2.3 Pricing with Service Level Agreements
3.1.4 Finding steady state probability
Let P rt(s1, s2, t) be the probability of service provider 1 in state s1 and service provider 2 in state s2 at time t. Let dt be an infinitely small time such that the probability of two or more events to take place in time dt is negligible. The method of findingP rt(s1, s2, t+dt)is as follows. The probability of service providers being in
some state at timet, P rt(s3, s2, t)is multiplied by the probability of state transition in time dt to state s1 from state s3.It is similar for the case when the state is P rt(s1, s3, t). As only one event can occur in time dt, we need to only consider single state changes. If the current stae is P rt(s1, s2, t) then it is multiplied by the probability of no state change in time dt. Probability of no state change in timedt is (1-probability of state change in time dt). All these terms are summed for each possible value of P rt(..., t) and the result is P rt(s1, s2, t+dt). P rt(s1, s2, t+dt) is written in terms of P rt(..., t)as given below.
P rt(s1, s2, t+dt) = X
s3,s16=s3
{P rt(s3, s2, t)× (Probability of state change from s3 to s1 in time dt)}
+ X
s3,s26=s3
{P rt(s1, s3, t)× (Probability of state change from s3 to s2 in time dt)}
+P rt(s1, s2, t) (1− X
s3,s16=s3
{ (Probability of state change from s1 to s3 in time dt)}
− X
s3,s26=s3
{
(Probability of state change from s2 to s3 in time dt)}) (3.3) Probability of state change froms3 tos1 and s3to s2depends on the event due to which the change took place and it is the rate of change multiplied by dt. If the number of events is two or more, the probability is zero because the time dt is so small that the probability of more than one event is negligible. If one event can change state from s3 to s1 or s3 to s2, this probability is the event rate multiplied by dt.
If the event is an arrival then it isλ×dt when all clients come to service provider 1 or λ2 ×dt when half the clients come to service provider 1, otherwise it is zero.
Which alternative will hold depends on the values of the decision matrix C(), as can be seen in the equation above. If s3is(m3, n3),n3is less than B1 and the event is a departure, the departure probability isS(1)×d×n3×dt. The probability of a client opening a session is S(0)×(m3−n3)×dt and the probability that a client closes a
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session and then becomes idle is S(1)×(1−d)×n3×dt. Ifn3is greater thanB1, it is replaced by B1. For example the departure probability becomesS(1)×d×B1×dt.
The same ways3tos2probabilities are written. Similarly probability of state change froms1tos3means the probability of state change froms1to any other state. Ifs1 is (m, n), it is the sum of three terms: λ×dt or λ2 ×dt or zero, S(0)×(m−n)×dt and S(1)×n×dt or S(1)×B1×dt .
The equation is rearranged such that P rt(s1,s2,t+dt)−P rt(s1,s2,t)
dt comes to the left hand side. Probability of state change from s1 and s2 to s1 and s2 when divided by dt, becomes the rate of state change from s1 and s2 tos1 and s2. Since we are finding a solution for the steady state, the left hand side becomes zero and time is removed from the equation. The following equation is then obtained.
0 = X
s3,s36=s1
{P r(s3, s2)×(Rate of change of state of service provider 1 from s3 to s1)}
+ X
s3,s36=s2
{P r(s1, s3)×(Rate of change of state of service provider 2 from s3 to s2)}
+P r(s1, s2)×
(− X
s3,s36=s1
{Rate of change of state of service provider 1 from s1 to s3}
− X
s3,s36=s2
{Rate of change of state of service provider 2 from s2 to s3})
The expanded equation is given below. Let states1be the state (m, n). And the state s2 be the state (m0, n0). The expanded equation contains three sets of terms and these are given separately inside three sets of curly brackets: {}. The first set contains all the possible transitions of service provider 1 to the present state (m, n).
The second set contains all the possible transitions of service provider 2 to the present state (m0, n0). The third set represents the possibility of both the service provider being in the present state (m, n, m0, n0) and rate at which any service provider leaves its present state.
The explanation of the first set is as follows. P r(m − 1, n −1, m0, n0) is the probability that service provider 1 is in state (m−1, n−1) and service provider 2 is
in state (m0, n0). It is multiplied by the mean arrival rate with which an arrival takes place for service provider 1. This rate is λ when the arrival is accepted by service provider 1 (given byC(1, m−1, n−1)≤0) and the price charged by service provider 2 is more than the price charged by service provider 1 (given by C(1, m−1, n−1)>
C(2, m0, n0)) or service provider 2 rejects an arriving client (given byC(2, m0, n0) = 1).
