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Annals of OperationsResearch ISSN 0254-5330Volume 181Number 1 Ann Oper Res (2010)181:503-514DOI 10.1007/s10479-010-0781-3

Adjustment of the moments of the projectcompletion times when activity times areexponentially distributed

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Ann Oper Res (2010) 181: 503–514DOI 10.1007/s10479-010-0781-3

Adjustment of the moments of the project completiontimes when activity times are exponentially distributed

Yousry H. Abdelkader

Published online: 24 August 2010© Springer Science+Business Media, LLC 2010

Abstract The problem of determining the moment of the project completion time (MPCT)with stochastic activity durations is addressed. In the direct approach the MPCT is obtainedafter analyzing all possible network paths which is NP-hard problem. A recursive methodfor determining MPCT by proceeding iteratively over the nodes is the computationally leastdemanding and therefore the most efficient approach. Unfortunately, the moment methodswhich are available in literature ignore some aspects. In this paper, an adjustment of themoment completion time is discussed. Two illustrative examples are presented.

Keywords Stochastic activity networks · Order statistics · Moments

1 Introduction

Stochastic Activity Networks (SANs) which require a priori cumulative distribution function(c.d.f.) has wide applications in the filed of production, industrial and military activities. Thedefinition of the SANs can be expressed as follows: let G = (V ,E) where V = {1,2, . . . , n}be the ordered set of n nodes (events), E be the set of arcs (activities) E ⊂ V × V . G isa connected directed acyclic graph. The graph has one starting and one ending node. Foreach activity e in E, we associate a random variable (r.v.) Xe that is the duration of thecorresponding activity and Fe is the cumulative distribution function of the r.v. Xe . Thetriple (V ,E,F ) represents the stochastic activity networks. The main problem in SANs(longest and shortest path) is divided into two categories. The first is to find the distributionof

νn = max(X1,X2, . . . ,Xn) and ξn = min(X1,X2, . . . ,Xn)

that is, finding Hn(t) = P (νn ≤ t) and Ln(x) = P (ξn ≤ x) or the rth moments of νn and ξn

or approximating them, where X1,X2, . . . ,Xn be r.v.’s represent the project activity dura-tions. It is worth mentioning that the distribution of νn is of interest in the project networkswhereas the distribution of ξn is of interest in the reliability networks.

Y.H. Abdelkader (�)Department of Mathematics, Faculty of Science, Alexandria University, Alexandria, Egypte-mail: yabdelkader@yahoo.com

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504 Ann Oper Res (2010) 181: 503–514

The second category is to find the distribution of the sum of n independent nonidenti-cally r.v.’s. This sum can be achieved through the convolution operation. For more detail inconvolution operation see, Martain (1965) and Agrawal and Elmaghraby (2001).

The problem associated with computing the moments of the project completion time byproceeding iteratively over the nodes has been discussed extensively in literature (see Clark1961, 1962; Sculli 1983; Sculli and Wong 1985; Kamburowski 1985a, 1985b; Golenko-Ginzburg 1989; Magott and Skudlarski 1993; Abdelkader 2003, 2004a, 2004b, 2006 amongothers). A survey of recent developments and complexity in SANs can be found in El-maghraby (1989, 2000 and 2005), Tavares (2002), Bendavid and Golany (2009) and thereferences therein.Most of the above work estimated the moment of the project completion time (MPCT) ofnode j recursively by

W1 = 0(1)

Wj = h(Wi + Xij , i ∈ Bj), j = 2,3, . . . , n

where

h(Y1, Y2, . . . , Yn) = E(max(Y1, Y2, . . . , Yn))k, k = 1,2, . . .

Bj denotes the set of direct predecessors of node j , Xij denotes the random duration ofarc(i, j), Wi is the realization time of the node i and Yi = Wi + Xij .

Unfortunately the papers referenced above replaced the r.v.’s Xij + Wi by their expectedvalues E(Wi + Xij ), which is obviously incorrect. This compound an error committed, byconsidering Wi as a r.v., and resulted in a loose estimate for the project completion time.

The purpose of the paper is to present an adjustment to the recursive method to determinethe moments of the project completion times. In the calculations of the moments of theproject completion time of node j , one needs formulae for the moments of the maximumof independent nonidentically distributed random variables which are shifted by differentconstant values. Therefore, we can rewrite formula (1) as:

W�1 = 0

(2)W�

j = h(Ci + Xij , i ∈ Bj), j = 2,3, . . . , n

where

h(Y ′1, Y

′2, . . . , Y

′n) = E(max(Y ′

1, Y′2, . . . , Y

′n))

k, k = 1,2, . . .

