Seite - 176 - in Algorithms for Scheduling Problems

Algorithms 2018,11, 35 Thealgorithmsolves thisexampleas follows. UsingFormulas (3)–(6) compute: Theprobabilitiesof risk factordriversonthe layersL0,L1,L2, . . . p(0)=(0.6471 0.1560 0.1969) p(1)=(0.3495 0.5549 0.0956) p(2)=(0.3354 0.2910 0.3736) p(3)=(0.1682 0.4012 0.4306) p(4)=(0.3895 0.2069 0.4036) p(5)=(0.4348 0.2719 0.2933) p(6)=(0.2815 0.4967 0.2218) p(7)=(0.1880 0.5044 0.3075) p(8)=(0.2978 0.4199 0.2823) p(9)=(0.3470 0.4771 0.1759) p(10)=(0.4495 0.0624 0.4881) Thecomplete transitionmatrices Mˆ(1)(s)= ( pˆ(1)ij (s) ) 3×3 Mˆ(1)(s)= ( pij(s) ) , s=1,2, . . . ,10 Mˆ (:, :,1)= 0.4124 0.4227 0.1649 0.8076 0.0130 0.1794 0.5732 0.0114 0.4153 Mˆ (:, :,2)= 0.1978 0.6567 0.1455 0.1029 0.8599 0.0372 0.6779 0.2258 0.0963 Mˆ (:, :,3)= 0.6331 0.3176 0.0494 0.1494 0.0275 0.8230 0.3925 0.5261 0.0814 Mˆ (:, :,4)= 0.1104 0.8743 0.0153 0.1262 0.0455 0.8284 0.2454 0.1270 0.6276 Mˆ (:, :,5)= 0.4019 0.3097 0.2884 0.7173 0.2444 0.0382 0.0672 0.0197 0.9131 Mˆ (:, :,6)= 0.1399 0.6222 0.2379 0.4887 0.1686 0.3427 0.6885 0.0584 0.2531 176