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Table 9-12 tj. A B C D E 8 12 0 12 4 12 8 22 1 5 a x2 jk 15 3 6 3 10 2356 18 5 17 4 8 7 2 15 10 7 10 0 60 20 20 70 t2 j. 100 0 3600 400 400 4500

Superposition. In the circuit of Fig. 4-50 nd the contribution of each source to I1 , I2 , I3 , and show that they add up to values found in Problem 4.43. From the source on the left: Ans. From the source on the right: From both sources: I1 36=51 I1 4=51 I1 32=51 I2 9=51 I2 18=51 I2 9=51 I3 27=51 I3 20=51 I3 7=51 (All in A)

Subtract a suitable number, say, 60, from all the data to obtain Table 9-12. Then v vb We now form Table 9-13. 2356 1 (4500) 5 (70)2 (5)(4) (70)2 (5)(4) 2356 900 245 245 2111 655

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1. Given the following code, generate test cases that satisfy the criteria of MM testing and every function pair testing: class Threes { char cout; public: Threes(){count = a ;} void PlusOne(){ if(count == a ) count = b ; if(count == b ) count = c ; if(count == c ) count = a ; } char* IsDiv() {if(count == a ){return yes ;}else{return no ;}} } class Num { Threes* SumMod3; int last; int number; void Digit(int newnum){int j; for (j=1; j<=newnum; j++) SumMod3->PlusOne();} public: Num(){SumMod3 = new Threes;} void Reduce() {while (number > 0){last = number (number/ 10)*10; Digit(last); number = number/10;} char* IsDivisibleBy3(int newnum) {number = newnum; Reduce; return SumMod3->IsDiv();} } Main() { Num* Test = new Num; int value; char* answer; cin >> value; answer = test->IsDivisibleBy3(value); cout << answer;

Node Voltage Method. In the circuit of Fig. 4-51 write three node equations for nodes A, B, and C, with node D as the reference, and nd the node voltages.

Table 9-13

1 1)

Fig. 4-51

4 5(4) 20

Ans:

For 4, 20 degrees of freedom we have F0.95 2.87. Therefore, we cannot reject the null hypothesis at a 0.05 level and therefore certainly cannot reject it at a 0.01 level.

Modifications for unequal numbers of observations 9.9. Table 9-14 shows the lifetimes in hours of samples from three different types of television tubes manufactured by a company. Using the long method, test at (a) the 0.05, (b) the 0.01 significance level whether there is a difference in the three types.

1. How is functional testing of object-oriented software done Functional testing of object-oriented software is no di erent from functional testing of conventional software. 2. Is statement coverage of object-oriented software useful Yes, statement coverage of object-oriented software should be done. It is probably the most minimal acceptable coverage. 3. Does MM testing subsume statement coverage (See Section 10.3.1.) No, MM testing requires that every method call be tested. However, a section of source code could not contain any method calls and thus not be tested by a set of test cases that achieve MM testing. 4. What is the advantage of function pair coverage Function pair coverage ensures that combinations of calls be executed. In the stack example (13.3), if the stack is called by an user interface, there may be only one call of each function. Thus, a simple sequence of create, push, and pop might achieve MM testing. Function pair coverage subsumes MM testing.

8 > Node A: < Node B: > : Node C:

Table 9-14

CHAP. 4]

Table 9-15 Total Sample 1 Sample 2 Sample 3 7 4 10 11 6 8 x # 9 8 6 5 8 grand mean 84 12 2 27 25 32 7 Mean 9 5 8

It is convenient to subtract a suitable number, say, 400, obtaining Table 9-15. In this table we have indicated the row totals, the sample or group means, and the grand mean. We then have v vb a (xjk

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