**6. Numerical example**

#### **6.1 Statement**

In a referral Hospital, an ophthalmologist doctor consults patients on odd days each week from 10h00 to 13h30. The patients arrive there following a Poisson distribution with parameter ~*λ* and the doctor's consultation following an expo-negative distribution with parameter *μ* The fuzzy parameters ~*λ* and *μ* are such that <sup>~</sup>*<sup>λ</sup> <sup>μ</sup>*<sup>~</sup> is approximately 0.4. We note that <sup>~</sup>*<sup>ρ</sup>* <sup>¼</sup> <sup>~</sup>*<sup>λ</sup> <sup>μ</sup>*<sup>~</sup> is the fuzzy traffic intensity. We further warn that this traffic intensity is a triangular fuzzy number and is denoted by ~*ρ* ¼ ð Þ 0*:*3j0*:*4j0*:*5 .

#### **6.2 Questions**


#### **6.3 Solution**

A careful reading of our example reveals that it is a fuzzy Markovian queue noted FM/FM/1 with a single server and infinite capacity. The fuzzy traffic intensity being about 0.4 implies that the fuzzy rates ~*λ* and *μ*~ are about 2 and 5, respectively. By assumption, since the fuzzy traffic intensity ~*ρ* is a triangular fuzzy number, the rates ~*λ* and *μ*~ are also triangular fuzzy numbers and can be written (cf. Remark 1, item a):

$$
\tilde{\lambda} = (\mathbf{1}|\mathbf{2}|\mathbf{3}) \quad \text{and} \quad \tilde{\mu} = (4|\mathbf{5}|\mathbf{6}).
$$

Thus, in fuzzy model, the rates ~*λ* and *μ*~ are fuzzy variables and the performance measures *N*~ *<sup>S</sup>* and *T*~ *<sup>S</sup>* are fuzzy time functions defined by:

$$\tilde{N}\_S(t) = \tilde{f}\_1(t, \tilde{\lambda}, \tilde{\mu}) = \frac{\tilde{\lambda} \left(1 - \exp\left(-(\tilde{\mu} - \tilde{\lambda})t\right)\right)}{\tilde{\mu} - \tilde{\lambda}} \tag{33}$$

$$\tilde{T}\_{\rm S}(t) = \tilde{f}\_{\rm 2}(t, \tilde{\lambda}, \tilde{\mu}) = \frac{\tilde{\lambda} - \tilde{\lambda} \exp\left(-(\tilde{\mu} - \tilde{\lambda})\right)}{\tilde{\mu} - \tilde{\lambda} \left[\tilde{\lambda} + (\tilde{\mu} - \tilde{\lambda}) \exp\left(-(\tilde{\mu} - \tilde{\lambda})t\right)\right]} \tag{34}$$

To evaluate these performance parameters by L–R method, we proceed as follows:

	- <sup>~</sup>*<sup>λ</sup>* <sup>¼</sup> ð Þ <sup>1</sup>j2j<sup>3</sup> and *<sup>μ</sup>*<sup>~</sup> <sup>¼</sup> ð Þ <sup>4</sup>j5j<sup>6</sup> and we have according to (21):

$$
\tilde{\lambda} = \langle \mathbf{2}, \mathbf{1}, \mathbf{1} \rangle\_{L-R} \quad \text{and} \quad \tilde{\lambda} = \langle \mathbf{5}, \mathbf{1}, \mathbf{1} \rangle\_{L-R} \tag{35}
$$

ii. Let us substitute the expressions of (35) into (33) and (34), and use the operations in (14)–(17) to obtain successively:

