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Posted (edited)

Hi,

I am having problem in understanding the following text of the book:

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 by zak100
Posted (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 by Ghideon
Posted
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]

Posted

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.

Posted

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.

Posted (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.

Untitled.png.1ee007abf38e6ba24c2aad1577c0f2a0.png

Edited by Strange
Stupid copy
Posted
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.

 

Posted
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!

Posted (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.

 

image.thumb.png.a0ffcc15ba9bbd2235c0e270f281c03b.png

Edited by zak100

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