MCS-284 Chapter 6 Homework

• 6.3 on page 455.
• 6.4 on page 455.
• 6.x1, 6.x2, 6.x3 The following code computes the maximum of two memory locations, whose addresses are in $a1 and $a2, and places the result back into the memory location specified by $a3.

      lw $t1, 0($a1)
      lw $t2, 0($a2)
      slt $t0, $t1, $t2
      add $t3, $t1, $zero
      beq $t0, $zero, done
      add $t3, $t2, $zero
  done:
      sw $t3, 0($a3)
  

  Assume the code is executed using the pipelined architecture is used as described in Chapter 6 with the following assumptions: Forwarding is used wherever possible. Branch hazards are dealt with as discussed on pages 418 and 419, i.e., the pipeline is optimized for branches that are not taken and that we moved the branch execution to the ID stage.
  • 6.x1: Identify above all data and branch hazards in the code. Identify which hazards can be helped by forwarding and which may lead to stalls.
  • 6.x2: Indicate in the table below which stages of each instruction are in the pipeline during execution of the code assuming the branch is not taken. Each entry should contain a IF, ID, EX, MEM or WB, or should be left blank.

    INSTRUCTION	CYCLE NUMBER
    	01	02	03	04	05	06	07	08	09	10	11	12	13	14	15
    lw $t1, 0($a1)
    lw $t2, 0($a2)
    slt $t0, $t1, $t2
    add $t3, $t1, $zero
    beq $t0, $zero, done
    add $t3, $t2, $zero
    sw $t3, 0($a3)

  • 6.x3: Indicate in the table below which stages of each instruction are in the pipeline assuming the branch is taken.

    INSTRUCTION	CYCLE NUMBER
    	01	02	03	04	05	06	07	08	09	10	11	12	13	14	15
    lw $t1, 0($a1)
    lw $t2, 0($a2)
    slt $t0, $t1, $t2
    add $t3, $t1, $zero
    beq $t0, $zero, done
    add $t3, $t2, $zero
    sw $t3, 0($a3)