MARIE & ISA - Assignment 2 for ITC544

This is the solution for Assignment 2 of ITC544 which covers topics like MARIE & ISA. It includes solutions to questions related to opcodes, memory, arithmetic expressions, and mnemonics.

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Full Name
Student ID No.
ITC544
Assignment 2: MARIE & ISA

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Question 1
1. 122 different opcodes cater for 122 different operations .
2x=122, x= log 122
log 2
¿ 6.93
Sinceisnot practically convenient deal with decimal bits , we round off the answer .
Thus x=7 Therefore the number bits needed for the opcode=7 bits
2. T he number of bits ¿ Total Number of bits−The number used for opcodes
¿ 16−6.93
¿ 9.07
≈ 9
Hence the number of bits ¿ the address part if the instruction is9.
3. T he maximum allowable ¿ memory =29
¿ 512 Bytes
4. The largest unsigned binary number that can be accomodated∈one word of memory
¿ 216
¿ 65536
Thus 65536 is the largest unsigned binary number that can be accommodated∈one word memory .
Question 2
Given that theaccumulator ( AC )=500 ,
1. Immediate=1000+500=1500
2. Direct =1400+500=1900
3. Indirect=1300+500=1800
4. Indexed =1000+500=1500
Question 3
Using the arithmetic expression S=(A+B)-(C+D)
Load A
Add B
Store Temp1
Load C
Add D
Store Temp2
Load Temp1

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Subt Temp2
Store Sum /Total of nine memory accesses; given that we execute C+D first, the process can
be carried out with 7 memory accesses.
Given that the processor has more than three registers:
Load R1, A
Load R2, B
Add R1, R2
Load R3, C
Load R4, D
Add R3, R4 // the operation does not required memory access
Subt R1, R4 //The operation does not required memory access
Store Sum // total of five memory accesses.
Question 4
Mnemonic Hex Description
0 JnS X Store the PC address X and jump X+1
1 Load X Load Contents of address X into AC
2 Store X Store the content of AC at address X
3 Add X Add the content of address X to AC
4 Subt X Subtract the content of address X from AC
5 Input Input a value from the keyboard into AC
6 Output Output the value in AC to the display
7 Halt Terminate program
8 Skipcond Skip next instruction on Condition
9 Jump X Load the value of X into PC
A Clear Put all zeros in AC
B Addl X
Use the value of X as true address of the data operand to add to
AC Indirect
C Jumpl X Use the value at X as the address to jump to
D Loadl X Use the value at X as the address of the value to the Load
IndirectE Storel X Use the value at X as the address of the location of storing value

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a)
Hexadecimal
Code
100 Start LOAD A 1108
101 ADD B 3109
102 STORE D 210B
103 CLEAR A000
104
OUTPU
T 6000
105 ADDl D B10B
106 STORE B 2109
107 HALT 7000
108 A, HEX 00FC 00FC
109 B, DEC 14 000E
10A C, HEX 108 108
10B D, HEX 0000 0000
b)
Symbol
Locatio
n
A 108
B 109
C 10A
D 10B
Start 100
c) The value stored in the ANC when the program terminates is 0108.