Greedy and Local Search Advanced Algorithms Nanjing University, - PowerPoint PPT Presentation

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 1 Since every 𝑓 ∈ 𝑁 must consume a subset to cover 𝑇 1 ∀𝐷, ∀𝑁: 𝑁 ≤ |𝐷| 𝑓 2 As a result, ∀𝑁: 𝑁 ≤ OPT primal = min |𝐷| 𝑇 2 𝑓 3 𝑇 3 𝑓 4 𝑇 4 𝑓 5

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 1 Since every 𝑓 ∈ 𝑁 must consume a subset to cover 𝑇 1 ∀𝐷, ∀𝑁: 𝑁 ≤ |𝐷| 𝑓 2 As a result, ∀𝑁: 𝑁 ≤ OPT primal = min |𝐷| 𝑇 2 𝑓 3 𝑇 3 GreedyMatchingCover ： 𝑓 4 Find arbitrary maximal 𝑁 for the dual problem. 𝑇 4 Return 𝐷 = {𝑗: 𝑇 𝑗 ∩ 𝑁 ≠ ∅} . 𝑓 5

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 1 Since every 𝑓 ∈ 𝑁 must consume a subset to cover 𝑇 1 ∀𝐷, ∀𝑁: 𝑁 ≤ |𝐷| 𝑓 2 As a result, ∀𝑁: 𝑁 ≤ OPT primal = min |𝐷| 𝑇 2 𝑓 3 𝑇 3 GreedyMatchingCover ： 𝑓 4 Find arbitrary maximal 𝑁 for the dual problem. 𝑇 4 Return 𝐷 = {𝑗: 𝑇 𝑗 ∩ 𝑁 ≠ ∅} . 𝑓 5 Since 𝑁 is maximal, returned 𝐷 must be a cover.

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 1 Since every 𝑓 ∈ 𝑁 must consume a subset to cover 𝑇 1 ∀𝐷, ∀𝑁: 𝑁 ≤ |𝐷| 𝑓 2 As a result, ∀𝑁: 𝑁 ≤ OPT primal = min |𝐷| 𝑇 2 𝑓 3 𝑇 3 GreedyMatchingCover ： 𝑓 4 Find arbitrary maximal 𝑁 for the dual problem. 𝑇 4 Return 𝐷 = {𝑗: 𝑇 𝑗 ∩ 𝑁 ≠ ∅} . 𝑓 5 Since 𝑁 is maximal, returned 𝐷 must be a cover. 𝐷 ≤ 𝑔 𝐽 ⋅ 𝑁 ≤ 𝑔 𝐽 ⋅ OPT primal

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 1 Since every 𝑓 ∈ 𝑁 must consume a subset to cover 𝑇 1 ∀𝐷, ∀𝑁: 𝑁 ≤ |𝐷| 𝑓 2 As a result, ∀𝑁: 𝑁 ≤ OPT primal = min |𝐷| 𝑇 2 𝑓 3 𝑇 3 GreedyMatchingCover ： 𝑓 4 Find arbitrary maximal 𝑁 for the dual problem. 𝑇 4 Return 𝐷 = {𝑗: 𝑇 𝑗 ∩ 𝑁 ≠ ∅} . 𝑓 5 Since 𝑁 is maximal, returned 𝐷 must be a cover. 𝐷 ≤ 𝑔 𝐽 ⋅ 𝑁 ≤ 𝑔 𝐽 ⋅ OPT primal GreedyMatchingCover has approximation ratio 𝑔 𝐽 .

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Set Cover: Find smallest 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . What if the frequency of each element is exactly 2?

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Set Cover: Find smallest 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . What if the frequency of each element is exactly 2? 𝑓 1 𝑓 1 𝑤 1 𝑤 1 𝑤 2 𝑓 2 𝑓 3 𝑓 2 𝑤 2 𝑤 3 incidence graph 𝑓 3 𝑤 3 𝑓 4 𝑓 4 𝑤 4 𝑤 4

Vertex Cover Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Set Cover: Find smallest 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . What if the frequency of each element is exactly 2? 𝑓 1 𝑓 1 𝑤 1 𝑤 1 𝑤 2 𝑓 2 𝑓 3 𝑓 2 𝑤 2 𝑤 3 incidence graph 𝑓 3 𝑤 3 𝑓 4 𝑓 4 𝑤 4 𝑤 4 Instance: An undirected simple graph 𝐻 = (𝑊, 𝐹) . Vertex Cover: Find smallest 𝐷 ⊆ 𝑊 s.t. ∀𝑓 ∈ 𝐹: 𝑓 ∩ 𝐷 ≠ ∅ .

