Improved results for a Improved results for a memory allocation memory allocation problemproblem

Rob van SteeUniversity of KarlsruheGermany

Leah EpsteinLeah EpsteinUniversity of HaifaUniversity of Haifa

IsraelIsrael

WADS 2007WAOA 2007

BackgroundBackground Allocate memory to parallel Allocate memory to parallel

processorsprocessors Goals:Goals:

– Each processor has sufficient Each processor has sufficient memorymemory

– Not too much memory is wastedNot too much memory is wasted

BackgroundBackground ProblemProblem: memory requirement : memory requirement

may vary over timemay vary over time If each processor can access only If each processor can access only

one memoryone memory, this is inefficient, this is inefficient If If allall processors share a single processors share a single

memory, there is heavy memory, there is heavy contentioncontention

Suggested modelSuggested model Chung et al. [SPAA 2004] suggest Chung et al. [SPAA 2004] suggest

a new architecturea new architecture Each memory may be accessed Each memory may be accessed

by at most by at most twotwo processors processors This avoids the mentioned This avoids the mentioned

problemsproblems

Bin packingBin packing They abstract this problem as a They abstract this problem as a

bin packing problembin packing problem Bins = memoriesBins = memories Items = memory requirements of Items = memory requirements of

processorsprocessors Thus, items may be larger than 1Thus, items may be larger than 1

Bin packingBin packing Items may be split and distributed Items may be split and distributed

among more than one binamong more than one bin Condition: each bin contains at most Condition: each bin contains at most

twotwo parts of items parts of items Corresponds to: each memory may be Corresponds to: each memory may be

accessed by at most two processorsaccessed by at most two processors We also consider the model where We also consider the model where

each bin may contain each bin may contain kk parts of items parts of items

Bin packingBin packing

Processor12345

Requirement

Bins

Assignment1,23,42

3,4,51

NotesNotes There may be many bins which There may be many bins which

are not fullare not full An item may be split more than An item may be split more than

onceonce No upper bound on how often No upper bound on how often

each item is spliteach item is split

ResultsResults Chung et al. [SPAA 2004]:Chung et al. [SPAA 2004]:

– (k=2) NP-hardness(k=2) NP-hardness– (k=2) 3/2-approximation(k=2) 3/2-approximation

Epstein & vS [WADS 2007]:Epstein & vS [WADS 2007]: – (k=2) 7/5-approximation(k=2) 7/5-approximation– For general k:For general k:

Next Fit is (2-1/k)-approximationNext Fit is (2-1/k)-approximation NP-hardness in the strong senseNP-hardness in the strong sense

ResultsResults

Epstein & vS [WAOA 2007]:Epstein & vS [WAOA 2007]: A polynomial time approximation A polynomial time approximation

scheme (PTAS) for every (constant) scheme (PTAS) for every (constant) kk– For every constant For every constant >0, a polynomial >0, a polynomial

time algorithm that uses at most 1+ time algorithm that uses at most 1+ times the minimal number of bins times the minimal number of bins

– at most at most OPT additional bins of size 1OPT additional bins of size 1

ResultsResults

Epstein & vS [WAOA 2007]:Epstein & vS [WAOA 2007]: A dual approximation scheme for A dual approximation scheme for

every (constant) kevery (constant) k– For every constant For every constant >0, a polynomial >0, a polynomial

time algorithm that uses time algorithm that uses bins of size bins of size 1+ 1+

– Larger bins, but no additional binsLarger bins, but no additional bins

Next Fit Next Fit in this contextin this context If current bin contains If current bin contains kk parts of parts of

items items oror is fulll is fulll, open a new bin, open a new bin Put as much as possible of the Put as much as possible of the

current item in the current bincurrent item in the current bin Open additional bin(s) if needed Open additional bin(s) if needed

(for large items)(for large items)

Next Fit: lower boundNext Fit: lower bound

One item of size One item of size Mk-Mk-11– Next Fit uses Next Fit uses Mk-Mk-11 bins bins

M M ((k-k-1)1)kk very small items very small items– Next Fit puts Next Fit puts kk items in one bin, items in one bin,

uses uses M M ((k-k-1)1) bins bins

Mk -1 M (k -1)

Next Fit: lower boundNext Fit: lower bound

Optimal solution needs only Optimal solution needs only MkMk bins bins Ratio: (Ratio: (MM (2 (2kk-1)-1) /-1)-1) /MkMk 2 – 1/2 – 1/kk

Mk -1 M (k -1)

