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2011_HW3_Solution

# 2011_HW3_Solution - 4.2(Random Test Generation Recall that...

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ECE255A – HW3 solutions P. 1/2 4.2 (Random Test Generation) Recall that the detection probability is n f f T d 2 = , where f T is the set of vectors that can detect fault f, and n is the number of inputs in the circuit. In this circuit, we have 3 inputs, so there is a total of 8 possible vectors. (b) For e/1, one vector, abc = 010 can detect it. Thus f d = 0.125 4.3 (Boolean Difference) (a) The set of vectors that can detect e/0 is () () () () () () () () () bc bc a bc b c a e b c a e b g e b g g b g g e b g g e e z e z e d d e e z = + = + = = = + + = + = = = = = m 1 0 1 Thus, the set of vectors is {011, 111}. 4.8 (D Algorithm) Initially, we place a D on b. The D-frontier at this time includes {d, e}. Next, we pick a D-frontier to propagate the fault effect across. Suppose we pick d. Then, the decision a=0 is made. At this time, the D-frontier becomes {x, e}. We pick the D-frontier that is closest to a PO. Thus, we pick x. The next decision is e1=0. This decision implies y=1

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2011_HW3_Solution - 4.2(Random Test Generation Recall that...

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