} usually takes cycles each due to overlapping but I1 (PO) takes 2 cycles to add “1” cycle extra (2 + 1) ⇒7+8+1=10
1
2
3
4
5
6
7
8
9
10
I1
IF
ID
OF
PO
PO
PO
WB
I2
IF
ID
OF
OF
OF
PO
PO
WB
I3
IF
ID
ID
ID
OF
OF
PO
PO
IF
ID
OF
PO
WB
I1
1
1
1
3
1
I2
1
1
1
3
1
I3
1
1
1
2
1
I4
1
1
1
2
1
I5
1
1
1
1
1
I6
1
1
1
1
1
If Type 1 – 2 instruction Type 2 – 2 instruction Type 3 – 2 instruction I::1+1+1+3+1=7 I2:1+2 extra cycle =3 I3:1+1 extra cycle =2 I4:1+1 extra cycle =2 I5:1 extra cycle =1 I6:1 extra cycle =1 =16 In this way, if we compute 100 instruction number of cycles required is If ID OF PO WB