Art of Multiprocessor Programming1 Futures, Scheduling, and Work Distribution Companion slides for The Art of Multiprocessor Programming by Maurice Herlihy.
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Art of Multiprocessor Programming
1
Futures, Scheduling, and Work Distribution
Companion slides forThe Art of Multiprocessor Programming
by Maurice Herlihy & Nir Shavit
Modified by Rajeev Alur forCIS 640 University of Pennsylvania
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How to write Parallel Apps?
• How to– split a program into parallel parts– In an effective way– Thread management
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Matrix Multiplication
BAC
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Matrix Multiplication
cij = k=0N-1 aki * bjk
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Matrix Multiplication class Worker extends Thread { int row, col; Worker(int row, int col) { this.row = row; this.col = col; } public void run() { double dotProduct = 0.0; for (int i = 0; i < n; i++) dotProduct += a[row][i] * b[i][col]; c[row][col] = dotProduct; }}}
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Matrix Multiplication class Worker extends Thread { int row, col; Worker(int row, int col) { this.row = row; this.col = col; } public void run() { double dotProduct = 0.0; for (int i = 0; i < n; i++) dotProduct += a[row][i] * b[i][col]; c[row][col] = dotProduct; }}}
a thread
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Matrix Multiplication class Worker extends Thread { int row, col; Worker(int row, int col) { this.row = row; this.col = col; } public void run() { double dotProduct = 0.0; for (int i = 0; i < n; i++) dotProduct += a[row][i] * b[i][col]; c[row][col] = dotProduct; }}}
Which matrix entry to compute
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Matrix Multiplication class Worker extends Thread { int row, col; Worker(int row, int col) { this.row = row; this.col = col; } public void run() { double dotProduct = 0.0; for (int i = 0; i < n; i++) dotProduct += a[row][i] * b[i][col]; c[row][col] = dotProduct; }}}
Actual computation
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Matrix Multiplication void multiply() { Worker[][] worker = new Worker[n][n]; for (int row …) for (int col …) worker[row][col] = new Worker(row,col); for (int row …) for (int col …) worker[row][col].start(); for (int row …) for (int col …) worker[row][col].join();}
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Matrix Multiplication void multiply() { Worker[][] worker = new Worker[n][n]; for (int row …) for (int col …) worker[row][col] = new Worker(row,col); for (int row …) for (int col …) worker[row][col].start(); for (int row …) for (int col …) worker[row][col].join();}
Create n x n threads
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Matrix Multiplication void multiply() { Worker[][] worker = new Worker[n][n]; for (int row …) for (int col …) worker[row][col] = new Worker(row,col); for (int row …) for (int col …) worker[row][col].start(); for (int row …) for (int col …) worker[row][col].join();}
Start them
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Matrix Multiplication void multiply() { Worker[][] worker = new Worker[n][n]; for (int row …) for (int col …) worker[row][col] = new Worker(row,col); for (int row …) for (int col …) worker[row][col].start(); for (int row …) for (int col …) worker[row][col].join();}
Wait for them to finish
Start them
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Matrix Multiplication void multiply() { Worker[][] worker = new Worker[n][n]; for (int row …) for (int col …) worker[row][col] = new Worker(row,col); for (int row …) for (int col …) worker[row][col].start(); for (int row …) for (int col …) worker[row][col].join();}
Wait for them to finish
Start them
What’s wrong with this picture?
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Thread Overhead
• Threads Require resources– Memory for stacks– Setup, teardown
• Scheduler overhead• Worse for short-lived threads
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Thread Pools
• More sensible to keep a pool of long-lived threads
• Threads assigned short-lived tasks– Runs the task– Rejoins pool– Waits for next assignment
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Thread Pool = Abstraction
• Insulate programmer from platform– Big machine, big pool– And vice-versa
• Portable code– Runs well on any platform– No need to mix algorithm/platform
concerns
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ExecutorService Interface
• In java.util.concurrent– Task = Runnable object
• If no result value expected• Calls run() method.
– Task = Callable<T> object• If result value of type T expected• Calls T call() method.
