Table of Contents
Computer Organization GATE Questions and Answers
Practice Set 6
Practice These Questions
1. Consider the following statements.
S1: The RISC processor has CPI always 1.
S2: In horizontal instruction control signals are always in encoded form.
S3: In vertical instruction control signals are always in encoded form.
S4: In terms of speed vertical instruction is slower than horizontal instruction.
Which of the above statements are true?
(a) Only S1 and S2 (b) Only S2 and S3
(c) Only S1, S3 and S4 (d) None of the above
Answer : (C) S1 is correct because this is the property of RISC processor that CPI is 1. S2 is not correct because horizontal instructions controls signals are in decoded form. S3 is true vertical instruction have signals in encoded form. Speed of vertical instruction is less as compare to horizontal instruction.
2. Computer uses addressing mode technique for __________.
(a) giving program versatility to user by providing facilities as a pointer to memory counters for
loop control
(b) reducing number of bits in the field of instruction
(c) provide rule or specification for calculating address
(d) All of the above
Answer : D.
3. Which of the following is added to the page table in order to track whether a page of cache has been modified since it was read from the memory?
(a) Reference bit (b) Dirty bit
(c) Tag bit (d) Valid bit
Answer (b) : Dirty Bit is the right answer because dirty bit is used to represent the status of the cache whether it has been modified or not after copying the data from main memory to cache. If dirty bit value is zero then it represent no modification has occurred if dirty bit value is 1 then it represents then modification has taken place.
4. Select the appropriate match of list 1 with list 2
A. Pointer 1. Indirect addressing mode
B. Position independent code 2. Immediate addressing mode
C. Constant operand 3. Relative addressing mode
Codes:
A B C
(a) 1 2 3
(b) 3 2 1
(c) 1 3 2
(d) 2 3 1
Answer : C
For making use of pointer in program we use indirect addressing mode because ponter store the address of a variable and same thing happen in indirect addressing mode store the address of effective address.
In relocation concept position independent code is used and to implement this relative addressing mode which use relocation register and set the difference of logical and physical address.
Immediate addressing mode provides the value directly in the instruction which implies the constant operand.
5. A 50 kbps device is connected to the processor. The interrupt overhead is 50 μsec. what is the performance achieved if interrupt initiated and data transferred is performed using programmed I/O is __________.
(a) 2.4 (b) 0.4
(c) 3 (d) 3.5
Answer: B
1 sec → 50 kbyte
1 byte → 1/150=20×10−6 =20
For interrupt driven mode it takes 50 μsec
So performance achieved when interrupt driven used over programmed I/O
= − / −
=20/50=0.4
6. Consider the following sequence of micro-operations.
MBR ‹ PC
MAR ‹ SP
M[MAR] ‹ MBR
PC ‹ Vector address.
Which of the following operation performed by this sequence?
(a) Instruction fetch (b) Operand fetch
(c) Interrupt subprogram initialization (d) Conditional branch
Answer: C
Program Counter holds the value of next instruction to be executed. nWe store the value of PC to MBR and value of stack pointer to MAR. Then store the value of PC which is available in MBR to location addressed by MAR.
Atleast vector address return to the PC. This can be done in interrupt subprogram initialization.
7. We have two designs P1 and P2 for a synchronous pipeline processor. P1 has 8 pipeline stages with execution time of 3 nsec, 2 nsec, 4 nsec, 8 nsec, 2 nsec, 5 nsec, 4 nsec and 1 nsec while design P2 has 5 stages each with 5 nsec each with 5 nsec execution time. How much time (in μsec) can be saved using design P2 over design P1 for executing 500 instructions? (upto 3 digit).
(a) 2.536 (b) 1.365
(c) 1.536 (d) 1.653
Answer: C
Execution time for pipeline = (k + n – 1) × tp
where k = Number of stages
n = Number of instruction
tp = Execution time = Max (all stages)
P1 = [8 + 500 – 1] × 8 = 4056
P2 = [5 + 500 – 1] × 5 = 2520
Time saved using P2 = 4056 – 2520 = 1536 nsec = 1.536 μsec
8. In which of the following addressing mode, the content of the program counter is added to the address part of the instruction to get the effective address?
