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1 B Solution COAL - Midterm

The document contains solutions to 4 questions from a midterm exam on machine architecture. Question 1 involves calculating MIPS rates for two machines, Machine A and Machine B. Machine B has the higher MIPS rate and is therefore better. Question 2 involves calculating the MIPS rate of a machine based on instructions performed and clock frequency. Question 3 shows stack and accumulator architectures for performing common arithmetic operations. Question 4 provides the step-by-step workings for two bitwise operations, AND and RCL, on two hexadecimal values.

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MUHAMMAD AQEEL ANWAR KAMBOH
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92 views2 pages

1 B Solution COAL - Midterm

The document contains solutions to 4 questions from a midterm exam on machine architecture. Question 1 involves calculating MIPS rates for two machines, Machine A and Machine B. Machine B has the higher MIPS rate and is therefore better. Question 2 involves calculating the MIPS rate of a machine based on instructions performed and clock frequency. Question 3 shows stack and accumulator architectures for performing common arithmetic operations. Question 4 provides the step-by-step workings for two bitwise operations, AND and RCL, on two hexadecimal values.

Uploaded by

MUHAMMAD AQEEL ANWAR KAMBOH
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Solution

Midterm Examination Fall 2022


Q. 1

Part (a)

For Machine A

(3*6 + 4*3 + 5*2 + 5*6) / 17 = 70 / 17 = 4.11

100 THz / ( 4.11 * 106 ) = 24.28 * 106

For Machine B

(2*10+4*6+6*2+4*2) / 20 = 64 / 20 = 3.2

100 THz / ( 3.2 * 106 ) = 31.25 * 106

Part (b)

Machine B is better because of its higher MIPS rate

Q. 2

Part (a)

(35*2 + 40*1 + 20*4 + 5*3) / 100 = 205 / 100 = 2.05

(100,000) * 2.05 * (1/f) = 69.87 * 106 / (50 *106 Hz)

Part (b)

1 / [(1 – 0.4) + (0.4/8)] = 1 / [(0.6) + (0.4/8)]

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Q. 3
F= R(C+D) – (RA+B)

Stack Architecture Accumulator Set General Purpose Register

PUSH A LDA R LOAD S, R


PUSH R MUL A MUL S, A
MUL ADD B ADD S, B
PUSH B
STA S LOAD F, C
ADD
PUSH D LDA C ADD F, D
PUSH C ADD D MUL F, R
ADD MUL R SUB F, S
PUSH R SUB S
MUL STA F
SUB
POP F

Q. 4
Initial Values:
DH = (6B)16
AH = (CB)16
CF = 1,
CL = 5
Instructions:
AND DH,AH
RCL DH, CL.

Solution:

CL CF DH
Starting 5 1 0 1 1 0 1 0 1 1
AND 4 0 0 1 0 0 1 0 1 1
RCL 3 0 1 0 0 1 0 1 1 0
2 1 0 0 1 0 1 1 0 0
1 0 0 1 0 1 1 0 0 1
0 0 1 0 1 1 0 0 1 0

***END**

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