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Future<T>Callable<T> task = …; …
Future<T> future = executor.submit(task);
…
T value = future.get();
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Future<T>Callable<T> task = …; …
Future<T> future = executor.submit(task);
…
T value = future.get();
Submitting a Callable<T> task returns a Future<T> object
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Future<T>Callable<T> task = …; …
Future<T> future = executor.submit(task);
…
T value = future.get();
The Future’s get() method blocks until the value is
available
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Future<?>Runnable task = …; …
Future<?> future = executor.submit(task);
…
future.get();
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Future<?>Runnable task = …; …
Future<?> future = executor.submit(task);
…
future.get();
Submitting a Runnable task returns a Future<?> object
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Future<?>Runnable task = …; …
Future<?> future = executor.submit(task);
…
future.get();
The Future’s get() method blocks until the computation is
complete
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Matrix Addition
00 00 00 00 01 01
10 10 10 10 11 11
C C A B B A
C C A B A B
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Matrix Addition
00 00 00 00 01 01
10 10 10 10 11 11
C C A B B A
C C A B A B
4 parallel additions
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Matrix Addition Taskclass AddTask implements Runnable { Matrix a, b; // multiply this! public void run() { if (a.dim == 1) { c[0][0] = a[0][0] + b[0][0]; // base case } else { (partition a, b into half-size matrices aij and bij) Future<?> f00 = exec.submit(add(a00,b00)); … Future<?> f11 = exec.submit(add(a11,b11)); f00.get(); …; f11.get(); … }}
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Matrix Addition Taskclass AddTask implements Runnable { Matrix a, b; // multiply this! public void run() { if (a.dim == 1) { c[0][0] = a[0][0] + b[0][0]; // base case } else { (partition a, b into half-size matrices aij and bij) Future<?> f00 = exec.submit(add(a00,b00)); … Future<?> f11 = exec.submit(add(a11,b11)); f00.get(); …; f11.get(); … }}
Base case: add directly
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Matrix Addition Taskclass AddTask implements Runnable { Matrix a, b; // multiply this! public void run() { if (a.dim == 1) { c[0][0] = a[0][0] + b[0][0]; // base case } else { (partition a, b into half-size matrices aij and bij) Future<?> f00 = exec.submit(add(a00,b00)); … Future<?> f11 = exec.submit(add(a11,b11)); f00.get(); …; f11.get(); … }}
Constant-time operation
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Matrix Addition Taskclass AddTask implements Runnable { Matrix a, b; // multiply this! public void run() { if (a.dim == 1) { c[0][0] = a[0][0] + b[0][0]; // base case } else { (partition a, b into half-size matrices aij and bij) Future<?> f00 = exec.submit(add(a00,b00)); … Future<?> f11 = exec.submit(add(a11,b11)); f00.get(); …; f11.get(); … }}
Submit 4 tasks
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Matrix Addition Taskclass AddTask implements Runnable { Matrix a, b; // multiply this! public void run() { if (a.dim == 1) { c[0][0] = a[0][0] + b[0][0]; // base case } else { (partition a, b into half-size matrices aij and bij) Future<?> f00 = exec.submit(add(a00,b00)); … Future<?> f11 = exec.submit(add(a11,b11)); f00.get(); …; f11.get(); … }}
Let them finish
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Dependencies
• Matrix example is not typical• Tasks are independent
– Don’t need results of one task …– To complete another
• Often tasks are not independent
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Fibonacci
1 if n = 0 or 1F(n)
F(n-1) + F(n-2) otherwise
• Note– potential parallelism– Dependencies
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Disclaimer
• This Fibonacci implementation is– Egregiously inefficient
• So don’t deploy it!
– But illustrates our point• How to deal with dependencies
• Exercise:– Make this implementation efficient!
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class FibTask implements Callable<Integer> { static ExecutorService exec = Executors.newCachedThreadPool(); int arg; public FibTask(int n) { arg = n; } public Integer call() { if (arg > 2) { Future<Integer> left = exec.submit(new FibTask(arg-1)); Future<Integer> right = exec.submit(new FibTask(arg-2)); return left.get() + right.get(); } else { return 1; }}}
Multithreaded Fibonacci
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class FibTask implements Callable<Integer> { static ExecutorService exec = Executors.newCachedThreadPool(); int arg; public FibTask(int n) { arg = n; } public Integer call() { if (arg > 2) { Future<Integer> left = exec.submit(new FibTask(arg-1)); Future<Integer> right = exec.submit(new FibTask(arg-2)); return left.get() + right.get(); } else { return 1; }}}
Multithreaded Fibonacci
Parallel calls
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class FibTask implements Callable<Integer> { static ExecutorService exec = Executors.newCachedThreadPool(); int arg; public FibTask(int n) { arg = n; } public Integer call() { if (arg > 2) { Future<Integer> left = exec.submit(new FibTask(arg-1)); Future<Integer> right = exec.submit(new FibTask(arg-2)); return left.get() + right.get(); } else { return 1; }}}
Multithreaded Fibonacci
Pick up & combine results
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Dynamic Behavior
• Multithreaded program is– A directed acyclic graph (DAG)– That unfolds dynamically
• Each node is– A single unit of work
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Fib DAGfib(4)
fib(3) fib(2)
submitget
fib(2) fib(1) fib(1)fib(1)
fib(1) fib(1)
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Arrows Reflect Dependenciesfib(4)
fib(3) fib(2)
submitget
fib(2) fib(1) fib(1)fib(1)
fib(1) fib(1)
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How Parallel is That?
