Explain the concept of segement memory if the 8086 excution unit calculates an effective address of 14A3H and DS contains 7000H what physical address will the BIU produce?

Segmented memory in the 8086/8088 means that the effective address is added to 16 times the segment address. In the question, an effective address of 14A3H with DS of 7000H, means the physical address is 714A3H. However, the BIU in the 8086 only addresses Words, not bytes, so the address generated is 714A2H. If the operation required more than one byte of data, subsequent even addresses of 714A4H, 714A6H, etc. would be generated.