Okay... let's redo q25 a i and ii again step by step:
The packets are: 0101100101 0110100111
Rules: the 1st digit and the last digit are start and end bits.
the 2nd digit is the direction X and Y.
In X direction: 1 is right, 0 is left
In Y direction: 1 is up, 0 is down
i. Spliting the digits away they become:
0 (start bit) 1 (right) 0110010 (mm) 1 (end bit)
0110010 = 50 base 10
therefore x direction: right 50 mm
similarly with the 2nd packet:
0 (start bit) 1 (up) 1010011 (mm) 1 (end bit)
1010011 = 83 base 10
therefore y direction: up 83 mm
ii. as for the checksum for the data above, the checksum bits are the sum of the two packets... In the exam i didn't include the direction into the calculation. I did 83 + 50 = 133 / 13 and get a reminder of 3 (sorry when i said i got 13 as the remainder).
If you included the directions into the addition, i.e:
10110010 + 11010011 then you should get 0 as the remainder.
The question is now... were u suppose to include them or not?