zak100 Posted April 23, 2020 Posted April 23, 2020 (edited) Hi, I am having problem in understanding the following text of the book: Quote The input consists of 6m items of size 1/7+ Absolon, followed by 6m items of size 1/3+ Absolon , followed by 6m items of size 1/2 + Absolon. One simple packing places one item of each size in a bin and requires 6m bins. If m = 1 then its possible to have 6 bins. All1/7 elements would go into 1 bin. All 6*1/3 items would go into 2 bins All 6 *1/2 items would go into 3 bins. Total = 6 bins It further says: Quote First fit requires 10m bins. Why First Fit would require 10 bins. I think First Fit would also require 6 bins. Somebody please guide me why First Fit would need 10 bins? Zulfi. Edited April 23, 2020 by zak100
zak100 Posted April 25, 2020 Author Posted April 25, 2020 Hi, I mean greek symbol like E. You can consider 'a' instead of absolon. Zulfi
Ghideon Posted April 25, 2020 Posted April 25, 2020 (edited) On 4/24/2020 at 1:07 AM, zak100 said: All 6*1/3 items would go into 2 bins But items are of size: On 4/24/2020 at 1:07 AM, zak100 said: 6m items of size 1/3+ Absolon They are larger than 1/3. So 6 of them does not fit in two bins. You need 3. Same for the 1/2+a size items, they need 6 bins 1+4+6=10 Edited April 25, 2020 by Ghideon 1
Strange Posted April 25, 2020 Posted April 25, 2020 6 hours ago, zak100 said: Hi, I mean greek symbol like E. You can consider 'a' instead of absolon. Zulfi Epsilon? [math]\frac{1}{7} + \epsilon[/math] 1
taeto Posted April 25, 2020 Posted April 25, 2020 Usually it is called \(\varepsilon.\) Whatever it is called, I suppose it is a positive number. In which case it is correct what Ghideon points out.
zak100 Posted April 25, 2020 Author Posted April 25, 2020 Hi my friend Ghideon, Quote 1+4+6=10 What you are saying is right but then we have 11m bins required not 10 m as the book says. And how the "simple packing" i.e. optimum requires 6m bins? Zulfi.
Strange Posted April 25, 2020 Posted April 25, 2020 (edited) 1 hour ago, zak100 said: What you are saying is right but then we have 11m bins required not 10 m as the book says. 1 + 4 + 6 = 10 1 + 3 + 6 = 10 1 hour ago, zak100 said: And how the "simple packing" i.e. optimum requires 6m bins? Because: On 4/24/2020 at 1:07 AM, zak100 said: One simple packing places one item of each size in a bin and requires 6m bins. Edited April 25, 2020 by Strange Stupid copy 1
Ghideon Posted April 25, 2020 Posted April 25, 2020 14 hours ago, Ghideon said: 1+4+6=10 I guess I can't type and/or count. 14 hours ago, Ghideon said: You need 3. 1 + 3 + 6 = 10 Sorry for confusing the discussion.
Strange Posted April 25, 2020 Posted April 25, 2020 2 minutes ago, Ghideon said: I guess I can't type and/or count. Doh. Same here. I worked it out myself, came up with 1 + 3 + 6 but then quoted your wrong numbers without noticing!
zak100 Posted April 27, 2020 Author Posted April 27, 2020 (edited) Hi my friends Ghideon and Strange, Thanks a lot. Your efforts helped me to understand the problem specially Strange's figure and Ghideon's calculation. I did my calculation: a =0.00795 == 1/7+0.00795=0.15075*6 = 0.9045 (1 bin) ½ +0.00795=0.50795= (6 bins) 1/3+.00795 = 0.3333+.00795= 0.34125 (2 can fit in one bin) and 6 will require 3 bin. Therefore total is 10 bins. 'a' is not perfect but its almost perfect. Its now clear to me. God bless you guys. Zulfi. Edited April 27, 2020 by zak100
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