 t1 1ms Transit time Master Slave Transit time = 2 ms t3=3ms t4=6ms Offset = -1 ms Slave is 1 ms slow Figure 7-3 Example Timestamps Between Master and Slave The assumption from Figure 7-3 is that the slave clock is 1 ms slow and that the forward and re- verse transit times are also 2 ms in each direction. So, you can start with t1 as 1 ms, t2 as 2 ms, t3 as 3 ms, and t4 as 6 ms (t2 and t3 are 1 ms less than they should be because of the offset). Filling in the values for the equation gives offset from master = (t2-t1t3-t4)/2=(2-1+3-6)/2-1 - Therefore, the equation shows that the slave clock does have a -1 ms offset from the master (mean- ing it is slow). Note that it does not matter how long the transit time is or how far the slave clock is wrong; the equation still gives an accurate solution to the value of the offset from the master. Should there be asymmetry in the forward and reverse paths, and that asymmetry is known and static (unchanging), then there are mechanisms to include that data in the equations to increase the accuracy of the recovered clock. This is covered in more detail in the upcoming section "Asym- metry Correction." Finally, there exists another mechanism to improve the recovered clock accuracy by using the cor- rectionField (correction field), which we address in the section "The Correction Field" and when talking about a transparent clock. To avoid overcomplicating the equations at this stage, those cor- rections are not included here. Suffice to say that there is a field available in the header of messages that allows PTP clocks to signal corrections arising from known errors in the timestamps or delays in the network (it is not used in Announce, signaling, and management messages). 7.3.1 Asymmetry and Message Delay The amount of time that the Sync and Delay_Req messages take to cross a packet path is not im- portant; it could be microseconds, milliseconds, or even minutes. This means that PTP still delivers time accurately, even over long-distance cable runs. Of course, a very long delay 