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The basic petri net model consists of condition nodes, arcs, event nodes, and tokens. If the input condition nodes for an event node all have tokens, then the event can re, the tokens are removed from the input nodes, and tokens are placed on all of the output nodes of the ring position. The condition nodes are usually represented by circles and the event nodes by horizontal lines or rectangles. In a petri net model, the condition nodes usually represent some required condition for instance, the existence of a test plan. A token at the condition means that the condition is met. An event node (the horizontal line) represents an event that can happen ( re) when all the requirements are met (tokens in all the condition nodes). Tokens are then placed at all the condition nodes that follow the event.

2 is 2) 1

10.10 Find the power factor for the circuit shown in Fig. 10-17.

where the sum is taken over all y corresponding to X E(Y u X 2)

With no voltage or current speci ed, P, Q, and S cannot be calculated. the cosine of the angle on the equivalent impedance. Zeq 3 j4 10 3:68 36:038

3.28. Find the conditional expectation of (a) Y given X, (b) X given Y in Problem 2.29, pages 58 59.

(a) (b) E(Y u X E(X uY x) 3 y)

Fig. 10-17

5 22

10.11 If the total power in the circuit Fig. 10-17 is 1100 W, what are the powers in the two resistors

3 `xf1(x u y) dx 2(1 3(1 y3) y2)

A petri net model of testing is shown in Fig. 2-5.

7 22

CHAP. 10]

19 11

P10

2(1 y y2) 3(1 y)

2x 3

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500 W.

m2)2f2(y u x) dy x)

Fig. 2-5.

10.12 Obtain the power factor of a two-branch parallel circuit where the rst branch has Z1 2 j4

2x>3 from Problem 3.28(a).

30 y

. To what value must the 6-

x2 18

Since the angle on the equivalent admittance is the negative of the angle on the equivalent impedance, its cosine also gives the power factor. Yeq 1 1 0:334 36:848 S 2 j4 6 pf cos 36:848 0:80 lagging

We shall present the proof for continuous random variables. A proof for discrete variables is similar if integrals are replaced by sums. If f(x) is the density function of X, then

The pf is lagging because the impedance angle is positive. Now, for a change in power factor to 0.90, the admittance angle must become cos 1 0:90 25:848. Then, 1 1 1 1 1 0 j Yeq 2 j4 R 10 R 5 requires 1=5 tan 25:848 1 1 10 R or R 3:20

m)2]

In object-oriented development ( 11), both the problem in the problem domain and the solution in the machine space are described in terms of objects In the solution, these objects normally become classes As the requirements and design phases of software development progress, the objects switch from being representations of the things in the problem domain to being programming structures in the software Object models represent entities and relationships between entities Each box represents a type of object, and the name, attributes, and the methods of the object are listed inside the box The top section of the box is for the name of the object, the second section is for the attributes, and the bottom section is for the methods An arc between two objects represents a relationship between the objects Arcs may be labeled in the center with a name of the association.

Since the integrand is nonnegative, the value of the integral can only decrease when the range of integration is diminished. Therefore, s2 3ux

10.13 A voltage, 28.28 608 V, is applied to a two-branch parallel circuit in which Z1 4 308 and Z1 5 608

P2 3

V V 7:07 308 A I2 5:66 08 A Z1 Z2 28:28 7:07 p 308 100 308 86:6 j50:0 S1 p 608 2 2 28:28 5:66 p 08 80:0 608 40:0 j69:3 S2 p 608 2 2 ST S1 S2 126:6 j119:3 174:0 43:38 VA I1 The power triangles and their summation are shown in Fig. 10-18.

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