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- If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the fetch cycle is 40% of the processor cycle time, what memory access speed is required to implement load operations with zero wait states and load operations with two wait states?SP=1239H, SS=9876H, the physical address is AAAFOH Non of them 1BC06H 0AAAFH 99999H if BX=1000, DS=0400, and AL=EDH, for the following instruction: MOV [BX] + 1234H, AL. the physical address is 6324H O 4244H 4234H 6234H 6243H OSuppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What is the Value of r0 and r1 after executing LDR r1, [r0, #2] -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]
- Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]ISA of a hypothetical CPU 1 Address Memory: Address Data - (8-bits) LOAD M 100 25 STORE M 101 90 ADD M 102 65 SUB M 103 36 MUL M 104 22 105 77 DIV M 106 89 Where: Note: all numbers are hexadecimal M-a memory address AC - accumulator Sample program: //line 1 /line 2 //line 3 //line 4 //line 5 //line 6 LOAD 100 ADD 101 STORE 106 LOAD 102 SUB 103 STORE 105 Answer the following questions based on the given information above. 1. What is the content of memory location 106 after executing line 3? 2. What is the content of memory location 105 after executing line 6? 3. Write a program segment that will multiply the content of memory location 105 with the content of AC and store the result at memory location 100. а. b. 4. For this CPU, what is the width of the program counter? (express answer in terms of bits, do not include the word "bits" in your answer)For the 8086 memory segment (011A) h with an offset (2B35) h, the memory address is
- Bus d'adresse Ox0 A7 0 A6 0 AS Oxo 0 A4 0 A3 0 A2 0 A1 Lecture Bus de contrôle Écriture AO Bus de données Mémoire d'Instructions Adr Adr Ctr Ctr Ctr Ox Mémoire de données Clavier Écran CD 0 EN Data Data EN Data EN Data EN Write the instruction STR R2, [R3] in hexadecimal (using 2 bytes). АО A1 DO D1 D d u r D3 D2 09Given the memory content below, fill in the table below the registers values. Initial values for PC=20A, AC=FFF8. (all values are in Hexadecimal) address Content 208 0209 209 000A 20A 7008 20B 1109 20C 9208Indirect Addressing Mode Instruction 002A J@ RETADR 3E2003 0030 RETADR RESW 1 Instruction Starts with 002A Opcode of J is 3C Object code is 3E2003 Show the detail work to get the above object code from the instruction. F9 F10 F11 F12 delet + ||
- For the instruction (0x6479), select all data paths that are used from the beginning of the Decode Instruction phase through the end of the Store Result phase. FYI: Be certain; Canvas deducts points for incorrect choices. OL tol OH to J OK to N OL to E OH to F OM to B OC to N OH to L A to F ON to O OM to N O to MConsider the simple program below: Memory Address Hex 100 101 102 103 104 105 106 107 203 204 205 206 207 Instruction Load 203 Add 204 Store 205 Add 100 Store 206 Halt Contents of Address (Hex) 1203 2204 2205 3100 2206 7000 0010 002B FFF7Complete the following table: MIPS Instruction op code rs rt rd shamt funct imm. /address Hexadecimal Representation add $t4, $s2, $s1 addi $s0, $t0, 123 lw $s6, -88($t7) Note: In MIPS register file, temporary registers $t0-$t7 have indices 8-15 (respec- tively). Also, the saved registers $s0-$s7 have indices 16-23 (respectively).