Upper bounds

Coffman, Garey and Johnson prove the following upper bounds on ${\displaystyle r_{n}}$:^[1]

• ${\displaystyle r_{n}\leq 8/7\approx 1.14}$ for n = 2;
• ${\displaystyle r_{n}\leq 15/13\approx 1.15}$ for n = 3;
• ${\displaystyle r_{n}\leq 20/17\approx 1.176}$ for n = 4,5,6,7;
• ${\displaystyle r_{n}\leq 122/100=1.22}$ for all n ≥ 8.

During the MultiFit algorithm, the lower bound L is always a capacity for which it is impossible to pack S into n bins. Therefore, ${\displaystyle L<r_{n}\cdot OPT(S,n)}$. Initially, the difference ${\displaystyle U-L}$ is at most sum(S) / n, which is at most ${\displaystyle OPT(S,n)}$. After the MultiFit algorithm runs for k iterations, the difference shrinks k times by half, so ${\displaystyle U-L\leq (1/2)^{k}\cdot OPT(S,n)}$. Therefore, ${\displaystyle U\leq (r_{n}+(1/2)^{k})\cdot OPT(S,n)}$. Therefore, the scheduling returned by MultiFit has makespan at most ${\displaystyle r_{n}+(1/2)^{k}}$ times the optimal makespan. When ${\displaystyle k}$ is sufficiently large, the approximation factor of MultiFit can be made arbitrarily close to ${\displaystyle r_{n}}$, which is at most 1.22.

Later papers performed a more detailed analysis of MultiFit, and proved that its approximation ratio is at most 6/5=1.2,^[2] and later, at most 13/11≈1.182.^[3] The original proof of this missed some cases; ^[4] presented a complete and simpler proof. The 13/11 cannot be improved: there is an instance with n=13 in which the approximation ratio is exactly 13/11.^[2]

Lower bounds

For n=4: the following^[5] shows that ${\displaystyle r_{n}\geq 20/17}$, which is tight. The inputs are 9,7,6,5,5, 4,4,4,4,4,4,4,4,4. They can be packed into 4 bins of capacity 17 as follows:

• 9, 4, 4
• 7, 6, 4
• 5, 4, 4, 4
• 5, 4, 4, 4

But if we run FFD with bin capacity smaller than 20, then the filled bins are:

• 9,7 [4 does not fit]
• 6,5,5 [4 does not fit]
• 4,4,4,4 [4 does not fit]
• 4,4,4,4
• 4

Note that the sum in each of the first 4 bins is 16, so we cannot put another 4 inside it. Therefore, 4 bins are not sufficient.

For n=13: the following^[2] shows that ${\displaystyle r_{n}\geq 13/11}$, which is tight. The inputs can be packed into 13 bins of capacity 66 as follows:

• 40,13,13 {8 times}
• 25,25,16 {3 times}
• 25,24,17 {2 times}

But if we run FFD with bin capacity smaller than 66*13/11 = 78, then the filled bins are:

• 40,25 {8 times}
• 24, 24, 17
• 17, 16, 16, 16
• 13, 13, 13, 13, 13 {3 times}
• 13

Note that the sum in each of the first 13 bins is 65, so we cannot put another 13 inside it. Therefore, 13 bins are not sufficient.

Performance with uniform machines

MultiFit can also be used in the more general setting called uniform-machines scheduling, where machines may have different processing speeds.^[6] When there are two uniform machines, the approximation factor is ${\displaystyle {\sqrt {6}}/2}$. When MultiFit is combined with the LPT algorithm, the ratio improves to ${\displaystyle {\sqrt {2}}+1/2}$.

Performance for maximizing the smallest sum

A dual goal to minimizing the largest sum (makespan) is maximizing the smallest sum. Deuermeyer, Friesen and Langston claim that MultiFit does not have a good approximation factor for this problem:^[7]

"In the solution of the makespan problem using MULTIFIT, it is easy to construct examples where one processor is never used. Such a solution is tolerable for the makespan problem, but is totally unacceptable for our problem _{[since the smallest sum is 0]}. Modifications of MULTIFIT can be devised which would be more suitable for our problem, but we could find none which produces a better worst-case bound than that of LPT."

