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Mean Project Completion Time in Dynamic Markov PERT Networks

Mean Project Completion Time in Dynamic Markov PERT Networks

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Amir Azaron, Hideki Katagiri, Kosuke Kato, Masatoshi Sakawa<br />

Proof. The expected length of arc (l,j) is<br />

774<br />

1<br />

1 2 N<br />

lj m , s ,...,<br />

1 m s<br />

2 mN<br />

λ ( s<br />

, because the<br />

)<br />

duration of activity (l,j) has exponential distribution with parameter<br />

1 2 N<br />

λ ( s , s ,..., s ) . The probability that the state vector of system<br />

lj<br />

m1 m2<br />

mN<br />

1 2 N<br />

1 2 N<br />

changes from ( s , s ,..., s ) <strong>in</strong> node l to ( s , s ,..., s ) , after do<strong>in</strong>g<br />

m1 m2<br />

mN<br />

the activity (l,j), would be<br />

1k1lj<br />

m m<br />

mN<br />

2k2lj<br />

m1m2<br />

... mN<br />

k1 k2<br />

kN<br />

Nk Nlj<br />

m1m<br />

mN<br />

P P P ...<br />

1 2...<br />

... 2 , because the<br />

cont<strong>in</strong>uous-time <strong>Markov</strong> processes correspond<strong>in</strong>g to the transitions of the social<br />

problems are <strong>in</strong>dependent. If the arc (l,j) belongs to the path with maximum<br />

expected length, which approximately considered as the longest path,<br />

then, the state vector of system <strong>in</strong> node j actually changes to<br />

1 2 N<br />

( s , s ,..., s ) with the <strong>in</strong>dicated probability. F<strong>in</strong>ally, by condition<strong>in</strong>g on<br />

k1 k2<br />

kN<br />

the state vector of system <strong>in</strong> node j, Theorem 1 is proved.<br />

Without los<strong>in</strong>g the generality, we assume that the nodes of the network are<br />

numbered from 1 to n, <strong>in</strong> which there should be no directed path from node j to<br />

node l for j>l. The follow<strong>in</strong>g algorithm can be used to approximate the mean<br />

project completion time <strong>in</strong> dynamic <strong>Markov</strong> <strong>PERT</strong> networks.<br />

Algorithm<br />

1 2 N<br />

Step 1. Beg<strong>in</strong> from node l=n. It is clear that V ( s , s ,..., s ) = 0<br />

m = 1,<br />

2,...,<br />

N , i = 1,<br />

2,...,<br />

N .<br />

for i<br />

i<br />

n m1<br />

m2<br />

mN

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