The original input was: 3  0 1 5 1  0 1 5 1  0 1 10 1 
The (sorted) input is:  3  0 1 5 1  0 1 5 1  0 1 10 1 

This detailed printout gives the state and remaining burst for each process

Before cycle    0:   unstarted  0  unstarted  0  unstarted  0.
Before cycle    1:     running  1      ready  0      ready  0.
Before cycle    2:     blocked  1    running  1      ready  0.
Before cycle    3:       ready  0    blocked  1    running  1.
Before cycle    4:     running  1      ready  0    blocked  1.
Before cycle    5:     blocked  1    running  1      ready  0.
Before cycle    6:       ready  0    blocked  1    running  1.
Before cycle    7:     running  1      ready  0    blocked  1.
Before cycle    8:     blocked  1    running  1      ready  0.
Before cycle    9:       ready  0    blocked  1    running  1.
Before cycle   10:     running  1      ready  0    blocked  1.
Before cycle   11:     blocked  1    running  1      ready  0.
Before cycle   12:       ready  0    blocked  1    running  1.
Before cycle   13:     running  1      ready  0    blocked  1.
Before cycle   14:  terminated  0    running  1      ready  0.
Before cycle   15:  terminated  0 terminated  0    running  1.
Before cycle   16:  terminated  0 terminated  0    blocked  1.
Before cycle   17:  terminated  0 terminated  0    running  1.
Before cycle   18:  terminated  0 terminated  0    blocked  1.
Before cycle   19:  terminated  0 terminated  0    running  1.
Before cycle   20:  terminated  0 terminated  0    blocked  1.
Before cycle   21:  terminated  0 terminated  0    running  1.
Before cycle   22:  terminated  0 terminated  0    blocked  1.
Before cycle   23:  terminated  0 terminated  0    running  1.
Before cycle   24:  terminated  0 terminated  0    blocked  1.
Before cycle   25:  terminated  0 terminated  0    running  1.
The scheduling algorithm used was Round Robbin

Process 0:
	(A,B,C,IO) = (0,1,5,1)
	Finishing time: 13
	Turnaround time: 13
	I/O time: 4
	Waiting time: 4

Process 1:
	(A,B,C,IO) = (0,1,5,1)
	Finishing time: 14
	Turnaround time: 14
	I/O time: 4
	Waiting time: 5

Process 2:
	(A,B,C,IO) = (0,1,10,1)
	Finishing time: 25
	Turnaround time: 25
	I/O time: 9
	Waiting time: 6

Summary Data:
	Finishing time: 25
	CPU Utilization: 0.800000
	I/O Utilization: 0.680000
	Throughput: 12.000000 processes per hundred cycles
	Average turnaround time: 17.333334
	Average waiting time: 5.000000