Zim Brightwood
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For when life gives you all the lemons

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Zim Brightwood
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CPSC313 Midterm Sample

CPSC313 Midterm Sample

Questão 1 de 31

1

1a) The classic RISC pipeline consists of 5 stages, what are they?

Selecione uma das seguintes:

  • Fetch, Decode, Execute, Memory, Write back

  • Fetch, Register, Execute, Memory, Write back

  • Load, Register, Execute, Memory, Write-read

  • Fetch, Register, Execute, Memory, Load-use

  • Load, Register, Execute, Memory, Control

Explicação

Questão 2 de 31

1

1b) Fetch is responsible for which actions? Check all that apply.

Selecione uma ou mais das seguintes:

  • Use pc to read next instruction from memory into instruction register

  • Determine instruction length and extract pieces of instruction

  • Advance the pc to the address of next instruction in sequence

  • Read values from register file

  • Update the pc for taken jump instructions

Explicação

Questão 3 de 31

1

1c) Which of the following is a decode action?

Selecione uma das seguintes:

  • read values from register

  • determine instruction length and extract pieces of instruction

  • access main memory

  • decode the difficulty of this course into a binary representation of tears and not tears

Explicação

Questão 4 de 31

1

1d) Which of the following is an execute stage action?

Selecione uma das seguintes:

  • perform ALU operations and determine whether jumps are taken

  • set the pc for taken jumps

  • access main memory

  • cry profusely into a bowl of alphabits hoping they'll arrange themselves into the proper values for the next stage

Explicação

Questão 5 de 31

1

1e) Which of the following is a memory stage action?

Selecione uma das seguintes:

  • access main memory, duhhhhhhhhhh

  • access main memory and write the values back to the register file, duhhhhhhhhhhh

  • when in doubt choose c

  • advance the pc to the next memory address in the sequence

Explicação

Questão 6 de 31

1

1f) Which of the following is a write back stage action?

Selecione uma das seguintes:

  • write values back to register file

  • write values back to main memory

  • write values back to the pc

  • forge your computer science degree when failing this course leads you into months of crippling depression

Explicação

Questão 7 de 31

1

2a) Why is the memory stage after the execute stage? Include a Y86 instruction that could not be implemented if you reversed the order.

Selecione uma das seguintes:

  • Execute needs to perform address calculations. mrmovl could not use base plus displacement addressing

  • Execute needs to perform math on register values. None of the op instructions would work (i.e. addl)

  • Memory needs to write a value from memory into a register. mrmovl would not work if the order was reversed.

  • Memory needs to read from memory in order for execute to work properly. ret would not work if the order was reversed.

Explicação

Questão 8 de 31

2

2b) Why is the execute stage after the decode stage? Include a Y86 instruction that could not be implemented if you reversed the order.

Selecione uma das seguintes:

  • Execute needs to be able to perform math on register values. If the order were reversed, none of the op instructions would work (i.e. addl)

  • Execute needs to perform address calculations. If the order were reversed, mrmovl would not work.

  • Decode needs to write a value from memory into a register. mrmovl would no longer work if the order was reversed.

  • Decode needs to read the values before execute stage to avoid stalling. popl would no longer work if the order was reversed.

Explicação

Questão 9 de 31

2

2c) Why is the write-back stage after the memory stage? Include a Y86 instruction that could not be implemented if you reversed the order.

Selecione uma das seguintes:

  • Memory needs to perform math on memory values. In the reverse order, none of the op instructions would work (i.e. addl)

  • Memory needs to perform address calculations. In the reverse order, mrmovl and rmmovl would not work.

  • Write back needs to be able to write a value from memory into a register. In the reverse order, mrmovl would not work.

Explicação

Questão 10 de 31

1

3a) RISC instruction sets do not allow ALU operations to read from memory. Explain how the structure of the pipeline leads this restriction.

Selecione uma das seguintes:

  • Since M comes after E, by the time we read the value from memory, it's too late.

  • Since M comes before E, by the time we read the value from memory, it's too late.

  • Since E comes after M, by the time we read the value from memory, it's too late.

