Adding on to that Zen 3 chiplet is 4.15B transistors. The consumer IO die is 2.09B transistors. The Threadripper IO dies is 8.34B transistors. M1 is 16B transi

Adding on to that Zen 3 chiplet is 4.15B transistors. The consumer IO die is 2.09B transistors. The Threadripper IO dies is 8.34B transistors. M1 is 16B transistors. A14 is 11.6B transistors. That difference is mostly 2 firestorm cores/cache, 4 GPU cores, another DDR4 controller, and some Thunderbolt/PCIe. Taking a ruler to a die shot and multiplying that area by transistor density (16B/119mm^2) gives us somewhere around 500M transistors per core plus 3mb of cache. Adjusting out to 4.5MB per core to match Zen 3 adds another 75-100M transistors. Zen 3 gives 518M transistors per core, but that includes interconnects. Taking a ruler to the core reveals about 14% of the die is interconnect, SMU, and wafer tests (about 580M transistors). Subtracting that from 4.15B and dividing into even core/cache gives 446M transistors per core. We'll assume an identical IO cost of 8.5B transistors and 150M transistors of interconnect per core. (64 * (600 + 150)) + 8500 is 56.5B transistors and 420mm^2 total area. A zen 3 chip with similar densities would be 310mm^2. No matter how you slice and dice that, Apple's chip is going to be significantly more expensive than AMD's chip. Thinking anything different is a pipe dream. EDIT: also note that formal verification adds a lot to chip pricing.