Model Answers HW2 - Chapter #3
1. The hypothetical machine of figure 3.4 also has two I/O instructions: 0011= Load AC fro 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 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 received from device 5 is 3 and that location 940 contains value of 2. Answer: Memory 300 3005 301 5940 302 7006 . . 940 0002 941 We will assume that the memory (contents in hex) as the previous table: 300: 3005; 301: 5940; 302: 7006 Therefore, the steps will be as the following: Step 1: 3005 → IR Step 2: 3 → AC Step 3: 5940 → IR Step 4: 3 + 2 = 5 → AC Step 5: 7006 → IR Step 6: AC → Device 6 ------------------------------------------------------------------------------------------------------2. Consider a hypothetical 32-bit microprocessor having 32-bit instructions composed of two fields: the first byte contains the opcode and the remainder the immediate operand or an operand address. a. What is the maximum directly addressable memory capacity (in bytes)? b. Discuss the impact on the system speed if the microprocessor bus has 1. a 32-bit local address bus and a 16-bit local data bus, or 2. a 16-bit local address bus and a 16-bit local data bus. c. How many bits are needed for the program counter and the instruction register? Answer: a. 2^(32-8) = 2^24 = 16,777,216 bytes = 16 MB ,(8 bits = 1 byte for he opcode). b.1. a 32-bit local address bus and a 16-bit local data bus. Instruction and data transfers would take three bus cycles each, one for the address and two for the data. Since If the address bus is 32 bits, the whole address can be transferred to memory at once and decoded there; however, since the data bus is only 16 bits, it will require 2 bus