Recent activity by Sahil91 / GATE Overflow for GATE CSE

3 answers

1

GATE CSE 2021 Set 1 | Question: 26
Consider the following grammar (that admits a series of declarations, followed by expressions) and the associated syntax directed translation $\text{(SDT)}$ ... used to type-check syntactically correct boolean variable declarations and boolean expressions. The actions will lead to an infinite loop

Consider the following grammar (that admits a series of declarations, followed by expressions) and the associated syntax directed translation $\text{(SDT)}$ acti...

10.5k views

commented Mar 1, 2021

3 answers

2

GATE CSE 2021 Set 1 | Question: 29
Assume that a $12$-bit Hamming codeword consisting of $8$-bit data and $4$ check bits is $d_8d_7d_6d_5c_8d_4d_4d_3d_2c_4d_1c_2c_1$ ... $0$ and $y$ is $1$ $x$ is $1$ and $y$ is $0$ $x$ is $1$ and $y$ is $1$

Assume that a $12$-bit Hamming codeword consisting of $8$-bit data and $4$ check bits is $d_8d_7d_6d_5c_8d_4d_4d_3d_2c_4d_1c_2c_1$, where the data bits and the check bits...

15.5k views

commented Feb 19, 2021

4 answers

3

GATE CSE 2021 Set 1 | Question: 45
Consider two hosts $P$ and $Q$ connected through a router $R$. The maximum transfer unit $\text{(MTU)}$ value of the link between $P$ and $R$ is $1500$ bytes, and between $R$ and $Q$ is $820$ bytes. A $\text{TCP}$ segment ... to resend the whole $\text{TCP}$ segment. $\text{TCP}$ destination port can be determined by analysing $\textit{only}$ the second fragment.

Consider two hosts $P$ and $Q$ connected through a router $R$. The maximum transfer unit $\text{(MTU)}$ value of the link between $P$ and $R$ is $1500$ bytes, and between...

10.4k views

commented Feb 19, 2021

7 answers

4

GATE CSE 2021 Set 1 | Question: 30
Consider the following recurrence relation. $T\left ( n \right )=\left\{\begin{array} {lcl} T(n ∕ 2)+T(2n∕5)+7n & \text{if} \; n>0\\1 & \text{if}\; n=0 \end{array}\right.$ Which one of the following options is correct? $T(n)=\Theta (n^{5/2})$ $T(n)=\Theta (n\log n)$ $T(n)=\Theta (n)$ $T(n)=\Theta ((\log n)^{5/2})$

Consider the following recurrence relation.$$T\left ( n \right )=\left\{\begin{array} {lcl} T(n ∕ 2)+T(2n∕5)+7n & \text{if} \; n>0\\1 & \text{if}\; n=0 \end{array}\r...

23.9k views

commented Feb 19, 2021

2 answers

5

GATE CSE 2021 Set 1 | Question: 41
An $articulation$ $point$ in a connected graph is a vertex such that removing the vertex and its incident edges disconnects the graph into two or more connected components. Let $T$ be a $\text{DFS}$ tree obtained by doing $\text{DFS}$ ... is a descendent of $u$ in $T$, then all paths from $x$ to $y$ in $G$ must pass through $u$.

An $articulation$ $point$ in a connected graph is a vertex such that removing the vertex and its incident edges disconnects the graph into two or more connected component...

14.0k views

commented Feb 19, 2021

2 answers

6

Test-Series
Suppose: TLB lookup time = 20 ns TLB hit ratio = 80% Memory access time = 75 ns Swap page time = 500,000 ns 50% of pages are dirty OS uses a single level page table What is the effective access time (EAT) if we assume the page fault rate is 10%? Assume the cost toupdate the TLB, the page table, and the frame table (if needed) is negligible. 1. 3777.5ns 2. 30220ns 3. 15110ns 4. 7555ns

Suppose: TLB lookup time = 20 nsTLB hit ratio = 80%Memory access time = 75 nsSwap page time = 500,000 ns50% of pages are dirtyOS uses a single level page tableWhat is the...

3.5k views

commented Jan 2, 2021

2 answers

7

OS:- Synchronzation
1. Progress Satisfied => Deadlock free 2. Starvation free => Deadlock free 3.Starvation freedom => Bounded wait Please explain these implications if valid or invalid

1. Progress Satisfied = Deadlock free2. Starvation free = Deadlock free3.Starvation freedom = Bounded waitPlease explain these implications if valid or invalid

1.4k views

commented Nov 16, 2020

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