You may assume that in this assembly language, each line of source code assembles to a single one-word (32-bit) instruction
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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?Will upvote! Find the memory address of the next instruction executed by the microprocessor, when operated in the real mode, for the following CS:IP and 80286 register combinations: a. DS=2F2E & DX=9D64 b. CS=9F7A & IP=AB27 c. ES=DE21 & DI=D75F d. SS=FF5C & BP=92B8 e. DS=DC67 & CX=2FE8The register content for an Intel 8086 microprocessor is as follows:CS = 1000H, DS = 2000H, SS = 5000H, SI = 2000H, DI = 4000HBX = 6783H, BP = 7000H, AX = 29FFH, CX = 8793H, DX = A297HCalculate the physical address of the memory where the operand is stored and the contents of the memory locations in each of the addresses shown below: a) MOV [SI], ALb) MOV [DI+6H], BXc) MOV [SI+BX–11], AXd) MOV [DI][BX]+28H, CXe) MOV [BP][SI]+17, DX
- A two-word instruction is stored in memory at an address designated by the symbol A. The address field of the instruction (stored at A + 1) is designated by the symbol Y. The operand used during the execution of the instruction is stored at an address symbolized by EA. An index register contains the value X. With the help of diagram, state how EA is calculated from the other addresses if the addressing mode of the instruction is (1)- direct (2)- indirect (3)- indexed (4)-Relative (5)- Register indirect1. T/F - if (B)=006000 (PC)=003600 (X)=000090, for the machine instruction 0x032026, the target address is 003000.2. T/F – PC register stores the return address for subroutine jump.3. T/F – S register contains a variety of information such as condition code.4. T/F – INPUT WORD 1034 – This means Operating system should reserve 1034 bytes in memory5. T/F - In a two pass assembler, adding literals to literal table and address resolution of local symbol are done using first pass and second pass respectively.If R0 = 0x20008000, after STMDA r0!, {r3, r9, r7, r1, r2} instruction is executed, register r7 will be stored in memory starting from which memory base address. A. R0 = 0x20007ff0 B. R0=0x20007fec C. R0 = 0x20007fff D. R0= 0x20007ff4 E. R0 = 0x20007ffef
- A SUB instruction stores a value 8467h at offset value BD3Fh. If the computed address is 5B68Eh, what will be the ending address? Assume real mode operation. O a. 5F94E O b.5F94E0 O c. 4F94F O d. 4F94F0You have a CPU which contains two processor cores, connected via a bus. Each core has its own 8 row, direct-mapped L1 cache and the two caches are coherent via snoopy cache coherent over the bus, with a write-invalidate mechanism. The block size in the cache is 8 bytes (two words). Core A: Load byte address 63 Core A: Load byte address 57 Core B: Store byte address 63 Core A: Store byte address 63 Core B: Load byte address 102 Core A: Load byte address 121 Core A: Load byte address 57 For each access from the list above, please indicate whether the access would be a hit, a compulsory miss, a conflict miss, a capacity miss, or a coherence miss. Before the sequence begins, both caches are empty. First access (Core A address 63): Second access (Core A address 57): Third access (Core B address 63): Fourth access (Core A address 63): Fifth access (Core B address 102): Sixth access (Core A address 121): Seventh access (Core A address 57):Single instruction computer (SIC) has only one instruction that for all operations our MIPS does. The instruction has the following format. sbn a, b, c # Mems[a]=Mem[a]- Mem[b]; if (Mem[a]<0) go to PC+c For example, here is the program to copy a number from location a to location b: Start: sbn temp, temp, 1 sbn temp, a, 1 sbn b, b, 1 sbn b, temp, 1 So build SIC program to add a and b, leaving the result in a and leaving b unmodified.
- Compute the physical address for the specified operand in each of the following instructions from previousproblem. The register contents and variables are as follows : (CS) = 0A0016, (DS) = 0B0016, (SI) =010016, (DI) = 020016 and (BX) = 030016.a) Destination operand of the instruction in (c)b) Source operand of the instruction in (d)c) Destination operand of the instruction in (e)d) Destination operand of the instruction in (f)e) Destination operand of the instruction in (g)Consider the following store instruction: SW R1, 0x000F(R0). Assume that the registers R0 and R1 are initialized with 0x00000001 and 0x53A78BC Frespectively. A section of the MIPS byte addressable data memory is shown. Give the memory word of the following memory locations after the SW operation: (a). 0x00000015. (b). 0x00000014. (c). 0x00000013. (d) 0x00000012.(e). 0x00000011. (f). 0x00000010.Assume that the state of the 8088’s registers and memory just prior to the executionof each instruction in problem 15 is as follows: * in photos*What result is produced in the destination operand by executing instructions (a)through (k)? *only h through k* (h) MUL DX(i) IMUL BYTE PTR [BX+SI](j) DIV BYTE PTR [SI]+0030H(k) IDIV BYTE PTR [BX][SI]+0030H