1. By Christos Giavroudis Dissertation submitted in partial fulfilment for the degree of Master of Science in Communication & Information Systems Department of Informatics & Communications TEI of Central Macedonia

2.  Suppose you need to place small objects in large containers with fixed size in order to have the minimum possible space, to use the fewest containers and the whole process can be done as soon as possible.  It is an optimization problem and belongs to combinatorial NP-hard problem

3.  There is a list of numbers called “weights”  These numbers represent objects that need to be packed into “bins” with a particular capacity  The goal is to pack the weights into the smallest number of bins possible

4. They have many applications, such as:  Objects coming down a conveyor belt need to be packed for shipping  A construction plan calls for small boards of various lengths, and you need to know how many long boards to order  Tour groups of various sizes need to be assigned to busses so that the groups are not split up  Placing computer files with specified sizes into memory blocks of fixed size.  the recording of a composer’s music where the length of the piece to be recorded are the weight and the bin capacity is the amount of time that can be stored on audio CD ( 80 minutes) and so on.

5.  There are many variations of this problem such as 1-D, 2-D, 3-D, linear programming packing by weight, packing by cost and so on. Bin Packing Problem, because of the high diversity, encloses many area of our lives.

6.  In literature there have been developed many heuristics algorithms to solve the Bin Packing Problem. Briefly, we mention that there are two major categories: the classified methods and the unclassified methods.  The classified are Next Fit (NF), First Fit (FF), Best Fit (BF) and the classified are Next Fit Decreasing (NFD), First Fit Decreasing (FFD), Best Fit Decreasing (BFD). Select to be presented in this thesis Best Fit Decreasing (BFD).

7.  Pack the weights 5, 7, 3, 5, 6, 2, 4, 4, 7, and 4 into bins with capacity 10

8.  There are many possible solutions

9.  Pack the weights 5, 7, 3, 5, 6, 2, 4, 4, 7, and 4 into bins with capacity 10  There are many possible solutions

10.  Pack the weights 5, 7, 3, 5, 6, 2, 4, 4, 7, and 4 into bins with capacity 10  There are many possible solutions

11.  We saw a solution with 5 bins  Is that the best possible solution?  If we add up all the weights, we get 5+7+3+5+6+2+4+4+7+4 = 47  Lower bound is 47/10 =4.7 where 10 is the capacity and 47 is the sum of all weights.  So, the best we can hope for is 5 bins.

12.  One heuristic method for packing weights is to look for these “best fits”  Consider all of the bins and find the bin that can hold the weight and would have the least room leftover after packing it.  One-at-a time algorithm  We have to decide what to do with each weight, in order, before moving on to the next one

13.  Let’s start over and use the best fit decreasing algorithm  Sort the list of weights from the biggest to the smallest

14.  This time we need to keep track of how much room is left in each bin. When we consider a weight, we look at all the bins that have room for it…

15.  …and put it into the bin that will have the least room left over  In this case, we only have one bin, so the 7 goes in there.

16.  For the next weight, we don’t have a bin that has room for it, so we make a new bin 3

17.  So, we make a new bin

18.  None of our bins have room for our next weight  So we make a new bin

19.  None of our bins have room for the next weight  So we make a new bin. 5

20.  Bin #4 has the least room left over, so that’s where we put our next weight

21.  Bin #3 has the least room left, so that’s where we put our next weight

22.  The next weight doesn’t fit into any of our bins, so we need to make a fifth bin

23.  The next weight fit into Bin#5  The next weight can go into either Bin#1or Bin#2.

24.  The last weight fits exactly into Bin#5.  The lower bound is:7+7+3+4+6+5+5+4+4+2=47/10=4.7  We got the best solution. This is called an optimal solution.

