
## 6.8 A Simple Example For The Qdsmp Transformation

We first recall that the qualitative extension of (22), denoted **qDSmP**ǫ(.) is given by **qDSmP**ǫ(∅) = 0 and
∀X ∈ GΘ **\ {∅}** by

$$qDSmP_{\epsilon}(X)=\sum_{Y\in\mathcal{C}^{0}}\frac{\sum_{\begin{subarray}{c}Z\in\mathcal{X}\cap Y\\ \mathcal{C}(Z)=1\\ \sum_{Z\in\mathcal{Y}}\ qm(Z)+\epsilon\cdot\mathcal{C}(Y)\end{subarray}}qm(Y)\tag{33}$$

where all operations in (33) are referred to labels, that is q-operators on linguistic labels and not classical operators on numbers.

Let's consider the simple frame Θ = {θ1, θ2} (here n = |Θ| = 2) with Shafer's model (i.e. θ1∩θ2 = ∅) and the following set of linguistic labels L = {L0, L1, L2, L3, L4, L5}, with L0 = Lmin and L5 = Lmax = Lm+1
(here m = 4) and the following qualitative belief assignment: qm(θ1) = L1, qm(θ2) = L3 and qm(θ1 ∪ θ2) =
L1. qm(.) is quasi-normalized since PX∈2Θ qm(X) = L5 = Lmax. In this example and with **DSmP**
transformation, qm(θ1 ∪ θ2) = L1 is redistributed to θ1 and θ2 proportionally with respect to their qualitative masses L1 and L3 respectively. Since both L1 and L3 are different from L0, we can take the tuning parameter ǫ = 0 for the best transfer. ǫ is taken different from zero when a mass of a set involved in a partial or total ignorance is zero (for qualitative masses, it means L0).

Therefore using (25), one has

 $\dfrac{x_{\theta_1}}{L_1}=\dfrac{x_{\theta_2}}{L_3}=\dfrac{L_1}{L_1+L_3}=\dfrac{L_1}{L_4}=L_{\frac{1}{4}\cdot5}=L_{\frac{5}{4}}=L_{1.25}$  s. 
and thus using (24), one gets

 $x_{\theta_1}=L_1\times L_{1.25}=L_{\frac{1\cdot(1.25)}{5}}=L_{\frac{1.25}{5}}=L_{0.25}$  $x_{\theta_2}=L_3\times L_{1.25}=L_{\frac{3\cdot(1.25)}{5}}=L_{\frac{3.75}{5}}=L_{0.75}$
$${\mathrm{Whence}}$$
$$q D S m P_{\epsilon=0}(\theta_{1}\cap\theta_{2})=q D S m P_{\epsilon=0}(\emptyset)=L_{0}$$
$$q D S m P_{\epsilon=0}(\theta_{1})=L_{1}+x_{\theta_{1}}=L_{1}+L_{0.25}=L_{1.25}$$ $$q D S m P_{\epsilon=0}(\theta_{2})=L_{3}+x_{\theta_{2}}=L_{3}+L_{0.75}=L_{3.75}$$
Naturally in our example, one has also
$$\begin{array}{l l}{{q D S m P_{\epsilon=0}(\theta_{1}\cup\theta_{2})=q D S m P_{\epsilon=0}(\theta_{1})+q D S m P_{\epsilon=0}(\theta_{2})-q D S m P_{\epsilon=0}(\theta_{1}\cap\theta_{2})}}\\ {{}}&{{=L_{1.25}+L_{3.75}-L_{0}=L_{5}=L_{\mathrm{max}}}}\end{array}$$
Since Hmax = log2 n = log2 2 = 1, using the qualitative extension of PIC formula (21), one obtains the following qualitative PIC value:

$$\begin{split}PIC=1+\frac{1}{1}\cdot[q D S m P_{\epsilon=0}(\theta_{1})\log_{2}(q D S m P_{\epsilon=0}(\theta_{1}))\\ +q D S m P_{\epsilon=0}(\theta_{2})\log_{2}(q D S m P_{\epsilon=0}(\theta_{2}))]\end{split}$$
$$+qDSmP_{\epsilon=0}(\theta_{2})\log_{2}(qDSmP_{\epsilon=0}(\theta_{2}))]$$  $=1+L_{1.25}\log_{2}(L_{1.25})+L_{3.75}\log_{2}(L_{3.75})\approx L_{0.94}$
since we considered the isomorphic transformation Li = i/(m + 1) (in our particular example m = 4 interior labels).

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