Tutorial 4: Computer Organization

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1 Tutorial 4: Computer Organization 1

2 Lecture Outcome A brief summary of flowcharts with an example. A recap of the central processing unit (CPU), the main memory, and the input/output subsystem. Address a bit pattern: Activity 1. Look at program execution: Connecting CPU and memory Fetch-decode-execute cycle, including its effect on registers. Activity 2. 2

Flowcharts Revision According to Lakin, Capon and Botten (1996:18), a flowchart is defined as a formalised graphic representation of a program logic sequence, work or manufacturing process,

3 Flowcharts Revision According to Lakin, Capon and Botten (1996:18), a flowchart is defined as a formalised graphic representation of a program logic sequence, work or manufacturing process, organisation chart or similar formalised structure. Figure 1: Flowcharts Common symbols used in process flow mapping (Adapted from Sharp & McDermott 2009:220). Lakin, R., Capon N., & Botten, N BPR enabling software for the financial services industry. Management Services. 40(3): Sharp, A. & McDermott, P Workflow modeling: tools for process improvement and application development. 2 nd ed. Norwood, Massachusetts, USA. Artech House. 3

4 Workflow Example Weather24 1. Draw a workflow diagram using flowchart symbols, based on the following scenario: To determine the weather, we first enter the Weather.news24 URL in our browser e.g., Firefox, Safari, or Internet Explorer. This request is sent to the Weather24 server. Weather24 server responses by sending us the home page of Weather24. Next, we click Gauteng, and read the temperature (Temp). Then the Temp is checked. If Temp < 20 C, the system prints Below Freezing. Else, if Temp > 20 C, the system prints Above Freezing. 4

5 Workflow Example Weather24 5

6 A recap of the central processing unit (CPU), the main memory, and the input/output subsystem Figure 5.1 Computer hardware (subsystems) 6

7 Address a bit pattern 1. Divide ( ) successively by 2 until the quotient is 0: /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = 62500, remainder is /2 = 31250, remainder is /2 = 15625, remainder is /2 = 7812, remainder is /2 = 3906, remainder is /2 = 1953, remainder is /2 = 976, remainder is 1 976/2 = 488, remainder is 0 488/2 = 244, remainder is 0 244/2 = 122, remainder is 0 122/2 = 61, remainder is 0 61/2 = 30, remainder is 1 30/2 = 15, remainder is 0 15/2 = 7, remainder is 1 7/2 = 3, remainder is 1 3/2 = 1, remainder is 1 1/2 = 0, remainder is 1 2: Read from the bottom to top as: Therefore, we need 25 bits to address each byte (Answer). 7

8 Address a bit pattern Activity 1 A computer has 12 MB (megabytes) of memory. How many bits are needed to address any single byte in memory? Remember to include all the steps you followed. Solution: Step 1: Divide ( ) successively by 2 until the quotient is 0: /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = , remainder is /2 = 93750, remainder is /2 = 46875, remainder is /2 = 23437, remainder is /2 = 11718, remainder is /2 = 5859, remainder is /2 = 2929, remainder is /2 = 1464, remainder is /2 = 732, remainder is 0 732/2 = 366, remainder is 0 366/2 = 183, remainder is 0 183/2 = 91, remainder is 1 91/2 = 45, remainder is 1 45/2 = 22, remainder is 1 22/2 = 11, remainder is 0 11/2 = 5, remainder is 1 5/2 = 2, remainder is 1 2/2 = 1, remainder is 0 1/2 = 0, remainder is 1 Step 2: Read from the bottom to top as Therefore, we need 24 bits to address each byte (Answer). The memory address space is 12 MB, or 2^24 (2^4 X 2^20). This means that we need log22^24, or 24 bits, to address each byte. 8

9 Program Execution Figure 5.19 The steps of a cycle We want our CPU repeatedly to FETCH the next instruction from memory into the instruction register DECODE the instruction (that is, work out which it is) EXECUTE the instruction 9

10 Connecting CPU and memory Fetch-decode-execute cycle

11 Fetch

12 From the previous slide: When an instruction is to be fetched, the program counter contains the address of the next instruction. In this case, the next instruction is our address [0000]. Thus, the program counter has [0000] stored inside of it. This address is then copied into the memory address register. The address is then sent along the address bus, where it waits to receive a signal on the control bus. Owing to us wanting to read the value in address [0000], the control unit (CU) sends a signal on the control bus to the memory controller to say, memory read. The contents in address [0000] can now be transferred along the data bus. 12

13 Fetch

14 From the previous slide: Owing to what we fetch being an instruction, the value from the RAM now goes into the current instruction register (CIR). At this point, the instruction has been fetched, so we can now increment the program counter to be ready to fetch the next instruction. 14

15 Fetch

16 From the previous slide: Here the program counter now contains the value [1], where it has been incremented by [1]. 16

17 Decode

18 From the previous slide: The instruction in the current instruction register (CIR) is now decoded by the decode unit. You will notice that there are two (2) pieces of information 1) opcode, and 2) operand the address or the data for the opcode to use. Let us look at [0101] in the list of instructions load. Now, go back to the RAM and look for [0101] and return it. 18

19 Execute

20 From the previous slide: Owing to the data that we need being at address [0101] we send [0101] to the memory address register (MAR). You will notice that the program counter (PC) and the memory address register (MAR) contain different values. This is because, the program counter (PC) stores the address of the next instruction after this one has finished execution and the MAR is storing the address of the data that we need in order to compete the execution of this instruction. At MAR the address is sent down the address bus, where we wait for a control signal on the control bus, to say that we want to read the memory. 20

21 Execute

22 From the previous slide: Here is address [0101] in the RAM. It is going to be sent down the data bus. 22

23 Execute

24 From the previous slide: Notice now how the memory data register (MDR) now contains the value that was in RAM. It is now ready to be passed to the accumulator (ACC). 24

25 Execute

26 From the previous slide: It is now ready to be passed to the accumulator (ACC). Now we are ready to fetch the next instruction. Conclusion This is the state of the register after one cycle. 26

27 Activity 2 Follow the pattern again from the previous slides and fetch-decode-execute. You need to fetch the next instruction which is it s address [0001], and see what effect the whole program has on the contents of the registers. We will discuss the homework in one of your tutorial classes. 27

28 Questions? 28

Outcomes. Lecture 13 - Introduction to the Central Processing Unit (CPU) Central Processing UNIT (CPU) or Processor

Outcomes. Lecture 13 - Introduction to the Central Processing Unit (CPU) Central Processing UNIT (CPU) or Processor Lecture 13 - Introduction to the Central Processing Unit (CPU) Outcomes What is a CPU? How are instructions prepared by the CPU before execution? What registers and operations are involved in this preparation