1- Convert the following binary numbers into signed decimal numbers in 2's complement representation: 00001110, 00110011, 10010010, 10001001. 2-List the number of words found in each memory device for the following numbers of address connections: (a) 8 (b) 11 (c) 14 3- Let a program have a portion fe of its code enhanced to run 4 times faster, to yield a system speedup 3.3 times faster. What is the fraction enhanced (fg)? 4 Draw memory and microprocessor contents before and after execution the following instruction: MOV AX, [1234H]; machine code for this instruction is ( AI 34 12)H. Suppose DS-400н, CS-6CОН, P- 200H, АХ-ID95H, [1234Н)-FAн, [1235H]-34H.
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- A.) What are the contents in hex of the memory location at the following address in binary: 0000 1100 0000 1111? (Enter hex like the following example: 0x2A3F) In all computer models, the contents of a memory location can be interpreted as data. B.) Interpret the contents at address 0xFFFF as a two's complement integer in base 10. C.) Interpret the contents at address 0x0C0C as two ASCII characters. D.) Interpret the contents at the same address as C.) above as an unsigned integer in base 10.A set of eight data bytes is stored in memory locations starting from XX70H. Write a program to add two bytes at a time and store the sum in the same memory locations, low order sum replacing the first byte and a carry replacing the second byte. If any pair does not generate a carry, the memory location of the second byte should be cleared. Data(H) F9, 38, A7, 56,98,52, 8F, F2Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to = 8 the third is equal to = 16 help me find the last word
- 49. Which of the following is/are restriction(s) in classless addressing ? a. The number of addresses needs to be a power of 2 b. The mask needs to be included in the address to define the block c. The starting address must be divisible by the number of addresses in the block d. All of the aboveThe following data segment starts at memory address 1000h (hexadecimal) .data printString BYTE "ASSEMBLY IS FUN",0 moreBytes BYTE 25(DUP)0 dateIssued DWORD ? dueDate DWORD ? elapsedTime Word ? What is the hexadecimal address of dueDate ? a. 1045h b. 1029h c.1010h d. 102DhBy assuming that X is the last digit of your student number and 3X is a two digitnumber, consider memory storage of a 64-bit word stored at memory word 3X ina byte-addressable memory(a) What is the byte address of memory word 3X?(b) What are the byte addresses that memory word 3X spans?(c) Draw the number 0xF1234567890ABCDE stored at word 3X in both big-endianand little-endian machines. Clearly label the byte address corresponding to eachdata byte value.
- Convert given code to LEGv8 code:int f, g, y //global 64-bit variablesint sum (int a, int b) { // at memory address X0+1000.return (a +b)} int main (void) // at memory address X0 + 800{f=2;g=3;y= sum (f, g);return y;}Convert this code, making valid assumptions about registers and register use. Notethat brackets and global variable declarations are not affecting the addresses of the instructionsin memory.76. ________ allows only committed data to be read, but does not require repeatable reads a. Read uncommitted b. Serializable c. Repeatable read d. Read committedSay you want to allocate an area in memory for storing any number in the range 0—.4,000,000,000. This memory area will start at location .2fffeb96. Give the addresses of each byte of memory that will be required.
- Q3. IF PC= 682 , AR=123 , DR=A2BF, IR= 5672, then execute the following M[AR] DR IR M[AR] Determine the value of IR after the execution and comment on the type of addressinWhat distinguishes entering the value 5 into cell number 6 from shifting the contents of cell number 5 into cell number 6 if the memory cell at address 5 presently has the value 8?Draw a picture illustrating the contents of memory, given the following data declarations: You need to mark all the memory addresses. Assume that your data segment starts at 0x1000 in memory. Name: .asciiz "Jim Bond!"Age: .byte 24Numbers: .word 11, 22, 33Letter1: .asciiz 'M' In this format HexadecimalAddress, Hex Value, Character/Number/Symbol, Binary Value, Decimal Value ALREADY HAVE! marking all the memory addresses. Assuming that the data segment starts at 0x1000 in memory. The Memory Layout looks like Byte by Byte Address Data 0x1000 4a 0x1001 61 0x1002 6d 0x1003 65 0x1004 73 0x1005 00 0x1006 18 0x1007 00 0x1008 0b 0x1009 00 0x100a 00 0x100b 00 0x100c 21 0x100d 00 0x100e 00 0x100f 00 0x1010 14 0x1011 00 0x1012 00 0x1013 00 0x1014 4d arrow_forward Step 2 In mips 1 word is equal to 4 bytes. Address Data…