In the table below, to have 8 memory addresses where each memory address holds one byte of data (byte-addressable), we need 3 bits to index every address. Memory Address Data Stored 000 A 001 010 011 1 100 2 101 110 M 111 N If you have 64 GiB of RAM memory, then how many bits are needed to have memory addresses for every byte-addressable memory location?
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In the table below, to have 8 memory addresses where each memory address holds one
byte of data (byte-addressable), we need 3 bits to index every address. (check image)
If you have 64 GiB of RAM memory, then how many bits are needed to have memory addresses for every byte-addressable memory location?
With n bits, we can identify different memory locations.
For example,
and so on...
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- (Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GBSuppose 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]The memory location at address of 0X003FB01 contains 1-byte memory variable J (0010_0001), and the memory location at the address of 0X003FB02 contains 1-byte memory variable K (0001 0010), see figure below. There is a 2-byte variable M which hold binary information M (1110 0101 0000 1i11). What is the address in hexadecimal format for 2-byte memory variable M, following little Endian computer? 7 Address in Data in Hex. Format Hex. Format 0X003FBF04 1110 0101 M OX003FBF03 0000 1111 0X003FBF02 0001 0010 0X003FBF01 0010 0001 J Its address in hexadecimal is 0X003FBF02. а. Its address in hexadecimal is 0×003FBF03. O b. Its address in hexadecimal is 0X003FBF04. Its address in hexadecimal is 0×003FBF01. d.
- Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): ADDRESS OxFFFF OXOCOE OXOCOD Ox0C0C OXOCOB OXOCOA 0x0C09 0x0000 CONTENTS 1111 1111 1111 1111 1111 1110 1101 1100 0001 1011 1100 0101 0110 0101 1000 0111 1100 0000 0100 0000 0011 0001 0101 0010 0000 1100 0000 1101 0000 0000 0000 0000 The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) What are the contents in hex of the memory location at following address in binary: 0000 1100 0000 1110? (Enter hex like the following example: Ox2A3F)Given a memory address, say 12A1H, find the Block number(i) the address belongs. Assume, the size of the Block is 8Bytes. Also, find the starting address of the block to which the address belongs. Find the capacity of RAM. Find the number of addressable locations in RAM. Find the total number of blocks RAM has. Assume RAM is byte-addressable.Given 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) 00000h
- The table below shows a segment of primary memory from a Von Neumann model computer Address Data 10101000 10001000 11001000 10011001 10100000 10101010 10110100 10111011 10001100 11001100 The program counter (PC) contains a value of 11001000. Find the value (in binary) that will be placed in MAR (memory address register)? MAR (in binray) %3D Find the value (in binary) that will be placed in MBR (memory buffer register)? MBR = (binray) %3DAccording 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 = OxC3ffffffBelow is a list of 32-bit memory address references, given as memory addresses. 12, 720, 172, 8, 764, 352, 760, 56, 724, 176, 744You would like to access a cache with the given memory addresses. The size of cache is 23 = 8-blocks. Your task is to: (1) find out the binary address, (2) fill out the tag and index for each memory address and (3) indicate whether the access is hit or miss in the following table:
- Microprocessor 8086 write An array called (A) has 300 unsigned byte numbers (chose your Owen data), write only one assemble to the following1-store the Maximum number of (A) in physical address AB200h. 2- store the Minimum number of (A) in the physical address CD100h. 3-rearrange the array given (A) ascending and save the new array in the memory starting at 3050e 4-save only the maximum odd number in the logical address 5200:0350Consider a memory implemented for 8086 microprocessor Draw the memory block diagram. Determine the values for A0 , /BHE ,address lines(A1..A19) and data lines(D0.. D15) in order to access: A byte at odd address [01FF3H] A byte at even address [01FFCH] A word at even address [01FFEH] A word at odd address [01ABFH]Consider a program consists of five segments: S0 = 600, S1 = 14 KB, S2= 100 KB, S3 =580 KB and S4 = 96 KB. Assume at that time, the available free space partitions of memory are 1200–1805, 50 – 150, 220-234, and 2500-3180.Find the following:d. What are the addresses in physical memory for the following logical addresses: 0.580, (b) 1.17 (c) 2.66 (d) 3.82 (e) 4.20?