Computer Science, asked by gautam6672, 9 months ago

Make the following changes to the basic computer.

Add a register to the bus system CTR( count register) to be selected with s2, s1, s0=000.
Replace the ISZ instruction with an instruction that loads a number into CTR.
LDC Address CTR <-M [Address]
Add a register reference instruction ICSZ: Increment CTR and skip next instruction if zero.
Discuss the advantage of this change

Answers

Answered by dhirajpatil30981
2

Explanation:

Add a Count register CTR to the bus system at the top so that the selection inputs .

6

Answered by Anonymous
1

Answer:

Changes to basic computer

Explanation:

There are four steps in the process.

Step 1 :

Stored Program Organization

Since there are 256K words in the memory unit,

Hence, there are 262144 words in the memory unit.

Number of bits in 1 word

Number of possible memory addresses

Number of registers

Step 2

Total number of registers

Therefore, number of bits in the register code part are 6.

Find number of bits in address part as follows:

Therefore, number of bits in the address part are 18.

Number of bits in the operation code can be calculated as follows:

Number of bits in the operand code Total number of bits – No. of bits in the address part – No. of bits in the register Code part – 1 (Indirect bit)

Therefore, number of bits in the operand code part

Therefore, number of bits in the operation code part are 7.

Step 3 of 4

b.

The Instruction word format is as follows:

0 to 17 denote address code part and it has 18 bits.

18 to 23 denote register code part and it has 6 bits.

24 to 30 denote operation code part and it has 7 bits.

31 denote indirect bit and it has 1 bit.

Step 4 of 4

Since the word size in this architecture is 32, so data input part will have 32 bits.

Therefore, number of bits in the data inputs of memory is 32.

Number of bits in the address input

Since, the computer’s memory unit has 256K words, which is, so the address inputs’ contains 18 bits.

Therefore, number of bits in the address inputs of memory is 18.

The correct answer would be 18.

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