**2. Analysis of CAC strategies in single tier networks: Two channel case**

We compare four basic CAC schemes by examining new call and handover call blocking probabilities:


We consider a *N*-channel cell without waiting positions and two Poisson call flows: handover call flow of intensity *A* and new call flow of intensity *B*. The holding times are exponentially distributed with mean value equal one. The exponential distribution simplifies the formulae and the fact that handover calls already has been served for some time before entering the cell considered, does not influence the remaining service time, as the exponential distribution is without memory. Our optimization criteria is the same for all schemes: to get the maximum revenue if each served *A*-call costs *K* units (*K >* 1) and each served *B*-call costs one unit.

## **2.1 Dynamic reservation strategy is better than fractional dynamic reservation strategy**

Let us start from the 2*nd* strategy: fractional dynamic reservation. This strategy seems to be more general than the dynamic reservation strategy, but we shall prove that such statement is not true. The system is modeled by a three-state Markov process (Fig. 1) having the following parameters: Number of channels *N* = 2, *A* and *B* = call flow intensities, *pi* = the probability of accepting *B*-calls for service in state *i*.

Fig. 1. Fractional dynamic reservation : two-channel state transition diagram.

The stationary state probabilities *Pi* are defined by equations (up to a normalization factor):

$$\begin{aligned} P\_0 &= 1\\ P\_1 &= A + B p\_0\\ P\_2 &= \frac{(A + B p\_0)(A + B p\_1)}{2} \end{aligned}$$

The average revenue *H* equals:

4 Will-be-set-by-IN-TECH

to take account of correlated streams. Sections 5.7 and 5.8 contain next step in ERT-method extension, namely the application of Neal's theory (Neal, 1971), and his formulas for

We compare four basic CAC schemes by examining new call and handover call blocking

• Strategy 1 - Dynamic reservation: the cutoff priority scheme is to reserve some channels for handover calls. Whenever a channel is released, it is returned to the common pool of

• Strategy 2 - Fractional dynamic reservation: the fractional guard channel scheme (the new call thinning scheme) is to admit a new call with a certain probability which depends on

• Strategy 3 - Static (fixed) reservation: all channels allocated to a cell are divided into two groups: one to be used by all calls and the other for handover calls only (the rigid

• Strategy 4 - New call bounding scheme: limitation of the number of simultaneous new

We consider a *N*-channel cell without waiting positions and two Poisson call flows: handover call flow of intensity *A* and new call flow of intensity *B*. The holding times are exponentially distributed with mean value equal one. The exponential distribution simplifies the formulae and the fact that handover calls already has been served for some time before entering the cell considered, does not influence the remaining service time, as the exponential distribution is without memory. Our optimization criteria is the same for all schemes: to get the maximum revenue if each served *A*-call costs *K* units (*K >* 1) and each served *B*-call costs one unit.

**2.1 Dynamic reservation strategy is better than fractional dynamic reservation strategy**

❥ A + B p<sup>0</sup>

✧✦

1

Fig. 1. Fractional dynamic reservation : two-channel state transition diagram.

*P*<sup>1</sup> = *A* + *Bp*<sup>0</sup>

*P*<sup>0</sup> = 1

❨

★✥

Let us start from the 2*nd* strategy: fractional dynamic reservation. This strategy seems to be more general than the dynamic reservation strategy, but we shall prove that such statement is not true. The system is modeled by a three-state Markov process (Fig. 1) having the following parameters: Number of channels *N* = 2, *A* and *B* = call flow intensities, *pi* = the probability

✧✦

2

❨

0 1 2

The stationary state probabilities *Pi* are defined by equations (up to a normalization factor):

*<sup>P</sup>*<sup>2</sup> <sup>=</sup> (*<sup>A</sup>* <sup>+</sup> *Bp*0)(*<sup>A</sup>* <sup>+</sup> *Bp*1) 2

✧✦

★✥

❥ A + B p<sup>1</sup>

★✥

**2. Analysis of CAC strategies in single tier networks: Two channel case**

covariance of correlated streams.

the number of busy channels.

division-based CAC scheme).

