The document provides a summary of Semi-Markov Decision Processes (SMDPs) in 10 points: 1. It describes the basic components of an SMDP including states, actions, rewards, policies, and value functions. 2. It discusses the concepts of optimal policies, average reward models, and discount factors in SMDPs. 3. It introduces the idea of transition times in SMDPs, which allows actions to take varying amounts of time. This makes SMDPs a generalization of Markov Decision Processes. 4. It notes that algorithms for solving SMDPs typically involve estimating the average reward per action to find an optimal policy.

1. 100 things I know.
Part I of III
Reinaldo Uribe M
Mar. 4, 2012

2. SMDP Problem Description.
1. In a Markov Decision Process, a (learning) agent is embedded
in an envionment and takes actions that affect that environment.
States: s ∈ S.
Actions: a ∈ As ; A = s∈S As .
(Stationary) system dynamics:
transition from s to s after taking a, with probability
a
Pss = p(s |s, a)
Rewards: Ra . Def. r(s, a) = E Ra | s, a
ss ss
At time t, the agent is in state st , takes action at , transitions to
state st+1 and observes reinforcement rt+1 with expectation
r(st , at ).

3. SMDP Problem Description.
2. Policies, value and optimal policies.
An element π of the policy space Π indicates what action,
π(s), to take at each state.
The value of a policy from a given state, v π (s) is the expected
cumulative reward received starting in s and following π:
∞
v π (s) = E γ t r(st , π(st )) | s0 = s, π
t=0
0 < γ ≤ 1 is a discount factor.
An optimal policy, π ∗ has maximum value at every state:
π ∗ (s) ∈ argmax v π (s) ∀s
π∈Π
π∗
v ∗ (s) = v (s) ≥ v π (s) ∀π ∈ Π

4. SMDP Problem Description.
3. Discount
Makes infinite-horizon value bounded if rewards are bounded.
Ostensibly makes rewards received sooner more desirable than
those received later.
But, exponential terms make analysis awkward and harder...
... and γ has unexpected, undesirable effects, as shown in Uribe
et al. 2011
Therefore, hereon γ = 1.
See section Discount, at the end, for discussion.

5. SMDP Problem Description.
4. Average reward models.
A more natural long term measure of optimality exists
for such cyclical tasks, based on maximizing the average
reward per action. Mahadevan 1996
n−1
1
ρπ (s) = lim E r(st , π(st )) | s0 = 0, π
n→∞ n
t=0
Optimal policy:
ρ∗ (s) ≥ ρπ (s) ∀s, π ∈ Π
Remark: All actions equally costly.

6. SMDP Problem Description
5. Semi-Markov Decision Process: usual approach, transition
times.
Agent is in state st and takes action π(st ) at decision epoch t.
After an average of Nt units of time, the sistem evolves to
state st+1 and the agent observes rt+1 with expectation
r(st , π(st )).
In general, Nt (st , at , st+1 ).
Gain (of a policy at a state):
n−1
π
E t=0 r(st , π(st )) | s0 = s, π
ρ (s) = lim
n→∞ n−1
E t=0 Nt | s0 = s, π
Optimizing gain still maximizes average reward per action, but
actions are no longer equally weighted. (Unless all Nt = 1)

7. SMDP Problem Description
6.a Semi-Markov Decision Process: explicit action costs.
Taking an action takes time, costs money, or consumes
energy. (Or any combination thereof)
Either way, real valued cost kt+1 not necessarily related to
process rewards.
Cost can depend on a, s and (less common in practice) s .
Generally, actions have positive cost. We simply require all
policies to have positive expected cost.
Wlog the magnitude of the smallest nonzero average action
cost is forced to be unity:
|k(a, s)| ≥ 1 ∀k(a, s) = 0

8. SMDP Problem Description
6.b Semi-Markov Decision Process: explicit action costs.
Cost of a policy from a state:
n−1
cπ (s) = lim E k(st , π(st )) | s0 = s, π
n→∞
t=0
So cπ (s) > 0 ∀π ∈ Π, s.
Nt = k(st , π(st )). Only their definition/interpretation
changes.
Gain
v π (s)/n
ρπ (s) =
cπ (s)/n

