2019-2020-1 Principles Of Computer Organization
Midterm – A
Ⅰ、Choose the only one correct answer and fill in the brackets (Total 10 points, 5 questions, 2 points per
question)
1. (
C
) are composed of hundred to thousand processors and terabytes of memory and having the
highest performance and cost.
(A) Desktop computers (B) Low-end servers (C) Supercomputers (D) Embedded computers
2. (
C
) Assume a color display using 8 bits for each of the primary colors (red, green, blue) per pixel
and a frame size of 1024 × 768, what is the minimum size in bytes of the frame buffer to store a frame?
(A) 768 K
3. (
D
(B) 1024K
(C) 2304K
(D) 3072K
) Computer X’s performance is 2 times as fast as the performance of computer Y, which runs a
given application in 10 seconds. How long will computer X take to run that application?
(A) 8 seconds
4. (
A
(B) 7 seconds
(C) 6 seconds
(D) 5 seconds
) Assume that registers $s0 and $s1 hold the values 0x80000000 and 0xD0000000, respectively.
What is the value of $t0 for the assembly code "add $t0, $s0, $s1" ?
(A) 0x50000000
5. ( C
(B) 0x80000000
(C) 0xD0000000
(D) 0xF0000000
) Suppose register $s0 has the number 0xF1234567 and that register $s1 has the binary number
0x12345678.What are the values of registers $t0 after this instruction?
slt $t0, $s0, $s1
(A) 3
(B)
2
(C) 1
1
(D) 0
II. Consider two different processors P1, and P2 executing the same instruction set with the clock rates and
CPIs given in the following table. (10 points)
Processor
Clock Rate
CPI
1.5
P1
2.1 GHz
P2
2.5 GHz
1.0
(1) Which processor has the highest performance expressed in instructions per second? (4 points)
(2) We are trying to reduce the time by 20% but this leads to an increase of 20% in the CPI. What clock rate
should we have to get this time reduction? (6 points)
(1)
P2 has the highest performance
performance of P1 (instructions/sec) = 2.1 × 109/1.5 = 1.4 × 109
performance of P2 (instructions/sec) = 2.5 × 109/1.0 = 2.5 × 109
(2)
Execution time = Instruction Count * CPI / Clock Rate
Execution time (new) / Execution time(old)
= ( CPI(new) / Clock Rate(new) ) / ( CPI(old) / Clock Rate(old) )
= CPI(new) * Clock Rate(old) / ( Clock Rate(new) * CPI (old) )
Execution time (new) = 0.8 * Execution time(old)
CPI(new) = 1.2 * CPI(old)
Clock Rate(new) = 1.2 * Clock Rate(old) / 0.8
(3 points) Clock Rate(P1 new) = 1.2 * 2.1 / 0.8 = 3.15 (GHz)
(3 points) Clock Rate(P2 new) = 1.2 * 2.5 / 0.8 = 3.75 (GHz)
2
III.
Suppose we have developed new versions of a processor with the following characteristics. (10 points)
Version
Voltage
Clock Rate
Version 1
5.0V
0.5 GHz
Version 2
3.3V
1.0 GHz
(1) How much has the capacitive load varied between versions if the dynamic power has been reduced by
10%? (2 points)
(2) How much has the dynamic power been reduced if the capacitive load does not change? (2points)
(3) Assuming that the capacitive load of version 2 is 80% the capacitive load of version 1, find the voltage for
version 2 if the dynamic power of version 2 is reduced by 40% from version 1. (6 points)
(1) Dynamic Power = V2 × clock rate × C / 2 .
DP2 = 0.9 DP1
C2/C1 = 0.9 × 52 × 0.5 × 109/3.32 × 1 × 109 = 1.033
(2) DP2/DP1 = 0.8712 => Reduction of 12.88%
(3)
DP2 = V22 × 1 × 109 × 0.8 × C1 = 0.6 × DP1
DP1 = 52 × 0.5 × 109 × C1
V22 × 1 × 109 × 0.8 × C1 = 0.6 × 52 × 0.5 × 109 × C1
V2 = ( (0.6 × 52 × 0.5 × 109)/(1 × 109 × 0.8) )1/2 = 3.062 (V)
3
IV. The table below shows the instruction type breakdown of a given application executed on 1, or 2
processors. Using this data, you will be exploring the speed­ up of applications on parallel processors.
Processors No. Instructions per Processor
CPI
Arithmetic Load/Store Branch Arithmetic Load/Store Branch
2560
1280
256
1
1
4
2
2
1
6
2
1350
800
128
(1) The table above shows the number of instructions required per processor to complete a program on a
multiprocessor with 1, or 2 processors. What is the total number of instructions executed per processor? (4
points)
(2) Given the CPI values on the right of the table above, find the total execution time for this program on 1,
and 2 processors. Assume that each processor has a 2 GHz clock frequency. (6 points)
(1)
Processors
Instructions per processor
4096
2278
1
2
(2)
Processors
Execution time (µs)
