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    I am new to operating system and i got some Questions about pipeling

    Firstly ,the question we have A processor has 24 registers*, uses 8-bit immediate*, and has 36 different instructions* (corresponding to 36 operation codes**) in its instruction set.These 36 instructions are classified into 4 types as listed belowi just going to give out type c)

    Type-C: takes 1 source register, 1 immediate, uses 1 destination register;

    i know how to do this but the problem here is i don't know how many bit do 24 registers and 36 opcode require

    (? = unknown)
    i know the answer will be ?opcode+ ?registers + 8bits immediates + ?registers = bits

    But i'd like to know how do i calculate the bits require

    Secondly, there's question asking me how much the pipeline speedup ratio:
    Assume a pipelined processor operates in four states: instruction fetch (IF), operand fetch (OF), operand execute (OE), and operand store (OS). A timing diagram of pipeline could be drawn as follows:
    we have the following code with the data dependency. 
    ADD R1,R3,R2	[R1] ← [R3]+[R2]
    SUB R4,R1,R5	[R4] ← [R1]-[R5]
    MUL R6,R7,R8	[R6] ← [R7]*[R8]
    This is the table :
    Cycle                  0 cycle 1 cycle 2 cycle 3 cycle 4 cycle 5 cycle 6 ......
    ADD R1,R2,R3            IF         OF          OE       OS      
    R1,R3,R2                               IF            OF        OE     OS
    R4,R1,R5                                              IF         OF      OE      OS
    R6,R7,R8                                                           IF       OF      OE      OS
    And this is all the information i get , but i don't know how many instruction in here also i don't know the Clock speed, how can i bring out the speed up ratio?

    Many thanks for reply or view
    my lecturer never reply my e-mail sigh...
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    I think is correct I'm not too sure but this is what I have done. Basically if the register was 32 it would be 2^5 and if the instructions were 64 it would 2^6
    so for c it would be:

    6opcode+ 5registers + 8bits immediates + 5registers = 24bits/3bytes

    If the numbers don't fit in base two exactly e.g 36 doesn't but
    the closest to it is 2^5 we use that

    Also if you get say 21bits you round it up to 24bits so its 3bytes bits should always be divided by 8.

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