7.3k views

Consider the following program segment for a hypothetical CPU having three user registers $R_1, R_2$ and $R_3.$

\begin{array}{|l|l|c|} \hline \text {Instruction}  &  \text{Operation }& \text{Instruction size (in Words)} \\\hline \text{MOV $R_1,5000$}
& \text{$R1$} \leftarrow  \text{Memory$[5000]$}& \text{$2$} \\\hline\text{MOV $R_2(R_1)$} & \text{$R2$} \leftarrow  \text{Memory$[(R_1)]$}& \text{$1$} \\\hline \text{ADD $R_2,R_3$} & \text{$R_2$} \leftarrow  \text{$R_2 + R_3$} & \text{$1$} \\\hline \text{MOV $6000,R_2$} & \text{Memory$[6000]$} \leftarrow  \text{$R_2$} & \text{$2$} \\\hline \text{Halt} & \text{Machine Halts} & \text{$1$} \\\hline \end{array}

Let the clock cycles required for various operations be as follows:

\begin{array}{|l|l|} \hline \text {Register to/from memory transfer}  &  \text{3 clock cycles } \\\hline  \text {ADD with both operands in register}  &  \text{1 clock cycles } \\\hline \text {Instruction fetch and decode}  &  \text{2 clock cycles }\\\hline \end{array}

The total number of clock cycles required to execute the program is

1. $29$
2. $24$
3. $23$
4. $20$

edited | 7.3k views
0
instead of register indirect in second instruction if memory indirect will be there then we should take 2 memory ref that means 6 cycles or 3 cycle ???

B.  $24 \text{ cycles}$

$$\begin{array}{|l|c|c|} \hline \text {Instruction} & \text{Size }& \text{Fetch and Decode + Execute} \\\hline \text{MOV} & \text{2} & \text{2} \times \text{2 + 3 = 7} \\\hline \text{MOV} & \text{1} & \text{2} \times \text{1 + 3 = 5} \\\hline \text{ADD} & \text{1} & \text{2} \times \text{1 + 1 = 3} \\\hline \text{MOV} & \text{2} & \text{2} \times \text{2 + 3 = 7} \\\hline \text{HALT} & \text{1} & \text{2} \times \text{1 + 0 = 2} \\\hline & \text{Total} & \text{24  Cycles} \\\hline \end{array}$$

by Boss
edited
0
Nice Explaination.
+1
One very sily clearification i need here is : we have not considered operand fetch for ADD instruction because we have already moved it to register R1, right? when to consider operand fetch seperaely in any question? will it be given directly in question?
+1
Operand fetch is not required if the content is already in registers, because processor operates directly on the registers. Think this way, where does the processor is going to keep the operands after fetching it from the registers. Instructions which use registers only can operate directly on the registers itself.
0
Why we are considering it as Non pipeline architecture as no. of clock cycles in pipeline will be less.
0
why execution time for HALT  is considered as 0?  won't it take 1 cycle in the end for execution?
+1

@Pawan Kumar 2 if nothing is mentioned in the question...shouldn't we consider it as 0.

0
Indirect addressing is present in second instruction, so there should be two memory references to get the data stored in the memory. therefore answer should be 27.

+1

In Indirect addressing also we consider only one register to/from memory transfer so that will take only 3 cycles.

+1
Why we are multiplying 2 to instruction size.
+2

@Vijay Kusumakar

CPU fetches one word at a time. For instructions with size 2 words, CPU has to fetch it two times

Hence, 2*clock cycles required to fetch and decode=2*2=4

0
Each instructions requires fetch and decode

So total instructions size is 7 words

Number of cycles = 7*2=14 cycles

Move     R1,5000 3 cycles for register /to memory transfer

Move     R2,(R1)  3 cycles

Move     6000,R2  3 cycle

Total = 14+10=24 cycles
by Boss
edited by
+1
very well explained
0
how instruction 4 will get 3 cycles is it go into instruction decode phase?
0
Memory transfer to register

The clock cycles are per block; if an instruction size is 2 then it requires twice no. of clock cycles.

 Instruction no.       size                                              no. of clock cycles 1                                2                                                                       3*2+2 2                                1                                                                       1*3+2 3                                1(add only)                                                      1 4                                2                                                                       3*2+2 5                                1                                                                       2(fetch and decode) Total                                                                  24

by Loyal
0
add also require fetch and decode
+4
Sir, one small doubt.

For the 2nd instruction , it's a register indirect addressing mode.

We need to consider 2 memory read for that instruction. Why are we considering only one memory read. Can you please explain?
0
correct xplaination ..nice
+1 vote
$3+3+1+3+7(2)=24$

Why the Instruction Fetch and Decode is counted 7 times when the number of instructions are 5?

Because the largest unit of memory that can be transferred to and from it, in a single operation is the word size. Since there are 7 words in total, to fetch them we need to devote 7 separate operations, and each such operation would take 2 clock cycles (given)
by Loyal
0
This explanation is not correct.
0

The equation I wrote here is literally equivalent to the one in the best answer, lol.

Still, please come up with the reasoning as to why is it not correct, I'd be happy to rectify myself :)

Btw, this is what Hamacher has to say: