What is the need of segmentation in microprocessor 8086?

At this point you may be wondering why Intel designed the 8086 family devices to access memory using the segment: offset approach rather than accessing memory directly with 20-bit addresses. The segment: offset, scheme requires only a 16-bit number to represent the base address for a segment, and only a 16-bit offset to access any location in a segment. This means that the 8086 has to manipulate and store only 16-bit quantities instead of 20-bit quantities. This makes for an easier interface with 8- and 16-bit-wide memory boards and with the 16-bit registers in the 8086.

The second reason for segmentation has to do with the type of microcomputer in which an 8086-family CPU is likely to be used. In a timesharing system, several users share a CPU. The CPU works on one user's program for perhaps 20 ms, then works on the next user's program for 20 ms. After working 20 ms for each of the other users, the CPU comes back to the first user's program again. Each time the CPU switches from one user's program to the next, it must access a new section of code and new sections of data. Segmentation makes this switching quite easy. Each user's program can be assigned a separate set of logical segments for its code and data. The user's program will contain offsets or displacements from these segment bases. To change from one user's program to a second user's program, all that the CPU has to do is to reload the four segment registers with the segment base addresses assigned to the second user's program. In other Words, segmentation makes it easy to keep users' programs and data separate from one another, and segmentation makes it easy to switch from one user's program to another user's program.