Q6: Assume total number of addresses a CPU can manage are 12 bits long. a. What is the maximum size of memory if byte addressable? b. What is the maximum size of memory if word (2-byte long) addressable? If CPU uses left most 2 bits to interleave across RAM chips. How many maximum number of RAM chips it can support. C.

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

Draw a picture illustrating the contents of memory, given the following data declarations: (You need to draw a diagram showing the memory addresses and their contents.) Assume that your data segment starts at 0x1000 in memory. Your answer should fill the table below and explain why. Addres Hexadecimal Hexadecimal S Value Value 0x1000 Ox |0x Ox Ox 0x Name: asciiz "Jones" Age: byte 48 Numbers: word 11, 20 Letter1: asciiz "Car" Letter2: .byte 25 Hexadecimal Value Hexadecimal Value

I have this problem from my textbook that I do not understand, despite re-reading the section on segmentation and paging. "The IBM system/370 architecture uses a two-level memory structure and refers to the two levels as segments and pages, although the segmentation approach lacks many of the features described in Chapter 8. For the basic 370 architecture, the page size may be either 2 KB or 4 KB, and the segment size is fixed at either 64 KB or 1 MB. For the 370/XA and 370/ESA architectures, the page size is 4 KB and the segment size is 1 MB. What advantages of segmentation does this scheme lack? What is the benefit of segmentation for the 370?"Can you help me undertand what they are looking for in this explanation?

Given a computer with 300 memory locations, each 18-bit assessable, how many bits of addressing are needed?  Given a computer with 10-bit addressability and 5 bits for the address, what's the maximum number of memory this computer can have in bits?

Let's double the size of the address field from 32b to 64b.  What's the largest decimal value now?  How might you organize this for readability?

How many bits are required to address 4G Bytes main memory if   a) Main memory is byte-addressable?     b)Main memory is word-addressable? (word size is 32 bits or 4 Bytes)     c)If the main memory will be built on 512M Bytes memory chips, how many RAM chips are needed?

Question 9

7.How many bits are stored in 1KByte of memory?

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…