- How many bits are in a frame?
- How many address bits are needed for each RAM chip?
- How many bits are required in the physical address?
- How many address bits are required for a 2048 bit memory?
- How many bits can a 2k memory unit store?
- How many address bits would be required for a memory that stores 64 4 bit words?
- How is virtual address space calculated?
- How do you find physical and logical addresses?
- How do I find my memory address range?
- How many bits are left for the address part of the instruction?
- What is the minimum number of address bits required to represent 1k byte memory?
- How do I calculate the number of address lines?
- How many bits are required for physical address for 128 KB?
- How many address lines are there in 8086?
- How many bits are needed in the Mar?
- How many address bits are required to represent a 32 K memory?
- What is address line and data line?

## How many bits are in a frame?

Finally, each frame includes a 32-bit CRC.

Like the HDLC protocol described in Section 2.3.

2, the Ethernet is a bit-oriented framing protocol.

Note that from the host’s perspective, an Ethernet frame has a 14-byte header: two 6-byte addresses and a 2-byte type field..

## How many address bits are needed for each RAM chip?

18 address bitsc) How many address bits are needed for each RAM chip? , so 18 address bits for each chip.

## How many bits are required in the physical address?

Answer: The physical address requires 4 bits for the frame number because there are 16 2 of them. Then, the physical address also requires 10 bits of the offset because there are 1024 2 of them.

## How many address bits are required for a 2048 bit memory?

11 bitsIt means that a memory of 2048 words, where each word is 4 bits. So to address 2048 (or 2K, where K means 2^10 or 1024), you need 11 bits, so 11 address lines.

## How many bits can a 2k memory unit store?

2048 bitsWith regard to memory, 2k usually means 2*(2^10) — that is, 2048. Now, depending on context, that may be 2048 bits, 2048 bytes ( 8*2048 bits), or 2048 words — (word-size-in-bits) * 2048 bits.

## How many address bits would be required for a memory that stores 64 4 bit words?

64MB = 67108864 Bytes/4 Bytes = 16777216 words in memory, and each single word can thus be addressed in 24 bits (first word has address 000000000000000000000000 and last has address 111111111111111111111111). Also 2 raised to 24 = 16777216, so 24 bits are needed to address each word in memory.

## How is virtual address space calculated?

A 36 bit address can address 2^36 bytes in a byte addressable machine. Since the size of a page 8K bytes (2^13), the number of addressable pages is 2^36 / >2^13 = 2^23.With 4 byte entries in the page table we can reference 2^32 pages. … 8 GB = 2^33 B. … Calculations are done in answer no.

## How do you find physical and logical addresses?

To compute the physical address:look up the page number in the page table and obtain the frame number.to create the physical address, frame = 17 bits; offset = 12 bits; then 512 = 29. 1m = 220 => 0 – ( 229-1 ) if main memory is 512 k, then the physical address is 29 bits.

## How do I find my memory address range?

The first address of memory 1 is 2000h = 8192. Memory 2 starts at 3000h, or 0011 0000 0000 0000 etc. So you can see that the binary digits in the 13th, 14th and 15th positions correspond to the starting addresses of the 4 memories: 0010 = memory 1, 0011 = memory 2, 0100 = memory 3 and 0101 = memory 4.

## How many bits are left for the address part of the instruction?

16 bits24-8=16 bits are left for the address part of the instruction.

## What is the minimum number of address bits required to represent 1k byte memory?

210 = 1024, so you need 10 bits to address every byte in a kilobyte. Likewise, you need 20 bits to address every byte in a megabyte, and 30 bits to address every byte in a gigabyte.

## How do I calculate the number of address lines?

If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 2 locations (0, 1, 2, and 3). As you can see, number of addressable locations = n^2. This means that n=log(1024) to the base 2.

## How many bits are required for physical address for 128 KB?

If it is 128 kilobyte, the 8 data lines will be enough for transferring 1 byte. 1024 bytes requires 10 bits of address, and 128 locations 7 bits, so a 17 bit address bus would be needed, assuming a separate address bus.

## How many address lines are there in 8086?

20 address lines8086 Microprocessor is an enhanced version of 8085Microprocessor that was designed by Intel in 1976. It is a 16-bit Microprocessor having 20 address lines and16 data lines that provides up to 1MB storage. It consists of powerful instruction set, which provides operations like multiplication and division easily.

## How many bits are needed in the Mar?

20 bitsThe MAR would be 20 bits long, since that is the size of a single address, 20 bits.

## How many address bits are required to represent a 32 K memory?

2e10 = 1024, so you need 10 bits to address every byte in a kilobyte. Likewise, you need 20 bits to address every byte in a megabyte, and 30 bits to address every byte in a gigabyte. 2e32 = 4294967296, which is the number of bytes in 4 gigabytes, so you need a 32 bit address for 4 GB of memory.

## What is address line and data line?

Data inside a computer is accessed by means of metallic conductors called address lines. Each of them carries a bit of information, the same as a data line. A group of address lines is called an address bus.