Q 5.1 [GATE 1987 : 2 Marks]Question: On receiving an interrupt from an I/O device the CPU
(a) halts for predetermined time
(b) hands over control of address bus and data bus to the interrupting device
(c) branches off to the interrupt service routine immediately
(d) branches off to the interrupt service routine after completion of the current instruction
Solution (Hinglish):Answer: (d)
CPU kabhi bhi beech instruction me interrupt handle nahi karta. Wo pehle current instruction complete karega, fir PC ko save karke ISR pe jump karega.
Q 5.2 [GATE 1990 : 2 Marks]Question: State True/False: The data transfer between memory and I/O devices using programmed I/O is faster than interrupt driven I/O.
Solution:Answer: FALSE
FALSE hai. Programmed I/O me CPU busy wait karta hai (polling), isliye slow hai. Interrupt driven me CPU free rehta hai, device interrupt dega tabhi kaam karega, to fast hai.
5.3 [GATE 1990 : 2 Marks]Question: State True/False: The flags are affected when conditional CALL or JUMP instructions are executed.
Solution: FALSE.
CALL/JUMP sirf PC change karte hain. Flags ALU ke operation se set hote hain jaise ADD, SUB.
5.4 [GATE 1990 : 2 Marks]Question: State True/False: Transferring data in blocks from the main memory to the cache memory enables an interleaved main memory unit to operate at its maximum speed.
Solution: TRUE. Interleaved memory me consecutive addresses alag-alag banks me hote hain. Block transfer se saare banks parallel kaam karte hain, speed max milti hai.
Q 5.5 [GATE 1991 : 2 Marks]Question: A certain moving arm disk storage, with one head, has specifications. Number of tracks/recording surface = 200, Disk rotation speed = 2400 rpm, Track storage capacity = 62,500 bits. The average latency of this device is P msec and the data transfer rate is Q bits/sec. Write the value of P and Q.
Solution:P = 12.5 msec, Q = 2.5 Mbps
2400 rpm = 40 rps. 1 Rotation = 25 msec.
P (Average Latency) = Half Rotation = 12.5 msec
Q (Data Rate) = Track Capacity _ rps = 62,500 _ 40 = 2,500,000 bits/sec
Q 5.6 [GATE 1995 : 2 Marks]Question: If the disk is rotating at 3600 rpm, determine the effective data transfer rate which is defined as the number of bytes transferred per second between disk and memory. Given size of track = 512 bytes.
Solution: 3600 rpm = 60 rps. Rate = 512 bytes _ 60 = 30,720 bytes/sec
Answer: 30,720 bytes/sec
Q 5.7 [GATE 1996 : 2 Marks]Question: For the daisy chain scheme of connecting I/O devices, which of the following statements is true?
(a) It gives non-uniform priority to various devices
(b)It gives uniform priority to various devices
(c)It is only useful for connecting slow devices to a processor device
(d)it requires a separate interrupt pin on the processor for each device.
Solution: Daisy chain me CPU ke sabse paas wale device ko highest priority milti hai. Jitna door, utni kam priority. Isliye non-uniform priority.
Answer: (a)
Q 5.8 [GATE 1996 : 2 Marks]Question: A micro program control unit is required to generate a total of 25 control signals. Assume that during any microinstruction at most two control signals are active. Minimum number of bits required in the control word is
(a) 2 (b) 25 (c) 10 (d) 12
Solution: (c) 10
25 signals ke liye 5 bits ( 25=32). At most 2 active hai to 5+5=10 bits. Combinational logic se 9 bits bhi enough hai par standard answer 10 hai.
Q 5.9 [GATE 1997 : 1 Mark]Question: Match: A. DMA I/O 1. High speed RAM, B. Cache 2. Disk, C. Interrupt I/O 3. Printer, D. Condition code Register 4. ALU
Solution: DMA->Disk (Fast), Cache->High speed RAM, Interrupt I/O->Printer (Slow), Condition Code->ALU
Answer: (b) A-2, B-1, C-3, D-4
Q 5.10 [GATE 1997 : 1 Mark]Question: I/O redirection
(a) implies changing the name of a file
(b) can be employed to use an existing file as input file for a program
(c) implies connection 2 programs through a pipe
(d) None
Solution:(c)
Redirection matlab program < file.txt. Existing file ko input bana dena. (c) wala pipe kehlata hai |.
Q 5.11 [GATE 1998 : 2 Marks]Question: Formatting of floppy disk refers to
(a) arranging data in contiguous fashion
(b) writing directory
(c) erasing system area
(d) writing identification information on all tracks and sectors
Solution:(d)
Formatting me har track/sector pe ID, header, sync marks likhe jate hain.
Q 5.12 [GATE 1999 : 2 Marks]Question: The main differences between a CISC and A RISC processor is/are that a RISC processor typically
(a) has fewer instructions
(b) has fewer addressing modes
(c) has more registers
(d) is easier to implement using hard-wired control
Solution:(d) All of the above
RISC me instruction kam, addressing mode kam, register zyada, aur hard-wired control hota hai jo fast hota hai. CISC me microprogrammed hota hai.
Q 5.13 [GATE 2004 : 2 Marks]Question: The microinstructions stored in the control memory of a processor have a width of 26 bits. Each microinstruction is divided into three fields: a micro-operation field of 13 bits, a next address field (X), and a MUX select field (Y). There are 8 status bits in the inputs of the MUX. How many bits are there in X and Y fields, and what is the size of control memory?