Minimal counterexamples

The upper bounds on ${\displaystyle r_{n}}$ are proved by contradiction. For any integers p ≥ q, if ${\displaystyle r_{n}>p/q}$, then there exists a (p/q)-counterexample, defined as an instance S and a number n of bins such that

• S can be packed into n bins with capacity q;
• FFD does not manage to pack S into n bins with capacity p.

If there exists such a counterexample, then there also exists a minimal (p/q)-counterexample, which is a (p/q)-counterexample with a smallest number of items in S and a smallest number of bins n. In a minimal (p/q)-counterexample, FFD packs all items in S except the last (smallest) one into n bins with capacity p. Given a minimal (p/q)-counterexample, denote by P₁,...,P_n the (incomplete) FFD packing into these n bins with capacity p, by P_n+1 the bin containing the single smallest item, and by Q₁,...,Q_n the (complete) optimal packing into n bins with capacity q. The following lemmas can be proved:

• No union of k subsets from {Q_1,...,Q_n} is dominated by a union of k subsets from {P_1,...,P_n+1} ("dominated" means that each item in the dominated subset is mapped to a weakly-larger item in the dominating subset). Otherwise we could get a smaller counterexample as follows. [1] Delete all items in the P_i. Clearly, the incomplete FFD packing now needs n-k bins, and still the smallest item (or an entire bin) remains unpacked. [2] In the optimal packing Q_i, exchange each item with its dominating item. Now, the k subsets Q_i are larger (probably larger than q), but all other n-k subsets are smaller (in particular, at most q). Therefore, after deleting all items in the P_i, the remaining items can be packed into at most n-k bins of size q.
• Each of Q_1,...,Q_n contains at least 3 items. This is because [a] each Q_i with a single item is dominated by the P_j that contains that item; [b] for each Q_i with two items x and y, if both x and y are in the same P_j, then Q_i is dominated by this P_j; [c] Suppose x≥y, x is in some P_j, and y is in some P_k to its right. This means that y did not fit into P_j. But x+y ≤ q. This means that P_j must contain some item z ≥ y. So Q_i is dominated by P_j. [d] Suppose x≥y, x is in some P_j, and y is in some P_k to its left. This means that there must be a previous item z ≥ x. So Q_i is dominated by P_k.
• Otherwise we had domination and, by the previous lemma, could get a smaller counterexample.
• Each of P_1,...,P_n contains at least 2 items. This is because, if some P_i contains only a single item, this implies that the last (smallest) item does not fit into it. This means that this single item must be alone in an optimal bundle, contradicting the previous lemma.
• Let s be the size of the smallest item. Then ${\displaystyle s>{\frac {n}{n-1}}(p-q)}$. Proof: Since s does not fit into the first n bundles, we have ${\displaystyle sum(P_{i})+s>p}$, so ${\displaystyle \sum _{i=1}^{n}sum(P_{i})+n\cdot s>n\cdot p}$. On the other hand, since all items fit into n bins of capacity q, we have ${\displaystyle \sum _{i=1}^{n}sum(P_{i})+s\leq n\cdot q}$. Subtracting the inequalities gives ${\displaystyle s>{\frac {n}{n-1}}(p-q)}$.
• The size of every item is at most ${\displaystyle q-2s}$. This is because there are at least 3 items in each optimal bin (with capacity q).
• The sum of items in every bin P_1,...,P_n is larger than ${\displaystyle p-s}$; otherwise we could add the smallest item.

5/4 Upper bound

From the above lemmas, it is already possible to prove a loose upper bound ${\displaystyle r_{n}\leq 5/4=1.25}$. Proof. Let S, n be a minimal (5/4)-counterexample. The above lemmas imply that -

• ${\displaystyle s>{\frac {n}{n-1}}(5-4)>1}$. Since the optimal capacity is 4, no optimal bin can contain 4 or more items. Therefore, each optimal bin must contain at most 3 items, and the number of items is at most 3n.
• The size of each item is at most ${\displaystyle 4-2s}$, and the size of each FFD bin is more than ${\displaystyle 5-s}$. If some FFD bin contained only two items, its sum would be at most ${\displaystyle 8-4s=5+(3-3s)-s<5-s}$; so each FFD bin must contain at least 3 items. But this means that FFD yields exactly n bins - a contradiction.

