If additionally to the usage and holding costs some fixed switching cost for chang- ing the servers will be introduced, we receive the model (see, [55]) for which the class of two-level hysteretic (q∗, q∗) optimal control rules exists, where 0 ≤ q∗ < q∗. Under these rules the slower server is switched on only when at an arrival epoch the number of jobs present in the system increases up to or above q∗ and the slower server is switched off when at a service completion epoch of the slower server the number of jobs left behind in the system is smaller or equal to q∗. But while the first steps in the numerical analysis of hysteretic optimal control rules have been done and despite the empirical evidence of the optimality of such rules, theoretical proof of this optimality is still lacking. Mentioned above hysteretic control rules are very similar to that we want to consider in this section for the PCM–problem without switching cost but with conditions (2.29) and (2.32). In view of the fixed set-up costs and these conditions, we can no longer restrict our- selves to the threshold policies with a single threshold level for the turning the slower server on, as in previous model.
As we have already mentioned numerical investigation lead to conjecture that if the servers are arranged as in (2.29) and condition (2.32) holds, the optimal control rule is characterized by threshold sequence depending on the system state x. With respect to this rule some server is used only if the current queue length is between the low and upper bounds. Moreover, it is not optimal at all to use always the fastest available server, i.e. Theorem (2.15) does not hold for the PCM–problem under condition (2.32). This can be easily shown by means of a simple system
without arrivals (λ = 0) and with two servers i, j, i > j, where ci µi > cj µj , ci ≥ cj and µi > µj. For the value function v(x) in the points Sixand Sjxwe get
v(Six) − v(Sjx) = ci µi − cj µj + c0q(x) 1 µi − 1 µj .
the last expression is nonnegative only if the queue length satisfies an inequality q(x) ≤ 1 c0 1 µj − 1 µi −1 c i µi − cj µj , where the expression in the right–hand side is greater than zero.
It is still an open problem to prove the theoretical optimality of this control rule. Nevertheless, the proposed numerical analysis confirms the assumptions which can be formulated as a following conjecture.
Conjecture 2.21 An optimal policy in the PCM–problem for the system with servers
arrangement (2.29) under condition (2.32) has such a structure, that at any state for idle server there exists two–level threshold switching rule that turns the faster but more expensive server on when the number of jobs in the system exceeds some prespecified upper level and turns such a server off when upon service completion by this server the number of jobs is below some prespecified lower level. Thus, some available in state x server j is used if q∗
j(x) ≤ q(x) < qj+1∗ .
We can rewrite Conjecture 2.21 in terms of the value function, i.e. we expect v(S0Sjx) ≤ v(S0Six) ⇒ v(Sjx) ≤ v(Six) for all j, i ∈ J0(x), j ≤ i or using the representation by means of the operators
(1 − S0)(Sj− Si)v(x) ≤ 0 for all i, j ∈ J0(x), j ≤ i. (2.46) Now we have to determine whether a server must be activated, that is we have to check the inequality (2.16). According to the optimal policy when the values of initial system parameters satisfy the conditions (2.29) and (2.32), the controller may use any of available server in some state, i.e. there exist no common rule to choose a server in each state as it was in previous cases. Therefore, the inequality
2.5. PROCESSING COST MINIMIZATION (PCM–PROBLEM) 63 (2.16) we can prove successfully only for the system with two servers and gen- eralization mainly follows from intuitive assumptions and numerical results. As before, it is necessary to show that if some function v(x) satisfies the condition (2.16) then the operator T0preserves this property.
Lemma 2.22 In the system with two servers the operator T0 preserves the prop-
erty (2.16) of the monotonicity increments of the function.
Proof: To prove that the operator T0 preserves the property (2.16) we have to prove that if this property holds for some function v(x) it holds also for the func- tion ˆv(x) = T0v(x).
Note, that for each point x it is possible to make only two decisions, namely, to send a job to the queue or to switch on some server. The last action depends on the queue length.
Using the definition (2.9) of the operator T0 it is necessary to check the in- equality
(1 − S0)(S0 − Sk)ˆv(x) = ˆv(S0x) − ˆv(Skx) − ˆv(S02x) + ˆv(S0Skx) = min{v(SlS0x) : l ∈ J0(x)} − min{v(SlSkx) : l ∈ J0(Skx)}− min{v(SlS02x) : l ∈ J0(x)} + min{v(SlS0Skx) : l ∈ J0(Skx)} ≤ 0, where for the system with two servers k = {1, 2}. As usual, we prove the result for any point x ∈ E and divide it into several cases.
