Consider the parallel circuit with probability as follows: (a) Compute the proba

Question

Consider the parallel circuit with probability as follows: (a) Compute the probability p of sending data successfully through this circuit as a function of p_1 and p_2. (b) Suppose we have 0.4 lessthanorequalto pi lessthanorequalto 0.9 and 0.6 lessthanorequalto p2 lessthanorequalto 0.8. Compute now the maximum q1 and minimum q2 of the probability p of sending data successfully through this circuit. (c) Generalize this to 3 parallel circuits with C3 with success prob = p3. Derive the probability p of sending data successfully through this circuit as a function of p_1, p_2, and p_3. (d) Suppose we have 0.4 lessthanorequalto pi lessthanorequalto 0.9 and 0.6 lessthanorequalto p2 lessthanorequalto 0.8, and 0.3 lessthanorequalto p3 lessthanorequalto 0.5. Compute now the maximum q1 and minimum q2 of the probability p of sending data successfully through this circuit with three components in parallel.

Explanation / Answer

(A)

   The probability p of sending data successfully through this circuit is

     P(Circuit with success) =P(C1 with success) + P(C2 with success) – P(C1 with success AND C2 with success)

Hence, p = p1 + p2 - p1*p2 ------------------(1)

(B)

   We know that, p+q = 1

For maximum q1, p1 is 0.4

then q1 = 1- p1 = 1- 0.4 = 0.6

For minimum q2, p2 is 0.8

then q2 = 1- p2 = 1- 0.8 = 0.2

Because, q = 1-p

by equation (1),

p = 0.4 + 0.8 - 0.4*0.8 = 0.88

and in parallel , p = 1- q1*q2

0.88 = 1 - q1*q2

q1*q2 = 0.12 = 0.6 * 0.2

Hence,  maximum q1 = 0.6 and minimum q2 = 0.2.

(C)

   The probability p of sending data successfully through this circuit is

    p' = p OR p3

p' = p + p3 - p*p3 ------------------(2)

p' = (p1 OR p2) OR p3

p' = (p1 + p2 - p1*p2) OR p3

     P'(Circuit with success) =P(C1 with success) + P(C2 with success) + P(C3 with success) – P(C1 with success AND C2 with success) – P(C2 with success AND C3 with success) – P(C3 with success AND C1 with success) + P(C1 with success AND C2 with success AND C3 with success)

Hence, p' = p1 + p2 + p3 - p1*p2 - p2*p3 - p3*p1 + p1*p2*p3

(D)

For maximum q1, p1 is 0.4

then q1 = 1- p1 = 1- 0.4 = 0.6

For minimum q2, p2 is 0.8

then q2 = 1- p2 = 1- 0.8 = 0.2

Because, q = 1-p

by equation (1),

p = 0.4 + 0.8 - 0.4*0.8 = 0.88

by equation (2) and each value of p3, we get p'

p' = (1/3)[ (0.88 + 0.3 - 0.88*0.3) + (0.88 + 0.4 - 0.88*0.4) + (0.88 + 0.5 - 0.88*0.5)]

= 0.928

and in parallel , p = 1- q1*q2*q3

0.928 = 1 - q1*q2*q3

q1*q2*q3 = 0.072 =0.6*0.2*0.6

Hence,  maximum q1 = 0.6 and minimum q2 = 0.2, and q3 = 0.6.

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