Mk

Next Fit: upper boundNext Fit: upper bound BlockBlock = set of full bins, followed = set of full bins, followed

by a bin with k itemsby a bin with k items Each bin >1 in a block was Each bin >1 in a block was

opened because the previous one opened because the previous one was fullwas full

WeightsWeights WeightWeight of item with size s = of item with size s = ss/k/k There are at least There are at least ss parts of item parts of item

with size s in any packingwith size s in any packing

OPT is at least the OPT is at least the total weight of total weight of all the itemsall the items

1 ii

i i

sOPT s

k k

Next Fit: upper boundNext Fit: upper bound Consider last bin in a block (Consider last bin in a block (not the not the

last blocklast block)) Has Has kk items, so at least items, so at least k k -1-1 items items

of weight of weight 1/1/kk (these items are not (these items are not split!)split!)

Block has 1 bin, then Block has 1 bin, then kk such items such items Else, consider all items in block Else, consider all items in block

except except the the k k -1 last items-1 last items

Next Fit: upper boundNext Fit: upper bound SS = all items in block except = all items in block except the the k k

-1 last items-1 last items How many items are in How many items are in S S ? ?

Unknown!Unknown! What is the minimal What is the minimal weight weight of of S S ?? Weight = , minimized for Weight = , minimized for

singlesingle item item

isk

Next Fit: upper boundNext Fit: upper bound SS = all items in block except = all items in block except the the k k -1 -1

last items last items For block of size For block of size bb, total size of items , total size of items

in S is strictly more than in S is strictly more than b b -1-1 Total weight is then at least Total weight is then at least b b //kk Weight in block at least Weight in block at least bb / /k k +(+(k k --

1)/1)/kk Note: also true for block of size 1Note: also true for block of size 1

The last blockThe last block Only one bin: weight at least 1/Only one bin: weight at least 1/kk

(has at least one item)(has at least one item) b b >1 bins: weight at least >1 bins: weight at least bb / /kk Let size of block Let size of block ii be be bbi i , then Next , then Next

Fit uses binsFit uses bins We know that OPT is at least the We know that OPT is at least the

total weighttotal weight

ib

Next Fit: upper boundNext Fit: upper bound Let m be the Let m be the amount of blocksamount of blocks

Also, OPT > NF – Also, OPT > NF – m m (in each block, at (in each block, at most one bin is most one bin is not completely fullnot completely full))

kkmNFk

kkbwOPT

i

m

ii

)1)(1(

)1()1(1

Next Fit: upper boundNext Fit: upper bound

Combining this we findCombining this we find

kkm

kkm

kk

kNFOPT

kk

kkm

kkNFOPT

kk

mNFOPT

1)1(1)1(1112

1)1(111

1,)1)(1(

mNFOPT

kkmNFOPT

2323

This talkThis talk We only discuss the algorithms We only discuss the algorithms

for k=2for k=2 General propertiesGeneral properties PTASPTAS Dual PTASDual PTAS

Representation of Representation of solutionssolutions A solution can be seen as a graphA solution can be seen as a graph Vertices = itemsVertices = items An edge between two items = An edge between two items = Parts of these two items share a binParts of these two items share a bin

– A loop means that a bin contains just A loop means that a bin contains just one part of an itemone part of an item

We can assume a forest (with loops)We can assume a forest (with loops)– Cycles can be easily removed without Cycles can be easily removed without

changing the cost (Chung et al.,2004).changing the cost (Chung et al.,2004).

ExampleExample

Items:

GraphGraph

Removal of cycleRemoval of cycle

Additional propertiesAdditional properties There exists an optimal packing forest There exists an optimal packing forest

where all items of size in (0,1/2] are where all items of size in (0,1/2] are leaves or singletonsleaves or singletons

If the forest representation is minimal If the forest representation is minimal (no tree can be split) then a packing is (no tree can be split) then a packing is implied from the treeimplied from the tree– Can be done iterativelyCan be done iteratively– Singletons are packed into separate binsSingletons are packed into separate bins– A leaf is packed with a maximum amount of A leaf is packed with a maximum amount of

its neighbor, and removed from the forestits neighbor, and removed from the forest

(Different) example(Different) example

1.5 0.4

0.21

0.9

0.2

0.3

Packing processPacking process

0.80.40.6

1 0.9

0.30.1

0.2

PTAS for k=2PTAS for k=2 We would like to use standard We would like to use standard

methods such asmethods such as– Linear grouping and roundingLinear grouping and rounding– Enumeration of patternsEnumeration of patterns

