Gate Exam

Consider a database table T containing two columns X and Y each of type integer. After the creation of the table, one record (X= 1, Y=l) is inserted in the table. Let MX and MY denote the respective maximum values of X and Y among all records in the table at any point in time. Using MX and MY, new records are inserted in the table 128 times with X and Y values being MX+1, 2*MY+1 respectively. It may be noted that each time after the insertion, values of MX and MY change. What will be the output of the following SQL query after the steps mentioned above are carried out? SELECT Y FROM T WHERE X=7;
option
(A) 127
(B) 255
(C) 129
(D) 257

Read Solution (Total 1)

ANS IS OPTION A BECAUSE X=1 Y=1, X=2 Y=3 , X=3 Y=7 , X=4 Y=15 , X=5 Y=31 ,X=6 Y=63 , X=7 Y =127
10 years agoHelpfull: Yes(1) No(0)

Gate Other Question

On a non-pipe lined sequential processor, a program segment, which is a part of the interrupt service routine, is given to transfer 500 bytes from an I/O device to memory.
Initialize the address register
Initialize the count to 500 LOOP:
Load a byte from device Store in memory at address given by address register
Increment the address register Decrement the count
If count != 0 go to LOOP
Assume that each statement in this program is equivalent to a machine
instruction which takes one clock cycle to execute if it is a non-load/store
instruction. The load-store instructions take two clock cycles to execute.
The designer of the system also has an alternate approach of using the DMA
controller to implement the same transfer. The DMA controller requires 20 clock
cycles for initialization and other overheads. Each DMA transfer cycle takes two
clock cycles to transfer one byte of data from the device to the memory.
What is the approximate speedup when the DMA controller based design is used
in place of the interrupt driven program based input-output?
option
(A) 3.4
(B) 4.4
(C) 5.1
(D) 6.7 Consider the following table of arrival time and burst time for three processes P0, P1 and P2. Process Arrival time Burst Time
P0 0 ms 9 ms
P1 1 ms 4ms
P2 2 ms 9ms
The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?
option
(A) 5.0 ms
(B) 4.33 ms
(C) 6.33 ms
(D) 7.33 ms