Now consider PT=0.5ms and TT=1ms for same GATE question.
P1 finished it's first TT=1ms and now ready to start it's first PT from S-R1 so source is free and now can transmit another packet P2. At T=1.5ms P1 has reached R1. now at this time P2 just finished it's half of first TT . When P1 has finished half if it's TT at R1 ,P2 has finished it's first TT and now ready to travel from S-R1.hope you can observe this.
So Total time to reach D for P1 = 1+1/2+1+1/2+1+1/2=4.5ms , so at T=4.5 P1 reached
Now as P2 is 1ms behind P1 so time to reach D for P2=5.5 , at T=5.5 ms P2 reached D.
at T=6.5 P3 reached.....
.......at T=1003.5 P1000 reached D.
Now consider PT=2ms and TT=1ms
so here at T=1ms P1 finished it's first TT and now read to travel form S-R1 and now S is free to transmit next packet P2 . At T=2ms P2 finished it's first TT and at that time P1 has completed half of it's travel time b/w S-R1.so we time to reach D for P1=1+2+1+2+1+2=9ms
At T=9ms P2 has finished half of it's last PT and it's in b/w the path R2-D.
so time to reach D for P2= at T=10 ms
for P3.....T=11 ms
....at T=1008 ms P1000 would reach D