime is greater than t0 . 18. The right-hand side of (4.5.4) is the probability that all components are up. Therefore the system is UP and vice versa. The right-hand side of (4.5.5) is 1 minus the probability that all components are down. Therefore it is the probability that at least one component is up and the whole system is UP.

18

5

Chapter 5.

1. The left-hand side of (5.2.4) is (∂R/∂p1 )q1 λ1 + (∂R/∂p2 )q2 λ2 + (∂R/∂p3 )q3 λ3 .

(5.1)

The coefficient at λ1 in the given expression is ψ1 ·λ1 +ψ3 ·λ1 . Comparing the coefficients at λ1 , we obtain that ∂R/∂p1 = (ψ1 + ψ3 )/q1 . Similarly, ∂R/∂p2 = (ψ1 + ψ2 )/q2 and ∂R/∂p3 = (ψ2 + ψ3 )/q3 . 2. The network is UP if at least two edges are up. This gives the following formula P (UP ) = p1 p2 (1 − p3 ) + p1 p3 (1 − p2 ) + p2 p3 (1 − p1 ) + p1 p2 p3 . Imp1 = ∂R/∂p1 = p2 + p3 − 2p2 p3 Imp2 = ∂R/∂p2 = p1 + p3 − 2p1 p3 Imp1 − Imp2 = (p2 − p1 )(1 − 2p3 ) > 0 since both brackets are negative. Similarly we conclude that Imp2 − Imp3 > 0 . Therefore, the most important is edge (a, b) and the less important is (a, c). 3. Let us number the components: the components in parallel have numbers 2, 3, 4 and the single-block component has number 1. The border states are the DOWN states which turn into UP state by replacing a single down component into an up component. The list of all DOWN states is the following: (0, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (0, 1, 1, 0), (0, 1, 0, 1), (0, 0, 1, 1), (0, 1, 1, 1), (1, 0, 0, 0)). All these states, except (0, 0, 0, 0) are border states: the states with zero on the first position turn into UP state by replacing component 1 by an up one: state (1, 0, 0, 0) ⇒ UP if any one of the components 2,3 or 4 becomes up. 4. Obviously, system reliability is R = p1 [1 − (1 − p2 )(1 − p3 )(1 − p4 )]. Then ∂R/∂p1 = Imp1 = [taking derivative and setting pi = p] = 1 − (1 − p)3 = 3p − 3p2 + p3 . Similarly, ∂R/∂p2 = p(1 − p)2 . After simple algebra, we see that 19

Imp1 − Imp2 ≥ 0. Thus the first component is more important than the second. The proof for the third and fourth components are similar. By symmetry, the result is the same. So, 1 is the most important component, the remaining components have equal importance. 5. The states with exactly two components up (we list only the up edges) are: [(s, a), (b, t)], [(s, a), (s, b)],[(a, t), (b, t)], [(a, t), (s, b)]. 6. We found in Problem 4 that R = p1 [1 − (1 − p1 )(1 − p2 )(1 − p3 )]. (Argument t is omitted at each pi (t)). It follows that ∂R/∂p1 = 1 − (1 − p1 )(1 − p2 )(1 − p3 ). Now from the list of all system border states (Problem 3) we see that the border states turning into UP state by means of ”activating” component 1 are the following: (0, 1, 1, 1), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (0, 0, 1, 1), (0, 1, 0, 1) and (0, 1, 1, 0). The probability to be in one of these states is (1 − p1 )[1 − (1 − p1 )(1 − p2 )(1 − p3 )]. If we multiply it by λ1 , we get exactly the first component of the ∇R vector multiplied by λ1 (1 − p1 ). Similarly, ∂R/∂p2 = p1 (1 − p3 )(1 − p4 ). Multiply it by (1 − p2 ) and we get the probability of the unique border state (1, 0, 0, 0) which goes into UP by ”activating” component 2. 7. If (s, t) is absolutely reliable, the network is UP and its reliability is 1. if (s, t) is eliminated, then the network has four min cuts of size 2. By B-P approximation, the failure probability is F ≈ 4α2 and the importance of (s, t) is Imp(s, t) ≈ 1 − (1 − 4α2 ) = 4α2 . If edge (s, a) is absolutely reliable, we can compress s and a, and our network will have 2 min cuts of size 3 and reliability R1 ≈ 1 − 2α3 . If edge (s, a) is eliminated, we have two min cuts of size 2 and the reliability R2 ≈ 1 − 2α2 . Thus, the Imp(s, a) ≈ 1 − 2α3 − (1 − 2α2 ) = 2α2 − 2α3 . Obviously Imp(s, t) > Imp(s, a). 8. Let 1, 2, 3 be the numbers of first series subsystem components, and 4, 5 - of the second. Denote by pi reliability of component i. System reliability is R = [1 − (1 − p1 )(1 − p2 )(1 − p3 )] · [1 − (1 − p4 )(1 − p5 )] = R1 · R2 . The importance of component 1 is Imp1 =