(a) Indexed addressing mode (b) Implied addressing mode
(c) Relative addressing mode (d) Register addressing mode
Answer :C
We know that in relative addressing mode content of the program counter is added to the address part of the instruction to get the effective address
9. Suppose that a cache is 20 times faster than main memory and cache memory can be used 80% of the time. What is the speed up factor ?
(a) 2.16 (b) 3.16
(c) 4.20 (d) 4.16
Answer : D
This can be solved Amdhal’s law
F = 80%, S = 20, overall speed-up = [(1−)+]−1
=[(1−0.8)+0.820]−1 = 4.16
10. If the last operation performed on a computer with an 8-bit word has an addition in which the two operands were 00000010 and 00000011, what would be the value of the Overflow, Sign and Half-Carry flags respectively?
(a) 0, 0, 0 (b) 0, 1, 0
(c) 1, 0, 1 (d) 0, 1, 1
Answer: A
A is the right option
0 0 0 0 0 0 1 0
0 0 0 0 0 0 1 1
———————–
0 0 0 0 0 1 0 1
———————–
Overflow = 0, sign = 0, Half carry = 0
Half carry indicate addition of packed decimal numbers. When carry takes out of the lower digit order, this flag is set. Auxiliary carry is also known as half carry
11. A 4 byte long PC-relative branch instruction is fetched from memory address 51210 and while its execution, the branch is made to location 88510. Find out the displacement in relative mode.
Answer : 369
Effective address = PC + Relative value
Relative value = EA – PC
= 885 – 516
12. Consider the following program segment:
Instruction Meaning Size (words)
I1 LOAD r0, 500 r0 ⃪ [500] 2
I2 MOV r1, r0 r1 ⃪ [r0] 1
I3 ADD ro, r1 r0 ⃪ r0 + r1 1
I4 INC r0 r0 ⃪ r0 + 1 1
I5 INC r1 r1 ⃪ r1 + 1 1
I6 ADD r0, r1 r0 ⃪ r0 + r1 1
I7 Store r1, r0 M[(r1)] ⃪ r0 2
I8 Halt Stop 1
Suppose that memory is word addressable and word size is 32 bits. Program is loaded into memory location (3000)10 on wards. What is the value of program counter register at the end of the program?.
Answer : 3009
Explanation:
(3009)
Word addressable storage
3000 – 3001
3002
3003
3004
3005
3006
3007 – 3008
3009
Valid program counter value after program is 3009.
13. A branch mark program is running on a 40 MHz processor. The executed program consists of 100,000 instruction executions, with the following instruction mix and clock cycle count.
Instruction Type Instruction Count Cycles per Instruction
Integer arithmetic 45000 1
Date transfer 32000 2
Floating point 15000 2
Control transfer 8000 2
The execution time in msec is __________.
Answer: 3.87 (3.86-3.88)
14. A hypothetical control unit supports 5 groups of mutually exclusive control signals. The numbers of bits that can be saved using vertical approach compared to horizontal are __________.
Solution: 22
Explanation:
(22)
In horizontal: 1 + 5 + 7 + 15 + 8 = 36
In vertical: log 1 + log 5 + log 7 + log 15 + log 8 = 1 + 3 + 3 + 4 + 3 = 14
Total saved bits = 36 – 14 = 22
15. Consider the micro-programmed control unit which support 256 instructions, each of which on an average takes 16 micro operations. The system support 16 flag conditions and 52 control signals. If vertical microprogramming control is used in the system then total length of control word is __________ (bits/word).
Answer: 22
Explanation:
(22)
Number of words in control memory = 256 × 16 = 4096 words
⇒ Address field = 12 bits
16 = 24 and 52 < 26
16. Consider a non-pipeline processor has clock rate of 25 MHz and CPI of 6, another processor designed with same clock rate and 8 stage instruction pipeline. If program has 500 instructions to be executed on both processors what will be the value of speed up factor ?
Answer: 5.91 (5.90-5.92)
Explanation:
5.91 (5.90-5.92)
Since we know that
Speed_Up = NonPipleline/Pipeline = n * tn / K + n -1
Speed_UP= (500 * 6) / (500+8-1)
n = 500
tn = 6 (for non-pipeline)
K = 8 (for pipeline) =500×6500+8−1=3000507=5.91
17. Consider the following statements:
S1: Compulsory miss can be reduced.
S2: Conflict miss can be reduced.
S3: Capacity miss can be reduced.
Which of the above statements are True?