$$\begin{split} \bar{N}\_{S}(t) &= \frac{\bar{\lambda}(1 - \exp\left(-(\bar{\mu} - \bar{\lambda})t\right))}{\bar{\mu} - \bar{\lambda}} \\ &= \frac{\langle 2, 1 \rangle\_{L-R} - \langle 2, 1 \rangle\_{L-R} \exp(-X)}{\langle 5, 1 \rangle\_{L-R} - \langle 2, 1 \rangle\_{L-R}} \qquad \text{with} \quad X = -\langle 3, 2, 2 \rangle\_{L-R} t \\ &= \frac{\langle 2 - 2 \exp(X), 1 + \exp(X), 1 + \exp(X) \rangle\_{L-R}}{\langle 3, 2 \rangle\_{L-R}} \\ &\approx \left\langle \frac{2 - 2e^X}{3}, \frac{(2 - 2e^X)X2}{3(3 + 2)} + \frac{1 + e^X}{3} - \frac{(1 + e^X)X2}{3(3 + 2)}, \frac{(2 - 2e^X)X2}{3(3 + 2)} + \frac{1 + e^X}{3} + \frac{(2 - 2e^X)X2}{3(3 + 2)} \right\rangle\_{L-R} \\ &= \left\langle \frac{2 - 2e^X}{3}, \frac{7 - e^X}{15}, \frac{7 - e^X}{3} \right\rangle\_{L-R} \\ &= \langle 0.67 - 0.67e^{-\frac{1}{3}t}, 0.47 - 0.1e^{-\frac{1}{3}t}, 2.3 - 0.3e^{-\frac{1}{3}t} \rangle\_{L-R}, \quad X = -3t \end{split}$$

*Computing the Performance Parameters of the Markovian Queueing System FM/FM/1… DOI: http://dx.doi.org/10.5772/intechopen.110388*

since *X* ¼ �h i 3,2,2 *tL*�*<sup>R</sup>* ¼ �3*t* when we change the L–R writing of *X* to the *α*-cut writing, with *α* ¼ 0ð Þ *t* ≥0

