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When both multiplexers a and b combine four 100 kbps channels using a time slot of 4 bits each frame generated from a and b has the size of 16 bits the frame at c is 32 bits what is the frame rate at?
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Most modern operating systems are either 32 bit or 64 bit but it mostly has to do with memory coding. 32bit can utilize a maximum or 4gb or RAM due to the amount of nu…mbers it uses to identify a location on the RAM (32 numbers). 64 bit can use far more memory (over 128gb) while 16 bit can use far less (not sure of exact numbers) Anything you buy today should be at least 32bit if not 64 bit
Frames of 1000 bits are sent over a 1-Mbps channel using a geostationary satellite whose propagation time from the earth is 270 msec?
Let t=0 denote the start of transmission. At t=1 msec, the first frame has been fully transmitted. At t=271 msec, the first frame has fully arrived. At t=272 msec, t…he frame acknowledging the first one has been fully sent. At t=542 msec, the acknowledementbearing frame has fully arrived. Thus, the cycle is 542 msec. A total of k frames are sent in 542 msec, for an efficiency of k/542. Hence a) k=1, efficiency=1/542=0,18% b) k=7, efficiency=7/542=1,29% c) k=4, efficiency=4/542=0,74%
1 frame= 1000 bytes where, 1 byte= 8 bits therefore, 1 frame=8000 bits
data link layer
Consider a logical address space of eight pages of 1024words each mapped onto a physical memory of 32 frames how many bits are there in the logical address?
As was given for a 4 Page, 1024 words & 64 frames (shown below) 4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2…+10=12 bits Physical memory = 10 +6 =16 bits The answer for this problem is 13. 8 pages -> 2^3 bits 1024 bytes -> 2^10 bits 32 frames -> 2^5 bits Therefore: Logical memory = 3+10=13 bits (Page + Word) Physical memory = 10 + 5 =15 bits (Word + Frame)
A better question would be - What is the difference between 16bit and 24bit color? Colour is usually represented on computers, & displayed using 3 colour elements - Red…, Green & Blue. 16bit colour (known as "high colour") refers to the fact that only a total of 16 binary bits are used to represent each colour. This usually results in 5 bits being used for red, 5 bits for blue, and 6 bits for green (due to the fact that we are apparently more visually sensitive to green) This gives 32 shades of red and blue available, and 64 shades of green. This results in 65536 different possible shades. 24bit colour (known as "true colour") results in 8 bits being used for each colour, allowing 256 shades of each, & a total of 4.2 Million colours. Pretty much all digital displays use 24bit colour & consider it to be the full or "true" colour mode. On a home computer 32bit means 24bit is used for red, green & blue. The extra 8 bits are usually used for storing internal information (like transparency or stencil information). 32bit is often used because memory can be read easier in steps of 32bits (4 bytes), than it can in 24bits (3 bytes). Visually they will look identical on your monitor. 24bit colour will look visually "better" than 16bit colour, when looking @ high colour images like photography, or smooth gradients of colour. In 16bit colour you can often see the "steps" as the colour transitions from one to the next. In 24bit the colours are so close together that they are often indiscernible, & on most digital monitors, to most eyes, will appear perfectly smooth. Most modern computer games require your graphics card to be capable of displaying 24bit colour & will no longer support 16bit.
The length of commands a microprocessor can address per clock cycle is in part measured by the bit-rate and the operations per cycle. In your case, the bit rate, the literal s…tring length of commands is addressed: Falling back on my Bus width answer, the elemental difference between a 16-bit and 32-bit processor is pretty easy when drawn out for you. I'll use the band name "ABBA" for an example. A 16-bit processor would run the string "ABBA" like this: Receive command Instruction #1 ; (This is "A" "B" in binary) Instruction #2 ; (This is "B" "A" in binary) Output A 32-bit processor does lit like this: Receive commandInstruction #1 ;;; ("ABBA" in Binary) OutputA lot more goes into the actual process than this, and a 32-bit processor isn't necessarily twice as fast at the same clock speed as a 16-bit processor. In fact, real world differences are quite small, due to some under-the-hood architectural differences between 16, 32, and 64 bit processors. Just keep in mind a lower-bit processor isn't compatible with a higher-bit program, but most higher-bit processors ARE compatible with lower-bit programs.
Channel has data rate of 4 Kbps and propagation delay of 20mswhat range of frame size do stop and wait give an efficiency of atleast 50?
