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Direct-Mapped TLB - Computer Architecture and Engineering - Exams, Exams of Computer Architecture and Organization

Main points of this past exam are: Direct-Mapped Tlb, Valid Translation, Frame Number, Physical Address, Fault Types, Indicate Fault, Memory Accesses, Physical Frame, Frame Number, Page Number

Typology: Exams

2012/2013

Uploaded on 04/02/2013

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University of California at Berkeley
College of Engineering
Computer Science Division - EECS
CS 152 D. Patterson & R. Yung
Fall 1995
Computer Architecture and Engineering
Midterm II
You may use two pages of notes. You have 180 minutes.
Please write your name on this cover sheet and also at the top left of each page.
The point value of each question is indicated in brackets after it.
Please show your work. Write neatly and be well organized. Good luck!
Your Name:
SID Number:
Discussion TA(s):
Problem Score
1 / 27
2 / 13
3 / 30
4 / 20
Total / 90
pf3
pf4
pf5
pf8

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Download Direct-Mapped TLB - Computer Architecture and Engineering - Exams and more Exams Computer Architecture and Organization in PDF only on Docsity!

University of California at Berkeley

College of Engineering

Computer Science Division - EECS

CS 152 D. Patterson & R. Yung Fall 1995

Computer Architecture and Engineering

Midterm II

You may use two pages of notes. You have 180 minutes. Please write your name on this cover sheet and also at the top left of each page. The point value of each question is indicated in brackets after it. Please show your work. Write neatly and be well organized. Good luck!

Your Name:

SID Number:

Discussion TA(s):

Problem Score

1 / 27

2 / 13

3 / 30

4 / 20

Total / 90

Question 1: Virtual Memory

An eight-entry direct-mapped TLB is implemented in the current design. Both the virtual and physical addresses are 32 bits wide, and page size is 4kB. Address translation is performed by this TLB for every memory access. NOTE: All addresses are given in hexadecimal.

a) Label the virtual and physical address fields used in address translation. [4 pt]

Virtual Address

31 0

Physical Address 31 0

c) Fill in the content of the TLB after the four memory accesses. [7 pt]

Table 3: Final TLB state

Index Virtual Page Number Physical Frame Number Read Write Valid

0 1 2 3 4 5 6 7

Question 2: I/O

A 1993 3.5 inch IBM disk rotates at 4318 revolutions-per-minute (RPM), has a random seek time of 11ms, transfers at 4 MB/s, has a capacity of 1 GB, and the mean-time-to-failure (MTTF) is 400,000 hours. The SCSI controller overhead is 2ms.

A 1995 3.5 inch IBM disk rotates at 7200 RPM, has a random seek time of 8ms, transfers at 12 MB/s, has a capacity is 4.2 GB, and the MTTF is 1,000,000 hours. The SCSI controller overhead today is 1ms.

a) On average, how much faster is the new disk than the old disk for a read of 4 kB assuming random seeks? Assume the disks are idle so that there is no waiting time. [4 pt]

b) If the actual seek time is 25% of the random seek, how much faster is the new disk now? [4 pt]

c) Now assume the read size is 1 megabyte, with seeks being 25% of random time. How much faster is the new disk? [4 pt]

d) What does this performance change in just two years suggested in the design of computer systems? [1 pt]

b) Here is a sequence of ten one-byte memory references (in hex) to the caches: R 0x0, W 0x4, R 0x6, R 0x20, R 0x25, W 0x27, W 0x1, R 0x23, R 0x3, R 0x4. NOTE: R - read access, W - write access

Fill in hit or miss for each memory references for the four cache configurations. Compute the final cache miss rates. NOTE: each cache starts out empty. [22 pt]

Results for the above example, cache E, are shown in the last column.

Table 6: Results of the cache references

references cache configurations

A B C D E

0x00 R miss

0x04 W miss

0x06 R miss

0x20 R miss

0x25 R miss

0x27 W miss

0x01 W miss

0x23 R miss

0x03 R miss

0x04 R miss

miss rate (%) 100%

Question 4: Pipelining

The designers are concerned about stalls in the pipeline. (Un)fortunately, the chief architect is out teaching. You are asked to examine the pipeline and recommend necessary bypasses to make the pipeline functional and to minimize pipeline stalls.

Assume a five-stage pipeline (IF, ID, EX, MEM, WB), with pipeline registers between adjacent stages. Pipeline registers are labeled T1 (IF/ID), T2 (ID/EX), T3 (EX/MEM), T4 (MEM/WB).

Fill in your bypass recommendations in Table 7. The table has a <pipeline register, opcode> pair in each column heading, and a <pipeline stage, opcode> pair for each row heading. The columns represent the source of bypasses. For example, column <T3, SUBU> is the output of EX stage latched in pipeline register T3 after executing SUBU. The rows represent the sink of the bypasses. For example, row <EX, LW> is the input to the ALU in EX stage for LW. [20 pt]

You may only forward to the components listed below: IF ID EX MEM WB Instr. Memory Comparator ALU Data Memory Sign Extend PC Only add bypasses when it is needed for correctness or to prevent stalls. When no forwarding is needed, say so. You will lose points for adding unnecessary bypasses or leaving entries blank!

A few entries are labeled for you as examples.

Table 7: Pipeline bypass

<pipeline register, opcode>

<stage, opcode>

T

(ID/EX)

BEQ

T

(EX/MEM)

SUBU

T

(EX/MEM)

SW

T

(MEM/WB)

LW

T

(MEM/WB)

JAL

IF, BEQ No forwarding

ID, JR No forwarding

ID, BNEZ

EX, JR Not applicable

EX, LW Not applicable

T3 -> ALU