Exercise 10. (15%) Consider a variant of the Round-Robin algorithm where the entries in the ready queue are pointers to the PCBs. 1. What would be the effect of putting two pointers to the same process in the ready queue? 2. What would be the major advantage of this scheme? 3. How could you modify the basic Round-Robin algorithm to achieve the same effect without the duplicate pointers? (This is an open-ended question.)