∂R1 · R2 . ∂p1

20

After setting all pi ≡ p we obtain Imp1 = (1 − p)2 [1 − (1 − p)2 ]. Components 1,2,3 have the same BIM’s. In a similar way we obtain that Imp4 = Imp5 = (1 − p)[1 − (1 − p)3 ]. Obviously, Imp1 < Imp4 . So, the components of the smaller system are more important.

21

6

Chapter 6.

1. Bridge network has four nodes (s, a, b, t) and five edges e1 = (s, a), e2 = (s, b), e3 = (a, t), e4 = (b, t), e5 = (a, b). s, t are terminals, a, b do not fail, the edges do fail, and their lives are i.i.d. r.v.’s. Network fails if the terminals become disconnected. In the process of destruction, the network can not fail with the failure of one edge. Thus f1 = 0. Obviously, it can not survive failure of four edges. Thus f5 = 0. If the first two failures are of edges 1, 2 or 3, 4 then the network fails with the second failure. There are 2·2!·3! = 24 permutations leading to the the failure on the second step. The total number of permutations is 5! = 120 and thus f2 = 1/5. The network survives three failure and fails on the fourth if the edges fail, for example, in the following order: 1, 3, 5, 2, 4 or 2, 5, 4, 1, 3. There are 3! · 2! · 2 = 24 permutations of this type. Thus f4 = 24/120 = 1/5 and therefore f3 = 3/5. 2. Define a matrix A = [a(i, j)], with 15 rows and and 10 columns. i denotes the row, j -the column. Suppose you have an exponential random number generator. For example, in Mathematica it is ExponentialDistribution[λ]. Set λ = 1. a. For fixed i, generate 10 replicas of the random variable X ∼ Exp(λ), denote them as x(i, 1), ..., x(i, 10) and set a(i, j) := x(i, j). b. Order the values in the i-th row in ascending order. ¸. Repeat a,b 15 times for i = 1 to i = 15. The desired sample will be the values a(i, 7), i = 1, 2, ..., 15. 3. By the definition of the r-th order statistic in a sample of n, in order to satisfy the inequality X(r) ≤ t there must be at least r values of Xi smaller or equal t. Let r ≤ j ≤ n. Fix any j value. j replicas of Xi must be smaller or equal t, the remaining (n − j) replicas must be greater than t. For fixed candidates of these replicas, the probability of this event is [F (t)]j · [1 − F (t)]n−j . There are n!/j!(n − j)! different ways to choose j candidates from a collection of n. This explains (6.1.2). The smallest value in the sample is larger than t if and only if all values in the sample exceed t. This explains (6.1.3). The largest value in the sample is ≤ t if and only if all values in the sample are ≤ t. This explains (6.1.5). To prove (6.1.6), note that P (Xmax ≤ t) = P (all Xi ≤ t) =