(a) Only S1 (b) Only S1 and S2
(c) Only S2 and S3 (d) S1, S2 and S3
Answer: (d)
S2: Conflict miss are occur when too many blocks are mapped into same line or set. So by increasing the associativity i.e. increases the size of set and increases the number of sets.
S1: Compulsory miss can be reduced by increasing the line size i.e., reduce number of lines.
S3: Capacity miss can be reduced by increasing the cache memory size.
All of the three statements are true.
18. Consider the hypothetical processor which support 512 k words memory. It uses the memory mapped IO configuration. In which when 2 most significants bits of address are 1 then assigned to IO port.
How many numbers of I/O port address and memory addresses are possible in the processor respectively?
(a) 1×217, 3×217 (b) 3×217, 1×217
(c) 2×217, 1×217 (d) None of these
Answer:
19. Which of the following statements are True?
S1: Reference bit in page table entry used for page replacement.
S2: In hierarchical memory access organization, CPU perform read and write operation on only
level 1 memory.
S3: In simultaneous memory access organization, CPU perform read and write operation on any
level of memory.
(a) S1 and S2 only (b) S1 and S3 only
(c) S2 and S3 only (d) S1, S2 and S3
Answer : (d)
All the above statements are correct.
S1: Reference bit sometimes called access bit used in page table entry to show if page is replaced or not.
S2: In hierarchical memory access, CPU perform read and write operation only on level 1 memory. If miss occur then data is first transferred to level 1 then CPU access data.
S3: In simultaneous memory access, CPU perform read and write operation on any level of memory i.e. not necessary to take data first into level 1 memory than access it.
20. A computer has 32-bit instruction and 9-bit address. If there are 400 two address instructions then how many one address instructions can be formulated?
(a) 214 (b) 232 – 200
(c) 214 – 400 (d) (214 – 400) × 29
21. Find the true statement
(a) In write through protocol, cache location and main memory location area unit updated simultaneously.
(b) In write back protocol, cache location and main memory location area unit updated at the same time.
(c) changed or dirty bits area unit utilized by write through protocol.
(d) None of those.
22. A 16KB 4-way set associative write-back cache is organized as multiple blocks, every of size 64-bytes.The processor generates 32-bit addresses. The cache controller maintains the tag data for every cache block comprising with one valid bit and one changed bit.
As several bits because the minimum required to spot the memory block mapped within the cache. what’s the whole size of memory required at the cache controller to store meta-data (tags) for the cache?
(a) 5362 bytes (b) 5361 bytes
(c) 704 bytes (d) 176 bytes
23. Take into account a system with the most access time as two hundred ns and cache time interval as ten ns. Hit magnitude relation for browse request is zero.8 and eight0% of the memory requests area unit for browse. If write through policy is employed, then the common time considering each browse and write requests is a little more.
(a) 169.6 ns (b) 192.4 ns
(c) 78.4 ns (d) None of those
24. Consider the following instructions.
I1 : R1 = 100
I2 : R1 = R2 + R4
I3 : R2 = R4 + 25
I4 : R4 = R1 + R3
I5 : R1 = R1 + 30
Calculate sum of (WAR, RAW and WAW) dependencies the above instructions.
(a) 10 (b) 12
(c) 6 (d) 8
25. A 4-way set associative cache memory consists of 128 blocks. The main memory consists of 32768 memory blocks and each block contains 512 eight bit words. Find how many bits are needed to represent TAG, SET and WORD field respectively?
(a) 5, 9, 10 (b) 10, 6, 8
(c) 10, 9, 5 (d) 10, 5, 9
26. Suppose directed mapped cache with 2m lines 2p bytes per cache lines. Memory is byte addressable of 2n bytes. Compute the space required for storing tags (in bits)?
(a) 2n – (m+p) (b) 2m × (n – (m+p))
(c) 2n+m (d) (m+n) × 2m
27. Consider a small two-way set-associative cache memory, consisting of 4 blocks. For choosing the block to be replaced, use the least recently used (LRU) scheme. The number of cache misses for the sequence of block addresses 18, 22, 10, 22, 18 is __________.
(a) 2 (b) 3
(c) 4 (d) 5
28. Consider a pipelined system with four stages: IF, ID, EX, WB. Following chart shows the clock cycles required by each instruction to compete each stage
How many clock cycles are required to complete the above instructions?
(a) 15 (b) 9
(c) 14 (d) 13