Explicação

Questão 11 de 31

2

3b) Describe how the pipeline could be modified to lift this restriction. You may not change the number
of pipeline stages or their basic function. Your revised pipeline does not need to be able to implement
every Y86 instruction (in fact it would not be able to) but it must be able to execute instructions like:

addl (%eax), %ebx # r[ebx] = r[ebx] + m[r[eax]]

Selecione uma das seguintes:

  • Swap the order of E and M

  • Allow for M to forward data back to E

  • Stall E until M has the correct data

Explicação

Questão 12 de 31

2

3c) This revised pipeline must place some restrictions on the revised ISA that it implements. One of
the impacts is on mrmovl and rmmovl. Describe the problem and modify the instructions so that
they will work with the new pipeline. Note that the revised versions need not be as powerful as the
original ones, but they must still load and store between memory and a register.

Selecione uma das seguintes:

  • We wouldn't be able to handle base plus addressing because E would come after M. To solve this, we simply eliminate base plus addressing

  • Memory access is rendered useless. To solve this, we simply eliminate the memory stage.

  • The write back stage would no longer be able to determine the correct register to move data to. To solve this, we would simply add another stage input to account for that lost addressing.

Explicação

Questão 13 de 31

2

Consider the following five-stage pipeline with stage delays (including overheads) of 34 ps,
42 ps, 75 ps, 50 ps and 18 ps.

What is the maximum clock rate (i.e., fastest) acceptable for this pipeline?

Arraste e solte para completar o texto.

    1 cycle / 75ps = 1000/75 Ghz
    1 cycle / 34ps = 1000/34 Ghz
    1 cycle / 42ps = 1000/42 Ghz
    1 cycle / 50ps = 1000/50 Ghz
    1 cycle / 18ps = 1000/18 Ghz

Explicação

Questão 14 de 31

2

Consider the following five-stage pipeline with stage delays (including overheads) of 34 ps,
42 ps, 75 ps, 50 ps and 18 ps.

What is the maximum throughput of this pipeline?

Arraste e solte para completar o texto.

    1/75 instructions per picosecond
    1/34 instructions per picosecond
    1/42 instructions per picosecond
    1/50 instructions per picosecond
    1/18 instructions per picosecond

Explicação

Questão 15 de 31

2

4c) Consider the following five-stage pipeline with stage delays (including overheads) of 34 ps,
42 ps, 75 ps, 50 ps and 18 ps.

What, if anything, might cause the actual throughput of programs to be lower than this maximum?

Selecione uma das seguintes:

  • pipeline bubbles

  • nothing would cause that, the maximum is actually also the minimum

  • jumps and returns

  • stack depletion

Explicação

Questão 16 de 31

2

Consider the following five-stage pipeline with stage delays (including overheads) of 34 ps,
42 ps, 75 ps, 50 ps and 18 ps.

What is the minimum instructions latency of this pipeline?

Arraste e solte para completar o texto.

    75ps + 75ps + 75ps + 75ps + 75ps
    34ps + 42ps + 50ps + 75ps + 18ps
    18ps + 18ps + 18ps + 18ps + 18ps

Explicação

Questão 17 de 31

3

5a) Consider the following piece of Y86 assembly code:
[0] addl %eax, %ebx
[1] irmovl $1, %eax
[2] mrmovl %(ebx), %ebx
[3] addl %ebx, %eax

List all of the output dependencies present in this code.

Selecione uma ou mais das seguintes:

  • 0 and 2 on %ebx

  • 0 and 1 on %eax

  • 0 and 3 on %ebx

  • 1 and 2 on %ebx

  • 1 and 3 on %eax

  • 2 and 3 on %ebx

Explicação

Questão 18 de 31

1

5a) Consider the following piece of Y86 assembly code:
[0] addl %eax, %ebx
[1] irmovl $1, %eax
[2] mrmovl %(ebx), %ebx
[3] addl %ebx, %eax

List all of the anti dependencies present in this code.

