# sol5 - Homework#5 solutions IEOR4405 1 1 Let f(j t denote...

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Homework #5 solutions / IEOR4405 1 1. Let f ( j, t ) denote the maximum weight set of jobs that can be completed by time t , using jobs 1 , 2 , ..., j (with j = 1 , 2 , .., 4 and t = 1 , 2 , ..., 10). We use the recurrence f ( j, t ) = max( f ( j - 1 , t ) , f ( j, t - 1) , f ( j - 1 , t - p j ) + w j ) , with the initialization f (0 , t ) = 0 for all t and f ( j, 0) = 0 for all j . The algorithm generates the values for f ( j, t ) given in the table below. They are generated row-by-row and every row is calculated from left to right. The optimal value is 18 and the corresponding schedule is 1-2-4-3. j \ t 0 1 2 3 4 5 6 7 8 9 10 0 0 0 0 0 0 0 0 0 0 0 0 1 0 6 6 6 6 6 6 6 6 6 6 2 6 6 10 10 10 10 10 10 10 10 10 3 0 6 6 10 10 10 12 12 16 16 16 4 0 6 6 10 10 10 13 13 17 17 17 5 0 6 6 10 10 10 13 14 14 17 18 2 Let f ( j, w ) be the minimum deadline d by which you can schedule jobs 1 , 2 , ...j and have the weight of the jobs processed atleast w . In a given state ( j, w ), we have the option to either choose the j th job or skip it. If the j th job is chosen, then we incur a processing time of p j units of time, and we enter state f ( j - 1 , w - w j ). Otherwise, we enter state, f ( j - 1 , w ) since there was no processing involved. Hence, the recursion will have the particular structure, f ( j, w ) = min ( f ( j - 1 , w ) , p j + f ( j - 1 , w - w j )) While filling the table, one can use backward induction and can start filling from the last element of the matrix.(lower right corner). Read the element in the matrix which has the value closest to but less than the specified deadline. Find the corresponding w 0 j

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