1. György Dósa, University of Pannonia, Veszprém, Hungary
The tight bound of First Fit Decreasing bin-packing algorithm is FFD(I) ≤ 11/9 OPT(I) + 6/9.
György Dósa,
University of Pannonia, Veszprém, Hungary

2. What is bin-packing? FFD:{ We are given a set of items I (instance),
I={p1,p2,…pn} , 0 ≤ pi ≤ 1 (sizes),
and unit capacity bins.
The problem: pack the items into a minimum number of bins (NP-hard)
EXAMPLE:
while
OPT(I)=2
FFD(I)=3
FFD:{
First sort the items into their decreasing sizes
The next item is always put into the first bin where it fits.

3. This bound can not be smaller!!
From the classical example of [2] E.G. Coffman, Jr., M.R. Garey, D.S. Johnson: Approximation algorithms for bin packing: A survey, in: Approx. Algorithms for NP-hard Problems, D. Hochbaum, (ed.), PWS Publishing, Boston, (1996) page 16. 6N +ε
-2ε
+2ε 3N
FFD(I)=11N 2N
OPT(I)=9N
we get an instance I, for which FFD(I) = 11/9 OPT(I), follows that
FFD(I) ≤ 11/9 OPT(I) + ?
This bound can not be smaller!!

4. Special thanks to: Zhiyi Tan, who asked me whether:
Is there an instance (item-set) I, for which
OPT(I)=5, and FFD(I) ≥ 7,
(i.e. the items can be packed into 5 unit capacity bins, and FFD packs them into at least 7 bins)?
Or: Does exist a 7/5 counterexample?
The answer is used in
[1] Zhong, W., Dósa, Gy., Tan, Z., On the machine scheduling problem with job delivery coordination, European Journal of Operations Research, online published, 2006
If yes: the shceduling algorithm introduced in the paper has a weak competitive ratio.
If not: The algorithm is very competitive.

5. BUT what is the case with OPT(I)= 6? Since then
Does exist a 7/5 example? ( OPT(I)=5, FFD(I) ≥ 7 )
(If we knew that for all I FFD(I) ≤ 11/9 OPT(I) + 6/9 holds, then
from OPT(I)= FFD(I) ≤ 11/9 · 5 + 6/9 =61/9 < 7 follows.)
The question looks to be quite trivial (but it is not).
The answer is: NOT, but it was an open question. See Appendix of [1], (the proof requires 4 pages).
BUT what is the case with OPT(I)= 6? Since then
FFD(I) ≤ 11/9 · 6 + 6/9 =72/9 = 8.
Does exist an instance I for which OPT(I)= 6, and FFD(I)=8?
The answer is YES:

6. By simple modification of example from [2] we get 8/6 example:
-2ε
+2ε 6N +ε +ε
+2ε
-2ε
Nr:
OPT(I)=6
FFD(I)=8
This bound also can not be smaller!
Follows: FFD(I) ≤ 11/9 OPT(I) + 6/9

7. Corollary: It is also easy to get instances for which
OPT(I) = 9N+ 6 + j and FFD(I)= 11N j, j=1,2,…
OPT(I) = 9N+11+ j and FFD(I)= 11N j, j=1,2,...
Corollary:
If we know that FFD(I) ≤ 11/9 OPT(I) + 6/9, then
for all integer m we know the maximum integer n,
for which there exists such instance I, that
OPT(I) = m, and FFD(I) = n,
as follows:

8. a d s o OPT(I) = m 1 2 3 4 5 6 7 8 9 10 FFD(I) = n 11 12 n – m = k mm 13 14 15 16 17 18 19 n 20 21 22 23 k 24 25 26 27 28 30 31 32 33 34 a d s o
We have seen these cases

9. Some history (on the additive constant)
FFD(I) ≤ 11/9 OPT(I) (D.S. Johnson, Doctoral Thesis, 1973)
FFD(I) ≤ 11/9 OPT(I) (B.S. Baker, 1985)
FFD(I) ≤ 11/9 OPT(I) (M. Yue, 1991)
A very good result, but:
The proof is hard to follow,
Many gaps remain to be verified by the reader
FFD(I) ≤ 11/9 OPT(I) + 7/ (L. Rongheng, M. Yue, 1997)
It is only a draft, conjectures the upper bound

10. Suppose that the statement is not true.
It remained only the proof for FFD(I) ≤ 11/9 OPT(I) + 6/9 (The proof needed 30 pages, and one year)
Suppose that the statement is not true.
Let I be a minimal counterexample
(minimum number of items)
Let X be the last (thus least) item
in the decreasing order of the items. Then it is alone in the last FFD bin.
If X is too big, bigger than 1/4:
Each bins contains at most three items, we get soon contradiction