25.  If the list of weights is very long, or if the bin capacity is very large, this can be impractical.  The weights are tasks that need to be completed  The bins are “processors,” which are the agents (people, machines, teams, etc.) that will actually perform the tasks

26.  Let’s start over and use the best decreasing fit algorithm in parallel implementation and using two cores with the same example.  Each core uses a bin.  The data set is divided into two parts with a cyclic data partition.  Each piece of data is mapped into a core (bin).

27.  We have the same example  Sort the list of weights from the biggest to the smallest.

28.  This time we need to separate the data set into two parts.

29.  When we consider a weight, we look at all bins that have room for it, and put it into the bin that will have the least room left.

30.  Now, we look the next weight separate of our data parts. None of our bins have room for the next weight.

31.  So we make a new bin both of data partition.

32.  Now, we put it into new bins per core that will have the least room left.

33.  we examine next weights separate and simultaneous.

34.  We put them into Bin#3 and Bin#4 respectively because these bins will have the least room left.

35.  After that, we examine again next weights separate and simultaneous.

36.  So we make new bins and put them into.

37.  The last weights have row. Bin #1 has the least room left over for the first part, so that’s where we put our weight. Alike, Bin#2 has the least room left over for the second part, so that’s where we put our weight

38.  We have the final result below

39.  This is a different result from the best fit decreasing algorithm. This is called also an optimal solution. The lower bound for the first part is 7+5+5+4+3=24/10=2.4. Alike for the second part the lower bound is 7+6+4+4+2=23/10=2.3.  We notice that the parallel implementation of algorithm use one more bin than serial, but it is executed in less steps.

40.  Will the serial execution of algorithm or the parallelization of algorithm be beneficial to us? From the previous examples, we can consider that:  For small datasets is preferable to use the serial algorithm.  For huge datasets is preferable to use the parallel algorithm.

41.  But, you should run the algorithm serial and then in parallel on different datasets.  In parallel execution, we will execute each dataset to varying number of processors in order to compare our results.  Only then, we say whether parallel processing is a beneficial or not.

42.  Parallel computing is the use of multiple processors to execute different parts of the same program concurrently.  A parallel computer is a collection of processing elements, that can solve big problems quickly by means of well coordinated collaboration.

43.  The implementation and the execution of algorithms made in the laboratory "Parallel and Distributed Processing" at the premises of the department of Computer Engineering of the Technological educational institution of Central Macedonia.  For the run of Algorithm computers were used (total 32 Processors), each processor has a 2.4 GHz. We used the software package MATLAB (version 6.2).

44. To run applications in MATLAB Distributed Computing Server must be done the following steps:  Installation of the MDCE  Run of MDCE  Set up of the network with the admincenter

45.  First, we set the current directory as \MATLAB\R2010a\toolbox\distcomp\bin  Afterwards, we run the in command window of MATLAB the following command !mdce install  Finally, we enable the MDCE in command window of MATLAB with the following command !mdce start

46.  We run the command admincenter outside of the MATLAB environment.  This commands occurs in directory  \MATLAB\R2010a\toolbox\distcomp\bin

47.  We select the menu Add or Find to define the Hosts of the network.

48.  We create a Jobmanager by the menu Start in Jobmanager part, which is called “BinPackingMan”. In this JobManager, we set up eight hostnames with 4 cores per hostname (PC).

49.  Close the admincenter window and we set by the menu parallel the Jobmanager.

50.  Parallel Computing Toolbox™ lets us solve computationally and data-intensive problems using multicore processors, GPUs, and computer clusters.  Totally, we had 32 cores. Then, we could run and executed Best Fit Decreasing algorithm parallel. We took the following useful conclusions on the execution time, the total wastage, the total number of bins used, the speed up and the efficiency on different size of datasets.

