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Main points of this exam paper are: Column Number, Maximum Credit, Computer Engineering, Gbit Dram Chip, Decoder Required, Mux Required, Muxes Required, Column Number, Column Offset, Decoder Required
Typology: Exams
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4 problems, 5 pages Exam Three 19 November 2003
Instructions: This is a closed book, closed note exam. Calculators are not permitted. If you have a question, raise your hand and I will come to you. Please work the exam in pencil and do not separate the pages of the exam. For maximum credit, show your work. Good Luck!
Your Name ( please print ) ________________________________________________
1 2 3 4 total
4 problems, 5 pages Exam Three 19 November 2003
Problem 1 (3 part, 12 points) Instruction Formats
An instruction format has the following field lengths for R-type and I-type instructions. Answer the following questions:
opcode RD RS1 RS 12 bits 9 bits 9 bits 9 bits opcode RD RS1 immediate value 12 bits 9 bits 9 bits 28 bits
Part A (4 points) How many registers are there?
Part B (4 points) How many instruction types are there?
Part C (4 points) What is the range of immediate values?
Problem 2 (3 parts, 32 points) Memory Systems
Part A (12 points) Consider a 4 Gbit DRAM chip organized as 512 million addresses of 8 bit words. Assume both the DRAM cell and the DRAM chip are square. The column number and offset concatenate to form the memory address. Using the organization approach discussed in class, answer the following questions about the chip. Express all answers in decimal.
number of columns column decoder required ( n to m )
type of mux required ( n to m ) number of muxes required number of address lines in column number
number of address lines in column offset
Part B (10 points) Consider a 256 Mbyte memory system with 64 million addresses of 4 byte words using 16 Mbit DRAM chips organized as 4 million addresses by 4 bit words.
word address lines for memory system
chips needed in one bank banks for memory system
memory decoder required ( n to m ) DRAM chips required
4 problems, 5 pages Exam Three 19 November 2003
Problem 4 (6 parts, 35 points) Microcode Reverse Engineering
The microcode fragment below is a mysterious program that runs on the provided datapath handout. All values are in hexadecimal. Unfortunately, don’t care values (X) have been converted to zeros. For full credit, be as specific and concise as you can (e.g., list shift types, amounts, and directions, logical functions, memory addresses and operations, etc.)
cycle X Y Z rwe im en im va au en -a/s lu en lf su en st ld en st en r/-w msel 1 0 0 1 1 1 3E8 0 0 1 C 0 0 0 0 0 0 2 1 0 2 1 0 0 0 0 0 0 0 0 1 0 1 1 3 2 0 3 1 1 10 0 0 0 0 1 0 0 0 0 0 4 2 0 2 1 1 FFFF^0 0 1 8 0 0 0 0 0 5 3 2 2 1 0 0 1 0 0 0 0 0 0 0 0 0 6 2 0 2 1 1 1 0 0 0 0 1 1 0 0 0 0 7 1 2 0 0 0 0 0 0 0 0 0 0 0 1 0 1
Part A (5 points) Describe the operation that occurs during cycle 2 (be specific)?
For the remaining parts, assume R2 contains the value 0x80100204 (hexadecimal) following cycle 2. Part B (5 points) What is the value (in hexadecimal) in R3 when cycle 3 completes?
Part C (5 points) What is the value (in hexadecimal) in R2 when cycle 4 completes?
Part D (5 points) What is the value (in hexadecimal) in R2 when cycle 6 completes?
Part E (5 points) Describe the operation that occurs during cycle 7 (be specific).
Part F (10 points) Describe the operation of this microcode fragment (be specific).
5
5
5
rwe
X
Y
Z au en
-a/s
im va
im en
lu en
lf 4
addr data
r/-w
msel
st en ld en
shift types0 = logical1 = arithmetic2 = rotate+ count shifts right- count shifts left
logical functionsX
out
lf
0
lf
1
lf
2
lf
3
cycle
cycle number
X^
register driven onto X bus
Y^
register driven onto Y bus
Z^
register written from Z bus
rwe
register write enable
im en
immediate enable on Y bus
im va
immediate value
au en
arithmetic unit enable
-a/s
-add / sub (0 = add, 1 = subtract)
lu en
logical unit enable
lf^
logical function
su en
shift unit enable
st^
shift type
ld en
load enable
st en
store enable
r/-w
read/-write (0 = write, 1 = read)
msel
memory select
description
operation description
su en
st 2
count
16 32
