Here is class 12 computer science [Type B] Unit 12 solutions for Sumita Arora back exercise assignment. Below includes both textual and video solutions wherever required. View all the answers in assignment for chapter 12 and for all chapters here.Watch all tutorials for chapter 12.
Separate parity bits from transmitted data row parities 1 1 1 1 0 0 0 | 0 1 0 1 0 1 0 1 | 0 1 1 1 1 1 1 1 | 1 ----------------------- column parities 1 0 1 0 0 1 0 | 1in row 1 - 1 1 1 1 0 0 0 we have four 1 bits hence parity bit will be 0 to get even parityreceived parity bit is 0 hence no error in row 1in row 2 - 1 0 1 0 1 0 1 we have four 1 bits hence parity bit will be 0 to get even parityreceived parity bit is 0 hence no error in row 2in row 3 - 1 1 1 1 1 1 1 we have seven 1 bits hence parity bit will be 1 to get even parityreceived parity bit is 1 hence no error in row 3in col 1 - 1 1 1 we have three 1 bits hence parity bit will be 1 to get even parityreceived column parity bit is 1 hence no error in column 1in col 2 - 1 0 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 2in col 3 - 1 1 1 we have three 1 bits hence parity bit will be 1 to get even parityreceived column parity bit is 1 hence no error in column 3in col 4 - 1 0 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 4in col 5 - 0 1 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 5in col 6 - 0 0 1 we have one 1 bits hence parity bit will be 1 to get even parityreceived column parity bit is 1 hence no error in column 6in col 7 - 0 1 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 7Note: The two-dimensional parity checks can detect single, two and three errors that occur anywhere in the matrix but it fails to detect four error bits in the transmitted data.as all the received parity bits are correct there are two possibilities 1. received data is error-less2. data contains 4 erroneous bits
Separate parity bits from transmitted data row parities 1 1 1 1 0 0 0 | 0 1 0 1 0 1 0 1 | 0 1 1 0 1 1 1 1 | 1 ----------------------- column parities 1 0 1 0 0 1 0 | 1in row 1 - 1 1 1 1 0 0 0 we have four 1 bits hence parity bit will be 0 to get even parityreceived parity bit is 0 hence no error in row 1in row 2 - 1 0 1 0 1 0 1 we have four 1 bits hence parity bit will be 0 to get even parityreceived parity bit is 0 hence no error in row 2in row 3 - 1 1 0 1 1 1 1 we have six 1 bits hence parity bit will be 0 to get even parityreceived parity bit is 1 hence there is error in row 3 and it can be any bit out of sevenin col 1 - 1 1 1 we have three 1 bits hence parity bit will be 1 to get even parityreceived column parity bit is 1 hence no error in column 1in col 2 - 1 0 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 2in col 3 - 1 0 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 1 hence there is error in column 3 in col 4 - 1 0 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 4in col 5 - 0 1 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 5in col 6 - 0 0 1 we have one 1 bits hence parity bit will be 1 to get even parityreceived column parity bit is 1 hence no error in column 6in col 7 - 0 1 1 we have two 1 bits hence parity bit will be 0 to get even parityreceived column parity bit is 0 hence no error in column 7Note: The two-dimensional parity checks can detect single, two and three errors that occur anywhere in the matrix but it fails to detect four error bits in the transmitted data.with above analysis we found that erroneous bit is present in 3rd row and 3rd columnhence the 3rd bit present in 3rd row is damaged bit
When data is receivedthe received data unit is divided into segments of m bits where m is size of checksumAll the m bit segments are added along with the checksum value.The value so obtained is complemented and the result is checked. following two cases are possible-If the result is zero,Receiver assumes that there is no error in received data and receiver accepts the data.If the result is non-zero,Receiver assumes that there is error in received dataand receiver discards the data In this example we are considering m = 16 bitReceived data:1000 0110 0101 11101010 1100 0110 00000111 0001 0010 10011000 0001 1001 01011101 1010 0110 0000 1000 0110 0101 1110 + 1010 1100 0110 0000 ----------------------- 1 0011 0010 1011 1110 + 1 (add carry forward 1 to answer)---------------------------- 0011 0010 1011 1111 + 0111 0001 0010 1001---------------------------- 1010 0011 1110 1000 + 1 1000 0001 1001 0101---------------------------- 0010 0101 0111 1101 + 1---------------------------- 0010 0101 0111 1110 + 1101 1010 0110 0000----------------------------- 1111 1111 1101 1110complement: 0000 0000 0010 0001as complement is not zero receiver will discard the data.
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