MCS 312: NP Completeness and Approximation algorthms

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MCS 312 NP Completeness and Approximation
algorthms

• Instructor
• Neelima Gupta
• ngupta_at_cs.du.ac.in

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Table of Contents

• Factor 2 algorithm for Bin Packing
• Factor 2 algorithm for Minimum Makespan
  Scheduling
• Reference Vazirani.

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Bin Packing Problem Definition

• Given n items of sizes x1 , xn between 0 and
  1, excluding 0, but including 1. Find a packing
  in unit-sized bins that minimizes the number of
  bins used.

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Greedy Approach factor 2

• Consider items in any arbitrary order.
• At any iteration I, we have a set of partially
  filled bins B1 . Bk
• To insert the next element xi , we check if it
  can be put in any of the already used bin. If not
  then we open another bin Bk1

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Analysis 2 factor approximation

• Prove by induction that if at any iteration we
  have m bins then at least m-1 of them are at
  least half full.
• Thus (m-1)/2 lt Sum_i,n xi
• Since bins are of unit length, sum of item sizes
  is a lower bound on OPT. i.e Sum_i,n xi lt
  OPT
• Thus m 1 lt 2 OPT i.e m lt 2OPT

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Minimum Makespan Scheduling

• Given n jobs with processing times p1 . Pn
  and an integer m , find an assignment of the jobs
  to m identical machines so that the completion
  time, called the makespan is minimized.

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Greedy Approach factor 2

• Consider jobs in any arbitrary order.
• Schedule the next job on the machine with the
  minimum makespan so far.

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Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
M1
M2
M3
t0
9
Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
2
M1
J1
M2
M3
t0
10
Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
2
M1
J1
3
M2
J2
M3
t0
11
Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
2
M1
J1
3
M2
J2
8
M3
J3
t0
12
Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
5
2
M1
J1
J4
3
M2
J2
8
M3
J3
t0
13
Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
5
2
M1
J1
J4
3
1
M2
J2
J5
8
M3
J3
t0
14
Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
5
2
M1
J1
J4
3
1
M2
J2
J5
7
8
M3
J3
J6
t0
15
Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
5
2
M1
J1
J4
3
1
M2
J2
J5
7
8
M3
J3
J6
t0
MAKESPAN OF THE SOLUTION IS 15
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Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
5
2
M1
J1
J4
3
1
M2
J2
J5
7
8
M3
J3
J6
t0
MAKESPAN OF THE SOLUTION IS 15
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Machines
Jobs Processing time J1
2 J2 3 J3
8 J4 5 J5
1 J6 7
7
2
M1
J1
J6
3
5
1
M2
J2
J4
J5
8
M3
J3
t0
MAKESPAN OF THE SOLUTION IS 15 BUT
OPT IS 9

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Analysis 2 factor approximation

• Average mount of time for which a machine runs is
  
• (sum_i pi ) /m
• Clearly OPT gt (sum_i pi ) /m
• Proof Suppose ti denotes the completion time
  of the ith machine in OPT. Then
• (sum_i pi ) /m is nothing but (sum_i
  ti ) /m
• Thus, OPT maxti gt (sum_i ti ) /m
  (sum_i pi ) /m
• Also, OPT gt maxpi

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Contd..

• Consider the machine M whose makespan is maximum
  in our solution. Let pl be the processing time of
  the last job jl scheduled on M. Let sl be the
  start time of jl on M.
• Then, makespan of our solution sl pl
• Let j1 .jr be the jobs scheduled just before jl.
• Then sl minTi lt (sum_i j1 .jr Ti )/m
  
• (sum_i j1 .jr pi )/m
• lt (sum_i
  pi ) /m
• Thus sl pl lt (sum_i pi ) /m
  maxpi
• lt 2 OPT.

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Thank You!