f available address lines are 36, then what will be the possible addressable memory bytes?
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If available address lines are 36, then what will be the possible addressable memory bytes?
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- 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 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]Suppose the RAM for a certain computer has 256M words, where each word is 16 bits long. a. What is the capacity of this memory expressed in bytes? b. If this RAM is byte addressable, how many bits must an address contain? c. If this RAM is word addressable, how many bits must an address contain?
- the 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 algorithmsGiven the following memory maps, determine the size of available memory space (ie the unused or free memory space). Convert your result to Kilobytes a) Memory 1 $0000 DIRECT PAGE REGISTERS $005F $0060 RAM 512 BYTES $025F $1800 HIGH PAGE REGISTERS $184F SE000 FLASH 8192 BYTES $FFFF b) Memory 2 |FFFFFH EPROM (128K) Flash Memory (128K) E0000h C0000h Unused 72000h Zilog SCC 70000h Unused 20000h SRAM (128K) 00000hAddressing Memory Suppose that there are 226 bytes in memory. In binary the lowest address is 00...00, and the highest address is 11...11. a. What is the lowest address (in decimal) if memory is byte addressable? b. What is the highest address (in decimal) if memory is byte addressable? c. Suppose that a word is 16-bits long. What is the lowest address (in decimal) if memory is word addressable? d. Suppose that a word is 16-bits long. What is the highest address (in decimal) if memory is word addressable? e. Suppose that a word is 32-bits long. What is the lowest address (in decimal) if memory is word addressable? f. Suppose that a word is 32-bits long. What is the highest address (in decimal) if memory is word addressable?
- According to the memory view given below, if RO = 0x20008000, then LDRSB r1, r1 = ?(data overlay big endian)? Memory address Data Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 ØxC3 ØX20007FFE ØXD4 ØX20007FFE OXE5 (Ctrl) A-R1 = 0XC3 B-R1 = 0x000000C3 C-R1 = OXC3000000 D-R1 = 0xffffffC3 E-R1 = OxC3ffffff7.How many bits are stored in 1KByte of memory?Assume a 2 byte memory. What are the lowest and highest addresses if memory is byte addressable?
- physcal addresses are 4s ng 4 Ame dat in a cetain compe, te addresses can be translaled without y TLB entries At most how many ditina vid the address translation peh has 12 vld The Translation Look aside Bulfer (TLB)i sine is kB and the word size iby The memory is word addresible. The pe virtual addresses are 64 bea long d th sine is miss?A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word ofmemory.a. How many bits are needed for the opcode?b. How many bits are left for the address part of the instruction?c. What is the maximum allowable size for memory?d. What is the largest unsigned binary number that can be accommodated in one word of memory?Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…