n ∏ i=1

22

Fi (t),

because all Xi are i.r.v.’s. 4. For n = 4 and r = 2, F(2) (t) = 6q 2 p2 + 4q 3 p + q 4 ; F(3) (t) = 4q 3 p + q 4 ; F(4) (t) = q 4 . By (6.3.2), Fs (t) = F(2) (t)f2 + F(3) (t)f3 = 4q 2 (1 − q)2 + 4q 3 (1 − q) + q 4 . After some boring algebra, it gives the desired result Fs (t) = (1 − p2 )2 . Now consider a parallel connection of two two-component series systems. System structure function ϕ(X) = 1−(1−X1 X3 )(1−X2 X4 )), where P (Xi = 1) = p. Therefore, system reliability R = 1 − (1 − p2 )2 . Finally, Fs (t) = 1 − R = (1 − p2 )2 . 5. If the system remains UP when the first r − 1 failures take place and goes DOWN with the r-th failure, the random instant when this happens is the r-th order statistic in the sample of n. (6.3.3) follows from the formula of the total probability. The event τs ≤ t may happen under one of n conditions, P (τs ≤ t|system fails with the r-th failure) = F(r) (t), r = 1, 2, ...n. 6. Suppose that the network has n edges numbered 1, 2, ..., n and the order in which they are born is given by some permutation of these numbers i1 , i2 , ..., ir , ..., in . Suppose that the network is DOWN at the instant of the first r − 1 births and gets UP at the r-th birth. Denote by br the probability of this event. Then the collection {b1 , b2 , ..., bn } is called the C-spectrum (construction spectrum) of the network. For example, if in the network on Figure 6.3 the births appear in the order 3, 2, 1, 4 the system gets UP on the third birth. Denote by θ the random time of system birth. Given that the system birth takes place at the r-th birth of an edge, the system birth time coincides with the value of the r-th order statistic of the birth times. Denote by Bi ∼ G(T ) the random time of the birth of edge i and by B(r) (t) the corresponding r-th order statistic in the sample of {Bi , i = 1, 2, ..., n}. Thus, applying the total probability formula, we arrive at the expression P (θ ≤ t) =

n ∑

br B(r) (t).

r=1

7. It follows from the problem description that the evaluation completion time τ is distributed as the third order statistic in the population of 5 from 23

the CDF F (t). Therefore P (τ ≤ t) =

5 ∑

5! [F (t)]r [(1 − F (t)]5−r . r!(5 − r)! r=3

8. Three nodes b, c, e can fail in 6 different orders: (b, c, e), (b, e, c), (c, b, e), (c, e, b), , (e, b, c), (e, c, b). In the first and the third sequences disconnection appears with the third failure. In other cases - on the second. So, f2 = 2/3, f3 = 1/3. 9. The C-spectrum is b2 = 1/3, b3 = 2/3. Consider, for example, permutation (1, 2, 3, 4). Suppose it gives the order of edge failures. s − t becomes disconnected with the second failure. Now read this permutation from right to left and suppose, it determines the order of edges births. Obviously, the s − t connection appears on the third step. Therefore, b3 = f2 , b2 = f3 . In general fk = bn+1−k . (3)

10. The lifetime of the first block is F1 (t) = F(2) (t), which is the CDF of the second order statistic in a sample of n = 3. The lifetime of the second block (4) has CDF F2 (t) = F(2) (t), which is the second order statistic in a sample of n = 4. Since the system is a series connection of these blocks, P (τs ≤ t) = [see (6.1.4)] = 1 − (1 − F1 (t))(1 − F2 (t)).

24

7

Chapter 7.

1. X has density 1/T in the interval [0, T ]. Thus P (X ≤ t) =

∫ 0

t

(1/T )dx = t/T, t ∈ [0, T ].

For t ∈ [0, T1 ] P (X ≤ t|X ≤ T1 ) =

P ((X ≤ t) ∩ (t ≤ T1 )) = t/T1 , P (X ≤ T1 )

which is the CDF of X ⋆ ∼ U (0, T1 ). 2. It is easier to prove that for t ∈ [0, T ] P (X1 > t) =

Exp[−λ1 t] − Exp[−λ1 T ] . 1 − Exp[−λ1 T ]

P (X1 > t) = P (1 − ξ(1 − e−λ1 T ) ≤ e−λ1 t ) = P (ξ > (1 − e

−λ1 t

1 − (1 − e

)/(1 − e

−λ1 t

−λ1 T

)) =

−λ1 T

)/(1 − e

(7.1)

).

