Parallel Traveling Salesman PhD Student: Viet Anh Trinh Advisor: - - PowerPoint PPT Presentation

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Parallel Traveling Salesman PhD Student: Viet Anh Trinh Advisor: Professor Feng Gu

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Agenda

• 1. Traveling salesman introduction
• 2. Genetic Algorithm for TSP
• 3. Tree Search for TSP

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Travelling Salesman

• Set of N cities
• Find the shortest closed non-looping

path that covers all the cities

• No city be visited more than once

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Travelling Salesman First Parallel Approach: Genetic Algorithm

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Travelling Salesman -Sequential Genetic Algorithm

Initialization 0123, 0231,1320, 0321 Fitness Evaluation (0123) = 1/(1 + 2 + 10 + 7) = 0.050 (0231) = 1/(3 + 10 + 4 + 5) =0.045 (1320) =1/(6 + 12 + 1 + 1) = 0.050 (0321) = 1/(8 + 12 + 18 + 5) = 0.023 Selection Cross-over Mutation Termination

Sequential GA Travelling Salesman

• Individuals
• Closed non-looping paths across all cities
• Initial

Population

• Set of randomly generated paths
• Evaluation
• Assess the fitness of the individual. Fitness is 1/

total distance of a given path

• Selection
• Select the fittest individuals ( biggest fitness,

smallest distance)

• Offspring

production

• Cross-over + mutation

Selection

Roulette Wheel Selection: P(choice – (0123)) = 0.05/(0.05+ 0.045 + 0.05 + 0.023) = 0.3 P(choice – (0231) = 0.27 P(choice – (1320)) = 0.3 P(choice – (0321)) = 0.13 If random number fall in: 0 ≤ 𝑠 < 0.3 choose 0123 0.3 ≤ 𝑠 < 0.57 choose 0231 0.57 ≤ 𝑠 < 0.87 choose 1320 0.87 ≤ 𝑠 < 1 choose 0321

1st path 2nd path 3rd path 4th path

Specialized Crossover Operator

Order Crossover(OX): No invalid part Normal Crossover: Invalid path appear

Mutation

• Select 2 random point and swap
• Ensure the valid path

Sequential Genetic Algorithm

Parallel Genetic Algorithm

Master: Initialization 0123, 0231,1320, 0321 Fitness Evaluation (1320) =1/(6 + 12 + 1 + 1) = 0.050 (0321) = 1/(8 + 12 + 18 + 5) = 0.023 Selection Cross-over Mutation Termination Fitness Evaluation (0123) = 1/(1 + 2 + 10 + 7) = 0.050 (0231) = 1/(3 + 10 + 4 + 5) =0.045 Selection Cross-over Mutation Termination Slave 1 Slave 2

Parallel Travelling Salesman - Master

• Master
• Initializes population
• Sends path to slaves
• Examine the best paths from slaves’ return results
• Slave
• Signals the master that it is ready for work
• Waits for paths to be sent by the master until a termination

message is received

• Evaluates the paths fitness
• Selection
• Crossover
• Mutation
• Sends the best c paths to nearby neighbors after k generations
• When finish, send best paths and their lengths to master

Time Complexity Sequential

Time Complexity: n :population size l : length of a path, number of cities g : number of generation Sequential Genetic Algorithm: Initialization : O(n) Evaluation: O(nl) Selection: O(nl) Crossover: c1x O(nl) Mutation: c2 x O(nl) Time: O(nl) + gO(nl) =O(gnl)

Time Complexity Parallel

• Isolated subpopulations
• Stepping model model: only send best individuals to neighbor

processor

• Communication time

Master send data to slave using scatter: tcomm1= O(nl/p) Slave send best c paths to neighbor processor after k generations: tcomm2= g/kO(cl) = g/kO(l) Slave send their c best paths and their length value to Master: tcomm3= O(cl)

• Computation time

Master Initialization : tcomp1 =O(n) Slave evaluation, selection, crossover, mutation: tcomp2 = O(gnl/p) Master final evaluation : tcomp3 =O(pc)

• Parallel time : tp = O(gnl/p)
• Speed up = ts/tp = p
• Efficiency = ts/ptp =1

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Travelling Salesman Second Parallel Approach: Tree Search

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Travelling Salesman

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Travelling Salesman – Tree Search

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Travelling Salesman Sequential

Algorithm

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Travelling Salesman Sequential

Algorithm

• City count: examines the partial tour if there are n

cities on the partial tour.

