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  2. Multiple subset sum - Wikipedia

    en.wikipedia.org/wiki/Multiple_subset_sum

    For separate items: the price-of-fairness of max-min fairness is unbounded. For example, suppose Alice has two items with values 1 and e, for some small e>0. George has two items with value e. The capacity is 1. The maximum sum is 1 - when Alice gets the item with value 1 and George gets nothing. But the max-min allocation gives both agents ...

  3. Subset sum problem - Wikipedia

    en.wikipedia.org/wiki/Subset_sum_problem

    The variant in which all inputs are positive, and the target sum is exactly half the sum of all inputs, i.e., = (+ +). This special case of SSP is known as the partition problem . SSP can also be regarded as an optimization problem : find a subset whose sum is at most T , and subject to that, as close as possible to T .

  4. List of mathematical series - Wikipedia

    en.wikipedia.org/wiki/List_of_mathematical_series

    This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums. Here, is taken to have the value

  5. First-fit bin packing - Wikipedia

    en.wikipedia.org/wiki/First-fit_bin_packing

    Here is a proof that the asymptotic ratio is at most 2. If there is an FF bin with sum less than 1/2, then the size of all remaining items is more than 1/2, so the sum of all following bins is more than 1/2. Therefore, all FF bins except at most one have sum at least 1/2. All optimal bins have sum at most 1, so the sum of all sizes is at most OPT.

  6. Knapsack problem - Wikipedia

    en.wikipedia.org/wiki/Knapsack_problem

    The target is to maximize the sum of the values of the items in the knapsack so that the sum of weights in each dimension does not exceed . Multi-dimensional knapsack is computationally harder than knapsack; even for D = 2 {\displaystyle D=2} , the problem does not have EPTAS unless P = {\displaystyle =} NP. [ 32 ]

  7. Bin covering problem - Wikipedia

    en.wikipedia.org/wiki/Bin_covering_problem

    The easy case is =, that is, all final items are smaller than 1/2. Then, the sum of every filled is at most 3/2, and the sum of remaining items is at most 1, so the sum of all items is at most / +. On the other hand, in the optimal solution the sum of every bin is at least 1, so the sum of all items is at least ().

  8. Running total - Wikipedia

    en.wikipedia.org/wiki/Running_total

    In fact, we would not even need to know the sequence at all, but simply add 6 to 18 to get the new running total; as each new number is added, we get a new running total. The same method will also work with subtraction, but in that case it is not strictly speaking a total (which implies summation) but a running difference; not to be confused ...

  9. Sum of normally distributed random variables - Wikipedia

    en.wikipedia.org/wiki/Sum_of_normally...

    This means that the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances (i.e., the square of the standard deviation is the sum of the squares of the standard deviations). [1]