Now P (X1 > t) = 1 − P (X1 ≤ t) = R1 (t) and fX1 (t) = − dRdt1 (t) . This gives the left-hand side of (7.4.2). 3. P (X ≤ t) = P (F −1 (V ) ≤ t) = |!| = P (F (F −1 (V )) ≤ F (t)) = F (t). 4. P (Z ≤ t) = P (Y β ≤ t) = P (Y ≤ t1/β ) = 1 − e−λt , Q.E.D. 5. We already know that if X ∼ U (0, 1), then Y = −λ−1 log(1 − X) ∼ Exp(λ). Using the result of the previous problem we conclude that Y 1/β ∼ W (λ, β). Therefore the generator is the following: Z = [−λ−1 log(1 − Random[ ])]1/β . 25

6. Set I = 0, N = 10, 000. a. For i = 1 to i = 3 generate Xi = [− log(1 − Random[ ])]1/β . b. Compute S = X1 + X2 + X3 . c. If S ≤ T , set I := I + 1. Repeat steps a, b, c N times. Estimate Pˆ (X1 + X2 + X3 ≤ T ) = I/N. z}|{

Estimate V ar[Pˆ ] = Pˆ (1 − Pˆ /N . Comment. Note that I ∼ B(N, P ), V ar[I] = N · P · (1 − P ) and V ar[I/N ] = P (1 − P )/N . Replacing P by its estimate Pˆ = I/N we obtain the estimate of the variance. 7. The question is, in fact, how many operations we have to perform in order to compute one replica of the estimator Bm (T ). Suppose we are on the j-th step. We already know the realizations of X1 = v1 , X2 = v2 , ..., Xj−1 = vj−1 and have the product of first j terms in (7.3.5). To get one replica vj of Xj we must generate one replica of the r.v. with density (7.3.4). Suppose it takes H arithmetic operations. Then we must find Fj (T −v1 −...−vj−1 ) and multiply by this quantity the expression (7.3.5). Suppose it takes another L operations. Thus, for simulating one replica of the unbiased estimator of r convolutions we have to perform r(H + L) = O(r) operations. Our method, therefore, is linear in r.

26

8

Chapter 8.

1. a). We will call the edges by their lifetimes. So, the edge connecting the left terminal with v1 will be called edge 8. Following Kruskal algorithm, we construct the maximal spanning tree. It consists of edges 13, 12, 11, 10, 9, 7, (note that edge 8 is omitted !) and 4. Edges 9,10 and 4 are ”hanging” edges with respect to the terminals. Eliminate them. The remaining tree has minimal edge 7, and the network lifetime is 7. When this edge fails, the terminal in the middle gets isolated. b). Because of node v1 failure, edges 8,9 will get lifetimes 6.5. Edge v2 failure changes edge 12 and 11 lifetimes to 10.5. Now the maximal spanning tree has edges 13, 10.5, 10.5, 10, 7, 6.5 and 4. Edges 4, 6.5 and 10 are ”hanging” and must be erased. The minimal edge in the remaining tree is 7. So, the network fails at time instant t = 7. 2. Replace the numerator by the first term in (8.4.6) and delete the multiples (1−fr ). The variance will only increase. Now replace fr by the largest value of {fr }, denote it maxk fk . This will only increase the numerator. In the denominator, replace all fr by their smallest value minr fr . In numerator ∑ ∑ we have ( [F(r) ]2 )0.5 . It is smaller then the F(r) . So, the fraction will only increase if the ratio of these sums is replaced by 1. This completes the proof. 3. Take the expression (6.1.2) for the r-th order statistic and set there F (t) = α. Afterwards, substitute F(r) (t) into (6.3.3). You will observe that α(1 − α)m−1 m!/(1!(m − 1)!) is multiplied only by f1 , α2 (1 − α)m−2 m!/(2!(m − 2)!) is multiplied by (f1 + f2 ), α3 (1 − α)m−3 m!/(3!(m − 3)!) is multiplied by (f1 + f2 + f3 ), and so on. In general, αx (1 − α)m−x m!/(x!(m−)!) has multiple (f1 + f2 + ... + fx ). On the other hand, there is an equivalent expression for system failure probability (8.5.5). Here the term αx (1−α)m−x is multiplied by the number of system failure states with exactly x components being down and (m − x) being up. From this follows (8.5.4). For the upper network in Fig. 8.3,c we have m = 15, x = 7 and f5 = 0.00205, f6 = 0.010155, f7 = 0.02982. C(7) = (f5 + f6 + f7 )15!/(7!8!) = 270.43. We conclude that the network has 270 failure states with 7 edges down and 8 edges up. 27

