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+//Example 20.3//
+a=1.21;//dollar/kg
+b=0.70;// fabrication yield rate
+phenolic=a/b
+mprintf("phenolic =$ %f /kg ",phenolic)
+a1=4.30;//dollar/kg
+b1=0.95;// fabrication yield rate
+polyester=a1/b1
+mprintf("\npolyester $%f /kg",polyester)
+//Then the net materials cost per part is
+c=2.9;//g/part
+d=1;//kg //kilogram
+e=1000;//g //gram
+p=phenolic*c*d/e
+mprintf("\np = $%f /part =0.5cents/part",p)
+py=polyester*c*d/e
+mprintf("\npy =$%f /part =1.3cents/part",py)
+a1=10;//dollar per hour/operator
+b1=1;//operator
+c1=35;//s/cycle
+d1=4;//parts/cycle
+e1=1;//hour
+f1=3600;//s //seconds
+p1=a1*b1*(c1/d1)*(e1/f1)
+mprintf("\np1 = %f /part  = 2.4cents/part",p1)
+c2=20;//s/cycle
+g1=5;//operator
+py1=a1*(b1/g1)*(c2/d1)*(e1/f1)
+mprintf("\npy1 =$%f /part =0.3 cents/part",py1)
+//The total cost (materials+labour)is then
+a3=0.5;//cents/parts
+b3=2.4;//cents/parts
+phenolic1=a3+b3
+mprintf("\nphenolic1 = %f cents/part",phenolic1)
+a4=1.3;//cents/part
+b4=0.3;//cents/part
+polyester2=(a4+b4);// /part
+mprintf("\npolyester2 = %f cents/part",polyester2)
+//the greatly reduced labor cost have given a net economic advantage to the polyster
+