• Define work:– Total time on one processor
• Define critical-path length:– Longest dependency path– Can’t beat that!
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Fib Workfib(4)
fib(3) fib(2)
fib(2) fib(1) fib(1)fib(1)
fib(1) fib(1)
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Fib Work
work is 17
1 2 3
84 765 9
1410 131211 15
16 17
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Fib Critical Pathfib(4)
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Fib Critical Pathfib(4)
Critical path length is 8
1 8
2 7
3 64
5
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Notation Watch
• TP = time on P processors
• T1 = work (time on 1 processor)
• T∞ = critical path length (time on ∞ processors)
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Simple Bounds
• TP ≥ T1/P– In one step, can’t do more than P
work
• TP ≥ T∞
– Can’t beat infinite resources
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More Notation Watch
• Speedup on P processors– Ratio T1/TP
– How much faster with P processors
• Linear speedup– T1/TP = Θ(P)
• Max speedup (average parallelism)– T1/T∞
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Matrix Addition
00 00 00 00 01 01
10 10 10 10 11 11
C C A B B A
C C A B A B
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Matrix Addition
00 00 00 00 01 01
10 10 10 10 11 11
C C A B B A
C C A B A B
4 parallel additions
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Addition
• Let AP(n) be running time – For n x n matrix– on P processors
• For example– A1(n) is work
– A∞(n) is critical path length
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Addition
• Work is
A1(n) = 4 A1(n/2) + Θ(1)
4 spawned additions
Partition, synch, etc
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Addition
• Work is
A1(n) = 4 A1(n/2) + Θ(1)
= Θ(n2)
Same as double-loop summation
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Addition
• Critical Path length is
A∞(n) = A∞(n/2) + Θ(1)
spawned additions in parallel
Partition, synch, etc
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Addition
• Critical Path length is
A∞(n) = A∞(n/2) + Θ(1)
= Θ(log n)
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Matrix Multiplication Redux
BAC
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Matrix Multiplication Redux
2221
1211
2221
1211
2221
1211
BB
BB
AA
AA
CC
CC
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First Phase …
2222122121221121
2212121121121111
2221
1211
BABABABA
BABABABA
CC
CC
8 multiplications
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Second Phase …
2222122121221121
2212121121121111
2221
1211
BABABABA
BABABABA
CC
CC
4 additions
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Multiplication
• Work is
M1(n) = 8 M1(n/2) + A1(n)
8 parallel multiplications
Final addition
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Multiplication
• Work is
M1(n) = 8 M1(n/2) + Θ(n2)
= Θ(n3)
Same as serial triple-nested loop
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Multiplication
• Critical path length is
M∞(n) = M∞(n/2) + A∞(n)
Half-size parallel multiplications
Final addition
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Multiplication
• Critical path length is
M∞(n) = M∞(n/2) + A∞(n)
= M∞(n/2) + Θ(log n)
= Θ(log2 n)
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Parallelism
• M1(n)/ M∞(n) = Θ(n3/log2 n)
• To multiply two 1000 x 1000 matrices– 10003/102=107
• Much more than number of processors on any real machine
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Work Distributionzzz…
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Work Dealing
Yes!
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The Problem with Work Dealing
D’oh!
D’oh!
D’oh!
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Work Stealing No work…
zzzYes!
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Lock-Free Work Stealing
• Each thread has a pool of ready work
• Remove work without synchronizing
• If you run out of work, steal someone else’s
• Choose victim at random
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Local Work Pools
Each work pool is a Double-Ended Queue
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Work DEQueue1
pushBottom
popBottom
work
1. Double-Ended Queue
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Obtain Work
•Obtain work•Run task until•Blocks or terminates
popBottom
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New Work
•Unblock node•Spawn node
pushBottom
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Whatcha Gonna do When the Well Runs Dry?
@&%$!!
empty
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Steal Work from OthersPick random thread’s DEQeueue
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Steal this Task!
popTop
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Task DEQueue
• Methods– pushBottom– popBottom– popTop
Never happen concurrently
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Task DEQueue
• Methods– pushBottom– popBottom– popTop
These most common –
make them fast(minimize use
of CAS)
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Ideal
• Wait-Free• Linearizable• Constant time
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Compromise
• Method popTop may fail if– Concurrent popTop succeeds, or a – Concurrent popBottom takes last
work
Blame the victim!
