Least cost feasible assignment found

{}

v1=b1

v1=c1

v2=b2

v2=c2

v3=b3

v3=c3

v1=b1

v2=b2

v1=b1

v2=c2

v1=b1

v3=c3

v1=b1

v3=b3

v1=b1

v2=b2

v3=b3

v1=b1

v2=b2

v3=c3

0

f({v1=b1 , v2=b2}) is true

1

2

2

3

3

4

Feasible

Previous slide

Next slide

Back to first slide

View graphic version