Question

The hypothetical machine of Figure 3.4 has two I/O instructions:

0011 = Load AC from I/O

0111 = Store AC to I/O

In these cases, the 12-bit address identifies a particular I/O device. Show the program execution (using the format of Figure 3.5) for the following program:

1. Load AC from device 5.

2. Add contents of memory location 940.

3. Store AC to device 6.

Assume that the next value retrieved from device 5 is 3 and that location 940 contains a value of 2.

34 Opcode Address (a) Instruction format Magnitude (b) Integer format Program counter (PC) = Address of instruction Instruction register (IR) = Instruction being executed Accumulator (AC) - Temporary storage (c) Internal CPU registers 0001 - Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory Figure 3.4 (d) Partial list of opcodes Characteristics of a Hypothetical Machine Memory 300 1 9 4 0 301 5 9 4 1 302 2 9 4 1 CPU registers Memory 3 0 0 PC 300 1 9 4 0 AC 301 5 9 4 11 19 4 0 IR 302 2 9 4 11 CPU registers 30 1PC 000 3 AC 11 9 4 0 IR 940 0 0 0 3 941 0 0 0 2 940 0 0 0 31 941 0 0 0 2 Step 1 Memory 300 1 9 4 0 301 5 9 4 11 302 2 9 4 1 Step 2 CPU registers Memory 30 1PC 300 1 9 4 0 OOO 3 AC 301 5 9 4 1 S 5 9 4 1 IR 3022 9 4 1 CPU registers 3 O 2 PC 0 0 0 5 AC 5 9 4 1 IR 3+2 = 5 940 0 0 0 3 941 0 0 0 2 Step 4 940 0 0 0 3 941 0 0 0 2 Step 3 Memory 300 1 9 4 0 301 5 9 4 1 302 2 9 4 1 4 CPU registers Memory 3 0 2 PC 300 1 9 4 0 0 0 0 5 AC 301 5 9 4 1 2 9 4 1 IR 302 2 9 4 1 CPU registers 30 3 PC OOO 5 AC ( 2 9 4 1 IR 940 0 0 0 3 941 0 0 0 21 Step 5 940 0 0 0 3 941 0 0 0 5 Step 6 Figure 3.5 Example of Program Execution (contents of memory and registers in hexadecimal)

The answer is

Step1: 3005 -> IR

Step2: 3->AC

Step3: 5940->IR

Step4: 3+2=5->AC

Step5: 7006->IR

Step6: AC-> Device6

How did the step 3 5940 come from and step 5 how did 7006 come from?

Answer 1

ANSWER textbook: pacgiam execution as per figuue 3.5 from Stepl. instanuction Register (IR) <3005 step2: Accumulator (AC 4-3 Step 3: Snstuction Registeen. CIR 5940 stepu' Accumulatour (AC) 65 (3+2) Step5: Instruction Registos CIR)<7006 Stepb: Device 6 < Accumulator (AC) Explanation - * Instruction eregistes (IR)_is the one which stoves instancion ---cucuently being executed ou de codede pao. glam _counte cec) is the one that holds adduess of the instruction being executed_at_curent_time. 12 bits is used to specify the address and bits in the IR_indicates_accumalater to be loaded let_the_contents_of_memory location be +300 13005.-__ # 301; 5940 302; 7006 Stella In stere the propean countea" bolds the value of "300" and "Thsforustion register holds contents of adduess "300" that is "3005" Memory CPU Registe37 300 3005 300 PC. 301 5940 ACC 302 7006 3005 IR Steli Searned with CamScanner

stemmer som steps - The_ content of address locatlon "cos" is moved to the accumulatow and fosturuction. Aegistej__holds the value "5940 Hemony CPU Registe) | 0003A/301 Ip 0003 ACC 5940 ID 300 3005 300 | 5940 30 2 7006 quo L2 Stepose - Add adduers of contente of accumulatour with contents "qWosind Stoned back to coccumulatoor. Memory Nemory 005 10003 CPO Registes, 302pc 0003 LACCA 5940JIR CPU Reginez 1302 Pc 005) 0003 Ooas fresca 300 3o 30012005 5940 BAR 3005 इपम 1006 → 3+2-5- 300 5900 D3t2=5 302 7006 940 Iccl Scannedith_ CamScanner

Stele The hentinstuuston (100.6.2 is fetched _302_ and pc is incremented from location - Memory 005 0003 CPU Register 303 Pc 0005 ACC 7006 IR 006 300 30) 302 3005 5940 7006 940 Stella The contents of the Acoue_stored in location 206 CPU Register 005 003 PC 006 005 303 coas 7006 ACC IR 300 3005 301 502 5940 7006 940 Thus, executtion to figure 3.5 Steps for the given from textbook done problem according cs Scanned with ERP- CamScanner

Answer 2

Question 1: Describe the fetch and execution cycles by executing the following instructions. 0001 Load AC from memory 0010 Add AC from memory 0011 And AC from memory 0100 OR AC from memory 0101 Store AC to memory ( indirect) AC PC Note : Explain your answer by drawing h000 1001 h001 2002 h002 3003 h003 4004 h004 5005 h005 6006 h006 7007 h007 8008