1. Nanoprogramming Extending microprogramming

2. Microprogramming The microprogram counter contains The address of the next microinstruction to be executed.

3. Microprogramming The microprogram memory contains all the microinstructions. Each machine level instruction is interpreted by one or more microinstructions.

4. Microprogramming If there are n machine- level instructions and each instruction is interpreted by m microinstructions, the size of the microprogram ROM is n.m lines.

5. Microprogramming The microinstruction register holds the bits of the current microinstruction. If this is p bits wide, the total size of the microprogram memory in bits is n.m.p

6. Microprogramming This structure requires a lot of fast microinstruction storage. For example, if there are 512 machine-level instructions, and each instruction is interpreted by four 200-bit microinstructions, the size of the ROM is 512 x 4 x 200 = 409,600 bits (51,200 bytes)

7. Nanoprogramming Microinstructions are very long; for example 200 bits. This requires a large amount of storage. However, of all the possible different microinstructions, a typical Microprogram ROM contains only a tiny fraction of possible microinstructions. Nanoprogramming reduces the number of control bits require to interpret an instruction set.

8. Nanoprogramming The microprogram memory (control ROM) is now much narrower because it contains pointers to the actual microinstructions.

9. Nanoprogramming The microinstruction register contains a short pointer that points to the nanoinstruction memory.

10. Nanoprogramming The nanoinstruction memory contains the actual microinstructions and is very wide.

11. Nanoprogramming Suppose that a microinstruction is 200 bits wide and that the microprogram memory contains only 120 unique microinstructions. Thus, out of 2 200 possible microinstructions, only 120 are actually used. Each microinstruction in the control ROM cam be replaced by a pointer that requires only 7 bits. 2 7 = 128 < 120

12. Nanoprogramming The nanoinstruction memory contains the 120 unique 100-bit- wide microinstructions.

13. Nanoprogramming Let’s use the previous example, with 512 machine-level instructions, where each instruction is interpreted by four 200-bit microinstructions. The size of the control ROM without nanoprogramming is 512 x 4 x 200 = 409,600 bits (51,200 bytes) With nanoprogramming (and assuming 120 unique microinstructions) the control ROM now requires 512 x 4 x 7 = 2048 x 7 = 14,336 bits because the 200-bit microinstruction has been replaced by a 7-bit pointer. The nanoinstruction memory contains 120 200-bit microinstructions or 120 x 200 = 24,000 bits. The total size of the read-only memory is the sum of the microinstruction memory and nanoinstruction memories or 14,336 + 24,000 = 38,336 bits (4,792 bytes). Nanoprogramming has reduced the storage requirement by a factor of 10