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Initial Control flow graph problem:
  Start location: evalSequentialSinglestart
      0: evalSequentialSinglestart -> evalSequentialSingleentryin : [], cost: 1
      1: evalSequentialSingleentryin -> evalSequentialSinglebb1in : A'=0, [], cost: 1
      2: evalSequentialSinglebb1in -> evalSequentialSinglebb5in : [ A>=B ], cost: 1
      3: evalSequentialSinglebb1in -> evalSequentialSinglebb2in : [ B>=1+A ], cost: 1
      8: evalSequentialSinglebb5in -> evalSequentialSinglebb4in : [ B>=1+A ], cost: 1
      9: evalSequentialSinglebb5in -> evalSequentialSinglereturnin : [ A>=B ], cost: 1
      6: evalSequentialSinglebb2in -> evalSequentialSinglebb5in : [], cost: 1
      4: evalSequentialSinglebb2in -> evalSequentialSinglebbin : [ 0>=1+free ], cost: 1
      5: evalSequentialSinglebb2in -> evalSequentialSinglebbin : [ free_1>=1 ], cost: 1
      7: evalSequentialSinglebbin -> evalSequentialSinglebb1in : A'=1+A, [], cost: 1
     10: evalSequentialSinglebb4in -> evalSequentialSinglebb5in : A'=1+A, [], cost: 1
     11: evalSequentialSinglereturnin -> evalSequentialSinglestop : [], cost: 1

Removing duplicate transition: 4.

Simplified the transitions:
  Start location: evalSequentialSinglestart
      0: evalSequentialSinglestart -> evalSequentialSingleentryin : [], cost: 1
      1: evalSequentialSingleentryin -> evalSequentialSinglebb1in : A'=0, [], cost: 1
      2: evalSequentialSinglebb1in -> evalSequentialSinglebb5in : [ A>=B ], cost: 1
      3: evalSequentialSinglebb1in -> evalSequentialSinglebb2in : [ B>=1+A ], cost: 1
      8: evalSequentialSinglebb5in -> evalSequentialSinglebb4in : [ B>=1+A ], cost: 1
      6: evalSequentialSinglebb2in -> evalSequentialSinglebb5in : [], cost: 1
      5: evalSequentialSinglebb2in -> evalSequentialSinglebbin : [], cost: 1
      7: evalSequentialSinglebbin -> evalSequentialSinglebb1in : A'=1+A, [], cost: 1
     10: evalSequentialSinglebb4in -> evalSequentialSinglebb5in : A'=1+A, [], cost: 1


Applied simple chaining:
  Start location: evalSequentialSinglestart
      0: evalSequentialSinglestart -> evalSequentialSinglebb1in : A'=0, [], cost: 2
      2: evalSequentialSinglebb1in -> evalSequentialSinglebb5in : [ A>=B ], cost: 1
      3: evalSequentialSinglebb1in -> evalSequentialSinglebb2in : [ B>=1+A ], cost: 1
      8: evalSequentialSinglebb5in -> evalSequentialSinglebb5in : A'=1+A, [ B>=1+A ], cost: 2
      5: evalSequentialSinglebb2in -> evalSequentialSinglebb1in : A'=1+A, [], cost: 2
      6: evalSequentialSinglebb2in -> evalSequentialSinglebb5in : [], cost: 1

Eliminating 1 self-loops for location evalSequentialSinglebb5in
  Self-Loop 8 has the metering function: B-A, resulting in the new transition 12.
  Removing the self-loops: 8.

Removed all Self-loops using metering functions (where possible):
  Start location: evalSequentialSinglestart
      0: evalSequentialSinglestart -> evalSequentialSinglebb1in : A'=0, [], cost: 2
      2: evalSequentialSinglebb1in -> evalSequentialSinglebb5in : [ A>=B ], cost: 1
      3: evalSequentialSinglebb1in -> evalSequentialSinglebb2in : [ B>=1+A ], cost: 1
     12: evalSequentialSinglebb5in -> [9] : A'=B, [ B>=1+A ], cost: 2*B-2*A
      5: evalSequentialSinglebb2in -> evalSequentialSinglebb1in : A'=1+A, [], cost: 2
      6: evalSequentialSinglebb2in -> evalSequentialSinglebb5in : [], cost: 1


Applied chaining over branches and pruning:
  Start location: evalSequentialSinglestart
      0: evalSequentialSinglestart -> evalSequentialSinglebb1in : A'=0, [], cost: 2
     13: evalSequentialSinglebb1in -> evalSequentialSinglebb1in : A'=1+A, [ B>=1+A ], cost: 3
      2: evalSequentialSinglebb1in -> evalSequentialSinglebb5in : [ A>=B ], cost: 1
     14: evalSequentialSinglebb1in -> evalSequentialSinglebb5in : [ B>=1+A ], cost: 2
     12: evalSequentialSinglebb5in -> [9] : A'=B, [ B>=1+A ], cost: 2*B-2*A

Eliminating 1 self-loops for location evalSequentialSinglebb1in
  Self-Loop 13 has the metering function: B-A, resulting in the new transition 15.
  Removing the self-loops: 13.

Removed all Self-loops using metering functions (where possible):
  Start location: evalSequentialSinglestart
      0: evalSequentialSinglestart -> evalSequentialSinglebb1in : A'=0, [], cost: 2
     15: evalSequentialSinglebb1in -> [10] : A'=B, [ B>=1+A ], cost: 3*B-3*A
     12: evalSequentialSinglebb5in -> [9] : A'=B, [ B>=1+A ], cost: 2*B-2*A
      2: [10] -> evalSequentialSinglebb5in : [ A>=B ], cost: 1
     14: [10] -> evalSequentialSinglebb5in : [ B>=1+A ], cost: 2


Applied simple chaining:
  Start location: evalSequentialSinglestart
      0: evalSequentialSinglestart -> [10] : A'=B, [ B>=1 ], cost: 2+3*B
     12: evalSequentialSinglebb5in -> [9] : A'=B, [ B>=1+A ], cost: 2*B-2*A
      2: [10] -> evalSequentialSinglebb5in : [ A>=B ], cost: 1
     14: [10] -> evalSequentialSinglebb5in : [ B>=1+A ], cost: 2


Applied chaining over branches and pruning:
  Start location: evalSequentialSinglestart
     16: evalSequentialSinglestart -> evalSequentialSinglebb5in : A'=B, [ B>=1 && B>=B ], cost: 3+3*B
     17: evalSequentialSinglestart -> [11] : A'=B, [ B>=1 ], cost: 2+3*B
     12: evalSequentialSinglebb5in -> [9] : A'=B, [ B>=1+A ], cost: 2*B-2*A


Applied chaining over branches and pruning:
  Start location: evalSequentialSinglestart
     17: evalSequentialSinglestart -> [11] : A'=B, [ B>=1 ], cost: 2+3*B
     18: evalSequentialSinglestart -> [12] : A'=B, [ B>=1 && B>=B ], cost: 3+3*B


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: evalSequentialSinglestart
     17: evalSequentialSinglestart -> [11] : A'=B, [ B>=1 ], cost: 2+3*B
     18: evalSequentialSinglestart -> [12] : A'=B, [ B>=1 && B>=B ], cost: 3+3*B


Computing complexity for remaining 2 transitions.

  Found configuration with infinitely models for cost: 2+3*B
  and guard: B>=1:
  B: Pos

Found new complexity n^1, because: Found infinity configuration.


The final runtime is determined by this resulting transition:
  Final Guard: B>=1
  Final Cost:  2+3*B

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^1
  Complexity value: 1

WORST_CASE(Omega(n^1),?)