Solution: 26 = 13 + X + Y => X+Y=13. 8 status bits ke liye Y = log2(8)=3. So X=10. Size = 210=1024 words.
Answer: (a) 10, 3, 1024
Q 5.14 [GATE 2004 : 2 Marks]Question: A hard disk with transfer rate 10 M bytes/sec constantly transferring data to memory using DMA. Processor runs at 600 MHz, and takes 300 and 900 clock cycles to initiate and complete DMA transfer respectively. If size is 20 Kbytes, what is % of processor time consumed for the transfer operation?
(a)5.0%
(b)1.0%
(c)0.5%
(d)0.1%
Solution: Transfer Time = 20KB/10MB = 2.048 ms. CPU Time = (300+900)/600MHz = 2 usec. % = 2us/2.048ms = 0.1%
Answer: (d) 0.1%
Q 5.15 [GATE 2004 : 2 Marks]Question: The storage area of a disk has innermost diameter 10 cm and outermost 20 cm. Max storage density 1400 bits/cm. Disk rotates at 4200 RPM. Computer has memory cycle time 40 nsec. If cycle stealing is used for data transfer, percentage of memory cycles stolen for one word is?
Solution: (c) 5%
Disk Data Rate = 2 _ pi _ r _ density _ rps. Average. Calculation se approx 5% aata hai.
Q 5.16 [GATE 2005 : 2 Marks – Linked]Question: A disk has 8 equidistant tracks. Diameters innermost and outermost are 1 cm and 8 cm. Innermost track capacity 10 MB. What is total data if used with drive that rotates it with (i) Constant Linear Velocity (ii) Constant Angular Velocity?
Solution: Dia are 1,2,3,4,5,6,7,8 cm. CAV me sab tracks =10 MB => Total=80 MB. CLV me capacity dia ke proportional => Total=10_(36)=360 MB.
Answer: (d) (i) 360 MB (ii) 80 MB
Q 5.17 [GATE 2005 : 2 Marks – Linked]Question: If disk has 20 sectors per track and is currently at end of 5th sector of inner-most track and head can move at speed 10 m/sec and rotating at 6000 RPM, how much time will it take to read 1 MB contiguous data starting from sector 4 of outer-most track?
Solution: Seek = 3.5 cm /10 m/s =3.5 ms + Rotational Latency + Transfer 1MB (2 sectors =1 ms) = Total ∼13.5 ms
Answer: (a) 13.5 ms
Q 5.18 [GATE 2005 : 1 Mark]Question: What is swap space in disk used for?
(a) Saving temp html pages
(b) Saving process data
(c) Storing superblock
(d) Storing device drivers
Solution: (b) Saving process data
Virtual Memory ke liye. Process data jab RAM full ho jaye.
Q 5.19 [GATE 2005 : 2 Marks]Question: A device with data transfer rate 10 KB/sec is connected to CPU. Data transferred byte-wise. Let the Interrupt overhead 4 μsec. The Byte transfer time between the device interface register and CPU or memory is negligible. What is minimum performance gain of operating device under interrupt mode over operating it under program controlled mode?
(a15
(b)25
(c)35
(d)45
Solution:(b) 25
Programmed = 100 usec per byte (10KB/sec). Interrupt = 4 usec. Gain=100/4=25
Q 5.20 [GATE 2005 : 2 Marks]Question: Consider disk drive: 16 surfaces, 512 tracks/surface, 512 sectors/track, 1 KB/sector, rotation 3000 rpm. Disk operated in cycle stealing mode whereby whenever one byte word ready it is sent to memory; for writing, disk reads 4 byte word from memory in each DMA cycle. Memory cycle time 40 nsec. Max % time CPU gets blocked during DMA is
(a)10
(b)25
(c)40
(d)50
Solution: (b) 25
1 Track=512KB. 1 Rotation=20ms. Rate=25.6 MB/sec. 4 byte per DMA cycle => 6.4M cycles/sec 40nsec=0.256 sec=25.6%
5.21 GATE 2006 – 1 MarkQuestion:
Which of the following DMA transfer modes and interrupt handling mechanisms will enable the highest I/O band-width?
(a) Transparent DMA and Polling interrupts
(b) Cycle-stealing and Vectored interrupts
(c) Block transfer and Vectored interrupts
(d) Block transfer and Polling interrupts
Solution:DMA: Block Transfer me DMA poora bus ek block ke liye hold karta hai, isliye Cycle Stealing se fast hai. Bandwidth max.Interrupt: Vectored me direct ISR address milta hai, Polling me CPU ko check karna padta hai kaun sa device. Isliye Vectored fast hai.Highest bandwidth = Block + Vectored.Answer: (c) Block transfer and Vectored interrupts
5.22 GATE 2007 – 1 MarkQuestion:
Which of the following systems is a most likely candidate example of a pipe and filter architecture?
(a) Expert system
(b) DB repository
(c) Aircraft flight controller
(d) Signal processing
Solution:
Pipe and Filter me ek stage ka output agle stage ka input banta hai, stream processing hota hai. Signal Processing (jaise audio/video compiler) iska best example hai. Har filter signal ko process karke aage bhejta hai.
Answer: (d) Signal processing