9. SMDP Problem Description
7. Optimality of π ∗ :
π ∗ ∈ Π with gain
n−1
E t=0 r(st , π(st )) | s0 = s, π ∗ ∗
v π (s)
π∗ ∗
ρ (s) = ρ (s) = lim = π∗
n→∞ n−1
= s, π ∗ c (s)
E t=0 k(st , π(st )) | s0
is optimal if
ρ∗ (s) ≥ ρπ (s) ∀s, π ∈ Π,
as it was in ARRL.
Notice that the optimal policy doesn’t necessarily maximize v π or
minimize cπ . Only optimizes their ratio.

10. SMDP Problem Description
8. Policies in ARRL and SMDPs are evaluated using the
average-adjusted sum of rewards:
n−1
H π (s) = lim E (r(st , π(st )) − ρπ (s)) | s0 = s, π
n→∞
t=0
Puterman 1994, Abounadi et al. 2001, Ghavamzadeh & Mahadevan 2007
This signals the existence of bias optimal policies that, while
gain optimal, also maximize the transitory rewards received
before entering recurrence.
We are interested in gain-optimal policies only.
(It is hard enough...)

11. SMDP Problem Description
9. The Unichain Property
A process is unichain if every policy has a single, unique
recurrent class.
I.e. if for every policy, all recurrent states communicate
between them.
All methods rely on the unichain property. (Because, if it
holds:)
ρπ (s) is constant for all s.
ρπ (s) = ρπ
Gain and value expressions simplify. (See next)
However, deciding if a problem is unichain is NP-Hard.
Tsitsiklis 2003

12. SMDP Problem Description
10. Unichain property under recurrent states. Feinberg & Yang, 2010
A state is recurrent if it belongs to a recurrent class of every
policy.
A recurrent state can be found, or proven not to exist, in
polynomial time.
If a recurrent state exists, determining whether the unichain
property holds can be done in polynomial time.
(We are not going to actually do it–it requires knowledge of
the system dynamics–but good to know!)
Recurrent states seem useful. In fact, existence of a recurrent
state is more critical to our purposes that the unichain
property.
Both will be required in principle for our methods/analysis,
until their necessity is furher qualified in section Unichain
Considerations below.

13. Intermission

14. Generic Learning Algorithm
11. The relevant expressions under our assumptions simplify, losing
dependence on s0
The following Bellman equation holds for average-adjusted state
value:
H π (s) = r(s, π(s)) − k(s, π(s))(ρπ ) + Eπ H π (s ) (1)
Ghavamzadeh & Mahadevan 2007
Reinforcement Learning methods exploit Eq. (1), running the
process and substituting:
State for state-action pair value.
Expected for obseved reward and cost.
ρπ for an estimate.
H π (s ) for its current estimate.

15. Generic Learning Algorithm
12.
Algorithm 1 Generic SMDP solver
Initialize
repeat forever
Act
Do RL to find value of current π Usually 1-step Q-learning
Update ρ.

16. Generic Learning Algorithm
13.
Model-based state value update:
H t+1 (st ) ← max r(st , a) + Ea H t (st+1 )
a
Ea emphasizes that expected value of next state depends on
action chosen/taken.
Model free state-action pair value update:
Qt+1 (st , at ) ← (1 − γt ) Qt (st , at )+
γt rt+1 − ρt ct+1 + max Qt (st+1 , a)
a
In ARRL, ct = 1 ∀t

17. Generic Learning Algorithm
14.a Table of algorithms. ARRL
Algorithm Gain update
t
AAC r(si , π i (si ))
i=0
Jalali and Ferguson 1989
ρt+1 ←
t+1
t+1
R–Learning ρ ← (1 − α)ρt +
Schwartz 1993 α rt+1 + max Qt (st+1 , a) − max Qt (st , a)
a a
H–Learning ρt+1 ← (1−αt )ρt +αt rt+1 − H t (st ) + H t (st+1 )
αt
Tadepalli and Ok 1998 αt+1 ←
αt + 1
SSP Q-Learning ρt+1 ← ρt + αt min Qt (ˆ, a)
s
Abounadi et al. 2001 a
t
HAR r(si , π i (si ))
i=0
Ghavamzadeh and Mahadevan 2007
ρt+1 ←
t+1

