output from PTP commands. The main one is known as offsetFromMaster (offset from master) and is defined as offset from master = <time on the slave clock> - <time on the master clock> Note that IEEE 1588-2008 defines the offset to be slave time - master time; should you compute it the other way around, you would see the same value but with the sign swapped. However, sticking to this convention, a negative value means the slave clock is slow, and positive means it is fast. Looking back at the mathematics a little more, you can conclude the following from Figure 7-2: + tms + offset from master t4 = t3 + tsm - offset from master Because the offset from master is positive if the slave time is fast, we need to add that offset to obtain t2 be- cause the slave clock will be timestamping with a value more than would be expected if the offset was zero. Similarly, for t4 the value of the timestamp on the master clock will be less than expected as the slave clock (and thus t3) is ahead of the master. Therefore, we must subtract the offset to determine t4 from t3 values. What we have been calling transit time is known in IEEE 1588 as one-way propagation delay and is calculated as a value called meanPathDelay, where mean has the meaning of average-be- cause it is an average of the forward and reverse transit times. Now, applying the assumption that the transit time is the same in both directions (meaning that tms = tsm = one-way delay = meanPathDelay): t2 = meanPathDelay+offset from master t4 = t3 + meanPathDelay-offset from master If we combine these two equations and solve for meanPathDelay, the offset from master can- cels out and we get meanPathDelay : (t2 - t1- offset from master +t4 - t3+ offset from master)/ meanPathDelay = (t2 - t4 - t3), /2 If we combine these two equations and solve for the offset from master, the meanPathDelay cancels out and we get offset from master = (t2 - t1-meanPathDelay + t3 - t4 + meanPath.Delay)/2 offset from master = - t4)/2 For a better understanding, see Figure 7-3 as a practical example.