PROBLEM 4 (21 PTs) a) What is the minimum number of bits required to represent:
ID: 3584822 • Letter: P
Question
PROBLEM 4 (21 PTs) a) What is the minimum number of bits required to represent: (3 pts) 65,537 colors? 32678 memory addresses in a computer?Numbers between 0 and 2048? b) A microprocessor has a 32-bit address line. The size of the memory contents of each address is 8 bits. The memory space is defined as the collection of memory positions the processor can address. (6 pts) 8 bits What is the address rangeowest to highest, in hexadecimal) of the memory space for this microprocessor? What is the size (in bytes, KB, or MB) of the memory space? 1KB = 210 bytes, 1MB = 220 bytes, 1GB = 230 bytes A memory device is connected to the microprocessor. Based on the size of the memory, the microprocessor has assigned the addresses 0x40800000 to 0x40BFFFFF to this memory device Address 0x40800000 0x40800001 What is the size (in bytes, KB, or MB) of this memory device? What is the minimum number of bits required to represent the addresses only for this memory device? 0x40BFFFFF c) The figure below depicts the entire memory space of a microprocessor. Each memory address occupies one byte. (12 pts) What is the size (in bytes, KB, or MB) of the memory space? What is the address bus size of the microprocessor? If we have a memory chip of 8MB, how many bits do we require to address 8MB of memory? We want to connect the 8MB memory chip to the microprocessor. For optimal implementation, we must place those 8MB in an address range where every single address share some MSBs (e.g.: 0x0000000 to 0x07FFFFF). Provide a list of all the possible address ranges that the 8MB memory chip can occupy. You can only use any of the non-occupied portions of the memory space as shown below 3 - 9 3Explanation / Answer
Ans 1.
a. 65537 colors -
2x = 65537 => x = log2(65537) = 16.000022 => 17 bits.
b. 32768 memory address.
2x = 32768 => x = log2(32768) = 15 => 15 bits.
c. Number between 0 and 2048.
If including 0 and 2048, then total numbers will be 2049.
2x = 2049 => x = log2(2049) = 11.0007 => 12 bits.
If excluding 0 and including 2048 or If excluding 0 and 2048, then total numbers will be 2048 and 2047 respectively.
2x = 2048 => x = log2(2048) = 11 => 11 bits.
2x = 2047 => x = log2(2047) = 10.999 => 11 bits.
As per chegg policy, please ask other questions separately. Please upvote if you found first answer useful.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.