Structure of FFD packing

To prove tighter bounds, one needs to take a closer look at the FFD packing of the minimal (p/q)-counterexample. The items and FFD bins P₁,...,P_n are termed as follows:

• A regular item is an item added to some bin P_i, before the next bin P_i+1 was opened. Equivalently, a regular item is an item in P_i which is at least as large as every item in every bin P_j for j>i.
• A fallback item is an item added to some bin P_i, after the next bin P_i+1 was opened. Equivalently, a fallback item is an item in P_i which is smaller than the largest item in P_i+1.
• A regular k-bin is a bin that contains k regular items and no fallback items.
• A fallback k-bin is a bin that contains k regular items and some fallback items.

The following lemmas follow immediately from these definitions and the operation of FFD.

• If k₁<k₂, then all k₁-bins are to the left of all k₂-bins. This is because all bins have the same capacity, so if more regular items fit into a bin, these items must be smaller, so they must be allocated later.
• If P_i is a k-bin, then the sum of the k regular items in P_i is larger than ${\displaystyle {\frac {k}{k+1}}\cdot p}$, since otherwise we could add another item before opening a new bin.
• If P_i and P_i+1 are both k-bins, and then the sum of the k regular items in P_i is at least as large as in P_i+1 (this is because the items are ordered by decreasing size).
• All regular k-bins are to the left of all fallback k-bins. This is because all bins have the same capacity, so if more fallback items fit into a bin, these items must be smaller, so they must be allocated later.

In a minimal counterexample, there are no regular 1-bins (since each bin contains at least 2 items), so by the above lemmas, the FFD bins P₁,...,P_n are ordered by type:

• Zero or more fallback 1-bins;
• Then, zero or more regular 2-bins;
• Then, zero or more fallback 2-bins;
• Then, zero or more regular 3-bins;
• Then, zero or more fallback 3-bins;
• and so on.

1.22 upper bound

The upper bound ${\displaystyle r_{n}\leq 1.22}$^[1] is proved by assuming a minimal (122/100)-counterexample. Each item is given a weight based on its size and its bin in the FFD packing. The weights are determined such that the total weight in each FFD bin is at least x, and the total weight in almost each optimal bin is at most x (for some predetermined x). This implies that the number of FFD bins is at most the number of optimal bins, which contradicts the assumption that it is a counterexample.

By the lemmas above, we know that:

• The size of the smallest item satisfies s > p-q = 22, so s = 22+D for some D>0.
• Each optimal bin contains at most 4 items (floor(100/22)), and each FFD bin contains at most 5 items (floor(122/22)).
• The size of every item is at most q-2s = 56-2D.
• The sum in each FFD bin is larger than p-s = 100-D.
• There are no 1-bins, since in a 1-bin, the size of the regular item must be at least p/2=61, while here the size of every item is less than 56.

If D>4, the size of each item is larger than 26, so each optimal bin (with capacity 100) must contain at most 3 items. Each item is smaller than 56-2D and each FFD bin has a sum larger than 100-D, so each FFD bin must contain at least 3 items. Therefore, there are at most n FFD bins - contradiction. So from now on, we assume D≤4. The items are assigned types and weights as follows.

• The two items in each regular 2-bin except maybe the last one have a size larger than (100-D)/2 each. All such items are called type-X₂, and assigned a weight of (100-D)/2. The last 2-regular bin is a special case: if both its items have a size larger than (100-D)/2, then they are type-X₂ too; otherwise, they are called type-Z, and their weight equals their size.
• The two regular items in each fallback 2-bin have a total size larger than 2*122/3; they are called type-Y₂, and their weight equals their size minus D.
• The three items in each regular 3-bin except maybe the last one have a size larger than (100-D)/3 each. All such items are called type-X₃, and assigned a weight of (100-D)/3. The last 3-regular bin is a special case: if all items in it have a size larger than (100-D)/3, then they are type-X₃ too; otherwise, they are called type-Z and their weight equals their size.
• The three regular items in each fallback 3-bin have a total size larger than 3*122/4; they are called type-Y₃, and their weight equals their size minus D.
• The four items in each regular 4-bin except maybe the last one have a size larger than (100-D)/4 each. All such items are called type-X₄, and assigned a weight of (100-D)/4. The last 4-regular bin is a special case: if all items in it have a size larger than (100-D)/4, then they are type-X₄ too; otherwise, they are called type-Z and their weight equals their size.
• The remaining items (including all fallback items in fallback 2-bins and 3-bins, all fallback 4-bins, and all other 5-item bins) are all called type-X₅, and their weight equals 22 (if D ≤ 12/5) or (100-D)/4 (otherwise). The threshold 12/5 was computed such that the weight is always at most 22+D, so that the weight is always smaller than the size.

