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Partial Program Admission by Path Enumeration

Partial Program Admission by Path Enumeration. Michael Wilson Advisor: Jonathan Turner also with Ron Cytron. Problem Context. Virtualizing the network core Router hosting platforms will run third-party networking protocols Router platforms run at Internet Core speeds (5-10 Gbps)

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Partial Program Admission by Path Enumeration

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  1. Partial Program AdmissionbyPath Enumeration Michael Wilson Advisor: Jonathan Turneralso with Ron Cytron

  2. Problem Context • Virtualizing the network core • Router hosting platforms will run third-party networking protocols • Router platforms run at Internet Core speeds (5-10 Gbps) • High-Speed Networking platforms are usually non-preemptive, lack OS protections • Third-party code cannot be trusted! • Memory protection is largely a solved problem • How to ensure that untrusted network protocols don’t hog the CPU? • Each protocol must adhere to a very strict cycle budget

  3. Existing Solutions • Worst-Case Execution Time (WCET) • Longest path analysis • Just find the longest path from entry to exit • Arose from circuit analysis • Won’t work with loops • Instrumentation • Add run-time checks to ensure code stays under budget • Our budgets are very tight; adding a run-time check might push “safe” paths over budget! • Integer Linear Programming • Requires developers to provide branch constraints • Deals well with loops • We can’t trust the developers! • A new solution is needed….

  4. Partial Program Admission • Better method: partial program admission • We can determine which execution paths are within budget and admit just those paths • Use an “over budget” exception handler • Modify the program at load time to enforce good behavior • Modifications result in zero runtime overhead • Requires explicit path enumeration, computationally expensive • Dynamic Programming to make the problem tractable

  5. Partial Program Admission • Why partial program admission? • What good is it to just run part of your program? • Not really about running part of a program, but about transforming a program to make proofs easier • Developer may know program branch constraints that render some paths “impossible” if (A)do foo(); if (not A)do bar(); • Substitute code duplication for developer knowledge

  6. Solution Concept SafeProgram Program Emit Enumerate Bound Coalesce CFG BXG CFT Compile BXT • Process is a series of graph transformations • Start with a Control Flow Graph (CFG) • Enumerate all paths within the CFG to produce a Control Flow Tree (CFT) • Bound all paths in the CFT by our budget to produce a Bounded Execution Tree (BXT) • Coalesce identical subtrees to reduce code duplication, producing the Bounded Execution Graph (BXG) • Our actual algorithm shortcuts directly from CFG to BXG

  7. Digraph representation of a program Vertices represent pieces of code Edges represent flow of control Weights represent cycle counts to traverse parent vertex Artificial source, sink: S, T S 0 A 2 2 B C 1 3 D 1 1 E F 1 4 G 1 T Control Flow Graph (CFG) CFG to CFT S 0 A A;if (cond1) B;else C;D;if (cond2) E;else F;G; 2 2 B C 1 3 D 1 1 E F 1 4 G 1 T

  8. S S S S 0 0 0 0 A A A A 2 2 2 2 2 B C B C B 1 1 1 3 3 D2 D2 D1 D1 D1 1 1 1 1 1 1 E1 E1 E2 E2 F1 F1 F2 4 1 1 1 1 4 4 G1 G2 G1 G4 G1 G3 G3 1 1 1 1 1 1 1 T1 T2 T1 T3 T1 T3 T4 CFG to CFT CFT to BXT S 0 A 2 2 B C 1 3 D 1 1 E F 1 4 G 1 T

  9. CFT to BXT S 0 A 2 2 Let’s prune this path by aborting to the exception handler Now all paths are under budget. All previously over-budget paths generate exceptions. This program is safe to run! At a budget of 10, this path ran out of time B C 1 3 D2 D1 1 1 1 1 E1 E2 F1 F2 4 1 1 4 G1 G2 G4 G3 However, it’s very bloated. 7 vertices have been replaced with 11. 1 1 1 1 T1 T2 T3 T4 X 6 11 9 6 8