Selecione uma ou mais das seguintes:

  • 0 and 1 on %eax

  • 0 and 2 on %ebx

  • 0 and 3 on %ebx

  • 1 and 2 on %ebx

  • 1 and 3 on %eax

  • 2 and 3 on %ebx

Explicação

Questão 19 de 31

1

5a) Consider the following piece of Y86 assembly code:
[0] addl %eax, %ebx
[1] irmovl $1, %eax
[2] mrmovl %(ebx), %ebx
[3] addl %ebx, %eax

List all of the causal dependencies present in this code.

Selecione uma ou mais das seguintes:

  • 0 and 1 on %eax

  • 0 and 2 on %ebx

  • 0 and 3 on %eax

  • 1 and 2 on %ebx

  • 1 and 3 on %eax

  • 2 and 3 on %ebx

Explicação

Questão 20 de 31

3

5b) Consider the following piece of Y86 assembly code:
[0] addl %eax, %ebx
[1] irmovl $1, %eax
[2] mrmovl %(ebx), %ebx
[3] addl %ebx, %eax

List all of the data hazards present in this code for the Y86, five stage pipeline

Selecione uma ou mais das seguintes:

  • All the dependencies

  • All the causal dependencies

  • All the output dependencies

  • All the anti dependencies

  • Just the causal and output dependencies

  • Just the causal and anti dependencies

  • Just the anti and output dependencies

Explicação

Questão 21 de 31

2

5c) Consider the following piece of Y86 assembly code:
[0] addl %eax, %ebx
[1] irmovl $1, %eax
[2] mrmovl %(ebx), %ebx
[3] addl %ebx, %eax

For each hazard, indicate the total number of bubbles added by the Pipe-Minus implementation

Selecione uma ou mais das seguintes:

  • 2 for dependency between 0 and 2 on %ebx

  • 1 for dependency between 0 and 2 on %ebx

  • 3 for dependency between 0 and 2 on %ebx

  • 0 for dependency between 1 and 3 on %eax

  • 1 for dependency between 1 and 3 on %eax

  • 2 for dependency between 1 and 3 on %eax

  • 0 for dependency between 0 and 2 on %ebx

  • 3 for dependency between 1 and 3 on %eax

  • 3 for dependency between 2 and 3 on %ebx

  • 2 for dependency between 2 and 3 on %ebx

Explicação

Questão 22 de 31

2

5d) Consider the following piece of Y86 assembly code:
[0] addl %eax, %ebx
[1] irmovl $1, %eax
[2] mrmovl %(ebx), %ebx
[3] addl %ebx, %eax

For each hazard, indicate the total number of bubbles added by the Pipe implementation

Selecione uma ou mais das seguintes:

  • 1 for dependency between 2 and 3 on %ebx

  • 2 for dependency between 2 and 3 on %ebx

  • 3 for dependency between 2 and 3 on %ebx

  • 0 for dependency between 2 and 3 on %ebx

Explicação

Questão 23 de 31

10

Describe the implementation of the following new instruction for Y86 Seq as you did in
Homework 2. This instruction is similar to mrmovl except that it adds 4 to rB and does not have a static-displacement operand. It would be useful, for example, for iterating over an array of integers.

Syntax:
mrmovincl (rB), rA
Semantics:
r[rA] <= m[r[rB]]
r[rB] <= r[rB] + 4
Memory Layout:
| 5 | F | rA | rB |
Describe each stage using a relaxed syntax similar as shown below. The Fetch and PC Update stages are complete. List only the code that would be added for this instruction.

Fetch:
f.iCd = m[F.pc] >> 4
f.iFn = m[F.pc] & 0xf
f.rA = m[F.pc+1] >> 4
f.rB = m[F.pc+1] & 0xf
f.valP = F.pc + 2
Decode:
d.srcA =
d.srcB =
d.dstE = D.rB
d.dstM =
d.valB =
Execute:
e.aluA = 4
e.aluB =
e.aluFun =
e.valE =
Memory:
m.valM =
Write Back:
r[W.dstE] =
r[W.dstM] =
PC Update (pseudo stage):
w.pc = W.valP

Arraste e solte para completar o texto.

