In the decode portion the control unit examines the op-code of the instructon stored in the IR to decide what actions are required.
In the execute phase, the control unit sends signals to the other units within the CPU which facilitate the execution of the instruction.
I feel you really should reference the figure you have just drawn in your discussion.
BUS ← IR MAR ← BUS MDR ← M[MAR] BUS ← MDR ALU OP: ACC + BUS ACCC ← ALU
PC → BUS as the first step is wrong in 2 out of 3 portions.
A special purpose register is a register which is used for a specific (or special) purpose. The following registers in wombat are special purpouse: PC, IR, MAR, MDR, INReg, OUTReg.
The register file is a collection of registers, it is not a single register.
The ALU is NOT a reigster. No way no how.
ADD, ADDI, ... are INSTRUCTIONS not registers.
2GHz = 2x109 cycles/ 1 second. But in this question we want time so start with 1 second 1 cycle ------------- x -------------- = .5 x 10-9 seconds/instruction 2 x 109 cycles 1 instruction or 500x10-12 seconds x 1012 ps / sec = 500ps. The answer is 500ps.
If you told me 2GS/cycle think for a moment, that is about 63 years. Are you willing to wait your entire life for a single instruction or two?
Class | CPI | Frequency |
---|---|---|
A | 1 | 20% |
B | 2 | 20% |
C | 4 | 60% |
1 second 3 cycles 1x109 instructions ------------- x ------------- x ------------------- = 3 sec/prog 2 x 109 cycles 1 instruction 1 prog
1 x .25 + 2 x .35 + 4 x .4 = .25 + .7 + 1.6 = 2.55 NEW CPI = 2.55 1 second 2.55 cycles 2.3x109 instructions ------------- x ------------- x --------------------- = 2.93 sec/prog 2 x 109 cycles 1 instruction 1 progSince the new compiler produces a faster program it should be used.
There are four data input lines. There are two control input lines. There is a single output line. The value of the control line is used to select a single input line to pass through to the output. Assume control has a value of 0 ≤ n ≤ 3, then input line n will be output.
b1 | b0 | O |
---|---|---|
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
O = b1·b0 + b1· b0