Bits in powerlab

Related BrainMass Solutions

• Transmission overhead associated with a protocol stack

  At Transport Layer: Since input data (19800 bits) is less than maximum data payload (20000 bits) of transport layer, entire data can fit in single transport packet (data unit).

• Cache: Expected access time, Address reference format

  [1b] Offset bits = lg(size of cache block in bytes) = lg(64) = 6 bits (bits 0:5) Index bits = lg(size of direct-mapped cache in bytes) = lg(32 KB) = lg(32768) = 15 bits (bits 6:20) Tag bits = 32 - 6 - 15 = 11 bits (bits 21:31).

• Multiplexer, Stuff Bits

  bits available for transmitting information in each group of 22 bits.

• Transmission schemes and Time sequence diagrams

  ; 7 data bits, 2 start bits, 1 stop bit, 4 bit FCS (i) each bit overhead in bits = 8/7*7=8 bits (ii) total overhead in bits = 1000*8=8000 bits overhead in time=8000/100= 80 sec d)Synchronous transmission; 1000 bit data field, 8 bit preamble

• Format of Memory Address for Various Cache Organizations

  Direct Mapped Cache: word, block, tag (number of bits in decimal) B.

• Set Associative Cache: Main Memory Address Format

  Hence we will require (12+7 =) 19 bits to address a word in main memory, with bits 0..6 indicating the word address in a block and bits 7..18 indicating the block address.

• Capacity, Data and Transfer Rates, Access Time of Disk Drive

  If you have difficulty in conceiving it, you can look at it in terms of the radius.

• Destination Address & Data Field

  I am assuming the following: Length of the data field (in hexadecimal)? = 4 bits = 0.5 Byte What is the length of the data field (in decimal)?

• Bits and bytes on a web page

  195327 Bits and bytes on a web page Create a Web Page named bits, html that prompts the user to enter a number of bits and stores that number in a variable named N.