Vertex Cover Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Set Cover: Find smallest 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . What if the frequency of each element is exactly 2? 𝑓 1 𝑓 1 𝑤 1 𝑤 1 𝑤 2 𝑓 2 𝑓 3 𝑓 2 𝑤 2 𝑤 3 incidence graph 𝑓 3 𝑤 3 𝑓 4 𝑓 4 𝑤 4 𝑤 4 Instance: An undirected simple graph 𝐻 = (𝑊, 𝐹) . Vertex Cover: Find smallest 𝐷 ⊆ 𝑊 s.t. ∀𝑓 ∈ 𝐹: 𝑓 ∩ 𝐷 ≠ ∅ . • Vertex cover is also NP-hard. • Decision version is one of Karp’s 21 NP -complete problems.

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . 𝑓 1 𝑤 1 Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 2 𝑤 2 The frequency of each element is exactly 2 𝑓 3 𝑤 3 Instance: An undirected simple graph 𝐻 = (𝑊, 𝐹) . 𝑓 4 𝑤 4 Primal: Find 𝐷 ⊆ 𝑊 s.t. ∀𝑓 ∈ 𝐹: 𝑓 ∩ 𝐷 ≠ ∅ . (Vertex Cover) Dual: Find 𝑁 ⊆ 𝐹 s.t. ∀𝑤 ∈ 𝑊: 𝑤 ∩ 𝑁 ≤ 1 . (Matching)

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . 𝑓 1 𝑤 1 Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 2 𝑤 2 The frequency of each element is exactly 2 𝑓 3 𝑤 3 Instance: An undirected simple graph 𝐻 = (𝑊, 𝐹) . 𝑓 4 𝑤 4 Primal: Find 𝐷 ⊆ 𝑊 s.t. ∀𝑓 ∈ 𝐹: 𝑓 ∩ 𝐷 ≠ ∅ . (Vertex Cover) Dual: Find 𝑁 ⊆ 𝐹 s.t. ∀𝑤 ∈ 𝑊: 𝑤 ∩ 𝑁 ≤ 1 . (Matching) A 2-approximation algorithm for the vertex cover problem GreedyMatchingCover ： Find arbitrary maximal matching 𝑁 of the input graph. Return 𝐷 = {𝑤: 𝑤 ∈ 𝑊 and 𝑤 ∩ 𝑁 ≠ ∅} .

Instance: A collection of subsets 𝑇 1 , 𝑇 2 , ⋯ , 𝑇 𝑛 ⊆ 𝑉 . Primal: Find 𝐷 ⊆ [𝑛] such that ڂ 𝑗∈𝐷 𝑇 𝑗 = 𝑉 . 𝑓 1 𝑤 1 Dual: Find 𝑁 ⊆ 𝑉 such that 𝑇 𝑗 ∩ 𝑁 ≤ 1 for all 𝑗 ∈ [𝑛] . 𝑓 2 𝑤 2 The frequency of each element is exactly 2 𝑓 3 𝑤 3 Instance: An undirected simple graph 𝐻 = (𝑊, 𝐹) . 𝑓 4 𝑤 4 Primal: Find 𝐷 ⊆ 𝑊 s.t. ∀𝑓 ∈ 𝐹: 𝑓 ∩ 𝐷 ≠ ∅ . (Vertex Cover) Dual: Find 𝑁 ⊆ 𝐹 s.t. ∀𝑤 ∈ 𝑊: 𝑤 ∩ 𝑁 ≤ 1 . (Matching) A 2-approximation algorithm for the vertex cover problem GreedyMatchingCover ： Find arbitrary maximal matching 𝑁 of the input graph. Return 𝐷 = {𝑤: 𝑤 ∈ 𝑊 and 𝑤 ∩ 𝑁 ≠ ∅} . • There is no poly-time <1.36-approx. alg. unless P = NP. • Assuming the unique game conjecture, there is no poly-time (2- ε )-approx. alg.

