A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the
output bit-string after stuffing is 01111100101, then the input bit-string is

**A. ** 0111110100

**B. ** 0111110101

**C. ** 0111111101

**D. ** 0111111111

**Answer : ****Option B**

**Explaination / Solution: **

Given 8 – bit delimiter pattern of 01111110.

Output Bit string after stuffing is 011111__0__0101

⇓

StuffedBit

Now, Input String is 0111110101

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Host A (on TCP/IP v4 network A) sends an IP datagram D to host B (also on TCP/IP V4
network B). Assume that no error occurred during the transmission of D. When D reaches B,
which of the following IP header field(s) may be different from that of the original datagram
D? (i) TTL (ii) Checksum (iii) Fragment Offset

**A. ** (i) only

**B. ** (i) and (ii) only

**C. ** (ii) and (iii) only

**D. ** (i), (ii) and (iii)

**Answer : ****Option D**

**Explaination / Solution: **

While an IP Datagram is transferring from one host to another host, TTL, Checksum and Fragmentation Offset will be changed.

While an IP Datagram is transferring from one host to another host, TTL, Checksum and Fragmentation Offset will be changed.

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An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP
packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant
fields in the header of the third IP fragment generated by the router for this packet are

**A. ** MF bit: 0, Datagram Length: 1444; Offset: 370

**B. ** MF bit: 1, Datagram Length: 1424; Offset: 185

**C. ** MF bit: 1, Datagram Length: 1500; Offset: 370

**D. ** MF bit: 0, Datagram Length: 1424; Offset: 2960

**Answer : ****Option A**

**Explaination / Solution: **

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The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013. For
international comparison, the GDP is compared in US Dollars (USD) after conversion based
on the market exchange rate. During the period 2012-2013 the exchange rate for the USD
increased from Rs. 50/ USD to Rs. 60/ USD. India’s GDP in USD during the period 2012-
2013

**A. ** increased by 5 %

**B. ** decreased by 13%

**C. ** decreased by 20%

**D. ** decreased by 11%

**Answer : ****Option D**

**Explaination / Solution: **

Per 100 Rs final value 107 Rs

Per 100 Rs final value 107 Rs

⇒ Per 100/50 Dollars final value 107/60

for 100 dollars____?

= ((100 × 50)/100) × (107/60) = 89.16

Decreased by 11%.

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What is the optimized version of the relation algebra expression where
A1, A2 are sets of attributes in with A_{1 }⊂ A_{2 }and F1, F2 are Boolean expressions based on

**A. **

**B. **

**C. **

**D. **

**Answer : ****Option A**

**Explaination / Solution: **

No Explaination.

No Explaination.

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A prime attribute of a relation scheme R is an attribute that appears

**A. ** in all candidate keys of R.

**B. ** in some candidate key of R.

**C. ** in a foreign keys of R.

**D. ** only in the primary key of R.

**Answer : ****Option B**

**Explaination / Solution: **

A prime attribute or key attribute of a relation scheme R is an attribute that appears in any of the candidate key of R, remaining attributes are known as non-prime or non-key tribute

A prime attribute or key attribute of a relation scheme R is an attribute that appears in any of the candidate key of R, remaining attributes are known as non-prime or non-key tribute

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Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.

**A. ** Only S1 is conflict-serializable.

**B. ** Only S2 is conflict-serializable.

**C. ** Both S1 and S2 are conflict-serializable.

**D. ** Neither S1 nor S2 is conflict-serializable.

**Answer : ****Option A**

**Explaination / Solution: **

Precedence graph for S1 & S2 are as follows

T1 : r1 (X) ; r1 (z) ; w1 (X) ; w1 (z)

T2 : r2 (X) ; r2 (z) ; w2 (z)

T3 : r3 (X) ; r3 (X) ; w3 (Y)

S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z)

S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z)

Which one of the following statements about the schedules is TRUE?

Precedence graph for S1 & S2 are as follows

∴Only S1 is conflict serializable.

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Consider the relational schema given below, where eId of the relation dependentis a foreign
key referring to empId of the relation employee. Assume that every employee has at least one
associated dependent in the dependent relation.

**A. ** some dependent.

**B. ** all dependents.

**C. ** some of his/her dependents.

**D. ** all of his/her dependents

**Answer : ****Option D**

**Explaination / Solution: **

employee (__empId__, empName, empAge)

dependent (__depId, eId__, depName, depAge)

Consider the following relational algebra query

The above query evaluates to the set of empIds of employees whose age is greater than that of

Part A of the above given relational algebra query will give the set of empIds of those
employees whose age is less than or equal to the age of some of his/her dependents.
Now when set of empIds of all employees minus set of empIds obtained from part A is done,
then we get the set of empIds of employees whose age is greater than that of all of his/her
dependents.

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Consider the decision problem 2CNFSAT defined as follows:

**A. ** NP-Complete.

**B. ** solvable in polynomial time by reduction to directed graph reachability.

**C. ** solvable in constant time since any input instance is satisfiable.

**D. ** NP-hard, but not NP-complete.

**Answer : ****Option B**

**Explaination / Solution: **

2 SAT is in P. This we can prove by reducing 2 SAT to directed graph reachability problem which is known to be in P.

__Procedure for reducing 2 SAT to reachability problem:__

{ φ | φ is a satisfiable propositional formula in CNF with at most two literal per clause}

For example, is a Boolean formula and it is in 2CNFSAT.

The decision problem 2CNFSAT is

2 SAT is in P. This we can prove by reducing 2 SAT to directed graph reachability problem which is known to be in P.

1. Let ϕ be CNF with clauses of length 2 and let P be the set of propositional
variables(literals) in ϕ

2. Build a graph G=(V,E) with V= P { p|p P} ∪ ¬ ∈ and (x, y E )∈ iff there is a clause in ϕ
that is equivalent to x y → (all the clauses are converted to equivalent implications and
the graph built is called as implication graph)

3. Observe that ϕ is unsatisfiable iff there is a p P ∈ such that there is both a path from p to
to ¬p and from ¬p to p in G.

This condition can be tested by running the reachability algorithm several times.

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Which of the following statements are CORRECT?

**A. ** 1 and 2 only

**B. ** 2 and 3 only

**C. ** 3 and 4 only

**D. ** 1 and 3 only

**Answer : ****Option D**

**Explaination / Solution: **

To implement recursion, activation record should be implemented by providing dynamic memory allocation. This dynamic allocation is done from runtime stack. Heap is essential to allocate memory for data structures at run-time, not for recursion. So statement 1 and 3 are correction.

1) Static allocation of all data areas by a compiler makes it impossible to implement
recursion.

2) Automatic garbage collection is essential to implement recursion.

3) Dynamic allocation of activation records is essential to implement recursion.

4) Both heap and stack are essential to implement recursion.

To implement recursion, activation record should be implemented by providing dynamic memory allocation. This dynamic allocation is done from runtime stack. Heap is essential to allocate memory for data structures at run-time, not for recursion. So statement 1 and 3 are correction.

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