Example 13: Fusion of three sources Consider a set of labels L = {Lmin = L0, L1, L2, L3, L4, L5, Lmax = L6}, and a frame of discernment Θ = {**A, B, C**} in Shafer's model (*i.e.* all intersections empty). Let's take the three following qualitative sources of evidence:

| A           | B   | B ∪ C   | A ∪ B ∪ C   |    |
|-------------|-----|---------|-------------|----|
| qm1(.)      | L2  | L0      | L0          | L4 |
| qm2(.)      | L0  | L3      | L0          | L3 |
| qm3(.)      | L0  | L0      | L5          | L1 |
| TABLE XVIII |     |         |             |    |

QUALITATIVE INPUTS FOR EXAMPLE 13

## - Qualitative Conjunctive Rule:

If one applies the Qualitative Conjunctive Rule (QCR), one gets:
qm∧(A) = qm1(A)qm2(A ∪ B ∪ C ∪ D)qm3(A ∪ B ∪ C ∪ D) = L2L3L1 = L2·3 6 L1 = L2·3·1 6·6= L1 6 Similarly,

 $qm_{\wedge}(B)=L_4L_3L_1+L_4L_3L_5=L_{\frac{4\cdot3\cdot1}{6\cdot6}}+L_{\frac{4\cdot3\cdot5}{6\cdot6}}=L_{\frac{2}{6}}+L_{\frac{10}{6}}=L_{\frac{12}{6}}$  $\wedge(B\cup C)=L_4L_3L_5=L_{\frac{4\cdot3\cdot5}{6\cdot6}}=L_{\frac{10}{6}}$
$\huge-244323\quad-2\frac{1}{6\cdot6}\quad-3$  $\huge-L$ $\huge-L$ $\huge-L$ $\huge-L$ $\huge-L$ $\huge-L$ $\huge-L$
qm∧(A ∪ B ∪ C ∪ D) = L4L3L1 = L4·3·1
6·6= L26
The total conflict is:
qm∧(∅) = qm1(A)qm2(B)qm3(B ∪ C) + qm1(A)qm2(B)qm3(A ∪ B ∪ C ∪ D) + qm1(A)qm2(A ∪ B ∪ C ∪ D)qm3(B ∪ C)
= L2L3L5 + L2L3L1 + L2L3L5 = L2·3·5 6·6+ L2·3·1 6·6+ L2·3·5 6·6= L5 6
+ L1 6
+ L56
= L11 6

## - Qualitative Pcr:

Applying the proportional conflict redistribution for the first partial conflict qm1(A)qm2(B)qm3(B ∪ C), one gets:

$${\frac{x_{1,A}}{L_{2}}}={\frac{y_{1,B}}{L_{3}}}={\frac{z_{1,B\cup C}}{L_{5}}}={\frac{L_{2}L_{3}L_{5}}{L_{2}+L_{3}+L_{5}}}={\frac{L_{\frac{5}{6}}}{L_{10}}}=L_{({\frac{5}{6}}\div10)\cdot6}=L_{\frac{3}{6}}$$
$$\mathbf{b}_{*}$$

so,

$$x_{1,A}=L_{2}L_{\frac{3}{6}}=L_{\frac{2\cdot3}{6\cdot6}}=L_{\frac{1}{6}}$$
$$y_{1,B}=L_{3}L_{\frac{3}{6}}=L_{\frac{3\cdot3}{6\cdot6}}=L_{\frac{1.5}{6}}$$
$$z_{1,B\cup C}=L_{5}L_{\frac{3}{6}}=L_{\frac{5\cdot3}{6\cdot6}}=L_{\frac{2\cdot5}{6}}$$

Applying the proportional conflict redistribution for the second partial conflict qm1(A)qm2(B)qm3(A∪B ∪C ∪D), one gets:

$${\frac{x_{2,A}}{L_{2}}}={\frac{y_{2,B}}{L_{3}}}={\frac{w_{1,A\cup B\cup C\cup D}}{L_{1}}}={\frac{L_{2}L_{3}L_{1}}{L_{2}+L_{3}+L_{1}}}={\frac{L_{\frac{1}{6}}}{L_{6}}}=L_{({\frac{1}{6}}+6)\cdot6}=L_{\frac{1}{6}}$$
