(Solved) : Program Execution Figure 35 Described Using Six Steps Expand Description Show Use Mar Mbr Q44036511 . . .
The program execution of Figure 3.5 is described using six steps. Expand this description to show the use of the MAR and MBR. Memory 300 19 4 0 301 5 9 4 1 3022 9 4 1 CPU registers Memory 3 0 0 PC 300 1941 AC 301594 19 4 0 IR 302 2 9 4 CPU registers 3 0 1 PC 0 0 3 AC 1 9 4 0 IR 1 940 0 941 0 0 0 0 0 3 2 940 0 941 0 0 0 0 0 3 2 Step 1 Memory 300 19 4 0 301 5 9 4 302 2 9 4 CPU registers 30 PC 0 0 0 3 AC 59 4 1 IR Step 2 Memory 300 19 4 0 301 5 9 4 1 302 2 94 CPU registers 3 O 2 PC 0 0.0 5 AC 5 9 4 1 IR 1 940 0 941 0 Step 3 0 0 0 0 3 2 940 0 941 0 0 0 0 0 3 2 Step 4 Memory 300 | 1940 301 594 302 2 9 4 CPU registers 3 0 2 PC 0 0 0 S AC 2 9 4 1 IR Memory 300 1 9 4 301S 9 302 2 2 CPU registers 3 0 3 PC 0 0 0 5 AC 2 9 4 1 IR 3 940 0 941 0 Step 5 0 0 0 0 3 2 940 0 941 0 Step 6 0 0 0 0 Figure 3.5 Example of Program Execution (contents of memory and registers in hexadecimal) Show transcribed image text The program execution of Figure 3.5 is described using six steps. Expand this description to show the use of the MAR and MBR. Memory 300 19 4 0 301 5 9 4 1 3022 9 4 1 CPU registers Memory 3 0 0 PC 300 1941 AC 301594 19 4 0 IR 302 2 9 4 CPU registers 3 0 1 PC 0 0 3 AC 1 9 4 0 IR 1 940 0 941 0 0 0 0 0 3 2 940 0 941 0 0 0 0 0 3 2 Step 1 Memory 300 19 4 0 301 5 9 4 302 2 9 4 CPU registers 30 PC 0 0 0 3 AC 59 4 1 IR Step 2 Memory 300 19 4 0 301 5 9 4 1 302 2 94 CPU registers 3 O 2 PC 0 0.0 5 AC 5 9 4 1 IR 1 940 0 941 0 Step 3 0 0 0 0 3 2 940 0 941 0 0 0 0 0 3 2 Step 4 Memory 300 | 1940 301 594 302 2 9 4 CPU registers 3 0 2 PC 0 0 0 S AC 2 9 4 1 IR Memory 300 1 9 4 301S 9 302 2 2 CPU registers 3 0 3 PC 0 0 0 5 AC 2 9 4 1 IR 3 940 0 941 0 Step 5 0 0 0 0 3 2 940 0 941 0 Step 6 0 0 0 0 Figure 3.5 Example of Program Execution (contents of memory and registers in hexadecimal)
Expert Answer
STEP:1 The value of Memory Buffer Register isbrimming to Instruction register(IR). value in Instruction Registeris “1940”. . . . . .
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