18. Generic Learning Algorithm
14.b Table of algorithms. SMDPRL
Algorithm Gain update
SMART t
Das et al. 1999
r(si , π i (si ))
i=0
ρt+1 ← t
MAX-Q c(si , π i (si ))
Ghavamzadeh and Mahadevan 2001 i=0

19. SSP Q-Learning
15. Stochastic Shortest Path Q-Learning
Most interesting. ARRL
If unichain and exists s recurrent (Assumption 2.1 ):
ˆ
SSP Q-learning is based on the observation that
the average cost under any stationary policy is
simply the ratio of expected total cost and expected
time between two successive visits to the reference
state [ˆ]
s
Thus, they propose (after Bertsekas 1998) making the process
episodic, splitting s into the (unique) initial and terminal
ˆ
states.
If the Assumption holds, termination has probability 1.
Only the value/cost of the initial state are important.
Optimal solution “can be shown to happen” when H(ˆ) = 0.
s
(See next section)

20. SSP Q-Learning
16. SSPQ ρ update.
ρt+1 ← ρt + αt min Qt (ˆ, a),
s
a
where
2
αt → ∞; αt < ∞.
t t
But it is hard to prove boundedness of {ρt }, so suggested instead
ρt+1 ← Γ ρt + αt min Qt (ˆ, a) ,
s
a
with Γ(·) a projection to [−K, K] and ρ∗ ∈ (−K, K).

21. A Critique
17. Complexity.
Unknown.
While RL is PAC.
18. Convergence.
Not always guaranteed (ex. R-Learning).
When proven, asymptotic:
convergence to the optimal policy/value if all state-action
pairs are visited infinite times.
Usually proven depending on decaying learning rates, which
make learning even slower.

22. A Critique
19. Convergence of ρ updates.
... while the second “slow” iteration gradually guides
[ρt ] to the desired value.
Abounadi et al. 2001
It is the slow one!
Must be so for sufficient approximation of current policy value
for improvement.
Initially biased towards (likely poor) observed returns at the
start.
A long time must probably pass following the optimal policy
for ρ to converge to actual value.

23. Our method
20.
Favours an understanding of the −ρ term, either alone in
ARRL or as a factor of costs in SMDPs, not so much as an
approximation to average rewards but as a punishment for
taking actions, which must be made “worth it” by the rewards.
I.e. nudging.
Exploits the splitting of SSP Q-Learning, in order to focus on
the value/cost of a single state, s.
ˆ
Thus, also assumes the existence of a recurrent state, and
that the unichain policy holds. (For the time being)
Attempts to ensure an accelerated convergence of ρ updates.
In a context in which certain, efficient convergence can be
easily introduced.

24. Intermission

25. Fractional programming
21. So, ‘Bertsekas splitting’ of s into initial sI and terminal sT .
ˆ
Then, from sI
Any policy π ∈ Π has an expected return until termination
v π (sI ),
and an expected cost until termination cπ (sI ).
v π (sI )
The ARRL problem, then, becomes max π
π∈Π c (sI )
Lemma
v π (sI )
argmax = argmax v π (s) + ρ∗ (−cπ (s))
π∈Π cπ (sI ) π∈Π
For ρ∗ such that max v π (s) + ρ(−cπ (s)) = 0
π∈Π

26. Fractional programming
22. Implications.
Assume the gain, ρ∗ is known.
Then, the nonlinear SMDP problem reduces to RL.
Which is better studied, well understood, simpler, and for
which sophisticated, efficient algorithms exist.
If we only use (r − ρ∗ c)(s, a, s ).
Problem: ρ∗ is usually not known.

