So can someone clarify networks question 678 cause the answers I'm looking at are different for each of these
I'm on my phone so I'm switching between tabs as I work out the explanation, so I'll come back and edit.
Question 6
-There are two pathways from Q to P. this excludes option A.
- only vertex R has a loop. Excludes option E and D
- Q has two pays to R, hence C. (One direct path and there's another one that sort of doubles back, te fork in the road near P)
Question 7
*if someone sees something incorrect in this, please point it out... We all know how silly networks can be*
- the graph has to be planar: true. It seems like it is always possible to draw this graph with no lines intersecting. (I think this might generally be true for a graph where e<v, but I don't know that for sure)
- the graph always has >1 face: false. Easily disproven by drawing a graph with only one face.
- all vertices even degree: false. the first graph I drew has two odd degree vertices.
- sum of degrees must = 8: true. The rule for "sum if degrees" is count the edges, multiply by two. Unless I'm missing something (possible), this should be true.
- can't loop: true. If it looped, a node would be isolated, but the graph is connected.
Hence, 3 true, so C.
Question 8
The critical path is 24, edges 3+7+14.
There is a second path, 4+5+14 that = 23.
Crashing edge 7, reduces time taken by 1, costs 100$.
Now reducing any edge *except for e* will be useless if we only crash once, because there are two critics paths. We will have to crash twice in order to lower both critical paths.
Reducing edge 3 makes it 2+6+14 = 22
Reducing edge 4, so 3+5+14 = 22.
Currently our cost is $300. In order to reduce it Anymore (ignoring E), we would have to crash two activities of ABCD. But the question states we can only crash each activity one time. So we can't crash anymore, but E hasn't been crashed yet, so crashing E gives $400.
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