N010 G00 G71 G40 G80;

N020 M06 T01;

N030 G42 M07;

040 G00 X-5 Y-5 S800 M03;

N050 G01 Z-10 F480;

.

.

N_G01 X_ Y_;

N_G00 X0 Y0 M09;

N_G00 Z0 M05;

N_ M30;

[Note: If not given Cutting dia = Assume 10-12mm Cutting speed = 15 m/min (Assume) Feed, f=0.15 mm/tooth]

Cutting Dia - D,

Cutting speed - V

$\therefore $Spindle speed = $\frac{320V}{V}$rpm

Feed = f mm/tooth

Assume No. of flutes as 4

$\therefore$ Z = 4

F = spindle speed xzxf. (in rpm) = mm/min

Drilling cycle (Canned)

N010 G00 G71 G40 G80;

N020 M06 T02;

N030 G00 X25 Y25 S575 M03;

N040 G44 M07; $\quad$ G44-tool length composition

N050 G81 G99 X25 Y25 Z-14 R2 F115

N060 X_ Y_

N070 X_ Y_

.

.

N_ G80;

N_M05;

[Note: If not given Drill dia = Depends on size of hole Cutting speed = 15 m/min for C.I (Assume)]

Drilling

Drill Diameter = D,

Cutting speed = V m/min,

Spindle speed = $\frac{320V}{V}$rpm Feed = 0.2 mm/rev

$\quad$= 0.2 $\times$ spindle speed