27. Nudging
23. Idea:
Separate reinforcement learning (leave it to the pros) from
updating ρ.
Thus, value-learning becomes method-free.
We can use any old RL method.
Gain update is actually the most critical step.
Punish too little, and the agent will not care about hurrying,
only collecting reward.
Punish too much, and the agent will only care about finishing
already.
In that sense, (r − ρc) is like picking fruit inside a maze.

28. Nudging
24. The problem reduces to a sequence of RL problems.
For a sequence of (temporarily fixed) ρk
Some of the methods already provide an indication of the sign
of ρ updates.
We just don’t hurry to update ρ after taking a single action.
Plus the method comes armed with a termination condition:
As soon as H k (sI ) = 0 then π k = π ∗ .

29. Nudging
25.
Algorithm 2 Nudged SSP Learning
Initialize
repeat
Set reward scheme to (r − cρ)
Solve by any RL method.
Update ρ From current H π (sI )
until H π (sI ) = 0

30. w − l space
26. D
We will propose a method for updating ρ and show that it
minimizes uncertainty between steps. For that, we will use a
transformation that extends the work of our CIG paper. But First.
Let D be a bound on the magnitude of unnudged reward
D ≥ lim sup{H π (sI ) | ρ = 0}
π∈Π
D ≤ lim inf {H π (sI ) | ρ = 0}
π∈Π
Observe interval (−D, D) bounds ρ∗ but the upper bound is tight
only in ARRL if all of D reward is received in a single step from sI .

31. w − l space
27. All policies π ∈ Π, from (that is, at) sI have:
real expected value |v π (sI )| ≤ D.
positive cost cπ (sI ) ≥ 1
28.a w − l transformation:
D+v π (sI ) D−v π (sI )
w= 2cπ (sI ) l= 2cπ (sI )

32. w − l space
28.b w − l plane.
D
l
w D

33. w − l space
29. Properties:
w, l ≥ 0
w, l ≤ D
D
w+l = ≤D
cπ (s I)
v π (sI ) = D ⇒ l=0
v π (sI ) = −D ⇒ w=0
lim (w, l) = (0, 0)
cπ (sI )→∞
30. Inverse transformation:
π π
v π (sI ) = D wπ −lπ
w +l cπ (sI ) = D wπ1 π
+l

34. Intermission

35. w − l space
31. Value.
w π − lπ
v π (sI ) = D
D w π + lπ
Level sets are lines.
w–axis, expected D.
l–axis, expected −D.
w = l, expected 0.
5D
−D
−0.
l Optimization → fanning
from l = 0.
0
Not convex, but splits the
0.5D
space.
So optimizers are vertices of
D
w D convex hull of policies.

36. w − l space
32. Cost.
D
1
cπ (sI ) = D
wπ+ lπ
Level sets are lnes with slope
−1.
w + l = D, expected cost 1.
l
Cost decreases with distance
1
to the origin.
Cost optimizers (both max
2
and min) also vertices.
4
8
w D

37. w − l space
33. The origin.
Policies of infinite expected cost.
Mean the problem is not unichain or sI is not recurrent.
And are troublesome for optimizing value.
So under our assumptions, the origin does not belong to the
space

38. Nudged value in the w − l space
34. SMDP problem in w − l.
π π
v π (sI ) D wπ −lπ
argmax π = argmax w 1 +l
= argmax wπ − lπ
π∈Π c (sI π∈Π D wπ +lπ π∈Π
D
/2
D
−
l
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D
w D

39. Nudged value in the w − l space
35. Nudged value, for some ρ.
argmax v π (sI ) − ρcπ (sI )
π∈Π
w π − lπ 1
= argmax D π + lπ
− ρD π
π∈Π w w + lπ
w π − lπ − ρ
= argmax D
π∈Π w π + lπ

40. Nudged value in the w − l space
36. Nudged value level sets
ˆ
(For a set ρ and all policies π with a given h)
ˆ
ˆ
D−h π D
lπ =
ˆ
wˆ − ρ
ˆ
D+h D+hˆ
Lines!
ˆ
Slope depends only on h (i.e., not on ρ)

41. Nudged value in the w − l space
37. Pencil of lines
ˆ ˇ
For a set ρ, any two h and h level set lines have intersection:
ρ ρ
,−
2 2
Pencil of lines with that vertex.

