Quick Answer: How Many Bits Indicate The Frame Number?

What is the size of physical and logical addresses?

Physical memory has 32 frames and we need 32 (2^5) bits to address each frames, requiring in total 5+10=15 bits.

A logical address space of 8 pages requires 3 bits to address each page uniquely, requiring 13 bits in total..

How is virtual memory page size calculated?

Finding Optimal Page SizePage Table Size = number of page entries in page table X size of one page entry.Let’s consider an example,Virtual Address Space = 2 GB = 2 X 2 ^ 30 Bytes.Page Size = 2 KB = 2 X 2 ^ 10 Bytes.Number of Pages in Page Table = (2 X 2 ^ 30)/(2 X 2 ^ 10) = 1 M pages.

Is 200 frames per second good?

At 144 Hz, for example, you will be able to see many more frames each second so you’ll get a smoother and more responsive experience overall. But running at 200 FPS with Vsync off rather than 144 FPS with Vsync on will still give you a difference between 5ms and upwards of 7ms of input latency.

What is a 48 bit address?

A 48-bit memory address can directly address every byte of 256 tebibytes of storage. 48-bit can refer to any other data unit that consumes 48 bits (6 octets) in width. Examples include 48-bit CPU and ALU architectures are those that are based on registers, address buses, or data buses of that size.

Is 25 frames per second good?

Film is sometimes shot at 25 FPS when destined for editing or distribution on PAL video. This refers to the interlaced field rate (double the frame rate) of PAL. … This number is sometimes referred to as 60 FPS but it is best to use 59.94 unless you really mean 60 FPS.

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 do you calculate frame number?

The size of a frame is the same as that of a page, so the size of a frame is 1024 bytes (210 bytes). If the physical memory is 32MB (225 bytes), the number of frames is 225 / 210 = 215 and this is also the maximum number of pages that can be present in memory at the same time.

How many bits is the physical address?

Word size versus address size Modern processors, including embedded systems, usually have a word size of 8, 16, 24, 32 or 64 bits; most current general purpose computers use 32 or 64 bits. Many different sizes have been used historically, including 8, 9, 10, 12, 18, 24, 36, 39, 40, 48 and 60 bits.

What is frame length?

Length – Length is a 2-Byte field, which indicates the length of entire Ethernet frame. This 16-bit field can hold the length value between 0 to 65534, but length cannot be larger than 1500 because of some own limitations of Ethernet. Data – This is the place where actual data is inserted, also known as Payload.

Why is 24 frames per second better?

If you want to watch a movie watch it at 24 frames. Originally 24fps was chosen as the film frame rate as a compromise between having a frame rate fast enough to create fluid motion to the eye and keeping film stock costs down.

How many bits wide is a virtual address?

Therefore, you need 20 (8+12) bits for a virtual address. There are 128 page frames. That requires 7 bits. Therefore, you need 19 (7+12) bits for a physical address.

How does frames per second work?

Frame rate is the speed at which those images are shown, or how fast you “flip” through the book and it’s usually expressed as “frames per second,” or FPS. Each image represents a frame, so if a video is captured and played back at 24fps, that means each second of video shows 24 distinct still images.

Why is virtual address used?

First, it allows us to extend the use of physical memory by using disk. Second, it allows us to have memory protection, because each virtual address is translated to a physical address. … Less number of I/O would be needed to load or swap each user program into memory.

How many 32 bit addresses are stored in the page directory?

A page directory also consists of 1024 * 32-bit entries (again fitting into a single page), each pointing to a page table. We can see that now 1024 * 1024 * 4KB = 32-bits and with this 3-level structure we are able to map the entire 4GB virtual address space.

How many bits are required for physical address for 128 kB?

Consider an 128 kB (total data size), four-way set-associative cache with 16 B blocks and LRU block replacement. The cache is physically tagged and indexed. Physical memory is 32MB, byte-addressable, and words are 4 bytes each. Virtual addresses are 32 bits, and pages are 16kB.