Page - 174 - in Algorithms for Scheduling Problems

Algorithms 2018,11, 35 Step7.For s−1,2, . . . , compute H(s−1)−H(s)= = ( 1− c(s)c(s−1) ) H(s−1)−c(s)(H∗(Ls)−HC∗(Ls−1))+c(s) N ∑ i=1 ( pi(s) · N ∑ j=1 pˆ(1)ij (s−1) · log ( pˆ(1)ij (s−1) )) Asthestoppingruleuse the followingrule: Stopat thecut s1 forwhich H(s1−1)−H(s1) H(0)−H(s1) < εholds. ThenthereducedSCmodelcontainsonly thecut (s1−1). 5.NumericalExample Inputdata: - thenumberorrisk factordrivers ineach layer,N=3; - levelofaccuracy ε=0.01; - theweight function c(s)= 1 (s+1)2 (selectedbythedecisionmaker). - probabilitiespfprime(s)=Pr (Af prime(s)) pf prime(0)=(0.3457 0.0835 0.0918) pf prime(1)=(0.1644 0.3017 0.0542) pf prime(2)=(0.1256 0.1602 0.2156) pf prime(3)=(0.0845 0.2001 0.3025) pf prime(4)=(0.2623 0.1056 0.2369) pf prime(5)=(0.2014 0.2032 0.1356) pf prime(6)=(0.1422 0.2258 0.1047) pf prime(7)=(0.1056 0.3241 0.2658) pf prime(8)=(0.1599 0.3056 0.1422) pf prime(9)=(0.2014 0.3068 0.0856) pf prime(10)=(0.2145 0.0241 0.2536) - probabilitiespfsecond(s)=Pr (Af second(s)) pf second(0)=(0.3014 0.0725 0.1051) pf second(1)=(0.1851 0.2532 0.0414) pf second(2)=(0.2098 0.1308 0.1280) pf second(3)=(0.0837 0.2011 0.1281) pf second(4)=(0.1272 0.1013 0.1667) pf second(5)=(0.2334 0.0687 0.1577) pf second(6)=(0.1393 0.2709 0.1171) pf second(7)=(0.0824 0.1803 0.0417) pf second(8)=(0.1379 0.1143 0.1401) pf second(9)=(0.1456 0.1703 0.0903) pf second(10)=(0.2350 0.0383 0.2345) transitionprobabilitymatricesM(2)(s)= ( p(2)ij (s) ) N×N , s=1,2, . . . ,10, M(2)(1)= 0.3124 0.3320 0.3556 0.3456 0.4158 0.2386 0.4258 0.0256 0.5486 174