# CS 484. Discrete Optimization Problems A discrete optimization problem can be expressed as (S, f) S is the set of all feasible solutions f is the cost.

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CS 484

Discrete Optimization Problems A discrete optimization problem can be expressed as (S, f) S is the set of all feasible solutions f is the cost function Goal: Find a feasible solution x opt such that f(x opt ) <= f(x) for all x in S

Discrete Optimization Problems Examples VLSI layout Robot motion planning Test pattern generation In most problems, S is very large S can be converted to a state-space graph and then can be reformulated to a search problem.

Discrete Optimization Problems NP-hard Why parallelize? Consider real-time problems robot motion planning speech understanding task scheduling Faster search through bigger search spaces.

Search Algorithms Depth First Search Breadth First Search Best First Search Branch and Bound Use cost to determine expansion Iterative Deepening A* Use cost + heuristic value to determine expansion

Parallel Depth First Search Critical issue is distribution of search space. Static partitioning of unstructured trees leads to poor load balancing.

Dynamic Load Balancing Consider sequential DFS

Parallel DFS Each processor performs DFS on a disjoint section of the tree. (Static load assignment) After the processor finishes, it requests unsearched portions of the tree from other processors. Unexplored sections are stored in the stack Pop off a section from the stack and give it to somebody else.

Parallel DFS Problems Splitting up the work How much work should you give to another processor? Determining a donor processor Who do you request more work from?

Work Splitting Strategies When splitting up a stack, consider Sending too little or too much increases work requests Ideally, rather than splitting the stack, you would split the search space. HARD Nodes high in tree --> big subtrees, & vice-versa

Work Splitting Strategies To avoid sending small amounts of work, nodes beyond a specified stack depth are not sent. Cut-off depth Strategies Send only nodes near bottom of stack Send nodes near cut-off depth Send 1/2 of nodes between bottom and cut-off

Load Balancing Schemes (Who do I request work from?) Asynchronous Round Robin each processor maintains target Ask from target then increment target Global Round Robin target is maintained by master node Random Polling randomly select a donor each processor has equal probability

Speedups of DFS

Best-First Search Heuristic is used to direct the search Maintains 2 lists Open  Nodes unsearched  Sorted by heuristic value Closed  Expanded nodes Memory requirement is linear in the size of the search space explored.

Parallel Best-First Search Concurrent processors pick the most promising node from the open list Newly generated nodes are placed back on the open list Centralized Strategy

Expand the node to generate successors Expand the node to generate successors Expand the node to generate successors at designated processor Global list maintained best node nodes Put expanded Get current Pick the best node from the list Place generated nodes in the list Pick the best node from the list Place generated nodes in the list Unlock the list Pick the best node from the list Place generated nodes in the list Unlock the list Lock the list

Centralized Best-First Search Termination condition A processor may find a solution but not the best solution. Modify the termination criteria (how?) Centralization leads to congestion Open list must be locked when accessed Extra work

Decentralizing Best-First Search Let each processor maintain its own open list Issues: Load balancing Termination (make sure it is the best)

Communication Strategies Random Periodically send some of the best nodes to a random processor Ring Periodically exchange best nodes with neighbors Blackboard Select best node from open list If l-value is OK then expand If l-value is BAD then get some from blackboard If l-value is GREAT then give some to blackboard

Ring Communication

Blackboard

What about searching a graph? Problem: node replication Possible solution: Assign each node to a processor Use hash function Whenever a node is generated, check to see if it already has been searched Costly

Speedup Anomalies Due to nature of the problem, speedup can vary greatly from one execution to the next. Two anomaly types: Acceleration Deceleration

Termination Detection Dijkstra's Token Termination Detection When idle, send idle token to next processor When idle token is received again, all done Tree-Based Termination Detection Associate a weight of 1 with initial work load Assign portions of the weight When finished give the weight portion back When processor 0 has weight of 1 --> all done.

Dijkstra’s Token Termination All processes are either active or inactive. inactive processes may not send messages other than the token active processes may turn inactive inactive processes may turn active if they receive a work message Termination can only occur if all processes are inactive We must determine if all processes are inactive and if there are no more messages in the system.

Dijkstra’s Token Termination Arrange the processes logically in a ring Since all processes must be inactive to terminate, designate process P0 as the process that can start termination detection When inactive, P0 sends a token traveling from process i to i + 1 The token only leaves a process if the process is inactive problem: an inactive process may receive a message and turn active after the token has already left. solution: introduce colors

Dijkstra’s Token Termination All processes are initially colored white. Any process i that sends a message to a process j such that j < i is suspect for reactivating a process: change that processes color to black If a black process receives a token, it colors the token black. If process 0 receives a white token, send poison pill

Dijkstra’s Token Termination 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 Active Inactive Token work

Dijkstra’s Token Termination Problem: Fast token and slow work Suppose process j sends work to process i < j Suppose the work message takes a long time to get there In the mean time, process j gets the token and changes it to black and sends it on. It then changes its state to white. P0 gets the black token and starts the process again Process i now receives the new white token before receiving the work message that is still in transit Process i passes on the white token Process j will also pass on a white token since it changed its state to white after sending on the black token The new white token will now arrive at P0 signaling termination  poison pill sent out

Dijkstra’s Token Termination Problem: Fast token and slow work Suppose process i sends work to process i + 4 Suppose the work message takes a long time to get there In the mean time, process i becomes idle. Process i now receives a white token Process i passes on the white token Process i + 4 receives the white token before the work message Process i + 4 will also pass on a white token The white token will now arrive at P0 signaling termination  poison pill sent out

Dijkstra’s Token Termination Solution: Message Counts Send message counts along with the token Initially, all processes are white and have a message count of 0 Whenever a process receives a message, it decrements its count and increments its count if it sends a message  sum of message counts will be zero iff all messages have been delivered Token sums message counts as it is passed.

Dijkstra’s Token Termination If P0 becomes inactive, it turns white and sends a white token with its message count to process 1 If a process sends or receives a message, it turns black Process i keeps the token as long as it is active. If it turns inactive: if process i is black, change token to black. Otherwise token color is unchanged add message count to the token forward the token change state to white

Dijkstra’s Token Termination If P0 receives a black token, try again. If P0 receives a white token Token has passed through only white processes However, a message may be in flight  token’s message count will be non-zero If message count is zero, send poison pill Otherwise try again

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