Scheduling 𝑛 identical machines

Scheduling 𝑛 identical machines 𝑜 jobs processing time 𝑞 𝑘 3 1 4 2 6 3 5 2 4 3

Scheduling 𝑛 machines 𝑜 jobs each with processing time 𝑞 𝑘

Scheduling 𝑛 machines 𝑜 jobs each with processing time 𝑞 𝑘 Completion time: 𝐷 𝑗 = ෍ 𝑞 𝑘 (of machine 𝑗 ) 𝑘: jobs assigned to machine 𝑗

Scheduling 𝑛 machines 𝑜 jobs each with processing time 𝑞 𝑘 makespan Completion time: 𝐷 𝑗 = ෍ 𝑞 𝑘 (of machine 𝑗 ) 𝑘: jobs assigned to machine 𝑗 𝐷 max = max 𝐷 𝑗 Makespan: 𝑗

Instance: 𝑜 jobs 𝑘 = 1,2, ⋯ , 𝑜 each with processing time 𝑞 𝑘 ∈ ℤ + . Problem: Find a schedule assigning 𝑜 jobs to 𝑛 identical machines so as the minimize the makespan.

Instance: 𝑜 jobs 𝑘 = 1,2, ⋯ , 𝑜 each with processing time 𝑞 𝑘 ∈ ℤ + . Problem: Find a schedule assigning 𝑜 jobs to 𝑛 identical machines so as the minimize the makespan. • “minimum makespan on identical machines” • Scheduling problem has many variations: machines could be different, jobs could have release- dates/deadlines, etc…

Instance: 𝑜 jobs 𝑘 = 1,2, ⋯ , 𝑜 each with processing time 𝑞 𝑘 ∈ ℤ + . Problem: Find a schedule assigning 𝑜 jobs to 𝑛 identical machines so as the minimize the makespan. • “minimum makespan on identical machines” • Scheduling problem has many variations: machines could be different, jobs could have release- dates/deadlines, etc… If 𝑛 = 2 , the scheduling problem can be used to solve the partition problem! Instance: 𝑜 positive integers 𝑦 1 , 𝑦 2 , ⋯ , 𝑦 𝑜 ∈ ℤ + . Problem: Determine whether there exists a partition of {1,2, ⋯ , 𝑜} into two sets 𝐵 and 𝐶 such that σ 𝑗∈𝐵 𝑦 𝑗 = σ 𝑗∈𝐶 𝑦 𝑗 .

Instance: 𝑜 jobs 𝑘 = 1,2, ⋯ , 𝑜 each with processing time 𝑞 𝑘 ∈ ℤ + . Problem: Find a schedule assigning 𝑜 jobs to 𝑛 identical machines so as the minimize the makespan. • “minimum makespan on identical machines” • Scheduling problem has many variations: machines could be different, jobs could have release- dates/deadlines, etc… If 𝑛 = 2 , the scheduling problem can be used to solve the partition problem! Instance: 𝑜 positive integers 𝑦 1 , 𝑦 2 , ⋯ , 𝑦 𝑜 ∈ ℤ + . Problem: Determine whether there exists a partition of {1,2, ⋯ , 𝑜} into two sets 𝐵 and 𝐶 such that σ 𝑗∈𝐵 𝑦 𝑗 = σ 𝑗∈𝐶 𝑦 𝑗 . • The partition problem is one of Karp’s 21 NP -complete problems. • Thus the considered scheduling problem is NP-hard.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Graham’s Li List Algorithm for Scheduling 𝑛 identical machines 𝑜 jobs each with processing time 𝑞 𝑘 List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time.

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio?

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio? OPT ≥ max 𝑞 𝑘 𝑘

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in the schedule, and last job on it is 𝑚 .

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in the schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in the schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in the schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 1 1 𝑛 σ 𝑘≠𝑚 𝑞 𝑘 ≤ 𝑛 σ 𝑘 𝑞 𝑘 ≤ OPT 𝐷 𝑙 − 𝑞 𝑚 ≤ 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘 since machine 𝑙 is least loaded when scheduling job 𝑚

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. What about the approximation ratio? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in the schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 ≤ 2 ⋅ OPT 1 1 𝑛 σ 𝑘≠𝑚 𝑞 𝑘 ≤ 𝑛 σ 𝑘 𝑞 𝑘 ≤ OPT 𝐷 𝑙 − 𝑞 𝑚 ≤ 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘 since machine 𝑙 is least loaded when scheduling job 𝑚

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. Algorithm List finishes within poly-time. Algorithm List has approximation ratio 2. Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 ≤ 2 ⋅ OPT 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 ≤ 1 𝑛 ෍ 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘 𝑘≠𝑚 𝑘

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. Algorithm List finishes within poly-time. Algorithm List has approximation ratio 2. Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 ≤ 2 ⋅ OPT 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 ≤ 1 𝑛 ෍ 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘 𝑘≠𝑚 𝑘 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 = 1 𝑞 𝑘 − 𝑞 𝑚 𝑛 ෍ 𝑛 ෍ 𝑛 𝑘≠𝑚 𝑘