  10. BXT to BXG S 0 A 2 2 Our bloat is greatly reduced! From our original 7 vertices, we now have 8. We can reduce this bloat by coalescing identical subtrees These two subtrees are identical. There’s no reason to keep them both. We have two remaining identical subtrees, so let’s merge them, too. These subtrees are not identical, so we can’t coalesce them. Let’s merge. B C 1 3 D2 D1 1 1 1 1 E1 E1/E2 E2 F1 Unfortunately, direct path enumeration is too slow to be useful. Fortunately, we can use Dynamic Programming. 1 1 4 G1/G2/G3 G1 G2 G1/G3 G3 1 1 1 T1 T2 T1/T3 T3 T1/T2/T3 X

  11. Dynamic Programming • Dynamic programming • Remember answers to previously solved problems, look them up later instead of recomputing them • This works when we have repeated subproblems or overlapping subproblems.

  12. Applying Dynamic Programming • Need to select repeated subproblems or overlapping subproblems • Our overlapping subproblem: • Given an execution subtree rooted at u and a cycle budget B, what is the bounded execution subtree bxtB(u)? • bxti(u) and bxtj(u) are often identical when i,j are close • We rely on the concept of Intervals of budgets • Interval [i,j] for execution tree rooted at vertex u such that bxti(u) = bxtj(u). • For all k such that i  k  j, bxti(u) = bxtj(u) impliesbxti(u)=bxtk(u)=bxtj(u) • From here on, we will only deal with maximal intervals

  13. Intervals have a recursive definition: • Given budget B and vertex u with • N children u1,…,uN • …and edge weight w(u,uk) • …each with interval [ik , jk] containing B-w(u,uk) • Then u’s interval is • The basis: for vertex T, intervals are [-∞,-1] and [0, ∞]. Intervals • Using intervals for dynamic programming • We want the interval for u, B=10 • Recurse for child intervals • Adjust for edge weights • Intersect child intervals • Now we want B=8 • 8 is in our interval [8,12]! No need to recompute! • Now we want B=6 • Still need to compute new intervals…. S Budget B=20 6 8 10 u [7,12] [8,12] [8,14] + + 4 1 v1 v2 [7,13] [3,8]

  14. The Algorithm • Preliminaries • Interval data type as described • Interval search object • Can associate a vertex and an interval • inserts, lookups • Assume we have done a reverse Dijkstra over the entire graph from T, storing the result in δ(v,T).

  15. The Algorithm interval bxg(integer R, vertex u) ifwe’ve already computed this subtree return the interval else if we’re over budget record an “exception” interval and return it else if we’re at the sink record a basis interval and return it else recursively analyze childrencombine results record and and return it • To create the bxg, we call bxg(B,S) where B is the budget and S is the root of the CFG. • The bounded execution graph is embedded in the intervals we have computed

  16. Evaluation Methodology • The duplication factor is crucial • High code bloat makes the algorithm useless • To test the algorithm, we created synthetic CFGs by applying graph transformations analogous to grammar production rules in a C-like language • Roughly 1000 vertices each, corresponding to code size of around 3600 instructions. • Our real programs of interest are much smaller(~50 vertices, <1000 instructions) • Acyclic graphs only for this examination • We also tested the algorithm on one real CFG from our problem area.

  17. Performance: Real CFG • Original code size: • 180 • Worst-case: • 296 at 85 cycles • (About 1.6x) • Longest path: • 108 cycles IPv4 Header Format Instructions Budget (cycles)

  18. Performance: Synthetic CFGs • Mean worst-case: • 1.6 • Median worst-case • 1.08 • Worst worst-case: • 23.4 Code Duplication Distribution Percentage of CFGs Maximum Duplication Required (Normalized)

  19. Future Work • Examine more real CFGs • Strategies to compact that BXG by adding run-time checks when they won’t impact safe paths • Incorporate program analysis for branch constraints • Detect loop iteration bounds • Detect mutually exclusive paths • Program analysis for the real WCET • Constant propagation over all branch constraints • Solves a decidable variation of the halting problem

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