YES(?, 15*b + 6)

Initial complexity problem:
1:	T:
		(1, 1)    evalspeedpldi4start(a, b) -> evalspeedpldi4entryin(a, b)
		(?, 1)    evalspeedpldi4entryin(a, b) -> evalspeedpldi4returnin(a, b) [ 0 >= a ]
		(?, 1)    evalspeedpldi4entryin(a, b) -> evalspeedpldi4returnin(a, b) [ a >= b ]
		(?, 1)    evalspeedpldi4entryin(a, b) -> evalspeedpldi4bb5in(a, b) [ a >= 1 /\ b >= a + 1 ]
		(?, 1)    evalspeedpldi4bb5in(a, b) -> evalspeedpldi4bb2in(a, b) [ b >= 1 ]
		(?, 1)    evalspeedpldi4bb5in(a, b) -> evalspeedpldi4returnin(a, b) [ 0 >= b ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb3in(a, b) [ a >= b + 1 ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb4in(a, b) [ b >= a ]
		(?, 1)    evalspeedpldi4bb3in(a, b) -> evalspeedpldi4bb5in(a, b - 1)
		(?, 1)    evalspeedpldi4bb4in(a, b) -> evalspeedpldi4bb5in(a, b - a)
		(?, 1)    evalspeedpldi4returnin(a, b) -> evalspeedpldi4stop(a, b)
	start location:	evalspeedpldi4start
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem:
2:	T:
		(1, 1)    evalspeedpldi4start(a, b) -> evalspeedpldi4entryin(a, b)
		(?, 1)    evalspeedpldi4entryin(a, b) -> evalspeedpldi4bb5in(a, b) [ a >= 1 /\ b >= a + 1 ]
		(?, 1)    evalspeedpldi4bb5in(a, b) -> evalspeedpldi4bb2in(a, b) [ b >= 1 ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb3in(a, b) [ a >= b + 1 ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb4in(a, b) [ b >= a ]
		(?, 1)    evalspeedpldi4bb3in(a, b) -> evalspeedpldi4bb5in(a, b - 1)
		(?, 1)    evalspeedpldi4bb4in(a, b) -> evalspeedpldi4bb5in(a, b - a)
	start location:	evalspeedpldi4start
	leaf cost:	4

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(1, 1)    evalspeedpldi4start(a, b) -> evalspeedpldi4entryin(a, b)
		(1, 1)    evalspeedpldi4entryin(a, b) -> evalspeedpldi4bb5in(a, b) [ a >= 1 /\ b >= a + 1 ]
		(?, 1)    evalspeedpldi4bb5in(a, b) -> evalspeedpldi4bb2in(a, b) [ b >= 1 ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb3in(a, b) [ a >= b + 1 ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb4in(a, b) [ b >= a ]
		(?, 1)    evalspeedpldi4bb3in(a, b) -> evalspeedpldi4bb5in(a, b - 1)
		(?, 1)    evalspeedpldi4bb4in(a, b) -> evalspeedpldi4bb5in(a, b - a)
	start location:	evalspeedpldi4start
	leaf cost:	4

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol evalspeedpldi4bb2in: X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ X_1 - 1 >= 0
  For symbol evalspeedpldi4bb3in: X_1 - X_2 - 1 >= 0 /\ X_2 - 1 >= 0 /\ X_1 + X_2 - 3 >= 0 /\ X_1 - 2 >= 0
  For symbol evalspeedpldi4bb4in: X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ -X_1 + X_2 >= 0 /\ X_1 - 1 >= 0
  For symbol evalspeedpldi4bb5in: X_1 - 1 >= 0


This yielded the following problem:
4:	T:
		(?, 1)    evalspeedpldi4bb4in(a, b) -> evalspeedpldi4bb5in(a, b - a) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ -a + b >= 0 /\ a - 1 >= 0 ]
		(?, 1)    evalspeedpldi4bb3in(a, b) -> evalspeedpldi4bb5in(a, b - 1) [ a - b - 1 >= 0 /\ b - 1 >= 0 /\ a + b - 3 >= 0 /\ a - 2 >= 0 ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb4in(a, b) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ a - 1 >= 0 /\ b >= a ]
		(?, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb3in(a, b) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ a - 1 >= 0 /\ a >= b + 1 ]
		(?, 1)    evalspeedpldi4bb5in(a, b) -> evalspeedpldi4bb2in(a, b) [ a - 1 >= 0 /\ b >= 1 ]
		(1, 1)    evalspeedpldi4entryin(a, b) -> evalspeedpldi4bb5in(a, b) [ a >= 1 /\ b >= a + 1 ]
		(1, 1)    evalspeedpldi4start(a, b) -> evalspeedpldi4entryin(a, b)
	start location:	evalspeedpldi4start
	leaf cost:	4

A polynomial rank function with
	Pol(evalspeedpldi4bb4in) = 3*V_2 - 2
	Pol(evalspeedpldi4bb5in) = 3*V_2
	Pol(evalspeedpldi4bb3in) = 3*V_2 - 2
	Pol(evalspeedpldi4bb2in) = 3*V_2 - 1
	Pol(evalspeedpldi4entryin) = 3*V_2
	Pol(evalspeedpldi4start) = 3*V_2
orients all transitions weakly and the transitions
	evalspeedpldi4bb5in(a, b) -> evalspeedpldi4bb2in(a, b) [ a - 1 >= 0 /\ b >= 1 ]
	evalspeedpldi4bb4in(a, b) -> evalspeedpldi4bb5in(a, b - a) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ -a + b >= 0 /\ a - 1 >= 0 ]
	evalspeedpldi4bb3in(a, b) -> evalspeedpldi4bb5in(a, b - 1) [ a - b - 1 >= 0 /\ b - 1 >= 0 /\ a + b - 3 >= 0 /\ a - 2 >= 0 ]
	evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb4in(a, b) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ a - 1 >= 0 /\ b >= a ]
	evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb3in(a, b) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ a - 1 >= 0 /\ a >= b + 1 ]
strictly and produces the following problem:
5:	T:
		(3*b, 1)    evalspeedpldi4bb4in(a, b) -> evalspeedpldi4bb5in(a, b - a) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ -a + b >= 0 /\ a - 1 >= 0 ]
		(3*b, 1)    evalspeedpldi4bb3in(a, b) -> evalspeedpldi4bb5in(a, b - 1) [ a - b - 1 >= 0 /\ b - 1 >= 0 /\ a + b - 3 >= 0 /\ a - 2 >= 0 ]
		(3*b, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb4in(a, b) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ a - 1 >= 0 /\ b >= a ]
		(3*b, 1)    evalspeedpldi4bb2in(a, b) -> evalspeedpldi4bb3in(a, b) [ b - 1 >= 0 /\ a + b - 2 >= 0 /\ a - 1 >= 0 /\ a >= b + 1 ]
		(3*b, 1)    evalspeedpldi4bb5in(a, b) -> evalspeedpldi4bb2in(a, b) [ a - 1 >= 0 /\ b >= 1 ]
		(1, 1)      evalspeedpldi4entryin(a, b) -> evalspeedpldi4bb5in(a, b) [ a >= 1 /\ b >= a + 1 ]
		(1, 1)      evalspeedpldi4start(a, b) -> evalspeedpldi4entryin(a, b)
	start location:	evalspeedpldi4start
	leaf cost:	4

Complexity upper bound 15*b + 6

Time: 0.388 sec (SMT: 0.368 sec)