Information system and software engineering miscellaneous

Information system and software engineering miscellaneous

Easy Questions

Information System and Software Engineering

Information system and software engineering miscellaneous
  1. The coupling between different modules of a software is categorized as follows
    1. Content coupling 2. Common coupling 3. Control coupling 4. Stamp coupling 5. Data coupling
    Coupling between modules can be ranked in the order of strongest (least desirable) to weakest (most desirable) as follows








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    Content coupling (high)
    Content coupling is the process in which one module relies on the internal workings of another module.
    Common coupling
    Common coupling is the process in which two modules share the same global data.
    Control coupling
    Control coupling is the process in which one module controls the flow of another, by passing it information. This information is basically what work is to be done.
    Stamp coupling
    Stamp coupling is the process in which modules share a composite data structure however use only a part of it.

    Correct Option: A


    Content coupling (high)
    Content coupling is the process in which one module relies on the internal workings of another module.
    Common coupling
    Common coupling is the process in which two modules share the same global data.
    Control coupling
    Control coupling is the process in which one module controls the flow of another, by passing it information. This information is basically what work is to be done.
    Stamp coupling
    Stamp coupling is the process in which modules share a composite data structure however use only a part of it.

  1. Which of the following statements are true?
    1. The context diagram should depict the system as a single bubble.
    2. External entities should be identified clearly at all levels of DFDs
    3. Control information should not be represented in a DFD.
    4. A data store can be connected either to another data store or to an external entity.









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    1. The context diagram always depicts the system as single bubble. Hence, it is true.
    2. External entities cannot be identified at all levels of DFD’s. Hence, it is false.
    3. DFD does not allow the representation of control information. Hence, it is true.
    4. A data store cannot be connected to other data stores or external entity. Hence, it is false.

    Correct Option: C

    1. The context diagram always depicts the system as single bubble. Hence, it is true.
    2. External entities cannot be identified at all levels of DFD’s. Hence, it is false.
    3. DFD does not allow the representation of control information. Hence, it is true.
    4. A data store cannot be connected to other data stores or external entity. Hence, it is false.

  1. The cyclomatic complexity of each of the modules A and B shown below is 10. What is the cyclomatic complexity of the sequential integration shown on the right hand side?









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    Individual cyclomatic complexity = 10
    Now, in the cases of sequential integration, the general behaviour is that the complexity is decreased than the two independent modules. This decrease is unitary and hence the answer is 19, (10 + 10 – 1 = 19).

    Correct Option: A

    Individual cyclomatic complexity = 10
    Now, in the cases of sequential integration, the general behaviour is that the complexity is decreased than the two independent modules. This decrease is unitary and hence the answer is 19, (10 + 10 – 1 = 19).

  1. A company needs to develop a strategy for software product development for which it has a choice of two programming languages L1 and L2. The number of Lines of Code (LOC) developed using L2 is estimated to be twice the LOC developed with L1. The product will have to be maintained for 5 years. various parameters for the company are given in the table below.

    Total cost of the project includes cost of development and maintenance. What is the LOC for L1 for which the cost of the project using L1 is equal to the cost of the project using L2?








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    Suppose, LOC of project L1 is x. The LOC of project L2 is 2x.
    Cost of project L1 = Development cost + Maintenance cost = Manyears × Development cost/year + Maintenance time × Maintenance cost/year

    =
    x
    		 × 1000000 + 5 × 100000
    10000

    = 100 x + 500000
    Cost of project L2 = Development cost + Maintenance cost
    =
    2x
    		 × 750000 + 5 × 50000
    10000

    = 100 × x + 500000 = 150 × x + 250000
    100 × x + 250000 = 150 × x
    50 x = 250000
    x = 5000

    Correct Option: B

    Suppose, LOC of project L1 is x. The LOC of project L2 is 2x.
    Cost of project L1 = Development cost + Maintenance cost = Manyears × Development cost/year + Maintenance time × Maintenance cost/year

    =
    x
    		 × 1000000 + 5 × 100000
    10000

    = 100 x + 500000
    Cost of project L2 = Development cost + Maintenance cost
    =
    2x
    		 × 750000 + 5 × 50000
    10000

    = 100 × x + 500000 = 150 × x + 250000
    100 × x + 250000 = 150 × x
    50 x = 250000
    x = 5000

  1. The following is comment written for a C function /* This function computes the roots of a quadratic equation. a.x2 + b.x + c = 0. The function stores two real roots. in *root1 and *root2 and returns the status of validity of roots. It handles four different kinds of cases.
    1. When coefficient a is zero irrespective of discriminant
    2. When discriminant is positive
    3. When discriminant is zero
    4. When discriminant is negative Only in cases 2 and 3, the stores roots are valid.
    Otherwise 0 is stored in the roots. the function returns 0
    when the roots are valid and – 1 otherwise.
    The function also ensures root 1 > = root 2.
    int get_QuadRoots float a, (float b, float c, float * root1, float * root2);
    */
    A software test engineer is assigned the job of doing Block Box testing. He comes up with the following test cases, many of which are redundant.

    Which one of the following options provides the set of nonredundant tests using equivalence class partitioning approach from input perspective for Black Box testing?








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    Test cases T1 and T2 both are examples of a = 0. Thus, T1 and T2 are redundant.
    For T3 discriminant b2 – 4ac = 4 – 4 = 0
    For T4 discriminant b2 – 4ac = (– 12)2 – 4 × 9 × 4 = 144 – 144 = 0
    For T5 discriminant b2 – 4ac = (– 2)2 – (– 3) × 4 × 1 = 4 + 12 = 16 > 0
    Thus T5 is example of discriminant > 0
    For T6 discriminant b2 – 4ac = (1)2 – 4 × 1 × 4 = 1 – 16 = – 13
    Thus T6 is example of discriminant < 0. means T3 and T4 are also redundant.

    Correct Option: C

    Test cases T1 and T2 both are examples of a = 0. Thus, T1 and T2 are redundant.
    For T3 discriminant b2 – 4ac = 4 – 4 = 0
    For T4 discriminant b2 – 4ac = (– 12)2 – 4 × 9 × 4 = 144 – 144 = 0
    For T5 discriminant b2 – 4ac = (– 2)2 – (– 3) × 4 × 1 = 4 + 12 = 16 > 0
    Thus T5 is example of discriminant > 0
    For T6 discriminant b2 – 4ac = (1)2 – 4 × 1 × 4 = 1 – 16 = – 13
    Thus T6 is example of discriminant < 0. means T3 and T4 are also redundant.