This set covers all five lessons in data transmission and checking: packets, serial and parallel transfer, parity, checksums, and detection versus correction. The questions go from easy to harder. Write every count and sum on paper.
Use the restricted pseudocode trace trainer and the safe Python reasoning sandbox to step through the checking loops. Record repeating slips in the mistake log and retest queue.
Questions
Q1. Which item in a packet header lets the receiver put packets back in order?
Show answer
The packet number (together with the total number of packets). Packets can take different routes and arrive out of order, so the receiver sorts them by this number.
Q2. A file of 3,000 characters is sent in packets that each carry 500 characters of data. How many packets are needed?
Show answer
3,000 ÷ 500 = 6. Check: 6 × 500 = 3,000. 6 packets.
Q3. A packet starts with a hop count of 4. The path has 6 routers. At which router is it discarded?
Show answer
After router 1 the count is 3, after router 2 it is 2, after router 3 it is 1 and after router 4 it is 0. The router that reduces it to 0 discards it, so router 4 discards the packet. It never reaches routers 5 and 6.
Q4. A link sends 1 bit per tick on each wire. How many ticks are needed to send 12 bytes (a) over a serial link and (b) over an 8-wire parallel link, ignoring skew?
Show answer
12 × 8 = 96 bits.
(a) Serial: 96 bits at 1 bit per tick = 96 ticks.
(b) Parallel: 96 ÷ 8 = 12 ticks. Check: 12 × 8 = 96.
Q5. Give one reason parallel transmission is a poor choice over a long cable.
Show answer
Bits on separate wires can arrive at slightly different times (skew) and can interfere with each other (crosstalk). The receiver may read the bits wrongly, so serial is more reliable over distance.
Q6. Find the even parity bit for 1101011 and the odd parity bit for 0011000.
Show answer
1101011 has 1,1,0,1,0,1,1 = 5 ones. Five is odd, so the even parity bit is 1 (total 6).
0011000 has 2 ones. For odd parity the total must be odd, so the parity bit is 1 (total 3).
Q7. A system uses even parity. The byte 01101101 arrives. Is an error detected? Explain.
Show answer
Count the 1s: 0,1,1,0,1,1,0,1 = 5. Five is odd, but even parity needs an even count, so an error is detected. Parity cannot show which bit is wrong.
Q8. A checksum is the sum of the values MOD 100. The sender sends 23, 48, 71, 9 with the checksum. (a) Calculate the checksum. (b) The receiver gets 23, 48, 71, 19. What does it find?
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(a) 23 + 48 = 71, 71 + 71 = 142, 142 + 9 = 151. 151 MOD 100 = 51.
(b) 23 + 48 + 71 + 19 = 161. 161 MOD 100 = 61. This is not 51, so an error is detected.
Q9. The byte 11001100 was sent with even parity (the last bit is the parity bit). It arrives as 10001110. Does the parity check pass? What does this show?
Show answer
Sent: 1,1,0,0,1,1,0,0 has 4 ones, even. Received: 1,0,0,0,1,1,1,0 has 4 ones, also even. The check passes, even though two bits changed (the second and the seventh). This shows parity cannot detect an even number of flipped bits.
Q10. Each bit is sent three times. Groups 110, 001 and 111 arrive. (a) What message is recovered? (b) Is this detection or correction? (c) What is the cost?
Show answer
(a) 110 has two 1s so the bit is 1. 001 has one 1 so the bit is 0. 111 gives 1. The message is 1, 0, 1.
(b) It is correction, because the receiver repairs the bits without asking again, provided only one copy in a group is wrong.
(c) Three times as much data is sent.
If you got these wrong
- Q1 to Q3 (packets, headers, hop count): go back to tracing a packet through a network.
- Q4 and Q5 (timing and skew): revisit serial and parallel transfer.
- Q6, Q7 and Q9 (parity): redo parity with a worked example, especially the count of 1s.
- Q8 (checksum): return to the checksum lesson. Write the running total after each value.
- Q10 (correction): read detection versus correction.
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