Note that the weight of each item is at most its size (the weight can be seen as the size "rounded down"). Still, the total weight of items in every FFD bin is at least 100-D:

• For regular 2-bins, regular 3-bins and regular 4-bins:
  • For the non-last ones, this is immediate.
  • The last such bins contain only Z-type items, whose weight equals their size, so the total weight of these bins equals their total size, which is more than 100-D.
• Fallback 2-bins contain two type-Y₂ items with total weight larger than 2*122/3-2D, plus at least one type-X₅ item with weight at least 22 (if D ≤ 12/5) or (100-D)/4 (otherwise). In both cases the total weight is more than 100-D.
• Fallback 3-bins contain three type-Y₃ items with total weight larger than 3*122/4-3D, plus at least one type-X₅ item with weight at least 22. So the total weight is more than 3*122/4+22-3D = 113.5-3D ≥ 105.5-D > 100-D, since D≤4.
• 5-item bins contain 5 items with size at least 22+D and weight at least 22, so their total weight is obviously more than 100-D.

The total weight of items in most optimal bins is at most 100-D:

• This is clear for any optimal bin containing a type-Y₂ item or a type-Y₃ item, since their weight is their size minus D, the weights of other items is at most their size, and the total size of an optimal bin is at most 100.
• For optimal bins containing only type-X₂, type-X₃, type-X₄ and type-X₅ items, it is possible to check all possible configurations (all combinations that fit into an optimal bin of size 100), and verify that the total weight in each configuration is at most 100-D.
• Optimal bins containing type-Z items might have a total weight larger than 100-D. Since the total weight is at most 100, there is an "excess weight" of at most D for each such bin. However, the number of type-Z items is limited:
  • If D > 12/5, then there are at most 5 type-Z items (2 in the last regular 2-bin and 3 in the last regular 3-bin; the items in the last regular 4-bin are all type-X₄). Therefore, the excess weight is at most 5D. Comparing the total weight of FFD vs. optimal bins yields s < 5D ≤ 20 < 22, a contradiction.
  • Otherwise, there are at most 9 type-Z items (2+3+4). Therefore, the excess weight is at most 9D. Comparing the total weight of FFD vs. optimal bins yields s < 9D ≤ 108/5 < 22, a contradiction.

13/11 upper bound

The upper bound ${\displaystyle r_{n}\leq 13/11\approx 1.182}$^[3] is proved by assuming a minimal ((120-3d)/100)-counterexample, with some d<20/33, and deriving a contradiction.

Example

Here is an example.^{[5]: Ex.5.2} Suppose there are four agents, and they have valuations of two types:

Both types can partition the chores into 4 parts of total value 75. Type A:

• 51, 12, 12
• 27.5, 27.5, 10, 10
• 27.5, 27.5, 10, 10
• 25, 10, 10, 10, 10, 10

Type B:

• 51, 24
• 27.5, 27.5, 20
• 27.5, 27.5, 20
• 8.33 {9 times}

If all four agents are of the same, then FFD with threshold 75 fills the 4 optimal bins. But suppose there is one agent of type B, and the others are of type A. Then, in the first round, the agent of type B takes the bundle 51, 24 (the other agents cannot take it since for them the values are 51,25 whose sum is more than 75).In the following rounds, the following bundles are filled for the type A agents:

• 27.5, 27.5, 12 [the sum is 67 - there is no room for another 10]
• 27.5, 27.5, 12 [the sum is 67 - there is no room for another 10]
• 10, 10, 10, 10, 10, 10, 10 [the sum is 70 - there is no room for another 10]

so the last two chores remain unallocated.

Optimal value guarantee

Using a more sophisticated threshold calculation, it is possible to guarantee to each agent at most 11/9≈1.22 of his optimal value if the optimal value is known, and at most 5/4≈1.25 of his optimal value (using a polynomial time algorithm) if the optimal value is not known.