    R_NONE
    D.rB
    D.rA
    r[d.srcB]
    E.valB
    A_ADD
    A_SUB
    A_INC
    4 + E.valB
    m4[M.valB]
    m4[M.srcB]
    W.valE
    W.valM
    R_ADD
    R_SUB

Explicação

Questão 24 de 31

6

Write Y86 assembly code that computes the sum of an array of integers where the address of
the array is stored in %ebx and the length of the array is in %ecx. Place the sum in %eax.

irmovl $0, #SET SUM TO 0
irmovl $, %edi #HOW MUCH WILL THE LENGTH DECREMENT BY AS YOU ITERATE?
irmovl $, %esi #AN ARRAY OF INTEGERS HAS ELEMENTS WITH A SIZE OF?
%ecx, %ecx #IS THE LENGTH OF THE ARRAY 0?
jle L1
L0: , %edx #STORE ADDRESS OF ARRAY INTO %edx
addl , %eax #ADD VALUE OF ELEMENT IN THE ARRAY TO SUM
addl , %ebx #INCREMENT THE ARRAY POINTER
subl , %ecx #DECREMENT THE LENGTH OF THE ARRAY
L0 #CHECK IF LENGTH IS STILL GREATER THAN ZERO
L1:

Arraste e solte para completar o texto.

    %eax
    1
    2
    3
    4
    5
    andl
    mrmovl
    rmmovl
    rrmovl
    (%ebx)
    %ebx
    %edx
    %esi
    %edi
    jg
    jge
    jle
    jl

Explicação

Questão 25 de 31

2

8a) What is spatial locality and why is it important?

Selecione uma ou mais das seguintes:

  • Spatial locality exists when memory accesses are clustered together to nearby memory addresses

  • Caches exploit spatial locality by storing data in multi-byte blocks.

  • Spatial locality exists when the same memory location is accessed repeatedly.

  • Caches exploit spatial locality by retaining recently accessed blocks in the cache.

  • Spatial locality exists between a pair of instructions when there are no dependencies between them and thus their execution order does not matter.

  • Pipelined (and super-scalar) processor architectures exploit spatial locality.

  • Spatial locality exists when programs explicitly indicate that threads of execution can be executed in parallel by either directly or indirectly creating multiple threads that can execute con-currently.

  • Multi-core processors (and hyper-threading) exploit spatial locality.

Explicação

Questão 26 de 31

2

8b) What is temporal locality and why is it important?

Selecione uma ou mais das seguintes:

  • Temporal locality exists when memory accesses are clustered together to nearby memory addresses.

  • Caches exploit temporal locality by storing data in multi-byte blocks.

  • Temporal locality exists when the same memory location is accessed repeatedly.

  • Caches exploit temporal locality by retaining recently accessed blocks in the cache.

  • Temporal locality exists between a pair of instructions when there are no dependencies between them and thus their execution order does not matter.

  • Pipelined (and super-scalar) processor architectures exploit temporal locality.

  • Temporal locality exists when programs explicitly indicate that threads of execution can be executed in parallel by either directly or indirectly creating multiple threads that can execute con-currently.

  • Multi-core processors (and hyper-threading) exploit temporal locality.

Explicação

Questão 27 de 31

2

8c) What is instruction-level parallelism and why is it important?

Selecione uma ou mais das seguintes:

  • Instruction-level parallelism exists between a pair of instructions when there are no dependencies
    between them and thus their execution order does not matter.

  • Pipelined (and super-scalar) processor architectures exploit instruction-level parallelism.

  • Instruction-level parallelism exists when programs explicitly indicate that threads of execution can be executed in parallel by either directly or indirectly creating multiple threads that can execute con-currently.

  • Multi-core processors (and hyper-threading) exploit instruction-level parallelism.

  • Instruction-level parallelism exists when the same memory location is accessed repeatedly.

  • Caches exploit instruction-level parallelism by storing data in multi-byte blocks.

  • Instruction-level parallelism exists when memory accesses are clustered together to nearby memory addresses.