4. Similar to (8.5.2), the probability to reveal in a single experiment a min cut of size 3 equals ϵ = 3!87!/90! = 6/(90 · 89 · 88) = 8.512 · 10−6 . For large M , (1 − x/M )M ≈ e−x . If ϵ is the probability to reveal the min cut in a single experiment, the probability to miss it in M experiments is (1 − ϵ)M . It must be less than 0.001. From e−x = 0.001 it follows that x = − log[0.001] = 6.91. So, from x/M = ϵ we obtain M = 6.91/ϵ ≈ 812, 000. Thus we must simulate at least 812,000 permutations. 5. a. There are 24=4! various orders in which 4 nodes can fail. Failure of any single node does lead to network failure. Thus f1 = 0. Network fails at the instant of the second failure if node 1 fails first, or anyone of nodes 2, 3, 4 fail first and node 1 fails next. This amounts to 12 permutations and thus f2 = 12/4! = 1/2. The network fails with the fourth failure if node 1 fails last. This gives 3!=6 permutations and f4 = 1/4. Therefore f3 = 1/4. Thus, Sp = [0, 0.5, 0.25, 0.25]. b. Following (8.4.4) we find, after routine algebra, that FN = 3F 2 − 3F 3 + F 4 . (In (8.4.3) we set m = 4, r = 2, 3, 4). To find the maximal permissible value of F we can use the operator FindRoot of Mathematica and solve the equation FN = 0.05. The result is F = 0.139. If node failure probability exceeds this value, network failure probability exceeds 0.05. Note also that the same formula for FN could be derived from the fact that the network is a parallel connection of two series systems, one with one component, another one - with three components. 6. Suppose that operator DiscUnif[m] generates random integer from the set of integers 1, ..., m. a. Define Matrix A[i,j] with 6 rows and n columns, n > 3. Set A[1,1]=1,A[1,2]=2,A[1,3]=3, Set A[2,1]=1,A[2,2]=3,A[2,3]=2, Set A[3,1]=2,A[3,2]=1,A[3,3]=3, Set A[4,1]=2,A[4,2]=3,A[4,3]=1, Set A[5,1]=3,A[5,2]=1,A[5,3]=2, Set A[6,1]=3,A[6,2]=2,A[6,3]=1. b. Let x=DiscUnif[6] c. Set j=4. d. Define A[x,j]=j; e. Set y=DiscUnif[j-1];a=A[x,y];b=A[x,j]; f. Set A[x,y]=b; A[x,j]=a; 28

g. j:=j+1; h. If j > n STOP ELSE GOTO to d; i. Read the permutation from the x-th row of A. This algorithm creates a random permutation of n > 3 integers in a recurrent way starting from a random permutation of 3 integers 1, 2, 3. These permutations are positioned into first 6 rows of matrix A. b. chooses randomly the x-th row of A containing one of these six permutations. c. for the first time (j=4) puts 4 on the last position number in the x-th row. e,f pick up a random element among the first j − 1 elements of the permutation; remembers it (A[x,y]=a) and exchanges it with the last element in the x-th row of matrix A. g, h create a loop. 7. It is seen from the figure, that there are 5 min cuts of size 4. The same result is provided by (8.5.1), with r = 4, m = 10. The B-P approximation is F ≈ 5ϵ4 . Equating it to 0.01 we obtain ϵ ≤ 0.211. 8. We present a small program written on Mathematica. Define m, the number of edges in the network; Define q, edge failure probability; Define a one-dimensional array Dsp containing the D-spectrum: Dsp=[f(1),f(2),...,f(n)]; Fsyst=Sum[Dsp[[r]]Sum[q j (1−q)(m−j) m!/(j!(m−j)!),(j, r, m)], (r, 1, m)] Print[Fsyst] The internal sum computes the r-th order statistic according to (8.4.3). The external sum computes (8.4.4). Dsp[[r]] picks up the r-th term of the spectrum, f(r). Fsyst is the system failure probability. For example, for the network shown on Fig. 8.3, b, we have m=10, Dsp=[0, 0, 0, 1/42, 4/42, 12/42, 25/42, 0, 0, 0]. Taking q=0.1, we obtain Fsyst=0.00833.

29

9

Chapter 9.

