solution to weighted intervals problem

(Latest Revision: Nov 07, 2021)

Solution to Weighted Intervals Problem

The Info You Had To E-Mail:

#   start    finish    weight    P    OPT
1    10        13        6       0     6
2    12        14        7       0     7
3    11        16        8       0     8
4    17        20       10       3    18
5    15        23        5       2    18
6    19        24        4       3    18
7    22        25        5       4    23
8    21        27        8       4    26

Backtracking finds that intervals 8, 4, and 3  
form an optimal solution, with total weight 26.

DETAILS SHOWING HOW TO DO THE PROBLEM

The Assigned Set of Intervals and Weights:

start   finish   weight
  12      14      7
  17      20     10
  22      25      5
  21      27      8
  19      24      4
  10      13      6
  15      23      5
  11      16      8

To solve the problem, we sort the intervals by increasing finish time, and compute the p-values.


# start   finish   weight    10     11     12     13     14     15     16     17     18     19     20     21     22     23     24     25     26     27
1  10       13       6        --------------------- p=0
2  12       14       7                      -------------- p=0
3  11       16       8               ------------------------------------ p=0
4  17       20      10                                                         ---------------------- p=3
5  15       23       5                                           --------------------------------------------------------- p=2
6  19       24       4                                                                       ----------------------------------- p=3
7  22       25       5                                                                                            --------------------- p=4
8  21       27       8                                                                                      ----------------------------------------- p=4

We then compute the OPT values from row 1 to row 8.

#   s-f    w    p   OPT
0   0-0    0    0    0
1  10-13   6    0    6   = max {OPT(0), w1+OPT(p(1))} = max { 0,  6 +  0} =  6
2  12-14   7    0    7   = max {OPT(1), w2+OPT(p(2))} = max { 6,  7 +  0} =  7
3  11-16   8    0    8   = max {OPT(2), w3+OPT(p(3))} = max { 7,  8 +  0} =  8
4  17-20  10    3   18   = max {OPT(3), w4+OPT(p(4))} = max { 8, 10 +  8} = 18
5  15-23   5    2   18   = max {OPT(4), w5+OPT(p(5))} = max {18,  5 +  7} = 18
6  19-24   4    3   18   = max {OPT(5), w5+OPT(p(6))} = max {18,  4 +  8} = 18
7  22-25   5    4   23   = max {OPT(6), w5+OPT(p(7))} = max {18,  5 + 18} = 23
8  21-27   8    4   26   = max {OPT(7), w5+OPT(p(8))} = max {23,  8 + 18} = 26

Then we do the backtracking.

OPT(8) > OPT(7), so we need interval 8, and we resume backtracking at p(8), which is interval 4
OPT(4) > OPT(3), so we need interval 4, and we resume backtracking at p(4), which is interval 3
OPT(3) > OPT(2), so we need interval 3, and we resume backtracking at p(3), which is interval 0, (but 0 means stop).

Note: Generally, when backtracking, if OPT(j) = OPT(j-1), it may or may not be possible to use interval j in an optimal solution. However interval j is not needed for an optimal solution, and probably the easiest thing to do is resume backtracking at row j-1. You will always find an optimal solution this way.