This arrival rate is λ2 when the arrival is accepted by service provider 1 (given by C(1, m−1, n−1)≤0) and the price charged by both the service providers are the same (given by C(1, m−1, n−1) = C(2, m0, n0)). Similarly, (m+ 1, n+ 1, m0, n0) is the state at which service provider 1 had an extra connected client and the rate at which this client departs is S(1)×(n+ 1) ×d when n + 1 is greater than the bandwidth B1 of service provider 1 or S(1)×B1 ×d otherwise. (m, n+ 1, m0, n0) represents the state where service provider 1 has one extra client in session and this client closes its session with the rateS(1)×(n+1)×(1−d)whenn+1is greater than the bandwidth of service provider 1 orS(1)×B1×(1−d)otherwise. (m, n−1, m0, n0) represents the possibility that service provider 1 has one client less in session (is idle) and this client opens a session with the rate S(0)×(m−n+ 1).
The second set of terms are identical to the first set of terms and therefore, the explanation is the same. As already mentioned, the third set represents the possibility of both the service providers being in the present state (m, n, m0, n0) and rate at which any service provider leaves its present state. The rate at which an arrival takes place is λ when a service provider accepts an arriving client (given by C(1, m, n) ≤ 0 or C(2, m0, n0)≤0). The rate at which an idle client opens a session for service provider 1 isS(0)×(m−n)and for service provider 2 isS(0)×(m0−n0). The rate at which a client in session closes a session for service provider 1 isS(1)×
(n , n≤B1 B1 , otherwise
!
and for service provider 2 is S(1)×
(n0 , n0 ≤B2 B2 , otherwise
! .
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0 = {
P r(m−1, n−1, m0, n0) , m6= 0, n6= 0
0 , otherwise
×
λ C(1, m−1, n−1)≤0
,{C(1, m−1, n−1)> C(2, m0, n0) or C(2, m0, n0) = 1}
, m6= 0, n6= 0
λ
2 , C(1, m−1, n−1)≤0
, C(1, m−1, n−1) =C(2, m0, n0) , m6= 0, n6= 0
0 , otherwise
+
P r(m+ 1, n+ 1, m0, n0)×S(1)×(n+ 1)×d , m < mmax , n+ 1 ≤B1 P r(m+ 1, n+ 1, m0, n0)×S(1)×B1×d , m < mmax
, n+ 1 > B1
0 , otherwise
+
P r(m, n+ 1, m0, n0)×S(1)×(n+ 1)×(1−d) , n < m , n+ 1 ≤B1 P r(m, n+ 1, m0, n0)×S(1)×B1×(1−d) , n < m
, n+ 1 > B1
0 , otherwise
+
P r(m, n−1, m0, n0)×S(0)×(m−n+ 1) , n >0
0 , otherwise
+{
P r(m, n, m0−1, n0 −1) , m0 6= 0, n0 6= 0
0 , otherwise
λ C(2, m0 −1, n0−1)≤0
,{C(2, m0−1, n0 −1)> C(1, m, n) or C(1, m, n) = 1}
, m0 6= 0, n0 6= 0
λ
2 , C(2, m0−1, n0−1)≤0
, C(2, m0−1, n0−1) =C(1, m, n) , m0 6= 0, n0 6= 0
0 , otherwise
+
P r(m, n, m0+ 1, n0+ 1)×S(1)×(n0+ 1)×d , m0 < mmax , n0+ 1≤B2 P r(m, n, m0+ 1, n0+ 1)×S(1)×B2×d , m0 < mmax
, n0+ 1> B2
0 , otherwise
+
P r(m, n, m0, n0+ 1)×S(1)×(n0+ 1)×(1−d) , n0 < m0 , n0+ 1≤B2 P r(m, n, m0, n0+ 1)×S(1)×B2×(1−d) , n0 < m0
, n0+ 1> B2
0 , otherwise
+
P r(m, n, m0, n0−1)×S(0)×(m0−n0+ 1) , n0 >0
0 , otherwise
}
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−{P r(m, n, m0, n0)
×(
λ , C(1, m, n)≤0 or C(2, m0, n0)≤0 0 , otherwise
+S(0)×(m−n) +S(0)×(m0−n0)
+S(1)×
(n , n≤B1 B1 , otherwise
!
+S(1)×
(n0 , n0 ≤B2 B2 , otherwise
!
)
} (3.4)
Since the sum of probabilities is 1, another equation is : X
s1,s2
P r(s1, s2) = 1 (3.5)
The values ofP r()are found by solving equations 3.4 and 3.5. The method used is LU Decomposition [7].