1. The super-state σ23 has edges 2 and 3 up, all other edges are down. Transition into it can be from σ11 and σ14 . 2. Using the properties of the Exponential distribution (Chapter 3) it is easy to find out the following. ∑

The sitting time in σ0 is τ0 ∼ Exp( 51 λi ). ∑ E[τ0 ] = 1/ 51 λi . The transition into σ1 has probability ∑5 λ1 / 1 λi . The sitting time in σ11 is τ11 ∼ Exp(λ1 + λ2 + λ4 + λ5 ). ∑ ∑ E[τ11 ] = 1/ i̸=3 λi . The transition into σ22 has probability λ4 / i̸=3 λi . The sitting time in σ23 is τ23 ∼ Exp(λ1 + λ4 + λ5 ). E[τ23 ] = 1/(λ1 + λ4 + λ5 ). The transition into U P has probability 1. The mean transition time along this trajectory ( call it u) is 1/

5 ∑

λi + 1/

1

∑

λi + 1/(λ1 + λ4 + λ5 ).

i̸=3

The probability of trajectory u is P (u) = (λ3 /

5 ∑ 1

λi ) · λ2 /

∑

λi .

i̸=3

3. If the sitting time in σ0 plus the sitting time in σ11 plus the sitting time in σ23 is less or equal t, then the trajectory is in UP at time t. Thus P0 = P (τ0 + τ11 + τ23 ≤ t|u). Now P1 = P (τ0 ≤ t|u) − P (τ0 + τ11 ≤ t|u), because the first term is the probability that the trajectory is not in σ0 at time t + 0, and the second term is the probability that it is in σ23 or in UP at time t. 4. There are two trajectories leading to σ26 : u1 = σ0 → σ14 → σ26 ,

30

u2 = σ0 → σ13 → σ26 , λ2 λ4 . i=1 λi · (λ1 + λ2 + λ3 + λ5 )

P (u1 ) = ∑5

λ5 (λ4 + λ3 ) . i=1 λi · (λ1 + λ3 + λ4 + λ5 )

P (u2 ) = ∑5

The answer: P (u1 ) + P (u2 ) + P (u3 ). 5. Sitting time in σ0 is τ0 ∼ Exp(λ2 + λ4 + λ5 ) Sitting time in σ3 is τ3 ∼ Exp(λ2 + λ5 ) Sitting time in σ1 is τ1 ∼ Exp(λ2 + λ4 ) The transition σ0 → σ3 has probability λ4 /(λ2 + λ4 + λ5 ). The transition σ0 → σ1 has probability λ5 /(λ2 + λ4 + λ5 ). 6. 1). The turnip has the root σ0 , three states on the first level σ1 , σ2 , σ3 and one super-state UP on the second level. There are three trajectories: ui = (σ0 → σi → U P ), i = 1, 2, 3. 2). λi = − log(qi ), or qi = e−λi . The sitting time in in σ0 is τ0 ∼ Exp(λ1 + λ2 + λ3 ). The sitting time in σ1 is τ1 ∼ Exp(λ2 + λ3 ). 3). p(u1 ) = (λ1 /(λ1 + λ2 + λ3 )) · 1. 4).A = P (τ0 ≤ 1) = 1 − e−Λ0 , where Λ0 = λ1 + λ2 + λ3 . By formula (5) from Appendix B, B = P (τ0 + τ1 ≤ 1) = 1 − e−Λ0

Λ1 Λ0 − e−Λ1 , Λ1 − Λ0 Λ0 − Λ1

where Λ1 = λ2 + λ3 . After some algebra we find out that P ⋆ = A − B = Λ0 (e−Λ1 − e−Λ0 )/λ1 5) follows from the above formulas for p(σ1 ), p(u1 ) and P ⋆ . This problem gives an explanation for the formula (9.3.7) for Φ(N).

31

10

Chapter 10.

