Three-bus Structure – Add R4, R5, R6 (meaning $R4 \leftarrow R5 + R6$)

In three-bus organization, bus A and bus B supply operands and bus C carries result, so execution needs only one clock cycle (fetch/decode steps same as usual).

Sequence:

1. Fetch & Decode: PC_{out}, MAR_{in}, Read, WMFC; MDR_{out}, IR_{in}; PC_{inc}; Decode
2. Execute Add: R5_{outA}, R6_{outB}, ALU_{add}, R4_{inC}, End

Explanation: Contents of R5 placed on Bus A, R6 on Bus B, ALU adds them concurrently, result on Bus C is latched into R4.

Note: For single-bus it would need 3 steps: R5_{out}, Y_{in}; R6_{out}, Y+ R6 → Z; Z_{out}, R4_{in}. For two-bus: R5_{outA}, R6_{outB}, ALU → Z; Z_{out}, R4_{in}. Three-bus is fastest.