ButBut– Patterns are treesPatterns are trees– Items sizes are not bounded from Items sizes are not bounded from

aboveabove

New tricksNew tricks Optimal packing is adapted in order Optimal packing is adapted in order

to make it possible to imitate itto make it possible to imitate it Trees are “truncated” Trees are “truncated”

– Partitioned into smaller treesPartitioned into smaller trees Large items are truncatedLarge items are truncated

– Partitioned into several items that are Partitioned into several items that are still large, but boundedstill large, but bounded

The last two adaptations increase The last two adaptations increase the number of bins by a factor of the number of bins by a factor of 1+1+

Trying all treesTrying all trees Type of a tree is (j,E)Type of a tree is (j,E)

– j is number of edgesj is number of edges– E is set of j-1 edges (so that it is a tree)E is set of j-1 edges (so that it is a tree)

Pattern = type plus vector of length jPattern = type plus vector of length j– Element i in the vector indicates Element i in the vector indicates groupgroup

of node iof node i ConstantConstant number of patterns number of patterns Only question: how many of each?Only question: how many of each?

Dual PTAS for k=2 Dual PTAS for k=2 (briefly)(briefly) We use an integer parameter K, We use an integer parameter K,

based on based on .. Items can be rounded down to a Items can be rounded down to a

sizes of the form i/Ksizes of the form i/K We round some non-zero sizes We round some non-zero sizes

into zero!into zero! By scaling, we get items of By scaling, we get items of

integer sizesinteger sizes, and a bin size , and a bin size KK

CuttingCutting There is no reason to cut items at There is no reason to cut items at

non-integer point non-integer point For each item size (at most K), we For each item size (at most K), we

define cutting patternsdefine cutting patterns We “guess” how to cut the input We “guess” how to cut the input

items items – Using enumerationUsing enumeration

Finding a solutionFinding a solution We use a layered graph to find an We use a layered graph to find an

optimal packing for each optionoptimal packing for each option– If the cutting was applied, the problem If the cutting was applied, the problem

becomes much easierbecomes much easier– However, we did not cut items larger than However, we did not cut items larger than

the bin sizethe bin size– We cut a piece off each one of them, but We cut a piece off each one of them, but

do not decide on all cutting pointsdo not decide on all cutting points The best (smallest number of bins) The best (smallest number of bins)

option is chosenoption is chosen Increasing items back to original sizes Increasing items back to original sizes

increases bins slightlyincreases bins slightly

Open problemsOpen problems PTAS, dual PTAS for non-constant kPTAS, dual PTAS for non-constant k (FPTAS or dual FPTAS are most (FPTAS or dual FPTAS are most

likely impossible since the problem likely impossible since the problem is strongly NP-hard for every is strongly NP-hard for every constant value of k)constant value of k)

Generalize 7/5-approximation for Generalize 7/5-approximation for larger larger kk

7/5-approximation7/5-approximation We show an approximation We show an approximation

algorithm for k=2 (at most two algorithm for k=2 (at most two parts of items per bin allowed)parts of items per bin allowed)

First it sorts the itemsFirst it sorts the items Then it packs them, starting with Then it packs them, starting with

the most difficult ones: size the most difficult ones: size between ½ and 1between ½ and 1

We use We use Next FitNext Fit as a subroutine as a subroutine

Step 1: SortingStep 1: Sorting Items of size at most ½ are Items of size at most ½ are smallsmall

– Sort them in order of Sort them in order of increasingincreasing size size Items larger than 1 are Items larger than 1 are largelarge

– Sort them in order of Sort them in order of decreasingdecreasing size size Other items are Other items are medium-sizedmedium-sized

– Sort them in order of Sort them in order of decreasingdecreasing size size– We pack these items firstWe pack these items first

Step 2: medium-sized Step 2: medium-sized itemsitems Pack Pack medium-sizedmedium-sized items items

one by one:one by one:– If current item fits in one bin If current item fits in one bin

with the smallest unpacked with the smallest unpacked small item, pack them small item, pack them togethertogether

– Else, pack current item Else, pack current item together with two together with two largestlargest unpacked small items in two unpacked small items in two binsbins

Step 3: all small items Step 3: all small items are packedare packed We do this until we run out of We do this until we run out of

either the medium-sized or the either the medium-sized or the small itemssmall items

If all small items are packed, use If all small items are packed, use Next Fit on rest, start with Next Fit on rest, start with medium-sized itemsmedium-sized items