11. > 1 + 9/11 ( FFD(I) - 2) OPT(I) ≥ Σ pi = X + P( B1) + Σ P (Bi)
If X is too small, not bigger than 2/11:
Each bins but the last one has content more than 9/11: the sum of the sizes of the items is too much, contradiction comes easily again:
OPT(I)
FFD(I)-1
OPT(I) ≥ Σ pi = X + P( B1) + Σ P (Bi)
i=2
k=1
> /11 ( FFD(I) - 2)
= 9/11 FFD(I) - 7/11
≥ 9/11 ( 11/9 OPT(L)+7/9) -7/11 = OPT(I)
(where P (Bi) = the sum of the sizes of items being in bin Bi )

12. Remained: if 2/11 < X ≤ 1/4 We show that
FFD(I) ≤ 11/9 OPT(I) + 27/36
(since 27/36 < 28/36 = 7/9, and OPT(I), FFD(I) are integers, it is OK)
Two main cases:
2/11 < X ≤ 1/ and /5 < X ≤ 1/4
Case Case 1

13. Case 1, 1/5 < X ≤ 1/4 we do a classification as follows: (not the same as in Yue’s paper)
NAME
CLASS
WEIGHT
Or simply
Large
1/2 < L
23/36(1-X) 23
Big
(1-X)/2 < B ≤ 1/2
18/36(1-X) 18
Medium
1/3 < M ≤ (1-X)/2
15/36(1-X) 15
Small
(1-X)/4 < S ≤ 1/3
12/36(1-X) 12
qUite small
1/4 < U ≤ (1-X)/4
9/36(1-X) 9
Very small
X ≤ V ≤ 1/4
Only the numerator differs here
the numerators

14. (1-X)FFD(I) ≤ Σk w( Bk) = w(I) = Σk w(Bk*) ≤ 11/9(1-X) OPT(I)
If all optimal bin’s weight is at most w(B*)≤11/9 (1-X), and all FFD bin has weight not more than w(B) ≤1-X, easily we get:
(1-X)FFD(I) ≤ Σk w( Bk) = w(I) = Σk w(Bk*) ≤ 11/9(1-X) OPT(I)
and we would be ready. Unfortunately, some FFD bins can have less weight. We say that that FFD bins have shortage.

15. Thus some definitions: let B be a bin,
B* be an optimal bin
w(B), w(B*), = the weight of the bin
The allowed weight of an optimal bin: 11/9 (1-X)
The required weight of an FFD bin: (1-X)
res(B*) = 11/9 (1-X) - w(B*), Reserve
shor(B) = (1-X) - w(B), (if it is positive) Shortage
sur(B) = w(B) - (1-X) (if it is positive) Surplus
rex(B) = 27/ REQuired
additive conStant

16. For example let B*=(2M,S) (an optimal bin)15 12
Allowed weight: 44
the real weight: 42
res(B*) = 2
w(S)=12
w(M)=15
The reserve of this optimal bin B*
w(B*) = 42
The allowed weight : 11/9 (1-X) = 44/36(1-X), or simply: 44
(only the numerator)

17. Two FFD bins: B1=(2M,S) B2=(S,U,V)15 12 M S U V 9 12
B w(B1) = 42
B w(B2) = 30
sur (B1) = 42 – 36 = 6
short(B2) = 36 – 30 = 6
(The required weight : (1-X) = 36/36(1-X), or simply: 36)

18. Short(I) ≤ res(I) + sur(I) + rex(I)
Then it is easy to see that the statement
FFD(I) ≤ 11/9 OPT(I) + 27/36
is equivalent with
Short(I) ≤ res(I) + sur(I) + rex(I)
(as M. Yue said: All shortage can be covered by all
reserve + surplus + the required additive constant.)

19. The all possible optimal bins in case 1/5 < X ≤ 1/4 are:
each optimal bin contains at least 3, at most 4 items L 1 B M 2 S U V 3 S 3 2 1 U V 4

20. The all possible FFD bins in case 1/5 < X ≤ 1/4 are:
(2, 3, or 4 items) L 1 B 2 M S U V M 1 S 3 2 U V 4

21. Then
Case analysis
Some FFD bins can not occur at the same time
For details please see the manuscript
In the other case 2/11 < X ≤ 1/5 the proof is similar, but there are much more optimal, and FFD bins.
(optimal, FFD bins)

22. Also thanks to Gouchuan Zhang, Leah Epstein and David Johnson, for their valuable help and comments, and THANK YOU! for your attention.