Average Completion Time on Multiple Machines

P ||PCj

j pjA 1

B 3

C 4

D 5

E 6

F 9

G 12

H 20

I 50

J 60

What is the right algorithm?

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Average Completion Time on Multiple Machines

P ||PCj

j pjA 1

B 3

C 4

D 5

E 6

F 9

G 12

H 20

I 50

J 60

What is the right algorithm?

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Average Completion Time on Multiple Machines

• P ||PCJ – SPT is optimal.

• P ||PwjCj – Is WSPT optimal?

Example

j wj pj1 1 1

2 1 1

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Average Completion Time on Multiple Machines

• P ||PCJ – SPT is optimal.

• P ||PwjCj – Is WSPT optimal?

Example

j wj pj1 1 1

2 1 1

3 100 99

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Average Completion Time on Multiple Machines

• P ||PCJ – SPT is optimal.

• P ||PwjCj – Is WSPT optimal?

Example

j wj pj1 1 1

2 1 1

3 100 99

• P ||PwjCj is NP-complete.

• WSPT is a (1 +p2)/2 -approximation for P ||PwjCj

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R|| PCj

• Can be solved as a matching problem.

• Left side node for each job j

• Right hand side node for the k th from last job on machine i

Example

J1 J2 J3 J4M1 6 4 1 3

M2 7 5 2 3

M3 3 8 5 3

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R|| PCj

• Can be solved as a matching problem.

• Left side node for each job j

• Right hand side node for the k th from last job on machine i

Example

J1 J2 J3 J4M1 6 4 1 3

M2 7 5 2 3

M3 3 8 5 3

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R|| PCj

• Can be solved as a matching problem.

• Left side node for each job j

• Right hand side node for the k th from last job on machine i

Example

J1 J2 J3 J4M1 6 4 1 3

M2 7 5 2 3

M3 3 8 5 3

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LP for the matching problem

Variable xijk = 1 if j is the k th from last job on Mi

minmX

i=1

nX

j=1

nX

k=1kpijxijk

s.t.

Each job runsmX

i=1

nX

k=1xijk = 1 j = 1 . . . n

Each machine/slot has at most 1 jobnX

j=1xijk 1 i = 1 . . .m; k = 1 . . . n

xijk 2 {0, 1} i = 1 . . .m; j = 1 . . . n; k = 1 . . . n

• Note that the may be unforced idleness e.g.

J1 J2M1 1 1

M2 10 10 it ③

Q|pmtn| PCj

• Algorithm is SRPT-FM. Shortest Remaining Processing Time on the

Fastest Machines.

• What about preemption in other models?

• P – doesn’t help

• R – NP-complete

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