## Solution to Sample Problems on Uncertain Reasoning

1. Suppose we have the following rule R1:

if (P1 and P2 and P3) or (P4 and not P5 then C1 (0.7) and C2 (-0.5)

and the certainty factors of P1, P2, P3, P4, P5 are as follows:

CF(P1) = 0.8,
CF(P2) = 0.7,
CF(P3) = 0.6,
CF(P4) = 0.9,
CF(P5) = -0.5,

What are the certainty factors associated with conclusions C1 and C2 after using rule R1?

Solution:
For P1 and P2 and P3, the CF is
min(CF(P1), CF(P2), CF(P3)) = min(0.8, 0.7, 0.6) = 0.6. Call this CFA.
For not P5, the CF is -CF(P5) = 0.5.
For P4 and notP5, the CF is min(0.9, 0.5) = 0.5. Call this CFB.
For (P1 and P2 and P3) or (P4 and notP), the CF is:
max(CFA, CFB) = max(0.6, 0.5) = 0.6.

Thus CF(C1) = 0.7 * 0.6 = 0.42
and CF(C2) = -0.5 * 0.6 = -0.30

2. The final answers to the previous question are CFR1(C1) = 0.42 and CFR1(C2) = -0.3. Suppose that we have, from another rule R2, the following certainty factors for C1 and C2:

CFR2(C1) = 0.7,
CFR2(C2) = -0.4

What are the certainty factors associated with C1 and C2 after combining the evidence from rules R1 and R2?

Solution:
For C1,

CF(C1) = CFR1(C1) + CFR2(C1) - CFR1(C1)*CFR2(C1)
= 0.42 + 0.7 - 0.42*0.7 = 1.12 - 0.294 = 0.826.

For C2,

CF(C2) = CFR1(C2) + CFR2(C2) + CFR1(C2)*CFR2(C2)
= -0.3 + -0.4 + (-0.3)*(-0.4) = -0.7 + 0.12 = -0.58

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Copyright (C) Bill Wilson, 2002, except where another source is acknowledged.