Crashing by hand.

Please see attachment for question: Answer below is what professor provided

Here is my answer. Please check my work. 

Initially, the critical path takes 10 weeks.  The critical path is B-F-H.  We choose B as it is only 500.  F is 650 H is 1200.

After redrawing the diagram, we see the project duration is 9 weeks and TWO critical paths. A-F-H and B-F-H.  I chose F as its only 650 and it is common to both paths.  The next closest is A and B (200 plus 500) which is 700.  H is 1200.

After redrawing the picture, we see project duration is 8 weeks and we still see THREE critical paths  A-E-H and A-F-H and B-F-H.

F cannot be crashed again.  While A is in common to two critical paths, crashing A will not shorten the 3rd critical path. So I choose A and B which is 700.  While H is in common to all three critical paths, H costs1200 which is more than 700.

After redrawing the diagram, we see the project duration is 7 weeks and THREE critical paths  A-E-H  A-F-H  and B-F-H

If you wanted to crash it to 6 weeks - we need to crash H despite its high cost.  (You do not need to crash to 6 weeks.)

F cannot be crashed again.  A cannot be crashed again.  While we can crash B, it will not save time on A-E-H and A-F-H.  The only alternative is to choose H for 1200.