Conversion of logical address and Physical address __ microcomputer principle

I have been learning 8086 recently, the class teacher suddenly asked this. For the problem "8086 CPU can provide 20-bit address information, can directly to 1M storage units to access, and the CPU can be used to provide address information registers are 16 bits, then how to use 16-bit registers to achieve 20-bit address addressing it" understand a lot.
About the logical address is allowed in the program in the address, 8086 of the logical address has a paragraph value and the offset within the paragraph two parts, the expression is "segment address: offset address within paragraph."

Conversion of logical and physical addresses. (Final Examination)
Previously said 20-bit address information can be accessed 1M memory units, that is, the address 00000h~fffffh, and the segment cs,ds,ss,es that store these addresses high 4 bits, such as 12345H, then a segment register will store 1234H high 4-bit information, which is the segment address.
The offset in the paragraph is the offset of the shift relative to the segment address.
The physical address is the 20-bit address information provided on the address bus select、read. Physical Address = segment Address *10h+ The offset address within the segment. Segment Address times 10H is because the segment address was taken high four digits, so after the restoration to let the segment address Left 4 digits (10H = 10000B), such as (cs) = 20a8h, (IP) = 2008H, then the physical address is 20a8h*10h+2008h = 22a88h.

Note: Segment address *16 (decimal) + offset address = Physical address.

My understanding: The Physical Address expression method is a hexadecimal *16 equivalent to moving the hexadecimal forward 1 bits,

FFFFF (10485575)/1024=1023kb (0~1023KB), "2 20 times equals 10485575/1024=1024KB"
1024KB is equal to 1MB, so the segment address is only to move one at the plus offset address, using the "segment Address: Offset Address" method can express 8086CPU index 1MB.

Italic is a reference.