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A Unified Framework for Schedule and Storage Optimization

A Unified Framework for Schedule and Storage Optimization. William Thies, Frédéric Vivien*, Jeffrey Sheldon, and Saman Amarasinghe MIT Laboratory for Computer Science * ICPS/LSIIT, Université Louis Pasteur http://compiler.lcs.mit.edu/aov. Motivating Example.

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A Unified Framework for Schedule and Storage Optimization

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  1. A Unified Framework forSchedule and Storage Optimization William Thies, Frédéric Vivien*, Jeffrey Sheldon, and Saman Amarasinghe MIT Laboratory for Computer Science * ICPS/LSIIT, Université Louis Pasteur http://compiler.lcs.mit.edu/aov

  2. Motivating Example for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  3. Motivating Example t = 1 (i, j) = (1, 1) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  4. Motivating Example t = 2 (i, j) = (1, 2) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  5. Motivating Example t = 3 (i, j) = (1, 3) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  6. Motivating Example t = 4 (i, j) = (1, 4) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  7. Motivating Example t = 5 (i, j) = (1, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  8. Motivating Example t = 6 (i, j) = (2, 1) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  9. Motivating Example t = 7 (i, j) = (2, 2) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  10. Motivating Example t = 8 (i, j) = (2, 3) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  11. Motivating Example t = 9 (i, j) = (2, 4) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  12. Motivating Example t = 10 (i, j) = (2, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  13. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j

  14. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  15. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 1 (i, j) = (1, 1) init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  16. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 2 (i, j) = {(1, 2), (2, 1)} init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  17. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 3 (i, j) = {(1, 3), (2, 2), (3, 1)} init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  18. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 4 (i, j) = {(1, 4), (2, 3), (3, 2), (4, 1)} init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  19. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 5 (i, j) = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  20. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 6 (i, j) = {(2, 5), (3, 4), (4, 3), (5, 2)} init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  21. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 7 (i, j) = {(3, 5), (4, 4), (5, 3)} init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  22. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 8 (i, j) = {(4, 5), (5, 4)} init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  23. Motivating Example t = 25 (i, j) = (5, 5) for i = 1 to n for j = 1 to n A[j] = f(A[j], A[j-2]) j Array Expansion t = 9 (i, j) = (5, 5) init A[0][j] for i = 1 to n for j = 1 to n A[i][j] = f(A[i-1][j], A[i][j-2]) i j

  24. Parallelism/Storage Tradeoff • Increasing storage can enable parallelism • But storage can be expensive • Phase ordering problem • Optimizing for storage restricts parallelism • Maximizing parallelism restricts storage options • Too complex to consider all combinations • Need efficient framework to integrate schedule and storage optimization Cache RAM Disk

  25. Outline • Abstract problem • Simplifications • Concrete problem • Solution Method • Conclusions

  26. Abstract Problem • Given DAG of dependent operations • Must execute producers before consumers • Must store a value until all consumers execute • Two parameters control execution: • 1. A scheduling function  • Maps each operation to execution time • Parallelism is implicit • 2. A fully associative store of size m

  27. Abstract Problem • We can ask three questions: • Two parameters control execution: • 1. A scheduling function  • Maps each operation to execution time • Parallelism is implicit • 2. A fully associative store of size m

  28. Abstract Problem • We can ask three questions: 1. Given , what is the smallest m? • Two parameters control execution: • 1. A scheduling function  • Maps each operation to execution time • Parallelism is implicit • 2. A fully associative store of size m

  29. Abstract Problem • We can ask three questions: 1. Given , what is the smallest m? 2. Given m, what is the “best” ? • Two parameters control execution: • 1. A scheduling function  • Maps each operation to execution time • Parallelism is implicit • 2. A fully associative store of size m

  30. Abstract Problem • We can ask three questions: 1. Given , what is the smallest m? 2. Given m, what is the “best” ? 3. What is the smallest m that is valid for all legal ? • Two parameters control execution: • 1. A scheduling function  • Maps each operation to execution time • Parallelism is implicit • 2. A fully associative store of size m

  31. Outline • Abstract problem • Simplifications • Concrete problem • Solution Method • Conclusions

  32. Simplifying the Schedule • Real programs aren’t DAG’s • Dependence graph is parameterized by loops • Too many nodes to schedule • Size could even be unknown (symbolic constants) • Use classical solution: affine schedules • Each statement has a scheduling function  • Each  is an affine function of the enclosing loop counters and symbolic constants • To simplify talk, ignore symbolic constants: (i) = B• i

  33. Simplifying the Storage Mapping • Programs use arrays, not associative maps • If size decreases, need to specify which elements are mapped to the same location

  34. Simplifying the Storage Mapping • Programs use arrays, not associative maps • If size decreases, need to specify which elements are mapped to the same location

  35. Simplifying the Storage Mapping • Programs use arrays, not associative maps • If size decreases, need to specify which elements are mapped to the same location ?

  36. Simplifying the Storage Mapping Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  37. Simplifying the Storage Mapping Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  38. Simplifying the Storage Mapping Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  39. Simplifying the Store Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  40. Simplifying the Store Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  41. Simplifying the Store Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  42. Simplifying the Store Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  43. Simplifying the Store Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  44. Simplifying the Store Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  45. Simplifying the Store Occupancy Vectors (Strout et al.) • Specifies unit of overwriting within an array • Locations collapsed if separated by a multiple of v v = (1, 1) j i

  46. Simplifying the Store Occupancy Vectors (Strout et al.) • For a given schedule, v is valid if semantics are unchanged using transformed array • Shorter vectors require less storage v = (1, 1) j i

  47. Outline • Abstract problem • Simplifications • Concrete problem • Solution Method • Conclusions

  48. Answering Question #1 • Given (i, j) = i + j, what is the shortest valid occupancy vector v? i j

  49. Answering Question #1 • Given (i, j) = i + j, what is the shortest valid occupancy vector v? i j

  50. Answering Question #1 • Given (i, j) = i + j, what is the shortest valid occupancy vector v? • Solution: v = (1, 1) i j

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