MARIE and ISA: An Exploration of Computer Architecture

ITC544
Computer Organisation and Architecture
ASSESSMENT 4
MARIE and ISA
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Student ID:

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Contents
Task 1.........................................................................................................................................2
(a)...........................................................................................................................................2
(b)...........................................................................................................................................4
Task 2.........................................................................................................................................7
Task 3.........................................................................................................................................8
Appendix..................................................................................................................................10
List of Figures
Figure 1: Output 1......................................................................................................................4
Figure 2: Output 2......................................................................................................................5
Figure 3: Output 3......................................................................................................................6
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Task 1
(a)
Program of Fibonacci number using MARIE simulator is shown below:
ORG 000
INPUT USER /Value from user is going to be taken through this
SUBT TWO /2 is going to be subtracted from here
PRINT, STORE Z /All the value of Z is going to be stored in AC
CLEAR /Through this, the value stored in the AC is going to be zero
ADD N1 /In AC, N1 is going to be added
ADD N2 /In AC, N2 is going to be added
STORE TOTAL /Through this total values are going to be stored
LOAD N2 /Value of N2 is going to be stored in AC
STORE N1 /values are going to be stored in N1
LOAD TOTAL /Now in AC value of total is going to be stored
STORE N2 /Now, N2 is going to be stored in AC
LOAD Z /All the values of Z are going to be stored in AC
SUBT ONE /1 is going to be subtracted from this
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SKIPCOND 000 /If AC<0 then skip condition of 000
JUMP PRINT /Through this loop is going to be printed
LOAD TOTAL /In AC total is going to be added
OUTPUT /Output is going to be displayed through this
HALT /Process is going to be halted
/INITILIZATIONS
N1, DEC 0
N2, DEC 1
ONE, DEC 1
TWO, DEC 2
USER, DEC 0
Z, DEC 0
TOTAL, DEC 0
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(b)
If user enters input as 17:
Figure 1: Output 1
If user enters input as 20:
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Figure 2: Output 2
If user enters input as 4:
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Figure 3: Output 3
The maximum value up to which this program run correct value is 2147483647. This is due
to the fact that Fibonacci runs up to (2^31-1).
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Task 2
As per given case scenario, size of instruction is 11 bits, therefore,
Total number of instruction can be= 211 = 2048
Also the size of address field as per scenario = 4 bits
So the instructions having a number of 2 address = 6*24*24
= 6*16*16
= 1536
And the address associated with 1 instruction= 30*24
= 30*16
= 480
As a result, number of 0 addresses that are possible = (2048) - (1536+480)
= (2048) – (2016)
= 32
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Task 3
As, A= (B + C) * (D – E)
Code for 0 address
push B
push C
add
pop A
push D
push E
sub
push A
mpy
pop A
Code for 1 address
lda B
add C
sta A
lda D
sub E
mpy A
sta A
Code for 2 address
load A, B
add A, C
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load X, D
sub X, E
mpy A, X
Code for 3 address
add A,B,C
sub X,D,E
mpy A, A, X
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Appendix
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