From Jeff Bagby's "White Paper - Accurate In-Room Frequency Response to 10Hz" he gives an example for adjusting for one port:

*The smaller of the two, usually the port, will need to be lowered in SPL (dB) because its smaller area will apply more pressure on the microphone making it appear louder. Its output needs to be lowered by the ratio of their areas using this formula: Lower the port by 20 Log (Port Diameter / Cone Diameter). So, for our 6.5” woofer, with its 5” cone area, working with a 2” port, the port output will need to be lowered by 20 Log (2 / 5) = -7.96 dB, or roughly 8 decibels"*

For dual ports he give this example:

*for dual ports, find the total port area of both ports and determine the effective diameter for the combined area. Now, take the measurement at only one port, find the ratio of the total port diameter to the cone and make the calculation and then add 6dB. So, for our example, if there had been 2 – 2” ports then the port output would be lowered by -7.96 + 6 dB = -1.96 dB. For dual woofers and a single port we will need to do it the other way around and subtract another 6dB from the port output.*

However, I'm not understanding this. If I calculate the total port area of the two ports you get a combined area of 6.28sq inches or an effective diameter of 2.28". If I then calculate the SPL adjustment by using the ratio of diameters (effective port/cone) = 20 Log (2.82/5) = -4.97dB. If I then add the 6dB I get +1.03dB.

In my particular situation I have a woofer with an effective diameter of 4.375", two ports of 0.625 diameter (front), and one port of 1.25" diameter (rear). I am trying to apply the correct adjustment to a port reading so I can stitch the woofer and port measurements into one SPL graph. I think I can figure it out once I understand how to do it correctly.