جامعة بنها - كمية العموم قسم الرياضيات المستوي الرابع )علوم حاسب( يوم االمتحان: االحد تاريخ االمتحان: 1024 / 21 / 12 المادة :

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1 جامعة بنها - كمية العموم قسم الرياضيات المستوي الرابع )علوم حاسب( يوم االمتحان: االحد تاريخ االمتحان: 1024 / 21 / 12 م المادة : بنية الحاسب )124 رس( الممتحن: د/ مصعب عبد الحميد محمد حسان مدرس بقسم الرياضيات بكمية العموم االسئمة و اإلجابة نموذج كاممة ورقة

2 Answer the following question Question 1. (12 marks) A- Define microoperations, instruction code, common bus system, assembler? (4 marks) B- Consider 128K X 16 memory. How many address lines does it have? How many data lines it have? What is its total capacity in Kbytes? What will be the total capacity if the address lines are changed to 20 lines without changing the word size? (3 marks) C- Draw the bus system for two registers such that each register has eight bits? (5 marks) Question 2. (14 marks) A- List the registers in the basic computer and write a description of their function and the number of bits that they contain? (4 marks) B- Write the instruction cycle steps in details? (6 marks) C- What is the difference between a direct and indirect address instruction? How many references to memory are needed for each type of instruction to bring an operand into a processor register? (4 marks) Question 3. (12 marks) A- For each timing signal, show the contents in hexadecimal of registers PC, AR, DR, IR, and SC of the basic computer when an ISZ indirect instruction is fetched from memory and executed. The initial content of PC is 7FE. The content of memory at address 7FE is EA9F. The content of memory at address A9F is 0C35. The content of memory at address C35 is FFFF. (4 marks) -Page 1-

3 B- List the memory reference instructions and for each instruction write operation decoder and symbolic description? (5 marks) C- List the rules of the assembly language? (3 marks) Question 4. (10 marks) A- List the categories of programs written for a computer? (3 marks) B- Write assembly language program to calculate the expression a+b-c? (4 marks) C- Given the common bus system of the basic computer, the following register transfers are to be executed in the system? For each transfer, specify: (1) the binary value that must be applied to bus select inputs S 2, S 1, and S 0 ; (2) the register whose LD control input must be active (if any); (3) a memory read or write operation (if needed) (3 marks) (i) AR PC (ii) IR M[AR] (iii) M[AR] TR Best Wishes Dr. Mosab Abd El-Hameed -Page 2-

4 Answer of Question 1 A- microoperations are operations executed on data stored in one or more registers instruction code is a group of bits that instructs the computer to perform a specific operation (sequence of microoperations). It is divided into parts (basic part is the operation part) common bus system is a scheme for transferring information between registers in a multiple-register configuration The translation of symbolic program into binary is done by a special program called an assembler B- This memory has -17 address lines data lines. - Total capacity of (128 x 1024 x 16) / (1024 x 8) = 256 Kbytes - If the address lines are changed to 20 lines without changing the word size, the memory will be 1024K X 16. Then the total capacity will be (1024 x 1024 x 16) / (1024 x 8) = 2048 Kbytes C- Register B Register A B 7 B 6 B 5 B 4 B 3 B 2 B 1 B 0 A 7 A 6 A 5 A 4 A 3 A 2 A 1 A 0 B 7 A 7 B 6 A 6 B 5 A 5 B 4 A 4 B 3 A 3 B 2 A 2 B 1 A 1 B 0 A 0 Mux7 Mux6 Mux5 Mux4 Mux3 Mux2 Mux1 Mux0 S 1 S 0 S 1 S 0 S 1 S 0 S 1 S 0 S 1 S 0 S 1 S 0 S 1 S 0 S 1 S 0 8-Line Common Bus

