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ECEN 4243 Spring 2002 Exam 2 5 April 2002 |
Name_______Solution_____ |
Read the instructions carefully before completing the exam. All answers must be shown on these pages – no attached pages will be graded. This exam is open book, open notes.
1. (20pts). The following instructions are executed in the order in the table. Fill in the following table with the register or memory location contents (in hexadecimal) after each instruction is executed.
|
|
R1 |
R2 |
R5 |
h97 |
h123 |
|
Initial contents |
h0 |
h2 |
h5 |
h7 |
hFF |
|
Mov R7, R5 |
h0 |
h2 |
h5 |
h7 |
hFF |
|
LD R6, h97 |
h0 |
h2 |
h5 |
h7 |
hFF |
|
Add R1, R6, R2 |
h9 |
h2 |
h5 |
h7 |
hFF |
|
LD R3, h123 |
h9 |
h2 |
h5 |
h7 |
hFF |
|
Xor R5, R6, R3 |
h9 |
h2 |
hF8 |
h7 |
hFF |
|
And R2, R2, R5 |
h9 |
h0 |
hF8 |
h7 |
hFF |
|
Jmp 255 |
h9 |
h0 |
hF8 |
h7 |
hFF |
|
Store R1, h123 |
h9 |
h0 |
hF8 |
h7 |
h9 |
2. (15pts). The following instructions are executed in the order in the table. Mark (with a Yes or No) which of the computer blocks is affected as the instruction is executed. Notice this does not include the fetch and decode steps.
|
|
ALU |
Register File |
Main Memory |
Program Counter |
Instruction Register |
Destination bus |
|
Mov R7, R5 |
Yes |
Yes |
no |
no |
no |
Yes |
|
LD R6, h97 |
Yes |
Yes |
Yes |
no |
no |
Yes |
|
Add R1, R6, R2 |
Yes |
Yes |
no |
no |
no |
Yes |
|
LD R3, h123 |
Yes |
Yes |
Yes |
no |
no |
Yes |
|
Xor R5, R6, R3 |
Yes |
Yes |
no |
no |
no |
Yes |
|
And R2, R2, R5 |
Yes |
Yes |
no |
no |
no |
Yes |
|
Jmp hFF |
no |
no |
no |
Yes |
Yes |
no |
|
Store R1, h123 |
Yes |
Yes |
Yes |
no |
no |
Yes |
3. (6pts) What are the three types of cache organization? __Direct Mapped_, __Set_Associative___, and _____Fully Associative________
4. (5pts) Which of those types of cache organization stores the full address with each word of data? ____Fully Associative__
5. (5pts) How many data words can be stored in the cache if the index portion of the address is 10 bits of a 32 bit address? _______1024__________
6. (10pts) The following instructions are executed in the order in the table. Complete the following table showing the contents for the Program Counter during the fetch step, the Instruction Register during execution, and the Program counter after execution. These are 32 bit registers so show all 8 hex digits.
Instruction format reminder:
|
Op code 7 bits |
Destination reg address 3 bits |
Src A register address 3 bit |
Src B register address 3 bits |
Value 16 bits |
Use these binary Opcodes:
|
Add |
1000000 |
|
Sub |
1100000 |
|
LD |
0010000 |
|
Store |
0011000 |
|
And |
1010000 |
|
Xor |
1011000 |
|
Jmp |
0000111 |
|
Halt |
1111111 |
|
Nop |
0000000 |
|
Mov |
0010001 |
|
|
Program counter during fetch |
Instruction Register during execution |
Program Counter after execution |
|
Nop |
h0000E000 |
h00000000 |
h0000E001 |
|
Mov R7, R5 |
h0000E001 |
h23E80000 |
h0000E002 |
|
LD R6, h97 |
h0000E002 |
h21800097 |
h0000E003 |
|
Add R1, R6, R2 |
h0000E003 |
h80720000 |
h0000E004 |
|
LD R3, h123 |
h0000E004 |
h20C00123 |
h0000E005 |
|
Xor R5, R6, R3 |
h0000E005 |
hB1730000 |
h0000E006 |
|
And R2, R2, R5 |
h0000E006 |
hA0950000 |
h0000E007 |
|
Jmp 255 |
h0000E007 |
h0E0000FF |
h000000FF |
|
Store R1, h123 |
h000000FF |
h30080123 |
h00000100 |
7. (10pts) List all of the possible stuck-at faults for the following logic circuit:

D
Sub-stuck-at-0, Sub-stuck-at-1,
X-stuck-at-0, X-stuck-at-1,
Y-stuck-at-0, Y-stuck-at-1, Cin-stuck-at-0,
Cin-stuck-at-1,
A-stuck-at-0, A-stuck-at-1, B-stuck-at-0,
B-stuck-at-1,
D-stuck-at-0, D-stuck-at-1, T-stuck-at-0,
T-stuck-at-1,
Sum-stuck-at-0, Sum-stuck-at-1,
Cout-stuck-at-0, Cout-stuck-at-1.
8. (10 points) For the previous logic circuit fill out the following table
|
Input Vector |
Fault Free outputs |
Node T-stuck-at-0 outputs |
Check the box if the Stuck-at fault is detected for that vector |
|||||||||
|
X |
Y |
Cin |
Sub |
Sum |
Cout |
Sum |
Cout |
A-s-a-0 |
A-s-a-1 |
T-s-a-0 |
Sum-s-a-0 |
|
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
ü |
|
|
|
|
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
|
ü |
ü |
ü |
|
|
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
ü |
|
|
ü |
|
|
0 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
ü |
|
|
|
|
|
0 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
|
ü |
ü |
ü |
|
|
1 |
1 |
1 |
1 |
0 |
1 |
0 |
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