  • Caches exploit instruction-level paralellism by retaining recently accessed blocks in the cache.

Explicação

Questão 28 de 31

2

8d) What is thread-level parallelism and why is it important?

Selecione uma ou mais das seguintes:

  • Thread-level parallelism exists when programs explicitly indicate that threads of execution can be executed in parallel by either directly or indirectly creating multiple threads that can execute con-currently.

  • Multi-core processors (and hyper-threading) exploit thread-level parallelism.

  • Thread-level parallelism exists between a pair of instructions when there are no dependencies
    between them and thus their execution order does not matter.

  • Pipelined (and super-scalar) processor architectures exploit thread-level parallelism.

  • Thread-level parallelism exists when the same memory location is accessed repeatedly.

  • Caches exploit thread-level parallelism by retaining recently accessed blocks in the cache.

  • Caches exploit thread-level parallelism by storing data in multi-byte blocks.

  • Thread-level parallelism exists when memory accesses are clustered together to nearby memory addresses.

Explicação

Questão 29 de 31

4

Polymorphic dispatch common to object-oriented languages like Java is implemented using a
double-indirect call instruction that reads an address from memory and then jumps to it. Explain why it is
challenging to implement such an instruction without stalling in a pipelined processor.

Selecione uma das seguintes:

  • The address of the next instruction to execute after the call is not known to the pipeline until the call instruction exits the memory stage and so predicting which instructions should be F, D, and E at this point is impossible without retaining execution history.

  • If the predPC is overwritten with the address of a call at a stage earlier than memory, the instructions already in the pipeline will bubble.

  • Couldn't we all just down a bottle of bleach instead of do this midterm?

Explicação

Questão 30 de 31

4

Consider the following instruction-execution frequencies for a program running on the
standard Y86 Pipe processor. The table shows, for example, that 7% of all instructions executed read the value of a register immediately after the preceding instruction modified that register by writing into it a value that came from memory.
7% read register immediately after an instruction writes into that register a value it reads from memory
6% read register immediately after an instruction writes into that register a value computed in Execute
12% conditional jump that is taken
8% conditional jump that is not taken
5% call
5% ret
57% the remaining introduce no bubbles

What is the average cycles per instruction for this execution?

Selecione uma das seguintes:

  • CPI = 1 + 0.07 + 0.08 * 2 + 0.05 * 3 = 1.38
    Because load-use causes 1 bubble, not taken penalties cause 2 bubbles, and ret instructions cause 3 bubbles

  • CPI = 1 + 0.07 * 2 + 0.08 * 3 + 0.05 = 1.43
    Because load-use causes 2 bubbles, not taken penalties cause 3 bubbles, and ret instructions cause 1 bubble

  • CPI = 1 + 0.06 * 2 + 0.12 * 3 + 0.05 = 1.41
    Because forwarding causes 2 bubbles, taken penalties cause 3 bubbles, and call instructions cause 1 bubble

  • CPI = 1 + 0.06 + 0.12 * 2 + 0.05 * 3 = 1.45
    Because forwarding causes 1 bubble, taken penalties cause 2 bubbles, and call instructions cause 3 bubbles

Explicação

Questão 31 de 31

4

Consider the following instruction-execution frequencies for a program running on the
standard Y86 Pipe processor. The table shows, for example, that 7% of all instructions executed read the value of a register immediately after the preceding instruction modified that register by writing into it a value that came from memory.
7% read register immediately after an instruction writes into that register a value it reads from memory
6% read register immediately after an instruction writes into that register a value computed in Execute
12% conditional jump that is taken
8% conditional jump that is not taken
5% call
5% ret
57% the remaining introduce no bubbles

What is the throughput of this execution on a 3-Ghz processor (ie 3*10^9 cycles per second)? (You may go back one question to see what you chose for the answer.

Selecione uma das seguintes:

  • (3*10^9 / 1.38) instructions per second

  • (3*10^9 / 1.41) instructions per seconds

  • (3*10^9 / 1.43) instructions per seconds

  • (3*10^9 / 1.45) instructions per seconds

Explicação

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