1. For the case when the optimal policies at the points Skxand S02x(where ˆv(x) is involved with the negative sign) are equal, i.e. f(Skx) = f (S02x) = f we have
(1 − S0)(S0− Sk)ˆv(x) = ˆv(S0x) − ˆv(Skx) − ˆv(S02x) + ˆv(S0Skx) = ˆ
v(S0x) − v(SfSkx) − v(SfS02x) + ˆv(S0Skx) ≤ v(SfS0x) − v(SfSkx) − v(SfS02x) + v(SfS0Skx) = (1 − S0)(S0− Sk)v(Sfx) ≤ 0.
2. For the case, when f(Skx) = 0and f(S02x) = kwe get the inequality (1 − S0)(S0− Sk)ˆv(x) = ˆv(S0x) − ˆv(Skx) − ˆv(S02x) + ˆv(S0Skx) = ˆ
v(S0x) − v(S0Skx) − v(S02Skx) + ˆv(S0Skx) ≤ v(S0Skx) − v(S0Skx) − v(S02Skx) + v(S02Skx) = 0.
3. If f(Skx) = 0and f(S02x) = l 6= k, the chain of relations is the following (1 − S0)(S0− Sk)ˆv(x) = ˆv(S0x) − ˆv(Skx) − ˆv(S02x) + ˆv(S0Skx) = ˆ v(S0x) − v(S0Skx) − v(S02Slx) + ˆv(S0Skx) ≤ v(S0Slx) − v(S0Skx) − v(S02Slx) + v(S02Skx) = (1 − S0)(Sl− Sk)v(S0x) ≤ 0,
by virtue of the property (2.46).
4. If f(Skx) = l 6= 0and f(S02x) = 0, the result will be the following (1 − S0)(S0− Sk)ˆv(x) = ˆv(S0x) − ˆv(Skx) − ˆv(S02x) + ˆv(S0Skx) = ˆ
v(S0x) − v(SlSkx) − v(S03x) + ˆv(S0Skx).
To determine the sign of the last expression we take the operator under consider- ation at the points Skxand S0x, that is
(1 − S0)(S0− Sl)v(Skx) = v(S0Skx) − v(SlSkx) − v(S02Skx) + v(S0SkSlx) ≤ 0, (1 − S0)(S0− Sk)v(S0x) = v(S02x) − v(S0Skx) − v(S03x) + v(S02Skx) ≤ 0, respectively. Now summing these inequalities we get
v(S02x) − v(SlSkx) − v(S03x) + v(S0SkSlx) ≤ 0.
Using the last inequality we can prove the statement for this case, namely, ˆ
v(S0x) − v(SlSkx) − v(S03x) + ˆv(S0Skx) ≤ v(S02x) − v(SlSkx) − v(S03x) + v(S0SkSlx) ≤ 0. 5. Finally in the case when f(Skx) = l 6= 0and f(S02x) = kwe have (1 − S0)(S0− Sk)ˆv(x) = ˆv(S0x) − ˆv(Skx) − ˆv(S02x) + ˆv(S0Skx) = ˆ
v(S0x) − v(SlSkx) − v(S02Skx) + ˆv(S0Skx) ≤
v(S0Skx) − v(SkSlx) − v(S02Skx) + v(S0SkSlx) = (1 − S0)(S0− Sl)v(Skx) ≤ 0. For the boundary points q(x) = N and J0(x) = ∅, the inequality (2.16) is also satisfied because the shift operators do not drive the point x outside the admissible set of states. Finally, the operators Tjalso retain the (S0, Sk)-submodularity of the functions by virtue of the fact that they are defined by (2.9).
Now we can prove that the value function of the model v = {v(x) : x ∈ E} is (S0, Sk)-submodular. This assertion follows from the fact that the property of (S0, Sk)-submodularity is retained for linear operations defining the operator B as in (2.5.2), the function c(x) is (S0, Sk)-submodular and successive approxima- tions Bnc(x)converge monotonously to the value function v(x).