Assume that CS=3500, DS=4500, SS=5500, SI=2200, DI=4200, BX=7300, BP=8000, AX=3420 (all values are in hex). Calculate the physical address of the memory and show the contents in each of the following: a) MOV [BP]+10,AX b) MOV [SI],AX c) MOV [BX][DI]+20,AX
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Assume that CS=3500, DS=4500, SS=5500, SI=2200, DI=4200, BX=7300, BP=8000, AX=3420 (all values are in hex). Calculate the physical address of the memory and show the contents in each of the following:
- a) MOV [BP]+10,AX b) MOV [SI],AX c) MOV [BX][DI]+20,AX
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- Assume a computer that has 32-bit integers. Use a table to clearly show how each of the following values would be stored sequentially in memory, starting at address 0x100, assuming each address holds one byte. 0x123456BA 0xabcde -28 (decimal values, assume the machine uses 2's complement notation. 0xFEDCDraw 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…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
- Assume the following values are stored at the indicated memory addresses and registers Address Value 0x100 0xaaa 0x104 0x123 0x108 0x12 0x10c 0x10 Register Value %eax 0x100 %ecx 0x1 %edx 0x3 Fill up the following table: %eax 0x104 $0x108 (%eax) 4(%eax) 9(%eax,%edx) 260(%ecx,%edx) 0xFC(,%ecx,4) (%eax,%edx,4)1. 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 "James Bond!"Age: .byte 24Numbers: .word 11, 22, 33Letter1: .asciiz "M"instruction is in the first picture please give me only implementation of int L1lookup(u_int32_t address) and int L2lookup(u_int32_t address) cacheSim.h #include<stdlib.h>#include<stdio.h>#define DRAM_SIZE 1048576typedef struct cb_struct {unsigned char data[16]; // One cache block is 16 bytes.u_int32_t tag;u_int32_t timeStamp; /// This is used to determine what to evict. You can update the timestamp using cycles.}cacheBlock;typedef struct access {int readWrite; // 0 for read, 1 for writeu_int32_t address;u_int32_t data; // If this is a read access, value here is 0}cacheAccess;// This is our dummy DRAM. You can initialize this in anyway you want to test.unsigned char * DRAM;cacheBlock L1_cache[2][2]; // Our 2-way, 64 byte cachecacheBlock L2_cache[4][4]; // Our 4-way, 256 byte cache// Trace points to a series of cache accesses.FILE *trace;long cycles;void init_DRAM();// This function print the content of the cache in the following format for an N-way cache with M Sets// Set 0…
- 1. The hypothetical machine has two I/O instructions: 0011 = Load AC from I/O 0111 = Store AC to I/O In these cases, the 12-bit address identifies a particular I/O device. List the steps for every execution for the following program and illustrate using table that explain the process below : a. Load AC from device 5. b. Add contents of memory location 940. c. Store AC to device 6. d. Assume that the next value retrieved from device 5 is 3 and that location 940 contains a value of 2. Please pointing a, b,c ans. Because one I already upload this question and I didn't understand which one is and of a...please write ans a, b , c please6. Assume a computer has 32-bit integers. Show how the value 0x0001122 would be stored sequentially in memory, starting at address 0x000, on both a big endian machine and a little endian machine, assuming that each address holds one byte. Address Big Endian Little Endian0x000 0x001 0x002 0x0032. 2) You are required to write an Assembly Language program segment to perform theoperation Ci = where Ai and Bi represents a set of 50 memory locations each storing avalue such that the A values are stored starting from memory location 100 while the B valuesare stored starting from memory location 200. The results are to be stored starting frommemory location 300.
- 8Gbx32 ROM element is given. a) Specify the address line and the data number line. b) How many bits is the total storage capacity of the memory? c) The total storage capacity of the memory can be specified in Megabytes.d) If we have two 4Gbx16 ROMs, two 2Gbx16 and 2Gbx32 modules, use these elements to design the 8Gbx32 bit memory unit as block diagrams. Express it in a descriptive way.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, F2the available space list of a computer memory is specified as follows: 9 start address block address in words 100 50 200 150 450 600 1200 400 determine the available space list after allocating the space for the stream of requests consisting of the following block sizes: 25,100,250,200,100,150 use i) first fit ii) best fit and iii) worst fit algorithms