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. Algorithm List finishes within poly-time. Algorithm List has approximation ratio 2. 2 − 1 ≤ ⋅ OPT Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 ≤ 2 ⋅ OPT 𝑛 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 ≤ 1 𝑛 ෍ 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘 𝑘≠𝑚 𝑘 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 = 1 𝑞 𝑘 − 𝑞 𝑚 𝑛 ෍ 𝑛 ෍ 𝑛 𝑘≠𝑚 𝑘

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. Algorithm List finishes within poly-time. Algorithm List has approximation ratio 2. 2 − 1 ≤ ⋅ OPT Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 ≤ 2 ⋅ OPT 𝑛 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 ≤ 1 𝑛 ෍ 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘 𝑘≠𝑚 𝑘 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 = 1 𝑞 𝑘 − 𝑞 𝑚 𝑛 ෍ 𝑛 ෍ 𝑛 𝑘≠𝑚 𝑘 Algorithm List has approximation ratio 2 − Τ 1 𝑛 .

List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. Algorithm List finishes within poly-time. Algorithm List has approximation ratio 2. 2 − 1 ≤ ⋅ OPT Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 ≤ 2 ⋅ OPT 𝑛 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 ≤ 1 𝑛 ෍ 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘 𝑘≠𝑚 𝑘 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 = 1 𝑞 𝑘 − 𝑞 𝑚 𝑛 ෍ 𝑛 ෍ 𝑛 𝑘≠𝑚 𝑘 Algorithm List has approximation ratio 2 − Τ 1 𝑛 . This bound is tight in the worst case. [Almost tight example: 𝑛 2 unit jobs followed by a length 𝑛 job. List generates makespan of 2𝑛 while OPT = 𝑛 + 1 .]

Local Search for Scheduling Start with an arbitrary solution: Keep making improvements by locally adjusting the solution, until no further improvement can be made ( local optimum )

Local Search for Scheduling Start with an arbitrary solution: Keep making improvements by locally adjusting the solution, until no further improvement can be made ( local optimum ) LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 .

Local Search for Scheduling Start with an arbitrary solution: Keep making improvements by locally adjusting the solution, until no further improvement can be made ( local optimum ) LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 .

Local Search for Scheduling Start with an arbitrary solution: Keep making improvements by locally adjusting the solution, until no further improvement can be made ( local optimum ) LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 .

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 .

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!)

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!) The approximation ratio of this algorithm?

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!) The approximation ratio of this algorithm? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!) The approximation ratio of this algorithm? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 .

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!) The approximation ratio of this algorithm? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!) The approximation ratio of this algorithm? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑘

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!) The approximation ratio of this algorithm? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 = 1 𝑞 𝑘 − 𝑞 𝑚 𝑛 ෍ 𝑛 ෍ 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑛 𝑘 𝑘≠𝑚 𝑘

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . This algorithm finishes within poly-time. (No job is transferred twice!) (2 − Τ 1 𝑛) The approximation ratio of this algorithm? OPT ≥ max 𝑞 𝑘 𝑛 ⋅ OPT ≥ ෍ 𝑞 𝑘 𝑘 𝑘 Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . 2 − 1 Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 ≤ ⋅ OPT 𝑛 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑞 𝑘 = 1 𝑞 𝑘 − 𝑞 𝑚 𝑛 ෍ 𝑛 ෍ 𝑞 𝑚 ≤ max 𝑞 𝑘 ≤ OPT 𝑛 𝑘 𝑘≠𝑚 𝑘

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . LocalSearch finds a schedule with makespan 𝐷 max ≤ 2 − 1 𝑛 ⋅ OPT List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . LocalSearch finds a schedule with makespan 𝐷 max ≤ 2 − 1 𝑛 ⋅ OPT List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. The schedule returned by List must be a local optimum!

LocalSearch ： Start with an arbitrary schedule. Repeat until no job can be reassigned (i.e., local optimum reached): Let 𝑚 be a job that finished last. If exists machine 𝑗 s.t. assigning job 𝑚 to 𝑗 allows 𝑚 finish earlier: Transfer job 𝑚 to earliest such 𝑗 . LocalSearch finds a schedule with makespan 𝐷 max ≤ 2 − 1 𝑛 ⋅ OPT List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. The schedule returned by List must be a local optimum! List will find a schedule with makespan 2 − 1 𝐷 max ≤ ⋅ OPT 𝑛

𝑛 identical machines 𝑜 jobs List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

𝑛 identical machines 𝑜 jobs List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

Longest Processing Time (L (LPT) 𝑛 identical machines 𝑜 jobs List (Graham 1966) ： For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine.