Y ′i = Ci + Xij and Ci are constants for i = 1,2, . . . , n.

In order to introduce our approach and demonstrate the deficiency of formula (1) we haveselected the exponential distribution (with different parameters) as the distribution functionof the duration of all the activities of the project. In Sect. 2, the kth moment will be derivedfollowing (2), and the special case with Ci = 0 for all i determined. Section 3 is devoted tointroduce two example networks that illustrate the error committed by using (1). Section 4gives the conclusions. An Appendix gives some details relative to the derivation of someequations in the main body of the paper.

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2 Main results

Let X1,X2, . . . ,Xn be independent nonidentically r.v.’s each having exponential cumulativedistribution function

Fi(x) = 1 − e−λix, x ≥ 0, λi > 0, (3)

and let X1:n ≤ X2:n ≤ · · · ,Xn:n be the corresponding order statistics, where Xn:n =max(X1,X2, . . . ,Xn) and X1:n = min(X1,X2, . . . ,Xn).

To lay the groundwork of this study, we begin with the following Lemma.

Lemma 1 Let X be a r.v. with finite mean μ defined on the range (A,∞), the kth momentμ(k) = E(Xk) can be written as

μ(k) = Ak(1 − F(A)) + k

∫ ∞

A

xk−1G(x)dx, (4)

where G(x) = 1 − F(x) denotes the survival function and A is a constant.

Proof By definition

μ(k) =∫ ∞

A

xkf (x)dx = −∫ ∞

A

xkdG(x).

Integrating by parts treating xk for differentiation, we obtain (4). �

In the following, some conventions and notations will be used. Let Xi (i = 1,2, . . . , n)

be independent nonidentically distributed random variables, A = max(C1,C2, . . . ,Cn) andB = min(C1,C2, . . . ,Cn) where Ci are shifting constants. Let FZ(z) be the c.d.f. of the r.v.Zn = max(X1 + C1,X2 + C2, . . . ,Xn + Cn), then

FZ(z) = P (X1 + C1 ≤ z,X2 + C2 ≤ z, . . . ,Xn + Cn ≤ z),

=n∏

i=1

Fi(z − Ci).

Similarly, if we denote FV (v) be the c.d.f. of the r.v. Vn = min(X1 +C1,X2 +C2, . . . ,Xn +Cn), then

FV (v) = 1 −n∏

i=1

Gi(v − Ci).

In the case when X1,X2, . . . ,Xn are independent nonidentically exponentially distributed,the c.d.f.’s of FZ(z) and FV (v) are, respectively, given by

FZ(z) =n∏

i=1

(1 − e−λi (x−Ci)

), (5)

FV (v) = 1 −n∏

i=1

e−λi (x−Ci). (6)

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In view of Lemma 1 the kth moment of order statistics, E(Zkr:n) = μ(k)

r:n, can be written as

μ(k)r:n = Ak(1 − Fr:n(A − Ci)) + k

∫ ∞

A

zk−1Gr:n(z − Ci)dz.

In particular, when r = 1 and r = n, we have

μ(k)

1:n = Bk

n∏i=1

Gi(A − Ci) + k

∫ ∞

B

zk−1n∏

i=1

Gi(z − Ci)dz, (7)

and

μ(k)n:n = Ak

(1 −

n∏i=1

Fi(A − Ci)

)+ k

∫ ∞

A

zk−1

(1 −

n∏i=1

Fi(z − Ci)

)dz. (8)

In the following theorem the kth moments μ(k)n:n = E(max(X1 + C1,X2 + C2, . . . ,Xn +

Cn))k, and μ

(k)

1:n = E(min(X1 + C1,X2 + C2, . . . ,Xn + Cn))k, of the maximum and the

minimum of a sample of size n from the exponential distribution will be derived. The othermoments (for r = 2,3, . . . , n − 1) can be obtained by using the procedure which presentedin Barakat and Abdelkader (2004).

Theorem 1 Let X1,X2, . . . ,Xn be independent nonidentically r.v.’s that obey the exponen-tial distribution given by (3). The kth moments, μ(k)

n:n and μ(k)

1:n, for k = 1,2, . . . are givenby

μ(k)n:n =

n∑j=1

(−1)j+1I∗j , (9)

and

μ(k)

1:n = I∗n, (10)

where

I∗j =

∑· · ·

∑1≤i1<i2<···<ij ≤n

e−∑jr=1(A−Cir )λir

(Ak + kAk−1

αj

+ k(k − 1)Ak−2

α2j

+ · · · + k!αk

j

), (11)

αj = (λi1 + λi2 + · · · + λij ), for j=1,2, . . . , n and (i1, i2, . . . , in) are permutations of(1,2, . . . , n) for which 1 ≤ i1 < i2 · · · < ij ≤ n, here α1 = λi and αn = ∑n

i=1 λi .