1
2
4.096
3.203
V. The following table shows manufacturing data for various processors.
Wafer Diameter
Dies per Wafer
Defects per Unit Area
Cost per Wafer
15 cm
90
10
0.018 defects/cm2
(1) Find the yield. ( 3 points )
(2) Find the cost per die. (2 points )
(3) If the number of dies per wafer is increased by 10% and the defects per area unit increases by 15%, find
the die area and yield. (5 points)
(1)
Wafer area =  × 7.52 = 176.7 cm2
Die area = 176.7/90 = 1.96 cm2
Yield =1/(1+(0.018*1.96/2))2 = 0.965
(2)
Cost per die = 10/(0.965*90)=0.115
(3)
Dies per wafer = 1.1 × 90 = 99
Defects per area = 1.15 × 0.018 = 0.0207 defects/cm2
Die area = wafer area/Dies per wafer = 176.7/99 = 1.785 cm2
Yield =1/(1+(0.0207*1.785/2))2 = 0.964
4
VI. The following table shows results for SPEC CPU2006 benchmark programs running on an AMD
Barcelona. (10 points)
Name Intr. Count(109)
Execution Time (seconds) Reference Time (seconds)
a.
perl
2118
500
9770
b.
mcf
336
1200
9120
(1) Find the CPI if the clock cycle time is 0.333 ns. (6 points)
(2) Find the SPECratio. (2 points)
(3) For these two benchmarks, find the geometric mean of the SPECratio. (2 points)
(1)
CPI = clock rate × CPU time/instr. count
clock rate = 1/cycle time = 3 GHz
CPI(perl) = 3 × 109 × 500/2118 × 109 = 0.7
CPI(mcf) = 3 × 109 × 1200/336 × 109 = 10.7
(2)
SPECratio(perl) = 9770/5 00 = 19.54
SPECratio(mcf) = 9120/1200 = 7.6
(3)
(19.54 × 7.6)1/2 = 12.19
VII. Consider a computer running programs with CPU times shown in the following table. (9 points)
FP Instr.
INT Instr.
L/S Instr.
Branch Instr.
Total Time
35 s
85 s
50 s
30 s
200 s
(1) How much is the total time reduced if the time for FP operations is reduced by 20%? ( 3 points)
(2) How much is the time for INT operations reduced if the total time is reduced by 20%? ( 3 points)
(3) Can the total time can be reduced by 20% by reducing only the time for branch instructions? (3 points)
(1)
Tfp = 35 × 0.8 = 28 s, Tnew_total = 28 + 85 + 50 + 30 = 193 s. Reduction: 3.5%
(2)
Tnew_total = 200 × 0.8 = 160 s, Tfp + Tl/s + Tbranch = 115 s, Tnew_int = 45 s. Reduction time INT: 47%
(3)
Tgoal = 200 × 0.8 = 160 s, Tfp + Tint + Tl/s = 170 s. NO
5
VIII. In the following problem, we will be investigating memory operations in the context of an MIPS
processor. The table below shows the values of an array stored in memory. Assume the base address of the
array is stored in register $s6 and offset it with respect to the base address of the array. (9 points)
Address
12
08
04
00
Data
1
6
4
2
(1) For the memory locations in the table above, write C code to sort the data from lowest to highest,
placing the lowest value in the smallest memory location shown in the figure. Assume that the data shown
represents the C variable called Array, which is an array of type int, and that the first number in the array
shown is the first element in the array. Assume that this particular machine is a byte-addressable machine
and a word consists of four bytes. ( 5 points)
(2) For the memory locations in the table above, write MIPS code to sort the data from lowest to highest,
placing the lowest value in the smallest memory location. Use a minimum number of MIPS instructions.
Assume the base address of Array is stored in register $s6. ( 4 points)
(1) (5 points)
temp = Array[3];
Array[3] = Array[2];
Array[2] = Array[1];
Array[1] = Array[0];
Array[0] = temp;
(2) (4 points)
temp = Array[3];
Array[3] = Array[2];
Array[2] = Array[1];
Array[1] = Array[0];
Array[0] = temp;
lw
lw
sw
lw
sw
lw
sw
sw
6
$t0,
$t1,
$t1,
$t1,
$t1,
$t1,
$t1,
$t0,
12($s6)
8($s6)
12($s6)
4($s6)
8($s6)
0($s6)
4($s6)
0($s6)
IX. For this problem, the table holds some C code. You will be asked to translate these C code statements to
MIPS assembly code. Use a minimum number of instructions. Assume that the values of a, b, i, and j are in
registers $s0, $s1, $t0, and $t1, respectively. ( 6 points)
for(i=0; i<10; i++)
a += b;
addi $t0, $0, 0
LOOP:
TEST:
beq $0, $0, TEST
add $s0, $s0, $s1
addi $t0, $t0, 1
slti $t2, $t0, 10
bne $t2, $0, LOOP
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X. For the following problems, the table holds C code functions. Assume that the first function listed in the
table is called first. You will be asked to translate these C code routines into MIPS assembly. (16 points)
compare:
int compare( int a, int b) {
if( sub(a,b) >= 0 )
return 1 ;
else
return 0 ;
}
int sub(int a , int b ) {
return a-b ;
}
addi $sp, $sp, –4
sw $ra, 0($sp)
add $s0, $a0, $0
add $s1, $a1, $0
exit:
sub:
jal sub
addi $t1, $0, 1
beq $v0, $0, exit
slt $t2, $0, $v0
bne $t2, $0, exit
addi $t1, $0, $0
add $v0, $t1, $0
lw $ra, 0($sp)
addi $sp, $sp, 4
jr $ra
sub $v0, $a0, $a1
jr $ra
8