Inferring Expected Runtimes Using Sizes

KoAT2 Proof WORST_CASE( ?, 2*Arg_0+23 {O(n)})

Initial Complexity Problem (after preprocessing)

Start:f
Program_Vars:Arg_0
Temp_Vars:T
Locations:f, g, h
Transitions:
f(Arg_0) -> h(T) :|: 0<T
f(Arg_0) -> 3/4:g(Arg_0) :+: 1/4:g(Arg_0-1)
g(Arg_0) -> g(Geometric (1/2)) :|: Arg_0<10

G f f g g f->g t₁ ∈ g₁ p = 3/4 f->g t₂ ∈ g₁ p = 1/4 η (Arg_0) = Arg_0-1 h h f->h t₀ ∈ g₀ η (Arg_0) = T τ = 0<T g->g t₃ ∈ g₂ η (Arg_0) = Geometric (1/2) τ = Arg_0<10

Timebounds:

Overall timebound:3+max([-(Arg_0)+10, 0])+max([-(Arg_0)+11, 0]) {O(n)}
0,0: f->h: 1 {O(1)}
1,1: f->g: 1 {O(1)}
2,1: f->g: 1 {O(1)}
3,2: g->g: max([-(Arg_0)+10, 0])+max([-(Arg_0)+11, 0]) {O(n)}

Expected Timebounds:

Overall expected timebound: 2*Arg_0+23 {O(n)}
0: f->[1:h]: 1 {O(1)}
1: f->[3/4:g; 1/4:g]: 1 {O(1)}
2: g->[1:g]: 2*Arg_0+21 {O(n)}

Costbounds:

Overall costbound: inf {Infinity}
0,0: f->h: inf {Infinity}
1,1: f->g: inf {Infinity}
2,1: f->g: inf {Infinity}
3,2: g->g: inf {Infinity}

Expected Costbounds:

Overall expected costbound: 2*Arg_0+23 {O(n)}
0: f->[1:h]: 1 {O(1)}
1: f->[3/4:g; 1/4:g]: 1 {O(1)}
2: g->[1:g]: 2*Arg_0+21 {O(n)}

Sizebounds:

1,1: f->g, Arg_0: Arg_0 {O(n)}
2,1: f->g, Arg_0: (-1)+Arg_0 {O(n)}

ExpSizeBounds:

(0: f->[1:h], h), T: T {O(n)}
(1: f->[3/4:g; 1/4:g], g), T: T {O(n)}
(1: f->[3/4:g; 1/4:g], g), Arg_0: 1+Arg_0 {O(n)}
(2: g->[1:g], g), T: T {O(n)}
(2: g->[1:g], g), Arg_0: 1+Arg_0+2*(2*Arg_0+21) {O(n)}