You need to use two formulas - firstly a= propagation delay/transmission time. We'll denote this by tprop and ttrans. tprop is given in the question, it is 20 x 10-3 ttrans… is Distance/Data Rate so ttrans = L / 4 x 103 ttrans = 80/L (becasue the powers of ten cancel each other out so you just multiply 20 x 4. we use powers of ten because we need the same format as the propogation delay a = 80/L 2. Use the formula 1/1+2a >= .5 = 1/1+ 160/L>=.5 First 1>= .5(1+160/L) second 1>= .5 + 80/L third 80/L >= 1 x .5 fourth .5 >= 80/L L >= 80/.5 L >= 160
16 bit compilers compile the program into 16-bit machine code that will run on a computer with a 16-bit processor. 16-bit machine code will run on a 32-bit processor, but 32-b…it machine code will not run on a 16-bit processor. 32-bit machine code is usually faster than 16-bit machine code. -DJ Craig Note With 16 bit compiler the type-sizes (in bits) are the following: short, int: 16 long: 32 long long: (no such type) pointer: 16/32 (but even 32 means only 1MB address-space on 8086) With 32 bit compiler the object-sizes (in bits) are the following: short: 16 int, long: 32 long long: 64 pointer: 32 With 64 bit compiler the object-sizes (in bits) are the following: short: 16 int: 32 long: 32 or 64 (!) long long: 64 pointer: 64 [While the above values are generally correct, they may vary for specific Operating Systems. Please check your compiler's documentation for the default sizes of standard types] Note: C language itself doesn't say anything about "16 bit compilers" and "32 bit compilers"
16-bit Windows applications were designed to run under Windows 3.0 and 3.1, while 32-bit Windows applications were designed for Windows 95, 98, NT, and 2000. They are written… to two different Application Program Interfaces (APIs) called "Win16" and "Win32". The main differences between the Win16 and Win32 APIs are:Memory model: Win16 uses a segmented memory model (each memory address is referred to using a segment address, and the offset within that segment), while Win32 uses a flat 32-bit address space.Multitasking: Win16 uses cooperative multitasking. This means that the application must relinquish control before another application or program can run. Win32 uses preemptive multitasking, in which the operating system (Windows NT, 95, 98, or 2000) assigns time slices to each process.Multithreading: Unlike Win16, Win32 supports multithreading. This means that each program is broken up into many threads, which can run simultaneously. Windows 3.1 and Windows for Workgroups 3.11 can run a small subset of Win32 applications, mostly older ones, by using a subsystem called "Win32s". Win32s translates Win32 system calls to Win16. This process is called "thunking". Windows 95, 98, NT, and 2000 can run Win16 applications by running them cooperatively in a Win16 compatibility box. (In the case of Windows NT, this is called "WOW" - Windows on Windows). If a 32-bit application crashes, it will not affect any other 32-bit or 16-bit applications. However, if a 16-bit application crashes, it might affect other 16-bit applications (but not 32-bit applications). Both APIs contain the mechanisms used to link applications and documents together (e.g., OLE and OLE2).
This refers to how the CPU processors the information, 32 bit is more current than 16 bit and much faster. 16 bit is obsolete because we not gave 64 bit systems.
How many bits are in logical addresss. If a logical address space of eight pages of 1024 words each mapped onto a physical memory of 32 frames?
a. Logical address will have 3 bits to specify the page number (for 8 pages) . 10 bits to specify the offset into each page (210 =1024 words) = 13 bits. … b. For (25) 11 32 frames of 1024 words each (Page size = Frame size) We have 5 + 10 = 15 bits.
Frames of 1000 bits are sent over a 1-Mbps channel using a geostationary satellite whose propagation time from the earth is 270 msec Acknowledgements are always piggybacked onto data frames The head?
Let t=0 denote the start of transmission.At t=1 msec,the first frame has been fully transmitted.At t=271 msec,the first frame has fully arrived.At t=272 msec,the frame … acknowledging the first one has been fully sent.At t=542 msec,the acknowledgement-bearing frame has fully arrived.Thus, the cycle is 542 msec.A total of k frames are sent in 542 msec,for an efficiency of k/542.Hence (a)k=1,efficiency=1/542=0.18% (b)k=7,efficiency=7/542=1.29% (c)k=4,efficiency=4/542=0.74%
Which bit is set by a Frame Relay switch to inform the source station that there is congestion on the network?
In order to know how many bits/second there are in 1 frame/second, you need to know how many bits are in that frame. In a typical asychronous serial protocol with 8 bits per f…rame, the bit rate would be 0.125 bits/second. If you are talking the IP network layer of TCP/IP, then the frame size is very dependent on the underlying message payload and headers. The original question, by the way, is invalid. Its asks "how many bits does...", but it should have asked "how manys bits per second does...".
Telegrams are sent in data packets. Bits packet frame segment data mean data packet and they belongs to Data link layer. Bye.