# Assessment 4 Maria and ISA

Home Assignment Answers Assessment 4 Maria and ISA

Question 1

• 199 instructions are given. To compute opcode bit, given bits are compared with formula 2n. It is founded that replacing n with 9 is satisfying our current requirement due to which 8 bits are enough for opcode (28 >= 199).
• Subtract opcode bits from bits in the word to obtain address part bits i.e. Bits in a word (24) – (8) Opcode bits. Hence, 16 bits are left for address part.
• To compute maximum size, address part bits are used in 2n where n is addresseded part bits. Hence, total memory size is 216 = 65536.
• In One Word, 011111111111111111111111 i.e. 223 – 1 is the largest bit that can be accommodated.

Question 2

• For Immediate Mode, Accumulator will take value 900 (ADD 900).
• Immediate AC value is used as a memory address in order to obtain the AC value of Direct Mode i.e. 1000.
• Similarly, AC value in direct Mode used as the memory address for indirect addressing mode to obtain the AC value i.e. 500.
• AC value in indexed mode is 500 because base register value i.e. 100 is added in Immediate Mode AC value in order to identify the memory address. At 1000 memory address, 500 value is present which is stored as AC value in indexed mode.

Question 3

Given Equation is F = (A-B) * (C*D+E)

MOV M1, A   [: - ‘A’ value is moved in register M1]

SUB M1, B    [: - ‘B’ value is subtracted from value present in M1 i.e. ‘A’]

MOV F, M1    [: - ‘A - B’ value is moved to F register]

MOV M1, C    [: - ‘C’ value moved to M1 register]

MUL M1, D    [: - ‘D’ value is multiplied with M1 value i.e. ‘C’ value]

ADD M1, E    [: - ‘E’ value is added after multiplication in M1]

MUL F, M1    [: - In the last, total value of M1 is multiplied with F value i.e. ‘A – B’]

SUB B             [: - ‘B’ value is subtracted from ‘A’]

STORE M1     [: - Stored subtracted value in M1 Register]

MUL D                       [: - with ‘C’, ‘D’ value is multiplied]

MUL M1         [: - Final value obtained from above is multiplied with the stored value of M1]

PUSH B                      [: - ‘B’ Value is entered in machine]

PUSH A                      [: - ‘A’ value is entered in machine]

SUB                [: - ‘B’ value is subtracted from ‘A’ value]

PUSH C                      [: - ‘C’ value is entered in machine]

PUSH D                      [: - ‘D’ value is entered in machine]

MUL               [: - Multiply ‘D’ vale with ‘C’ value]

PUSH E                      [: - ‘E’ value is entered in machine]

MUL               [: - Multiplied ‘C*D+E’ value with ‘A – B’]

POP                 [: - Extracted Solution i.e. (A – B) * (C * D + E)

Question 4

Given Instruction Length is 10 bits.

Address Field Size is 3 bits.

To get 2-address instruction, we have used 2 address field of 3 bits: -

Opcode bits = Instruction length [10 bits] – address bits [6 bits]

Opcode bits = 4 bits.

Total instruction formed from 4 bits opcode are 24 = 16 in which 15 instructions are already used.

To get 1-address instruction, we have used 1 address field of 3 bits: -

Opcode bits = Instruction length [10 bits] – address bits [3 bits]

Opcode Bits = 7 bits

Total instruction formed from 7 bits opcode are 27 = 128 in which 7 instructions are already used.

To obtain opcode bits, no address bits are used.

Opcode bits = Instruction length [10 bits] – address bits [0 bits]

Opcode Bits = 10 bits

Still, 0-address instruction is able to accommodate possible instruction as shown below: -

There are 10 bits, so total instruction formed are 210 including 2-address instruction and 1-address instruction.

Total Instruction = 210 = 1024.

0-address instruction = 1024 – 15 – 7 = 1002.

Suitable total number of zero-address instruction is 1002.

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