1. Components 1 and 2 are in series, together they are in parallel to 1 and all this block is in series to component 4. It is elementary to check that the structure function is ϕ(X) = [1 − (1 − X2 X3 )(1 − X1 )]X4 . System reliability is E[ϕ(X)] = R(p) = [1 − (1 − p2 p3 )(1 − p1 )]p4 = |pi ≡ p| = [1 − [1 − p2 )(1 − p)]p. After setting all pi = p, we obtain R = p3 + p2 − p4 . 2. What is called the normalized C ⋆ -spectrum in this chapter is exactly the C-spectrum introduced in Chapter 8. Consider any permutation of n component numbers. Without loss of generality let this permutation be (1, 2, 3, ..., k, k + 1, ..., n) Suppose we do the destruction of the network, erasing components as they appear from left to right. Suppose that the network gets DOWN at the failure of component k, on the k-th step. Now consider the same permutation and do the construction process moving from right to left. Obviously, the network will be UP exactly at adding component k, not before and not later, i.e. on the (n − k + 1)-th step of the process. This proves that fk = gn−k+1 . 3. Since R(p) = [1 − (1 − p2 p3 )(1 − p1 )]p4 , R(0, 1, 1, 1) = p2 p3 p4 ; R(1, 0, 1, 1) = p1 p4 , R(1, 1, 0, 1) = p1 p4 ; R(1, 1, 1, 0) = 0. After setting pi ≡ p, we obtain F V IM1 = 1 −

p ; 1 + p − p2

F V IM2 = 1 −

1 = F V IM3 ; F V IM4 = 1. 1 + p − p2

32

4. Rewrite the formula (10.3.2) in the following form: BIMj =

n ∑ Zi,j pi−1 q n−i i=1

i!(n − i)!

−

n ∑ (Yi − Zi,j )pi q n−i−1 i=1

i!(n − i)!

,

(10.1)

where the first sum stands for R(p1 , ..., pj−1 , 1j , pj+1 , ..., pn ) and the second for R(p1 , ..., pj−1 , 0j , pj+1 , ..., pn ). Now, placing R(p1 , ..., pj−1 , 0j , pj+1 , ..., pn ) into the numerator of the formula (10.6.1) and replacing Yi − Zi,j by Vi,j we arrive at (10.6.. 5. The following 4 permutations of edges have anchor equal 2: (1, 2, 3, 4), (1, 2, 4, 3), (2, 1, 3, 4), (2, 1, 4, 3). Anchor equal 3 has the following 14 permutations: (1, 3, 2, 4), (1, 4, 2, 3), (2, 3, 1, 4), 2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 2, 1), (4, 1, 2, 3), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 2, 1). Finally, the following 6 permutations have anchor 4: (1, 3, 4, 2, (1, 4, 3, 2), (3, 1, 4, 2), (3, 4, 1, 2), (4, 1, 3, 2), (4, 3, 1, 2). From this we get the expression (10.2.1).

33

11

Chapter 11.

1. Subtract (11.3.3) from (11.3.2). After simple algebra, the result will be D = r2 (1 − p)2 − p2 (1 − p)(1 − r). Cancel out (1 − p). D > 0 because for r > p, r2 > p2 and 1 − p > 1 − r. 2. Consider the Taylor series for R(p1 + δ1 , p2 + δ2 , ..., pn + δn ) and take the first two terms. For small δi R(p1 + δ1 , ..., pn + δn ) ≈ R(p1 , ..., pn ) +

n ∑

(∂R/∂pi )δi .

i=1

If at our disposal is only one nonzero δi , the greatest reliability increment will be achieved for the component j which maximizes (∂R/∂pi )δi . Recall that the partial derivative is the component j BIM . 3. Let us use the following notation: Res -integer resource for reliability improvement. p1, p2, p3 - component reliability. f i(p1 , p2 , p3 ) - reliability gradient of component i = 1, 2, 3. R = R(p1, p2, p3) -system reliability. di = 0.2 · exp[−pi]]-reliability increase of component i after investing into it one unit of resource. Set pi := min(1, pi + di)- component i improved reliability. Qi = f i · di increment in system reliability after component i is was reinforced. Description of the Algorithm a. For i = 1, 2, 3, compute di, f i and Qi = f i · di,. b. If Q1 = max[Q1, Q2, Q3], set p1 := min(1, p1 + Q1), Else If Q2 = max[Q1, Q2, Q3], set p2 := min(1, p2 + Q2), Else set p3 := min(1, p3 + Q3). c. Calculate system reliability R = R(p1, p2, p3). d.Set Res := Res − 1. e. If Res > 0, GOTO a. Else STOP Print[R] 4. a. In a five-node network, locating the terminal in node 2 provides reliability 1 since there is a reliable path connecting all three terminals. (Network reliability is defined as all terminal connectivity). b. In a six-node network, nodes 2,3,5 and 6 by symmetry are equally good for locating the third terminal. 34