42. Nudged value in the w − l space
D 38. Zero nudged value.
D−0 π D
lπ =
ˆ
wˆ − ρ
D+0 D+0
l lπ = w π − ρ
ˆ ˆ
−D
Unity slope.
0 Negative values above, positive
below.
w D
D

ρ


ρ



If whole cloud above w = l, some
 ,− 
 

2 2

negative nudging is the optimizer.
(Encouragement)

43. Nudged value in the w − l space
D
l
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w D

44. Nudged value in the w − l space
40. Initial bounds on ρ∗ .
−D ≤ ρ∗ ≤ D
(Duh! but nice geometry)

45. Enclosing triangle
42. Nomenclature
D
41. Definition.
ABC such that:
ABC ⊂ w − l space.
(w∗ , l∗ ) ∈ ABC. l B
●
Slope of AB segment, unity. mβ
mα
wA ≤ wB 1
● C
wA ≤ wC A● mγ
w D

46. Enclosing Triangle
43. (New) bounds on ρ.
Def. Slope mζ projection of point X(wX , lX ) to w = −l line.
mζ wX − lX
Xζ =
mζ + 1
Bounds:
Aα = Bα ≤ ρ∗ ≤ Cα
wA − lA = wB − lB ≤ ρ∗ ≤ wC − lC
44. So, collinearity (of A, B and C) implies optimality.
(Even if there are multiple optima)

47. Right and left uncertainty
45. Iterating inside an enclosing triangle.
1 Set ρ to some value within the bounds
ˆ
(wA − lA ≤ ρ ≤ wC − lC ).
ˆ
2 Solve problem with rewards (r − ρc).
ˆ
46. Optimality.
If h(sI ) = 0
Done!
Optimal policy found for current problem solves SMDP and
termination condition has been met.

48. Right and left uncertainty
47.a If h(sI ) > 0
Right uncertainty.
l
B
●
●
S
T
●
● C
A●
w
y1

49. Right and left uncertainty
47.b Right uncertainty.
Derivation:
y1 = Sα − Tα
1
= ((1 − mβ )wS − (1 − mγ )wT − (mγ − mβ )wC )
2
Maximization:
∗ 2s ab(ρ/2 − Cβ )(ρ/2 − Cγ ) + a(ρ/2 − Cβ ) + b(ρ/2 − Cγ )
y1 =
c
s = sign(mβ − mγ )
a = (1 − mγ )(mβ + 1)
b = (1 − mβ )(mγ + 1)
c = (b − a) = 2(mγ − mβ )

50. Right and left uncertainty
48.a If h(sI ) < 0
Left uncertainty.
l
B
●
●
● C
A● R
●
y2
w

51. Right and left uncertainty
48. Left uncertainty.
Is maximum where expected.
(When value level set crosses B)
y2 = Rα − Qα
∗ (ρ/2 − Bα )
y2 = (Bα − Bγ )
(ρ/2 − Bγ )

52. Right and left uncertainty
49. Fundamental lemma.
As ρ grows, maximal right uncertainty is monotonically
ˆ
decreasing and maximal left uncertainty is monotonically
increasing, and both are non-negative with minimum 0.

53. Optimal nudging
50.
Find ρ (between the bounds, obviously) such that the
ˆ
maximum resulting uncertainty, either left or right, is min.
Since both are monotonic and have minimum 0, min max
when maximum left and right uncertainties are equal.
Remark: bear in mind this (↑) is the worst case. It can
terminate immediately.
ρ is gain, but neither biased towards observations (initial or
ˆ
otherwise), nor slowly updated.
Optimal nudging is “optimal” in the sense that with this
update the maximum uncertainty range of resulting ρ values is
minimum.