*<sup>T</sup>*<sup>~</sup> *<sup>S</sup>*ðÞ¼ *<sup>t</sup>* <sup>~</sup>*<sup>λ</sup>* � <sup>~</sup>*<sup>λ</sup>* exp � *<sup>μ</sup>*<sup>~</sup> � <sup>~</sup>*<sup>λ</sup> <sup>μ</sup>*<sup>~</sup> � <sup>~</sup>*<sup>λ</sup>* <sup>~</sup>*<sup>λ</sup>* <sup>þ</sup> *<sup>μ</sup>*<sup>~</sup> � <sup>~</sup>*<sup>λ</sup>* exp � *<sup>μ</sup>*<sup>~</sup> � <sup>~</sup>*<sup>λ</sup> <sup>t</sup>* <sup>¼</sup> h i 2,1,1 *<sup>L</sup>*�*<sup>R</sup>* � h i 2,1,1 *<sup>L</sup>*�*<sup>R</sup>* expð Þ �*<sup>X</sup>* h i 3,2,2 *<sup>L</sup>*�*<sup>R</sup>* h i 2,1,1 *<sup>L</sup>*�*<sup>R</sup>* � h i 3,2,2 *<sup>L</sup>*�*<sup>R</sup>* expð Þ �*<sup>X</sup>* with *<sup>X</sup>* ¼ �h i 3,2,2 *<sup>L</sup>*�*<sup>R</sup>* ¼ �3*<sup>t</sup>* <sup>¼</sup> <sup>2</sup> � <sup>2</sup>*e<sup>X</sup>*, 1 <sup>þ</sup> *eX*, 1 <sup>þ</sup> *<sup>e</sup><sup>X</sup> L*�*R* h i 3,2,2 *<sup>L</sup>*�*<sup>R</sup>* <sup>2</sup> � <sup>3</sup>*eX*, 1 <sup>þ</sup> <sup>2</sup>*eX*, 1 <sup>þ</sup> <sup>2</sup>*eX* h i*<sup>L</sup>*�*ReX* <sup>¼</sup> <sup>2</sup> � <sup>2</sup>*e<sup>X</sup>*, 1 <sup>þ</sup> *<sup>e</sup><sup>X</sup>*, 1 <sup>þ</sup> *eX L*�*R* h6 þ 9*eX*, 5 þ 8*eX*, 9 þ 16*eX* >*<sup>L</sup>*�*<sup>R</sup>* <sup>≈</sup> <sup>2</sup> � <sup>2</sup>*e<sup>X</sup>* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* , <sup>2</sup> � <sup>2</sup>*eX* <sup>9</sup> <sup>þ</sup> <sup>16</sup>*e<sup>X</sup>* ð Þ <sup>6</sup> <sup>þ</sup> <sup>9</sup> <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* <sup>þ</sup> <sup>9</sup> <sup>þ</sup> <sup>16</sup>*eX* ½ � <sup>þ</sup> <sup>1</sup> <sup>þ</sup> *<sup>e</sup><sup>X</sup>* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* � <sup>1</sup> <sup>þ</sup> *<sup>e</sup><sup>X</sup>* <sup>9</sup> <sup>þ</sup> <sup>16</sup>*e<sup>X</sup>* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* ð Þ <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* � <sup>5</sup> � <sup>16</sup>*eX* ½ �, <sup>2</sup> � <sup>2</sup>*e<sup>X</sup>* <sup>5</sup> <sup>þ</sup> <sup>8</sup>*e<sup>X</sup>* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* ð Þ <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* � <sup>5</sup> � <sup>8</sup>*eX* ½ � <sup>þ</sup> <sup>1</sup> <sup>þ</sup> *<sup>e</sup><sup>X</sup>* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* <sup>þ</sup> <sup>1</sup> <sup>þ</sup> *<sup>e</sup><sup>X</sup>* <sup>5</sup> <sup>þ</sup> <sup>8</sup>*e<sup>X</sup>* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* ð Þ <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* � <sup>5</sup> � <sup>8</sup>*eX* ½ � *L*�*R* <sup>¼</sup> <sup>2</sup> � <sup>2</sup>*eX* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* , <sup>24</sup> <sup>þ</sup> <sup>29</sup>*e<sup>X</sup>* � <sup>23</sup>*eX* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* ð Þ <sup>15</sup> <sup>þ</sup> <sup>25</sup>*eX* ð Þ, <sup>16</sup> <sup>þ</sup> <sup>21</sup>*e<sup>X</sup>* � <sup>7</sup>*e*<sup>2</sup>*<sup>X</sup>* <sup>6</sup> <sup>þ</sup> <sup>9</sup>*eX* ð Þ <sup>1</sup> <sup>þ</sup> *eX* ð Þ *L*�*R*

with *X* ¼ �h i 3,2,2 *<sup>L</sup>*�*Rt* ¼ �3*t* and *t*≥ 0.

#### **6.4 Supports and modals**

$$\begin{split} \text{supp}\left(\hat{N}\_{S}(t)\right) &= \left[\left(0.67-0.67e^{X}\right)-\left(0.47-0.1e^{X}\right), \left(0.67-0.67e^{X}\right)+\left(2.3-0.3e^{X}\right)\right] \\ &= \left[0.2-0.6e^{X}, 3-e^{X}\right], \quad X=-3t, t\geq 0 \\ \text{supp}\left(\hat{T}\_{S}(t)\right) &= \left[\frac{2-2e^{X}}{6+9e^{X}}-\frac{24+29e^{X}-23e^{X}}{(6+9e^{X})(15+25e^{X})}, \frac{2-2e^{X}}{6+9e^{X}}+\frac{16+21e^{X}-7e^{X}}{(6+9e^{X})(1+e^{X})}\right] \\ &= \left[\frac{1-3e^{X}}{15+25e^{X}}, \frac{3-e^{X}}{1+e^{X}}\right], \quad X=-3t, t\geq 0 \\ \text{ker}\left(\hat{N}\_{S}(t)\right) &= \quad 0.67-0.67e^{X}, \quad \text{where} \quad X=-3t, t\geq 0 \\ &\quad \text{ker}\left(\hat{T}\_{S}(t)\right) &= \quad \frac{2-2e^{X}}{6+9e^{X}}, \quad X=-3t, t\geq 0 \end{split}$$