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm?

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑘

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑘

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑘 W.l.o.g.: • # of jobs > # of machines (i.e., 𝑜 > 𝑛 ) • makespan is achieved by some job bigger than 𝑛 (i.e., 𝑚 > 𝑛 ) Otherwise, LPT returns an optimal solution already!

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑘 W.l.o.g.: • # of jobs > # of machines (i.e., 𝑜 > 𝑛 ) 𝑞 𝑛 + 𝑞 𝑛+1 ≤ OPT • makespan is achieved by some job bigger than 𝑛 (i.e., 𝑚 > 𝑛 ) Otherwise, LPT returns an optimal solution already!

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑛 ෍ 𝑞 𝑘 ≤ OPT 𝑘 W.l.o.g.: • # of jobs > # of machines (i.e., 𝑜 > 𝑛 ) 𝑞 𝑛 + 𝑞 𝑛+1 ≤ OPT • makespan is achieved by some job bigger than 𝑛 (i.e., 𝑚 > 𝑛 ) 𝑞 𝑚 ≤ 𝑞 𝑛+1 Otherwise, LPT returns an optimal solution already!

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 𝑞 𝑚 ≤ 𝑞 𝑛+1 ≤ 1 2 𝑞 𝑛 + 𝑞 𝑛+1 ≤ OPT 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑛 ෍ 𝑞 𝑘 ≤ OPT 2 𝑘 W.l.o.g.: • # of jobs > # of machines (i.e., 𝑜 > 𝑛 ) 𝑞 𝑛 + 𝑞 𝑛+1 ≤ OPT • makespan is achieved by some job bigger than 𝑛 (i.e., 𝑚 > 𝑛 ) 𝑞 𝑚 ≤ 𝑞 𝑛+1 Otherwise, LPT returns an optimal solution already!

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. This algorithm finishes within poly-time. The approximation ratio of this algorithm? Assume machine 𝑙 finishes last in final schedule, and last job on it is 𝑚 . ≤ 3 Makespan 𝐷 max = 𝐷 𝑙 = 𝐷 𝑙 − 𝑞 𝑚 + 𝑞 𝑚 2 ⋅ OPT 𝑞 𝑚 ≤ 𝑞 𝑛+1 ≤ 1 2 𝑞 𝑛 + 𝑞 𝑛+1 ≤ OPT 𝐷 𝑙 − 𝑞 𝑚 ≤ 1 𝑛 ෍ 𝑞 𝑘 ≤ OPT 2 𝑘 W.l.o.g.: • # of jobs > # of machines (i.e., 𝑜 > 𝑛 ) 𝑞 𝑛 + 𝑞 𝑛+1 ≤ OPT • makespan is achieved by some job bigger than 𝑛 (i.e., 𝑚 > 𝑛 ) 𝑞 𝑚 ≤ 𝑞 𝑛+1 Otherwise, LPT returns an optimal solution already!

LongestProcessingTime (LPT) ： Sort jobs so that 𝑞 1 ≥ 𝑞 2 ≥ ⋯ ≥ 𝑞 𝑜 . For each job 𝑘 = 1,2, ⋯ , 𝑜 do: Assign job 𝑘 to a currently least loaded machine. • We have shown LPT has approximation ratio (at most) Τ 3 2 . • By a more careful analysis, it can be shown LPT is actually a Τ 4 3 approximation algorithm. • The problem of “ minimum makespan on identical machines ” has a PTAS ( P olynomial T ime A pproximation S cheme). ∀𝜗 > 0, ∃ poly-time (1 + 𝜗) -approx. alg. for the problem

Online Scheduling 𝑛 identical machines Jobs arrive (revealed) one-by-one Schedule decision must be made once a job arrives, without seeing jobs in the future.

Online Scheduling 𝑛 identical machines Jobs arrive (revealed) one-by-one Schedule decision must be made once a job arrives, without seeing jobs in the future.

Online Scheduling 𝑛 identical machines Jobs arrive (revealed) one-by-one Schedule decision must be made once a job arrives, without seeing jobs in the future.

Online Scheduling 𝑛 identical machines Jobs arrive (revealed) one-by-one Schedule decision must be made once a job arrives, without seeing jobs in the future.