Proof From (5) and (8), we have

μ(k)n:n = Ak

(1 −

n∏i=1

(1 − e−λi (A−Ci))

)+ k

∫ ∞

A

zk−1

(1 −

n∏i=1

(1 − e−λi (z−Ci ))

)dz

= Ak∑

· · ·∑

1≤i1<i2<···<ij ≤n

(−1)j+1e−∑jr=1(A−Cir )λir + k

∫ ∞

A

zk−1J (z)dz, (12)

where

J (z) =n∑

i=1

e−(z−ci )λi −∑ ∑1≤i1<i2≤n

eλi1 Ci1 +λi2 Ci2 e−(λi1 +λi2 )z

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+∑ ∑ ∑

1≤i1<i2<i3≤n

eλi1 Ci1 +λi2 Ci2 +λi3 Ci3 e−(λi1 +λi2 +λi3 )z

+ · · · + (−1)n+1n∑

i=1

e−∑ni=1 λi (z−ci ).

Upon using the integral

∫unebudu = ebu

b

(un − nun−1

b+ n(n − 1)un−2

b2+ · · · + (−1)nn!

bn

), (13)

where “b” is a real constant. The second term in (12) can be written as

k

∫ ∞

A

zk−1J (z)dz =n∑

j=1

(−1)j+1∑

· · ·∑

1≤i1<i2<···<ij ≤n

e−∑jr=1(A−Cir )λir

×(

Ak + kAk−1

αj

+ k(k − 1)Ak−2

α2j

+ · · · + k!αk

j

). (14)

By summing the first term in (12) and the terms of (14), we get (9).Similarly, (10) can be obtained by using (6) and (7) and replacing “A” by “B” in (11)

which completes the proof. �

Corollary 1 Set Ci = A = 0 in Theorem 1, we get

μ′(k)n:n = k!

n∑j=1

(−1)j+1Ij , (15)

where

Ij =∑

· · ·∑

1≤i1<i2<···<ij ≤n

1

(λi1 + λi2 + · · · + λij )k.

Formula (15) is used only when there is a number of n activities in parallel from the sourceto the end node.

Some special formulae for the first and the second moments of the maximum of two andthree r.v.’s obtained by (9) are given in the Appendix.

Corollary 2 Let Yi = Wi + Xij be the sum of two r.v.’s defined by (1) and μ∗i = E(Yi), then

U(k)n:n = E(max(Y1, Y2, . . . , Yn))

k = μ(k)n:n = k!

n∑j=1

(−1)j+1Ij , (16)

where

Ij =∑

· · ·∑

1≤i1<i2<···<ij ≤n

1

(μ∗i1

+ μ∗i2

+ · · · + μ∗ij)k

.

The following Proposition 1 gives bounds for the expected value of the maximum of n r.v.’swith shifting constants.

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508 Ann Oper Res (2010) 181: 503–514

Proposition 1 Let X1,X2, . . . ,Xn be independent nonidentically r.v.’s follow the ex-ponential distribution with finite means μ1,μ2, . . . ,μn. For each n ≥ 1, let Un:n =E max(Y1, Y2, . . . , Yn), μn:n = E max(X1 + C1,X2 + C2, . . . ,Xn + Cn), and κn =max(μ1,μ2, . . . ,μn), then

Un:n ≥ μn:n ≥ κn. (17)

Proof The second part of the inequality, μn:n ≥ κn, can be verified by Jensen’s inequality:

ψ[E(X)] ≤ E[ψ(X)]

for each convex function ψ and each r.v. X. The first part of the inequality, Un:n ≥ μn:n,follows from the Theorem and consequent Corollary 2 since I ∗

j ≤ I j . Proposition 1 is alsovalid for k > 1. �

To sum up the computations for obtaining the kth moments of the project completion timein stochastic activity networks, we present the following algorithm:

INPUT: k (k = 1,2, . . . , ) and μij (the mean duration of activity (i, j)).

OUTPUT: The kth moment of the project completion time of node j (W�j , for j =

2,3, . . . n).Step 1. Set W�

1 = 0 (starting node).step 2. Compute W�

j , for j = 2,3, . . . n by using equations (2), (9) and (11).Step 3. Go to step 2.

