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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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In serial communication disconnection and continuing how start bit and stop bit recorganized by the receiving device for correct framing?
Answer The start bit in serial communication is opposite polarity puls than the stop bit depending on the system the start bit for instance is a positive puls (hi…gh) and the stop bit will be a negative puls (low)
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%
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
What type of construction is it when the internal body structure of a vehicle is used as its frame a Unibody b Body-frame c Integral or d Body-over-frame?
Consider a logical address space with 4 pages of 1024 bytes per pageeach mapped onto a physical memory of 64 frames. how many bits are there in logical address?
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
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.
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
32 bits. which is 4 bytes (8 bits in each byte).
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)
1 frame= 1000 bytes where, 1 byte= 8 bits therefore, 1 frame=8000 bits
Bit Interval: The time required to send one signal bit. Bit Rate: The number of bits that are conveyed or processed per unit of time. (Example: 100MB/sec)
8 bit registers cannot be used as 16 bit registers. The reverse works, however, as the 16 bit general purpose registers of the 8086 and 8088 can be used as pairs of 8 bit regi…sters. AX is divided into AH (high 8 bits) and AL (low 8 bits), and BX, CX, and DX are similarly divided. Operations on 16 bit and operations on 8 bit registers are similar. So you can do add ah, bl, just as you could do add ax, bx.
In Windows XP
Refers to native data size & processing speed, each of those potentially twice as fast provided you have Ram etc. to support.
The ISA slot started as 8-bit, and then evolved to 16-bits.
Theoretically you can use any slot, but practically use ports starting from 1st.
If you are using TC.EXE (or TCC.EXE) then it is 16-bit.
Which answer lists the correct number of bits associated with each term. 8 bits per double word 32 bits per word 64 bits per quadruple word or 4 bits per byte?
A person will not be able to know which answer lists the correct number of bits without knowing what the answer choices are. In order to know which of the answers is corre…ct the answer options should be given.