**2.3 Static reservation**

In the cases *R* = 0 and *R* = 2, this strategy does not differ from the strategy 1 above. Therefore, it remains to consider the case *R* = 1. This Markov model has four states (Fig. 2):


Fig. 2. Two channels, static reservation: the model and the state transition diagram.

From linear balance equations under the assumption of statistical equilibrium we obtain the state probabilities (up to a normalizing factor):

$$\begin{aligned} P\_{00} &= 2 + 2A + B \\ P\_{01} &= (A + B)^2 + 2B \\ P\_{10} &= A(2 + A + B) \\ P\_{11} &= A(\left(A + B\right)^2 + A + 2B) \end{aligned}$$

The average revenue takes the form:

$$H\_4 = \frac{A\left(2 + 4A + 3B + 2A^2 + 3AB + B^2\right) \cdot K + B\left(2 + 4A + B + A^2 + AB\right) \cdot 1}{2 + 4A + 3B + 3A^2 + 5AB + B^2 + A^3 + 2A^2B + AB^2}$$

Our aim is to prove that the static reservation strategy cannot be more profitable than the dynamic reservation strategy. That is, we must prove that for any values of *A*, *B* and *K* at least one of the values of *H*<sup>0</sup> and *H*<sup>1</sup> are not smaller than *H*4.

It is easy to verify that by replacing *K* in this formula by *K*<sup>1</sup> from expression (4) we obtain *H*<sup>4</sup> = 2. We get the same result by substituting this value of *K*<sup>1</sup> for *K* in expressions (2) and (3), i.e. the equalities *H*<sup>0</sup> = *H*<sup>1</sup> = *H*<sup>4</sup> = 2 are true for any *A*, *B* and given *K* = *K*1. Furthermore, we note that *H*0, *H*<sup>1</sup> and *H*<sup>4</sup> are linear functions of *K*, i.e. straight lines. Note that for *K < K*<sup>1</sup> the inequalities *H*<sup>0</sup> *< H*<sup>4</sup> *< H*<sup>1</sup> are true. Hence these three straight lines intersect at one point, and the straight line *H*<sup>4</sup> is located between the two others. This means that for any values of *A*, *B* and *K*, at least one of the values of *H*<sup>0</sup> and *H*<sup>1</sup> is not less than the value *H*4, q.e.d.

### **2.4 New call bounding scheme (strategy 4)**

In case of a 2-channel system, the only nontrivial variant of strategy 4 (Restriction on number of B-calls admitted) is: no more than one B-call. The state transition diagram of this model 6 Will-be-set-by-IN-TECH

In the cases *R* = 0 and *R* = 2, this strategy does not differ from the strategy 1 above. Therefore,

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10 11

A + B

A A 1 1

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B

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1

00 01

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<sup>2</sup> + *A* + 2*B*)

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it remains to consider the case *R* = 1. This Markov model has four states (Fig. 2):

**2.3 Static reservation**

(00) – both channels are free,

First channel

(01) – the common channel is engaged, (10) – the guard channel is engaged, (11) – both channels are engaged.

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state probabilities (up to a normalizing factor):

one of the values of *H*<sup>0</sup> and *H*<sup>1</sup> are not smaller than *H*4.

**2.4 New call bounding scheme (strategy 4)**

The average revenue takes the form:

A

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B

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Second channel

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Fig. 2. Two channels, static reservation: the model and the state transition diagram.

*P*<sup>00</sup> = 2 + 2*A* + *B P*<sup>01</sup> = (*A* + *B*)<sup>2</sup> + 2*B P*<sup>10</sup> = *A*(2 + *A* + *B*) *P*<sup>11</sup> = *A*((*A* + *B*)

From linear balance equations under the assumption of statistical equilibrium we obtain the

*<sup>H</sup>*<sup>4</sup> <sup>=</sup> *<sup>A</sup>* (<sup>2</sup> <sup>+</sup> <sup>4</sup>*<sup>A</sup>* <sup>+</sup> <sup>3</sup>*<sup>B</sup>* <sup>+</sup> <sup>2</sup>*A*<sup>2</sup> <sup>+</sup> <sup>3</sup>*AB* <sup>+</sup> *<sup>B</sup>*2) · *<sup>K</sup>* <sup>+</sup> *<sup>B</sup>* (<sup>2</sup> <sup>+</sup> <sup>4</sup>*<sup>A</sup>* <sup>+</sup> *<sup>B</sup>* <sup>+</sup> *<sup>A</sup>*<sup>2</sup> <sup>+</sup> *AB*) · <sup>1</sup> 2 + 4*A* + 3*B* + 3*A*<sup>2</sup> + 5*AB* + *B*<sup>2</sup> + *A*<sup>3</sup> + 2*A*2*B* + *AB*<sup>2</sup>

Our aim is to prove that the static reservation strategy cannot be more profitable than the dynamic reservation strategy. That is, we must prove that for any values of *A*, *B* and *K* at least

It is easy to verify that by replacing *K* in this formula by *K*<sup>1</sup> from expression (4) we obtain *H*<sup>4</sup> = 2. We get the same result by substituting this value of *K*<sup>1</sup> for *K* in expressions (2) and (3), i.e. the equalities *H*<sup>0</sup> = *H*<sup>1</sup> = *H*<sup>4</sup> = 2 are true for any *A*, *B* and given *K* = *K*1. Furthermore, we note that *H*0, *H*<sup>1</sup> and *H*<sup>4</sup> are linear functions of *K*, i.e. straight lines. Note that for *K < K*<sup>1</sup> the inequalities *H*<sup>0</sup> *< H*<sup>4</sup> *< H*<sup>1</sup> are true. Hence these three straight lines intersect at one point, and the straight line *H*<sup>4</sup> is located between the two others. This means that for any values of *A*, *B* and *K*, at least one of the values of *H*<sup>0</sup> and *H*<sup>1</sup> is not less than the value *H*4, q.e.d.

In case of a 2-channel system, the only nontrivial variant of strategy 4 (Restriction on number of B-calls admitted) is: no more than one B-call. The state transition diagram of this model looks similar to that in Fig. 2, and the expression for average revenue is:

$$H\_5 = \frac{2\left(A\left(A+B+1\right)\cdot K + B\left(A+1\right)\right)}{2\left(A+B+1\right) + A\left(A+2B\right)}$$

This strategy is similar to strategy 3 (Static reservation). As shown in Fig. 3, the straight lines *H*<sup>4</sup> and *H*<sup>5</sup> are close: up to point *K <* 1.25 strategy 4 is a little more profitable, and from 1.25 *< K* strategy 3 is more profitable. But for any *K*, strategy 4 is worse than the optimal strategy 1, since *H*<sup>5</sup> *< H*<sup>0</sup> up to *K <* 1.25 and *H*<sup>5</sup> *< H*<sup>2</sup> from 1.25 *< K*.

Fig. 3. Dependence of the revenue from handover call cost *K* for three models: *D* – dynamic reservation, *F* – static reservation, and *L* – restriction on number of admitted *B*-calls.

Conclusion. Hence it has been proved mathematically that the optimal service strategy in a two-channel system is dynamic reservation. Graphically, this fact is illustrated in Fig. 3 :


Problem 1. It is noteworthy that all four straights in Fig. 2 intersect at one point. This experimental fact deserves further study for systems with more than two channels.