• Best tour: check if the complete tour has a lower

cost than “best tour”

• Update best tour: replace the current best tour

with this tour

• Feasible: checks to see if the city or vertex has

already been visited.

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Travelling Salesman Sequential

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Travelling Salesman Sequential

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Travelling Salesman Sequential

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Travelling Salesman Parallel

Static load balancing (picture) à Imbalance load Solution à Dynamic load balancing

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Travelling Salesman Parallel

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Travelling Salesman Parallel

Terminologies

• Donor process: the process that sends work
• Recipient process: the process that requests/receives work
• Half-split: ideally, the stack is split into two equal pieces such

that the search space of each stack is the same

• Cutoff depth: to avoid sending very small amounts of work,

nodes beyond a specified stack depth are not given away

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Travelling Salesman Parallel

Some possible strategies

• 1. Send nodes near the bottom of the stack
• Works well with uniform search space; has low splitting cost
• 2. Send nodes near the cutoff depth
• Performs better with a strong heuristic (tries to distribute the

parts of the search space likely to contain a solution)

• 3. Send half the nodes between the bottom and the cutoff

depth

• Works well with uniform and irregular search space

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Travelling Salesman Parallel

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Travelling Salesman Parallel

The entire space is assigned to master

• When slave runs out of work, it gets more work from

another slave using work requests and responses

• Unexplored states can be conveniently stored as local stacks

at processors.

• Slave terminate when reaching final state

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Travelling Salesman Parallel

• Load balancing scheme: Random polling (RP)

When a processor becomes idle, it randomly selects a donor. Each processor is selected as a donor with equal probability, ensuring that work requests are evenly distributed.

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Travelling Salesman Parallel

• Let W be serial work and pWp be parallel work.
• Search overhead factor s is defined as pWP/W
• Quantify total overhead To in terms of W to compute

scalability. § To = pWp – W

• Upper bound on speed up is p×1/s.

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Travelling Salesman Parallel

Assumption:

• Search overhead factor = one
• Work at any processor can be partitioned into independent pieces

as long as its size exceeds a threshold ε.

• A reasonable work-splitting mechanism is available.

§ If work w at a processor is split into two parts ψw and (1–ψ)w, there exists an arbitrarily small constant α (0 < α ≤ 0.5),such that ψw > αw and (1–ψ)w > αw. § The constant α sets a lower bound on the load imbalance from work splitting.

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Travelling Salesman Parallel

• If processor Pi initially had work wi, after a single request by processor Pj

and split, neither Pi nor Pj have more than (1–α)wi work.

• For each load balancing strategy, we define V(P) as the total number of

work requests after which each processor receives at least one work request (note that V(p) ≥ p).

• Assume that the largest piece of work at any point is W.
• After V(p) requests, the maximum work remaining at any processor is less

than (1–α)W; after 2V(p) requests, it is less than (1–α)2W; …

• After (log1/1(1- α )(W/ε))V(p) requests, the maximum work remaining at any

processor is below a threshold value ε.

• The total number of work requests is O(V(p) log W).

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Travelling Salesman Parallel

• If tcomm is the time required to communicate a piece of work, then

the communication overhead TO is TO = tcommV(p)log W

• The corresponding efficiency E is given by:

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Travelling Salesman Parallel

• Random Polling

§ Worst case V(p) is unbounded. § We do average case analysis.

• Let F(i,p) represent a state in which i of the processors have been

requested, and p–i have not.

• Let f(i,p) denote the average number of trials needed to change from

state F(i,p) to F(p,p) (V(p) = f(0,p)).

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Travelling Salesman Parallel

• We have

As p becomes large, Hp ≃ 1.69 ln p. Thus, V(p) = O(p log p). To = O(p log p log W) Therefore W = O(p log2p).

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END OF PRESENTATION

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