3 Numerical results

In order to perform the comparison between the adjusted exponential distribution and ex-ponential distribution as well as PERT lower bound (see Devroye 1979), two example net-works are given in Figs. 1 and 2 with the mean indicated on each arc. The first networkcomposed of 24 activities and 13 nodes which initially proposed by Skudlarski (1991). Thesecond network was originally proposed by Kleindorfer (1971) and subsequently used byElmaghraby (2000). It contains of 38 activities and 20 nodes. All activities are assumed tobe exponentially distributed.

Example 1 The network in Fig. 1 is taken from Skudlarski (1991). He obtained the exactmean and variance project completion time to be 10.9 and 9.84, respectively. He assumedthat the duration of each activity follows the exponential distribution and using a generalizedstochastic Petri nets analyzer. Five estimates for the MPCT of node j are shown in Table 1.The exponential estimate (EE) and the second moment exponential estimate (SMEE) areobtained by using formula (1), the adjusted first moment exponential estimate (AFEE), theadjusted second moment exponential estimates (ASEE) and the PERT lower bound estimate(PLBE). Both AFEE and ASEE are obtained by using formula (2). The variance of theproject completion time using EE and the adjusted proposed estimate are 2546.7 and 2.23,respectively.

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Fig. 1 Skudlarski’s network with mean duration indicated on each arc

Table 1 Comparison of PERT (LB), AFEE, EE, ASEE and SMEE

Node First moment est. Second moment est.

PLBE AFEE EE ASEE SMEE

2 1 1.083 1.083 2.10 2.10

3 1.33 1.67 3.30 1.83 5.32

4 2 2.42 6.56 3.18 15.78

5 3 3.64 15.42 5.25 42.98

6 3.33 4.19 18.06 7.37 85.85

7 4 4.96 25.28 10.50 172.98

8 4.33 5.64 32.44 14.48 330.23

9 5.66 6.89 49.80 20.30 645.67

10 6 7.44 55.79 27.27 1174.25

11 6.66 8.21 68.06 36.88 2142.14

12 7 8.89 79.67 49.41 3851.81

13 8.33 10.15 105.2 66.57 6978.32

Example 2 The network which is depicted in Fig. 2 is taken from Kleindorfer (1971) andreconsidered by Elmaghraby (2000) as an example of a large network (38 activities and 20nodes). By using straightforward Monte Carlo Sampling (MCS) with 20,000 iterations asa reference, we obtained the mean and the variance of the project duration of 62.59 and32.6, respectively. The mean and the variance obtained by the exponential estimates are249.33 and 27481.05, respectively. While the mean and the variance obtained by the adjustedproposed estimates are 71.91 and 200.7, respectively. Table 2, summarizes the completiontime of each node j for the same five estimates which are mentioned in Example 1. SinceW1 = 0, W2 = 10, W3 = 11.5, W4 = 2.33, W6 = 15.5 and W10 = 15.33 are equal for thethree mean estimates (PLBE, AFEE and EE), Table 2 provides the other node estimates.

Form the above results and the examples we can conclude the following remarks:

Remark 1 The results of the above examples agree with the bounds which stated in theproposition.

Remark 2 There is a large gap between AFEE and EE. The percentage difference betweenAFEE and EE of the last node for the mean project completion time is equal to 84.75% inExample 1, while 71.15% in Example 2. This error occurs when considering Wi is a r.v., see

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Fig. 2 The num beron each arc represents the activity mean

(1); and it disappeared when Wi is regarded as a constant, see (2). This is logically consistentsince in (1) the expectation for Wi is considered more than once.

Remark 3 The straightforward method (or Direct Approach) which is proposed by Martain(1965), Hartley and Wartham (1966), Ringer (1969), Dodin (1985) and Fisher et al. (1985)required to solve, in Example 1, 23-fold integral over the region determined by 233 inequal-ities (the number of paths). In the second example these numbers are much larger sincethey have considered the maximum of path times which are the sums of the activity timesbelonging to each individual path.

4 Conclusions

In this paper, we have presented an adjustment of the recursive method for determiningthe moments of the project completion time in a stochastic activity network by proceedingiteratively over the nodes. In the direct approach the expected project completion time isobtained after analyzing all possible network paths, which is NP-hard problem even for

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Table 2 Comparison of PERT (LB), AFEE, EE, ASEE and SMEE

Node First moment est. Second moment est.

