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?
29 people found this useful
Consider a 3 different raster system with resolutin of 64048012801024 and 25602048 what size frame buffer is needee for each of theae system tostore 12 bits and 24 bit?
to store 12 bits per pixel . 1.for system with resolution 640 by 480. frame buffer size=(640*480*12)/8=0.46Mbyte. 2.for system with resolution 1280 by 1024. frame buffer s…izs=(1280*1024*12)/8=1.96Mbyte. 3.for system with resolution 2560 by 2048. frame buffer sizs=(2560*2048*12)/8=7.86Mbyte. to store 24 bits per pixel . 1.for system with resolution 640 by 480. frame buffer size=(640*480*24)/8=0.92Mbyte. 2.for system with resolution 1280 by 1024. frame buffer sizs=(1280*1024*24)/8=3.93Mbyte. 3.for system with resolution 2560 by 2048. frame buffer sizs=(2560*2048*24)/8=15.72Mbyte
Sliding window protocol using 3 bit sequence number and maximum window size of 4 Starting window positions at A and B before any frames are sent?
i need program C for sliding window protocl
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 acknowl…edging 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 t prop and t trans. . t prop is given in the question, it is 20 x 10 -3… t trans is Distance/Data Rate . so t trans = L / 4 x 10 3 . t trans = 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
Framing bit is a common practice in telecommunication. Framing bitis a bit that precedes a T1 or J1 frame. It is used for the purposeof synchronization of the incoming data wi…th the receiver.
Framing bit does not contain any dta however this bit is used for syncronization between two peers
If a tv screen has 525 scan line and an aspect ratio of 3 is to 4 and if each pixel contains 12 bits worth of intensity information how many bits per second are required to 30 frames each second?
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 fra…mes of 1024 words each (Page size = Frame size) We have 5 + 10 = 15 bits.
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)
data link layer
The year of 1972.
1 frame= 1000 bytes where, 1 byte= 8 bits therefore, 1 frame=8000 bits
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...".
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 firstframe has been fully transmitted. At t=271 msec, the first frame has fully arrived. Att=272 msec, the frame ack…nowledging the first one has been fully sent. At t=542 msec, theacknowledementbearing frame has fully arrived. Thus, the cycle is542 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%
PCI slot and 32-bit is 2 deffrent things PCI slot is able to have more extra stuff in ur computer like Graphics card to emprove gamming exp and 32-bit is ur computer Operating… System like Windows 7 there r 2 deffrent windows 7, 32 and 64 bit for example the 32 bit is limited on RAM now days if u have more then 3GB RAM u wont be able to run no more then 3GB.... 64-bit is better and faster able to run more up to date Hardware like havin 16GB or more RAM or what ever u please to have in ur computer