Step 4: some small Step 4: some small items remainitems remain Pack each small item in Pack each small item in its own its own

binbin Pack large items into these bins Pack large items into these bins

using Next Fitusing Next Fit Still use ordering from before:Still use ordering from before:

– Largest large items firstLargest large items first– Smallest small items firstSmallest small items first

Step 5: small items leftStep 5: small items left We have now run out of either the We have now run out of either the

large or the small itemslarge or the small items If bins with small items remain, If bins with small items remain,

repackrepack these items two per bin these items two per bin

Step 6: large items leftStep 6: large items left Pack remaining large items using Pack remaining large items using

Next FitNext Fit

Upper boundUpper bound We use the same weight definition as We use the same weight definition as

before: weight of item with size s isbefore: weight of item with size s is

We know that OPT is at least the We know that OPT is at least the total weighttotal weight

How much weight do we pack in a How much weight do we pack in a bin? bin?

2i is sk

Amount of weight Amount of weight packedpackedStep 2:Step 2: Pack medium-sized items one by one:Pack medium-sized items one by one:

– If current item fits in one bin with the If current item fits in one bin with the smallest unpacked small item, pack them smallest unpacked small item, pack them together. together. Weight ½+½ = Weight ½+½ = 11

– Else, pack current item together with two Else, pack current item together with two largest unpacked small items in two bins. largest unpacked small items in two bins. Three items of weight ½ in two bins: Three items of weight ½ in two bins: ¾¾

Amount of weight Amount of weight packedpackedStep 3:Step 3: If all small items are packed, use Next If all small items are packed, use Next

Fit on rest, start with medium-sized Fit on rest, start with medium-sized itemsitems

We assume We assume manymany small items small items existexist

Then, this step Then, this step does not occurdoes not occur

Amount of weight Amount of weight packedpackedStep 4:Step 4: Pack each small item in Pack each small item in its own binits own bin Pack large items into these bins using Pack large items into these bins using

Next FitNext Fit Still use ordering from before:Still use ordering from before:

– Largest large items firstLargest large items first– Smallest small items firstSmallest small items first

Amount of weight Amount of weight packedpacked Consider a large item which is Consider a large item which is

packed into packed into gg bins bins It is packed together with It is packed together with gg small small

items (total weight items (total weight g g /2)/2) The large item has size strictly The large item has size strictly

more than (more than (g g -1)/2, so weight at -1)/2, so weight at least least g g /4/4

Average weight per bin = Average weight per bin = 3/43/4

Amount of weight Amount of weight packedpackedStep 5:Step 5: We have now run out of either the We have now run out of either the

large or the small itemslarge or the small items If bins with small items remain, If bins with small items remain,

repack these items two per binrepack these items two per bin

Weight: ½ + ½ = Weight: ½ + ½ = 11

Packed weight: Packed weight: summarysummary Step 2: ½+½=1, or 3 times ½ in Step 2: ½+½=1, or 3 times ½ in

two bins, average ¾two bins, average ¾ Step 4: average ¾Step 4: average ¾ Step 5: ½+½=1Step 5: ½+½=1

Gives upper bound of 4/3! (Since Gives upper bound of 4/3! (Since OPT is at least total weight) OPT is at least total weight)

However…However… There may be only one small item There may be only one small item

left at the end of Step 5left at the end of Step 5– „„If bins with small items remain, If bins with small items remain,

repack these items repack these items two per bintwo per bin““ We may run out of small items in We may run out of small items in

Step 2, so that Step 3 is executedStep 2, so that Step 3 is executed We may run out of small items in We may run out of small items in

Step 4, so that Step 6 is executedStep 4, so that Step 6 is executed

Upper boundUpper bound It turns out that these cases are It turns out that these cases are

the most difficultthe most difficult Proof is a number of pagesProof is a number of pages Eventually we find an upper Eventually we find an upper

bound of bound of 7/57/5 in all cases in all cases

Open questionsOpen questions Improve this approximation ratioImprove this approximation ratio

– We can give a PTAS, but this is much We can give a PTAS, but this is much less efficient already for epsilon = 2/5less efficient already for epsilon = 2/5

Generalize this algorithm for larger kGeneralize this algorithm for larger k– We have the upper bound of Next Fit, We have the upper bound of Next Fit,

which is 2-1/k, and a PTASwhich is 2-1/k, and a PTAS– Can we improve on Next Fit efficiently?Can we improve on Next Fit efficiently?

Thank you!Thank you!