PLBE AFEE EE ASEE SMEE

5 14.16 15.81 21.12 264.22 694.24

7 19 22.36 24.94 709.64 984.05

8 16.16 17.81 23.12 321.2 1069.07

9 22.50 24.44 30.15 722.82 1431.20

11 24.50 27.67 49.69 777.02 3537.89

12 37 39.10 54.89 1736.98 4888.90

13 30 33.97 51.74 1270.92 3895.22

14 43.50 46.97 80.29 2560.44 10558.00

15 39.33 42.13 84.72 1784.55 11167.70

16 47 51.80 110.53 2709.20 18939.60

17 34.33 42.41 67.12 2003.71 7255.43

18 40.83 43.63 95.26 1906.03 15891.90

19 43.33 51.70 159.45 2719.51 36009.90

20 60 71.91 249.33 5371.75 89646.50

moderated network (for example see Hagstrom 1988). By adjusting the recursive method weavoid the error committed by considering Wi as a random variable in (1). The error has beenobviated by considering Wi as a constant in (2). But there is still a little errors committeddue to the non-stability of the exponential distribution with respect to the maximizationoperation and the independent assumption between the random variables. It is our intentionto extend (2) to other distributions such as the Normal; Weibull; Erlang which may be moreappropriate for describing the activity durations.

Acknowledgements The author is very grateful to anonymous referees for their valuable comments andsuggestions that improved the presentation of the paper. I would also like to thank Prof. Salah E. Elmaghrabyfrom North Carolina State University, USA for his remarks that improved the paper.

Appendix

Some special cases arises from (7) and (9).

Case 1 The first and the second moments of max(X1,X2 + C) are given by

μ∗2:2 = C + e−λ1C

λ1+ 1

λ2− e−λ1C

(λ1 + λ2). (18)

μ∗(2)

2:2 = B1e−λ1C + B2 − B3e

−λ1C, (19)

where

Bi = 2

λ2i

+ 2C

λi

+ C2, i = 1,2

B3 = 2

(λ1 + λ2)2+ 2C

(λ1 + λ2)+ C2.

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Case 2 The first and the second moments of max(X1 + C1,X2 + C2) are given by

μ∗2:2 = e−(A−C1)λ1

(A + 1

λ1

)+ e−(A−C2)λ2

(A + 1

λ2

),

− e−(A−C1)λ1−(A−C2)λ2

(A + 1

(λ1 + λ2)

). (20)

μ∗(2)

2:2 = B1e−(A−C1)λ1 + B2e

−(A−C2)λ2 − B3e−(A−C1)λ1−(A−C2)λ2 , (21)

where A = max(C1,C2), and

Bi = A2 + 2A

λi

+ 2

λ2i

, for i = 1,2,

B3 = A2 + 2A

(λ1 + λ2)+ 2

(λ1 + λ2)2.

Case 3 The first and the second moments of max(X1 +C1,X2 +C2,X3 +C3) are given by

μ∗3:3 = B1 + B2 + B3 − (D12 + D13 + D23) + D123, (22)

where

Bi = e−(A−Ci)λi

(A + 1

λi

),

Dij = e−(A−Ci)λi−(A−Cj )λj

(A + 1

(λi + λj )

), i < j for i, j = 1,2,3,

D123 = e−(A−C1)λ1−(A−C2)λ2−(A−C3)λ3

(A + 1

(λ1 + λ2 + λ3)

),

μ∗(2)

3:3 =3∑

i=1

E1iK1i −∑ ∑1≤i<j≤3

E2ijK2ij + E3K3,

where

E1i = e−(A−Ci)λi ,

K1i = A2 + 2A

λi

+ 2

λ2i

, for i = 1,2,3.

E2ij = e−(A−Ci)λi−(A−Cj )λj ,

K2ij = A2 + 2A

(λi + λj )+ 2

(λi + λj )2,

E3 = e−(A−C1)λ1−(A−C2)λ2−(A−C3)λ3 ,

K3 = A2 + 2A

(λ1 + λ2 + λ3)+ 2

(λ1 + λ2 + λ3)2,

where A = max(C1,C2,C3).

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Remark 4 Set A = C2 > 0, and C1 = 0 in Case 2, Case 1 is follow.

Remark 5 When C1 = C2 = 0 in Case 2, we get μ2:2 and μ(2)

2:2. Also, set C1 = C2 = C3 = 0in Case 3, we get μ3:3 and μ

(2)

3:3.

Remark 6 Var(Xn:n) = σ (2)n:n = μ(2)

n:n − μ2n:n.

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