54. Optimal nudging
51. Enclosing triangle into enclosing triangle.
52. Strictly smaller (both area and, importantly, resulting
uncertainty)

55. Obtaining an initial enclosing triangle
53. Setting ρ(0) = 0 and solving.
Maximizes reward irrespective of cost. (Usual RL problem)
Can be interpreted geometrically as fanning from the w axis
to find the policy with w, l coordinates that subtends the
smallest angle.
The resulting optimizer maps to a point somewhere along a
line with intercept at the origin.
54. Optimum of the SMDP problem above but not behind that
line.
Else, contradiction.

56. Obtaining an initial enclosing triangle
56. Either way, after iter. 0, uncertainty reduces in at least half.

57. Conic intersection
57. Maximum right uncertainty is a conic!
  
c −(b + a) −Cα c r
∗  −(b + a) ∗
r y1 1 c (Cβ a + Cγ b)   y1  = 0
2
−Cα c (Cβ a + Cγ b) Cα c 1
58. Maximum left uncertainty is a conic!
  
0 1 (Bγ − Cγ ) r
∗ ∗
r y2 1  1 0 −Bγ   y2  = 0
(Bγ − Cγ ) −Bγ −2Bα (Bγ − Cγ ) 1

58. Conic intersection
59. Intersecting them is easy.
60. And cheap. (Requires in principle constant time and simple
matrix operations)
61. So plug it in!

59. Termination Criteria
62.
We want to reduce uncertainty to ε.
Because it is a good idea. (Right?)
So there’s your termination condition right there.
63. Alternatively, stop when |h(k) (sI )| < .
64. In any case, if the same policy remains optimal and the sign of
its nudged value changes between iterations, stop:
It is the optimal solution of the SMDP problem.

60. Finding D
65. A quick and dirty method:
1 Maximize cost (or episode length, all costs equal 1).
2 Multiply by largest unsigned reinforcement.
66. So, at most one RL problem more.
67. If D is estimated too large, wider initial bounds and longer
computation, but ok.
68. If D is estimated too small (by other methods, of course),
points outside the triangle in w − l space. (But where?)

61. Recurring state + unichain considerations
69. Feinberg and Yang: Deciding whether the unichain condition
holds can be done in polynomial time if a recurring state exists.
70. Existence of a recurring state is common in practice.
71. (Future work) It can maybe be induced using ε–MDPs.
(Maybe).
72. At least one case in which no unichain is no problem: games.
Certainty of positive policies.
Non-positive chains.
73. Happens! (See experiments)

62. Complexity
74. Discounted RL is PAC (–efficient).
75. In problem size parameters (|S|, |A|) and 1/γ.
76. Episodic undiscounted RL is also PAC.
(Following similar arguments, but slightly more intricate
derivations)
77. So we call a PAC (–efficient) method a number of times.

63. Complexity
78. Most worstest case foreverest when choosing ρ(k) is not
reducing uncertainty.
79. Reducing it in half is a better bound for our method.
80. ... and it is a tight bound...
81. ... in cases that are nearly optimal from the outset.
82. So, at worst, log 1 calls to PAC:
ε
PAC!

64. Complexity
83. Whoops, we proved complexity! That’s a first for SMDP
(or ARRL, for that matter).
84. And we inherit convergence from invoked RL, so there’s
also that.

65. Tipically much faster
85. Worst case happens when we are ”already there.
86. Otherwise, depends, but certainly better.
87. Multi-iteration reduction in uncertainty way better than 0.5· ,
because it accumulates geometrically.
88. Empirical complexity better than the already very good upper
bound.

66. Bibliography I
S. Mahadevan. Average reward reinforcement learning: Foundations, algorithms, and empirical results. Machine
Learning, 22(1):159–195, 1996.
Reinaldo Uribe, Fernando Lozano, Katsunari Shibata, and Charles Anderson. Discount and speed/execution
tradeoffs in markov decision process games. In Computational Intelligence and Games (CIG), 2011 IEEE
Conference on, pages 79–86. IEEE, 2011.