Question

Consider a system that consists of 3 parameters, P1, P2, and P3. P1 has three values: 0, 1 and 2. P2 has two values: a and b. P3 has three values: x, y and z. Apply algorithm IPO to build a pairwise test set for this system. The parameters should be covered in the given order, i.e., P1, P2, and P3. That is, you should build a pairwise test set T for P1 and P2 first, and then extend T to cover P3. Use “-” to represent don’t care values, i.e., values that do not affect coverage. Clearly indicate your tie-breaking rules that may be needed in the test generation process. You must show intermediate steps to obtain full credits.

Answer 1

Solution:

given that: p1 (0,1,2)

p2(a,b)

p3(x,y,z)

Note: Every possible combination of values of the parameters must be covered

then all combinations coverage requires 18 tests:

Each Choice Coverage ! Each parameter value must be covered in at least one test case. !

Pairwise Coverage

T-Wise Coverage !

Given any t parameters, every combination of values of these t parameters must be covered in at least one test case. !

T-Wise Coverage

Given any t parameters, every combination of values of these t parameters must be covered in at least one test case.

Base Choice Coverage

For each parameter, one of the possible values is designated as a base choice of the parameter

A base test is formed by using the base choice for each parameter

Multiple Base Choices Coverage

At least one, and possibly more, base choices are designated for each parameter.

The notions of a base test and subsequent tests are defined in the same as Base Choice.

Subsumption Relation

The IPO Strategy

First generate a pairwise test set for the first two parameters, then for the first three parameters, and so on .

A pairwise test set for the first n parameters is built by extending the test set for the first n –

1 parameters –

Horizontal growth: Extend each existing test case by adding one value of the new parameter –'

Vertical growth: Adds new tests, if necessary

Algorithm IPO_H (T, pi )

Assume that the domain of pi contains values v1 , v2 , …, and vq ;  = { pairs between values of pi and values of p1 , p2 , …, and pi-1

if ( |T| <= q) for 1 <= j <= |T|, extend the j th test in T by adding value vj and remove from  pairs covered by the extended test

else for 1 <=j <= q, extend the j th test in T by adding value vj and remove from  pairs covered by the extended test;

for q < j <= |T|, extend the j th test in T by adding one value of pi such that the resulting test covers the most number of pairs in , and remove from  pairs covered by the extended test

Algorithm IPO_V(T, )

let T’ be an empty set;

for each pair in 

assume that the pair contains value w of pk , 1 ≤ k < i, and value u of pi ;

if (T contains a test with “-” as the value of pk and u as the value of pi )

modify this test by replacing the “-” with w

else

add a new test to T’ that has w as the value of pk , u as the value of pi , and “-” as the value of every other parameter;

T = T  T’

(here  symbal is is pie)

Any probblem please comment