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## Hybrid Algorithm for mean T

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**Hybrid Algorithm for mean T**Baker p.49 Hybrid Algorithm: Dominance matrix to decompose problem in to sub-problems. Branch and Bound Original problem for DP n! = C(n,1) + C(n,2) + ………….+ C(n,n) 30~35 jobs → out of computation time N! N1! N2!**Decomposition**Ex. Divide 50 persons into 4 subgroups by according to their height. Mutual exclusive S Decompose S1,S2,………,Sn and For any , put i in Then the objective function**Hybrid Algorithm**Notation: SPT EDD**Bj**tj Hybrid Algorithm Property 3.2: (i before j) Property 3.3: (j before i)**Steps:**Baker p.51**Final table**2-5-1-4-6-7-XXX-10 389**By DP**2-5-1-4-6-7-9-3-8-10**HW.**Use Hybrid to solve**Single Machine Heuristics**Sule p.14 • Earliest Due Date (EDD) Rule • Cost Over Time (COVERT) Rule • Shortest Processing Time (SPT) Rule • Largest Penalty per Unit Length (LPUL) Rule • Largest Weight (WT) and LPUL Rule • Shortest Processing Time and LPUL Rule • Shortest Weighted Processing Time (SWPT) Rule • Critical Ratio (CR) Rule**COVERT**Notations: TT - Total processing times of all jobs RT - Remaining total processing times of the set of unscheduled jobs ST - Starting time for the next scheduled job,0 for the first job CF - Coefficient PR - Priority**PR=0**PR=1 Di TT Pi ST+Pi ST COVERT Rules Step l: calculate PR for all jobs RT Step 2: calculate CF for job i, yet to be scheduled. Step 3: Job with the Max CF is scheduled next. Repeat step1**Ex.**Iteration 1. Step l : Caculate PR TT=93 RT=93**Ex.**Step 2: Step3: job 3 is scheduled. Iteration 2. . . . .**Backward & Forward Method**• Backward&ForwardMethod Initial scheduled Improve：Swapping to jump out the local optimal • Use any one of the Lower Bound Heuristic method • Not the optimum**BF Method**5 jobs X X X X X 5 4 3 2 1 4: 5 – 1 3: 5 – 2 , 4 – 1 2: 5 – 3 , 4 – 2 , 3 – 1 1: 5 – 4 , 4 – 3 , 3 – 2 , 2 –1 Lag =**Ex.**(1) Backward Phase: Step1. TT = 37+27+1+28 = 93 Calculate the penalty for each job if it is to be complete at T=93. Job1: (93-49)*1=44 Job2: (93-36)*5=285 Job3: (93-1)*1=92 Job4: (93-37)*5=280 • job 1 has the least cost, so it is scheduled in the last position**EX.**Step2. T= 93-37 = 56 Calculate the penalty for each job if it is to be complete at T=56. Job2: (56-36)*5=100 Job3: (56-1)*1=55 Job4: (56-37)*5=95 • job 3 has the least cost, so it is scheduled in position 3**EX.**Step3. T= 56-1 = 55 Calculate the penalty for each job if it is to be complete at T=55. Job2: (55-36)*5=95 Job4: (55-37)*5=90 • job 4 has the least cost, so it is scheduled in position 2 Step4. The final sequence 2-4-3-1 The total penalty for the sequence 2-4-3-1 is (18*5)+(55*1)+(44*1)=189**Ex.**(2) Forward Phase: Initial sequence : 2-4-3-1 penalty = 189 Lag = 3 : 1-4-3-2 penalty = 490 – no change Lag = 2 : 3-4-2-1 penalty = 144 – change New sequence : 3-4-2-1 penalty = 144 Lag = 3 : 1-4-2-3 penalty = 644 – no change Lag = 2 : 2-4-3-1 penalty = 189 – no change Lag = 2 : 3-1-2-4 penalty > 144 – no change Lag = 1 : 4-3-2-1 penalty = 172 – no change Lag = 1 : 3-2-4-1 penalty = 139 – change New sequence : 3-2-4-1 . . . We can get the final sequence is 3-2-4-1**HW.**• Use Forward + Covert to solve the following problem : (or you want to compare your solution with BF)