Download CAAP '86: 11th Colloquium on Trees in Algebra and by Eric G. Wagner (auth.), Paul Franchi-Zannettacci (eds.) PDF

By Eric G. Wagner (auth.), Paul Franchi-Zannettacci (eds.)

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Read Online or Download CAAP '86: 11th Colloquium on Trees in Algebra and Programming Nice, France, March 24–26, 1986 Proceedings PDF

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Additional info for CAAP '86: 11th Colloquium on Trees in Algebra and Programming Nice, France, March 24–26, 1986 Proceedings

Example text

Lastly, for each column replacement, we count the number N of working devices. 2) ans = 93 89 91 92 90 93 From the above, we see, for example, that replacing the third component with an ultrareliable component resulted in 91 working devices. The results are fairly inconclusive in that replacing devices 1, 2, or 3 should yield the same probability of device failure. 2) ans = 8771 8795 8806 >> 9135 8800 8796 9178 8886 8875 In both cases, it is clear that replacing component 4 maximizes the device reliability.

7 Solution For Y = aX + b, we wish to show that Var[Y ] = a2 Var[X]. 12 says that E[aX + b] = aE[X] + b. Hence, by the definition of variance. Var [Y ] = E (aX + b − (aE [X] + b))2 (1) = E a2 (X − E [X])2 (2) = a E (X − E [X]) (3) 2 2 Since E[(X − E[X])2 ] = Var[X], the assertion is proved. 9 Solution With our measure of jitter being σT , and the fact that T = 2X − 1, we can express the jitter as a function of q by realizing that Var[T ] = 4 Var[X] = 4q (1 − q)2 (1) Therefore, our maximum permitted jitter is σT = √ 2 q = 2 msec (1 − q) (2) Solving for q yields q 2 − 3q + 1 = 0.

Therefore D has the Pascal PMF PD (d) = d−1 j−1 [(1 − p)/2]j [(1 + p)/2]d−j 0 d = j, j + 1, . . 13 Solution (a) Let Sn denote the event that the Sixers win the series in n games. Similarly, Cn is the event that the Celtics in in n games. The Sixers win the series in 3 games if they win three straight, which occurs with probability (1) P [S3 ] = (1/2)3 = 1/8 The Sixers win the series in 4 games if they win two out of the first three games and they win the fourth game so that P [S4 ] = 3 (1/2)3 (1/2) = 3/16 2 (2) The Sixers win the series in five games if they win two out of the first four games and then win game five.

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