Sample questions
Nc Time Sync TickDifficulty 1
A client-server time exchange logs four timestamps in milliseconds: T1 (client send) = 2000, T2 (server receive) = 2035, T3 (server send) = 2042, T4 (client receive) = 2090. Using theta = ((T2-T1)+(T3-T4))/2, what is theta?
- a-6.5 ms✓
- b6.5 ms
- c-13 ms
- d41.5 ms
Explanation:(T2-T1) = 35, (T3-T4) = -48, sum = -13, divided by 2 = -6.5 ms. The +6.5 ms option flips the sign, -13 ms skips the final division by 2, and 41.5 ms comes from adding the absolute values of the two terms instead of their signed sum.
Nc Time Sync TickDifficulty 1
Using the same offset formula theta = ((T2-T1)+(T3-T4))/2 with T1=500, T2=460, T3=470, T4=530 (all in ms), what is theta?
- a-100 ms
- b-50 ms✓
- c50 ms
- d-40 ms
Explanation:(T2-T1) = -40, (T3-T4) = -60, sum = -100, divided by 2 = -50 ms. -100 ms skips the division, 50 ms flips the sign, and -40 ms only uses the first term while ignoring (T3-T4).
Nc Time Sync TickDifficulty 2
In question 1 the computed offset was theta = -6.5 ms, defined as theta = T(B) - T(A) where A is the client (initiator) and B is the server (responder). What does this negative value mean about the two clocks?
- aThe server's clock reads about 6.5 ms ahead of the client's clock.
- bThe round-trip delay is negative, so the measurement is invalid.
- cThe server's clock reads about 6.5 ms behind the client's clock.✓
- dThe client sent the packet 6.5 ms later than it should have.
Explanation:theta = T(B) - T(A) is negative exactly when B (server) reads earlier than A (client) at the same instant, i.e. the server clock trails the client clock by 6.5 ms. A negative theta says nothing about delta (round-trip delay) and nothing about send timing.
Nc Time Sync TickDifficulty 1
Using T1=2000, T2=2035, T3=2042, T4=2090 (ms) and delta = (T4-T1) - (T3-T2), what is delta?
Explanation:(T4-T1) = 90, (T3-T2) = 7, so delta = 90 - 7 = 83 ms. 90 ms forgets to subtract the second term, 97 ms adds instead of subtracting, and 7 ms reports only the second term.
Nc Time Sync TickDifficulty 2
In delta = (T4-T1) - (T3-T2), what does the subtracted term (T3-T2) represent, and why is it removed?
- aThe client's clock drift since the last sync; it is removed to correct the client's clock.
- bThe server's own processing time between receiving and replying; it is removed to isolate pure network delay.✓
- cThe one-way network delay from server to client; it is removed to avoid double-counting it.
- dThe transmission time of the packet's payload bytes; it is removed to ignore packet size.
Explanation:(T3-T2) is the time the server held the packet before replying. Subtracting it from the round-trip span (T4-T1) leaves only time spent in transit, so server processing delay does not inflate the measured network delay.
Nc Time Sync TickDifficulty 2
A measured round-trip delay delta comes out to 50 ms. Is it always safe to assume each one-way leg took 25 ms?
- aNo — on an asymmetric path the two legs can differ a lot.✓
- bYes — the delta formula already guarantees a symmetric path.
- cNo — delta is always exactly four times the true one-way delay.
- dYes, but only when the connection goes through a relay server.
Explanation:delta only measures the round trip, T(A->B->A). It carries no information about how that total splits between the forward and return legs, so dividing by two is an assumption, not a measured fact.