51. function bfd(numberOfPackages,binSize,totalBins,package,totalWastage) tic i=1; while i<=numberOfPackages currentPackage=0; while(currentPackage<binSize && i<=numberOfPackages) min=300; position=0; %position of minimun wastage for j=totalBins:-1:1 if wastage(j)-package(i)>=0 && wastage(j)-package(i)<min min=wastage(j)-package(i); position=j; end a=binSize-(currentPackage+package(i)); if a>=0 && a<min position=0; end if position~=0 && wastage(position)~=0 wastage(position)=wastage(position)-package(i); i=i+1; else if currentPackage+package(i)<=binSize currentPackage=currentPackage+package(i); i=i+1; else break end totalBins=totalBins+1; wastage(totalBins)=binSize-currentPackage; end

52. for i=1:totalBins totalWastage=totalWastage+wastage(i); end timerValue=toc; str=sprintf('Total wastage: %d', totalWastage); disp(str) str=sprintf('Total bins used: %d', totalBins); disp(str) str=sprintf('Total time elapsed: %f', timerValue); disp(str) str=sprintf('Average values: %f', mean(package)); disp(str) str=sprintf('Standard deviation: %f\n', std(package)); disp(str) end

53. clear; clc; binSize=100; % stable size of bin for i=[2^10 2^12 2^14 2^16] numberOfPackages=i; % number of package for packing packarray=randi(binSize- 1,numberOfPackages,1); %table of packages random size package=sort(packarray); %sort data table str=sprintf('dataset_%d',i); save(str); end

54. function a=data_partition(labs,package) n=length(package); for i=1:labs for j=1:n/labs k=j+1; a(i,j)=package(labs*(k-1- mod(i,labs)); end

55. function c=calc_wastage(binSize,data) for j=1:length(data)/labs k=j+1; min=300; c(i,j)=min+sub_package(labs*(k-1)- mod( i,labs)); end

56. str=sprintf('-Best Fit Decreasing Algorithm-\n------------------\n'); disp(str); str1=sprintf('dataset_%d',2^10); load(str1); totalBins=0; %total number bins used totalWastage=0; %total wastage bfd(numberOfPackages,binSize,totalBins,package,totalWastage) n=length(package); for tloop=1:10 for labs=[1,2,4,8,16,32] matlabpool('open', 'local', labs); tic totalBins=0; %total number bins used totalWastage=0; %total wastage sub_package=data_partition(labs,package); str=sprintf('----Distributed Best Fit Decreasing Alogorithm for clones:---%d',labs); disp(str) part=n/labs;

57. spmd(labs) for p=1:labs temp=sub_package(p,:); if (labindex==p) bfd(part,100,0,temp,0); end tt(tloop,labs)=toc; matlabpool('close'); end for labs=[1,2,4,8,16,32] at(labs)=(sum(tt(:,labs))-max(tt(:,labs))-min(tt(:,labs)))/8 end save results_2_10

58. The command matlabpool enables the parallel language features in the MATLAB language by starting a parallel job that connects this MATLAB client with a number of labs. matlabpool('open', 'local', labs); starts a worker pool using the local parallel configuration and the number of the available labs is labs. matlabpool('close'); stops the worker pool, destroys the parallel job, and makes all parallel language features revert to using the MATLAB client for computing their results.

59. Total wastage Datasetserialcore 1core 2core 4core 8core 16 2^10 dataset14560 14760146601496015160 2^12 dataset57980 58080 58580 2^14 dataset231264 231364231464231864231964 2^16 dataset933434 933534933634936934934434

61. Total used Bins Datasetserialcore 1core 2core 4core 8core 16 2^10 dataset665 667666669671 2^12 dataset2634 2635 2640 2^14 dataset10478 10479104801048410485 2^16 dataset42004 4200542006 42014

63. Total execution time Datasetcore 1core 2core 4core 8core 16 core 32 2^10 dataset0,10820,0930,1050,12380,15840,2307 2^12 dataset0,28830,14180,12570,14530,19530,2763 2^14 dataset3,10,860560,33770,2570,34590,5208 2^16 dataset47,755912,12523,4051,24791,05731,5077

65. Thank you!