I need help showing the following problem in a diagram(all 6 steps) the first step was already done.
1.The PC contains 300, the address of the first instruction. This instruction
(the value 1940 in hexadecimal) is loaded into the IR (Instruction set register), and the PC is
incremented. Note that this process involves the use of a memory address register and a
memory buffer register. For simplicity, these intermediate registers are ignored.
2. The first 4 bits (first hexadecimal digit) in the IR indicate that the AC is to be loaded.
00012 = 110 =Load AC from memory
00102 = 210= Store AC to memory
01012 = 510 = Add to AC from memory
The remaining 12 bits (three hexadecimal digits) specify the address (940) from which data are
to be loaded.
3. The next instruction (5941) is fetched from location 301, and the PC is incremented.
4. The old contents of the AC and the contents of location 941 are added, and the result is
stored in the AC.
5. The next instruction (2941) is fetched from location 302, and the PC is incremented.
6. The contents of the AC are stored in location 941
Step by stepSolved in 2 steps
- Show the following problem in a diagram(all 6 steps).1.The PC contains 250, the address of the first instruction. This instruction (the value 1940 inhexadecimal) is loaded into the instruction register IR, and the PC is incremented. Note that thisCPU Registers300 PC AC 1940 IR process involves the use of a memory address register and a memory buffer register. For simplicity, these intermediate registers are ignored.2. The first 4 bits (first hexadecimal digit) in the IR indicate the opcode.00012 = 110 =Load AC from memory00102 = 210= Store AC to memory01012 = 510 = Add to AC from memoryThe remaining 12 bits (three hexadecimal digits) specify the address from which data(saymemory location 941contains the data 0003) are to be loaded.3. The next instruction (5941) is fetched from location 251, and the PC is incremented.4. The old contents of the AC and the contents of location 941 (say the data is 0004)are added,and the result is stored in the AC.5. The next instruction (2941) is…arrow_forwardIn this case, every instruction gets its own data in a way that is completely separate from how the data of other instructions is obtained. We utilize: A Multiple Data/Instruction B Multiple Data/Instruction C Singular Data Single Inst D Singe Inst Multiplearrow_forward
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