Ex 1. Show how the following values would be stored by byte-addressable machines with 32-bit words, using little endian and then big endian format. Assume that each value starts at address 1016. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations. a) O×456789A1 b) O×0000058A
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- Show how the following values would be stored by byteaddressable machines with 32-bit words, using little endian and then big endian format. Assume that each value starts at address 10 . Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations.Q.) 0x0000058AAssume 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,AXDraw 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…
- Show how the following values would be stored by machines with 32-bit words, using little endian and then big endian format. Assume each value starts at address 1016. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations. 1234567816 ABCDEF1216 8765432116Show how the following values would be stored by byte addressable machines with 32- bit words, using the little endian and then big endian. Assume each value starts at address 10 base 16. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations. a. 456789A116 b. 0000058A16Question Show how the following values would be stored bybyte-addressable machines with 32- bit words, using little endianand then big endian format. Assume each value starts at address301816. Draw a diagram of memory for each, placing the appropriatevalues in the correct (and labeled) memory locations. a. 56789ABC16 b. 2014111910 The memory unit of a computer has256K words of 32 bits each. The computer has an instruction formatwith 4 fields: an opcode field; a mode field to specify 1 of 7addressing modes; a register address field to specify one of 16registers; and a memory address field. Assume an instruction is 32bits long. Answer the following: a. How large must the mode field be? b. How large must the register field be? c. How large must theaddress field be? d. How large is the opcode field? 4. In a computerinstruction format, the instruction length is 12 bits and the sizeof an address field is 4 bits. Is it possible to have: 13 2-address instructions 45 1-address instructions…
- 6. 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 0x003Giventhe following assignment of some program’s virtual pages to physical pages in a system with 4 KiB byte pages, what physical memory address corresponds to virtual address 20000? (All values are given in decimal.)Assume 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
- 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 pleaseCA_6 We study the properties of cache memory, and for reasons of easier design and efficient circuits, we assume that the cache capacity is 2i Bytes, and cache line size is 2j Bytes, with i and j being natural numbers: (a) How many bits should the tag field have? And can the tag field contain 0 bit (i.e., be empty)? Elaborate (b) Repeat the above for the index field. (c) Repeat the above for the byte-offset field. (d) Finally, depict a figure showing a cache line, indicate what fields it possibly has, state the possible sizes of these fields, and explain the uses of these fields.By assuming that X = 3, and 33 is a two digit number, consider memory storage of a 64-bit word stored at memory word 33 in a byte-addressable memory (a) What is the byte address of memory word 33? (b) What are the byte addresses that memory word 33 spans? (c) Draw the number 0xF1234567890ABCDE stored at word 33 in both big endian and little-endian machines. Clearly label the byte address corresponding to each data byte value.