UGC NET COMPUTER SCIENCE SOLVED PAPERS 2017-19 - UGC NET Computer Science November 2017 Paper 3

6. A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:

  • Option : B
  • Explanation :
    Microprocessor instruction format, which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations.ie:
    Nov2017 cs
    F1,F2,F3 each having seven distinct micro-operation. So, 3 bits are required for each. Condition field have four status, it needs 2 bits for four different condition. Branch field have four option so,it needs 2 bits for four option. Now there are 128 different memory location, So, there 7 bits atre required for 128 diffeent location. Instruction Field:

    Nov2017 cs
    Total bits are 20. So, option (B) is correct.
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9. If every non-key attribute is functionally dependent on the primary key, then the relation is in __________ .

  • Option : B
  • Explanation :
    Conditions for various normal forms:
    1. 1 NF - A relation R is in first normal form (1NF) if and only if all underlying domains contain atomic values only.
    2. 2 NF - A relation R is in second normal form (2NF) if and only if it is in 1NF and every non-key attribute is fully dependent on the primary key.
    3. 3 NF - A relation R is in third normal form (3NF) if and only if it is in 2NF and every non-key attribute is non-transitively dependent on the primary key.
    4. BCNF - A relation R is in Boyce-Codd normal form (BCNF) if and only if every determinant is a candidate key.

    Example: Relation R(XYZ) with functional dependencies {X -> Y, Y -> Z, X -> Z}. Notice here Y -> Z, in question it is not mention that non prime attribute is only dependent on primary key so this FD is perfectly valid. This relation is in 2NF but not in 3NF because of every non-key attribute is transitively dependent on the primary key. Here {X} will be candidate key. So, option (B) is correct.
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10. Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is atomic, and following functional dependencies exist.
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is __________ .

  • Option : A
  • Explanation :
    If we find closure of A: A+ → All atribute except D. Similarly for other keys we can find closure, but D can't be derived from any key and it must be added to all keys to be derived from. That's why this relation is in 1NF, since there is partial dependency so, this relation is not in 2NF. So, option (A) is correct
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