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Dreaded ABA Problem
topCAS
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Dreaded ABA Problem
top
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Dreaded ABA Problem
top
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Dreaded ABA Problem
top
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Dreaded ABA Problem
top
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Dreaded ABA Problem
top
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Dreaded ABA Problem
top
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Dreaded ABA Problem
topCAS
Uh-Oh …
Yes!
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Fix for Dreaded ABA
stamp
top
bottom
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Bounded DEQueuepublic class BDEQueue { AtomicStampedReference<Integer> top; volatile int bottom; Runnable[] tasks; …}
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Bounded DQueuepublic class BDEQueue { AtomicStampedReference<Integer> top; volatile int bottom; Runnable[] tasks; …}
Index & Stamp (synchronized)
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Bounded DEQueuepublic class BDEQueue { AtomicStampedReference<Integer> top; volatile int bottom; Runnable[] deq; …}
Index of bottom taskno need to synchronizeDo need memory barrier
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Bounded DEQueuepublic class BDEQueue { AtomicStampedReference<Integer> top; volatile int bottom; Runnable[] tasks; …}
Array holding tasks
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pushBottom()
public class BDEQueue { … void pushBottom(Runnable r){ tasks[bottom] = r; bottom++; } …}
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pushBottom()
public class BDEQueue { … void pushBottom(Runnable r){ tasks[bottom] = r; bottom++; } …} Bottom is the index to
store the new task in the array
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pushBottom()
public class BDEQueue { … void pushBottom(Runnable r){ tasks[bottom] = r; bottom++; } …}
Adjust the bottom index
stamp
top
bottom
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Steal Work
public Runnable popTop() { int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = oldTop + 1; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom <= oldTop) return null; Runnable r = tasks[oldTop]; if (top.CAS(oldTop, newTop, oldStamp, newStamp))
return r; return null; }
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Steal Work
public Runnable popTop() { int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = oldTop + 1; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom <= oldTop) return null; Runnable r = tasks[oldTop]; if (top.CAS(oldTop, newTop, oldStamp, newStamp))
return r; return null; } Read top (value & stamp)
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Steal Work
public Runnable popTop() { int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = oldTop + 1; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom <= oldTop) return null; Runnable r = tasks[oldTop]; if (top.CAS(oldTop, newTop, oldStamp, newStamp))
return r; return null; } Compute new value & stamp
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Steal Work
public Runnable popTop() { int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = oldTop + 1; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom <= oldTop) return null; Runnable r = tasks[oldTop]; if (top.CAS(oldTop, newTop, oldStamp, newStamp))
return r; return null; }
Quit if queue is empty
stamp
top
bottom
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Steal Work
public Runnable popTop() { int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = oldTop + 1; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom <= oldTop) return null; Runnable r = tasks[oldTop]; if (top.CAS(oldTop, newTop, oldStamp, newStamp))
return r; return null; } Try to steal the task
stamp
topCAS
bottom
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Steal Work
public Runnable popTop() { int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = oldTop + 1; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom <= oldTop) return null; Runnable r = tasks[oldTop]; if (top.CAS(oldTop, newTop, oldStamp, newStamp))
return r; return null; }
Give up if conflict occurs
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;}
Take Work
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;} 103
Take Work
Make sure queue is non-empty
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;}
Take Work
Prepare to grab bottom task
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;}
Take Work
Read top, & prepare new values
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } return null;} 106
Take Work
If top & bottom 1 or more apart, no conflict
stamp
top
bottom
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;} 107
Take Work
At most one item left
stamp
top
bottom
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;} 108
Take WorkTry to steal last task.Always reset bottombecause the DEQueue will be empty even if unsuccessful (why?)
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;}
Take Work
stamp
topCAS
bottom
I win CAS
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;}
Take Work
stamp
topCAS
bottom
If I lose CASThief must have won…
Art of Multiprocessor Programming
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Runnable popBottom() { if (bottom == 0) return null; bottom--; Runnable r = tasks[bottom]; int[] stamp = new int[1]; int oldTop = top.get(stamp), newTop = 0; int oldStamp = stamp[0], newStamp = oldStamp + 1; if (bottom > oldTop) return r; if (bottom == oldTop){ bottom = 0; if (top.CAS(oldTop, newTop, oldStamp, newStamp)) return r; } top.set(newTop,newStamp); return null;}
Take Work
failed to get last task Must still reset top
Art of Multiprocessor Programming
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Variations
• Stealing is expensive– Pay CAS– Only one thread taken
• What if– Randomly balance loads?
Art of Multiprocessor Programming
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Work Stealing & Balancing
• Clean separation between app & scheduling layer
• Works well when number of processors fluctuates.
• Works on “black-box” operating systems
top related