5 Answer of Question 2 A- List of BC registers: DR 16 Data Register Holds memory operand AR 12 Address Register Holds address for memory AC 16 Accumulator Processor register IR 16 Instruction Register Holds instruction code PC 12 Program Counter Holds address of instruction TR 16 Temporary Register Holds temporary data INPR 8 Input Register Holds input character OUTR 8 Output Register Holds output character B- Instruction Cycle Steps: 1- Fetch an instruction from memory 2- Decode the instruction 3- Read the effective address from memory if the instruction has an indirect address 4- Execute the instruction This cycle repeats indefinitely unless a HALT instruction is encountered Instruction Cycle Fetch and Decode 1- Initially, the Program Counter (PC) is loaded with the address of the first instruction in the program 2- The sequence counter SC is cleared to 0, providing a decoded timing signal T 0 3- After each clock pulse, SC is incremented by one, so that the timing signals go through a sequence T 0, T 1, T 2, and so on T 0 : AR PC (this is essential) The address of the instruction is moved to AR. T 1 : IR M[AR], PC PC+1 The instruction is fetched from the memory to IR, and the PC is incremented. T 2 : D 0,, D 7 Decode IR(12-14), AR IR(0-11), I IR(15) Next figure shows the instruction cycle flowchart. we can determine the type of instruction (see in the figure). For example: D'7IT3: AR M[AR] D'7I'T3: Nothing D7I'T3: Execute a register-reference instr. D7IT3: Execute an input-output instr.

6 C- Direct Address Instruction has its operand address specified directly in the instruction, but the Indirect Address Instruction has the address of the operand s address, i.e. not directly the address of the operand. Direct Address Instruction: needs 2 visits to memory. The 1 st one is to fetch the instruction, 2 nd is to bring the operand.

7 Indirect Address Instruction: needs 3 visits to memory. The 1 st one is to fetch the instruction, 2 nd is to bring the operand address. And the 3 rd is to bring the operand itself. Answer of Question 3 A- Given 7FE X X X 0 T0 7FE 7FE X X 1 T1 7FF 7FE X EA9F 2 T2 7FF A9F X EA9F 3 T3 7FF C35 X EA9F 4 T4 7FF C35 FFFF EA9F 5 T5 7FF C EA9F 6 T6 800 C EA9F 0 B- Symbol Operation Symbolic Description AND D 0 AC AC M[AR] ADD D 1 AC AC + M[AR], E C out LDA D 2 AC M[AR] STA D 3 M[AR] AC BUN D 4 PC AR BSA D 5 M[AR] PC, PC AR + 1 ISZ D 6 M[AR] M[AR] + 1, if M[AR] + 1 = 0 then PC PC+1 C- Each line of an assembly language program is arranged in three columns called fields: 1-The Label field: May be empty or specify a symbolic address. 2- The Instruction field: Specifies a machine instruction or pseudo instruction. 3- The Comment field: May be ORG Lab Lab, ADD op1 / this is an add operation. Label Instruction Comment Answer of Question 4 A- The categories of programs written for a computer: 1- Binary code. This is a sequence of instructions and operands in binary that list the exact representation of instructions as they appear in computer memory.

8 2- Octal or hexadecimal code. This is an equivalent translation of the binary code or hexadecimal. 3- Symbolic code. The user employs symbols (letters, numerals, or special characters) for the operation part, the address part, and other parts of the instruction code. Each symbolic instruction can be translated into one binary coded instruction. 4- High-level programming languages. These are special languages developed to reflect the procedures used in the solution of a problem rather than be concerned with the computer hardware behavior. For example, C++, Java, C#. B- Suppose we calculate ORG 0 /Origin of program is location 0 LDA A /Load operand from location A ADD B /Add operand from location B STA D /Store sum in location D LDA C /Load operand from location C CMA /Complement AC INC /Increment AC ADD D /Add operand from location D STA E /Store the result in location E HLT /Halt computer A, DEC 83 /Decimal operand B, DEC 23 /Decimal operand C, DEC 15 /Decimal operand D, DEC 0 /Sum stored in location D E, DEC 0 /The result stored in location E END /End of symbolic program C-

Chapter 5. Computer Architecture Organization and Design. Computer System Architecture Database Lab, SANGJI University

Chapter 5. Computer Architecture Organization and Design Computer System Architecture Database Lab, SANGJI University Computer Architecture Organization and Design Instruction Codes Computer Registers