calls admitted to the network.

of accepting *B*-calls for service in state *i*.

probabilities:

channels.

$$\begin{split}H &= \frac{A(P\_0 + P\_1) \cdot K + B(p\_0 P\_0 + p\_1 P\_1) \cdot 1}{P\_0 + P\_1 + P\_2} \\ &= \frac{2(AK + Bp\_0 + (A + Bp\_0)(AK + Bp\_1))}{2 + (A + Bp\_0)(2 + A + Bp\_1)}\end{split} \tag{1}$$

This expression is the ratio of two polynomials, each one with probabilities *p*<sup>0</sup> and *p*<sup>1</sup> included in the first power. Consider the expression (1) as a function of one of the probabilities *p*. After multiplying by the relevant constant we can get it into the form (*p* + *a*)/(*bp* + *c*). The derivative of this expression has the form (*ac*)/(*bp* + *c*)2. Consequently, the expression (1) has invariable sign in the range of values (0, 1), and its extreme values are located at the ends of the interval, i.e. probabilities *p*<sup>0</sup> and *p*<sup>1</sup> can only take values 0 or 1. Therefore, the dynamic reservation strategy has advantage over fractional dynamic reservation strategy.

### **2.2 Dynamic reservation strategy is better than static reservation strategy**

This model (strategy 1) is a special case of the previous one: you can reserve 0, 1 or 2 channels, corresponding to choice of probability (*p*0, *p*1) in the form of (1,1) (1,0) or (0,0), which, in own order, corresponds to the values of *R* of 0, 1 or 2. Accordingly, the revenue from formula (1) takes the form:

$$H\_0 = \frac{2(AK+B)(1+A+B)}{2(1+A+B)+(A+B)^2} \tag{2}$$

$$\begin{aligned} H\_1 &= \frac{2(B + AK(1 + A + B))}{2 + (A + B)(2 + A)} \\ H\_2 &= \frac{2AK(1 + A)}{2 + 2A + A^2} \end{aligned} \tag{3}$$

How many channels should be reserved? This depends on the parameters *K*, *A* and *B*. To find the optimal value of *R*, one should solve two equations pointing to the boundary of *K*:

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

$$H\_0 = H\_1 \quad \text{and} \quad H\_1 = H\_2 \dots$$

We get:

$$K\_1 = 1 + \frac{2 + A + B}{A(1 + A + B)}\tag{4}$$

$$K\_2 = 1 + \frac{2(A+1)}{A^2}$$

It is easy to verify that for any values of *A* and *B*, the inequality *K*<sup>1</sup> *< K*<sup>2</sup> is true since:

$$K\_2 - K\_1 = \frac{2 + 2A + 2B + A^2 + AB}{A^2(1 + A + B)} \dots$$

Hence we have the following solution for optimal reservation *R* at *N* = 2 channels:

*R* = 0 if *K < K*<sup>1</sup> *R* = 1 if *K*<sup>1</sup> *< K < K*<sup>2</sup> *R* = 2 if *K*<sup>2</sup> *< K*

looks similar to that in Fig. 2, and the expression for average revenue is:

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*.

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

Call Admission Control in Cellular Networks 117

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

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.

two-channel system is dynamic reservation. Graphically, this fact is illustrated in Fig. 3 :

• reservation *R* = 0, optimal for values of *K* ≤ 1.25, reservation line *D*(*R* = 0), • reservation *R* = 1, optimal for values of *K* ≥ 1.25, reservation line *D*(*R* = 1).

experimental fact deserves further study for systems with more than two channels.

**3. Comparison of four strategies in single tier network: Common case**

Hence it has been proved mathematically that the optimal service strategy in a

It is noteworthy that all four straights in Fig. 2 intersect at one point. This

For a *N*-channel loss system where *B*-calls are accepted with probability *pi*,

depending on the number of busy channels *i* (*i* = 0, 1, . . . , *N*), the optimal fractional dynamic

Conclusion.

Problem

Theorem.

 1.

**3.1 Fractional dynamic reservation**

reservation is limited to probabilities *pi* equal to 0 or 1.
