Yeah because say one of the 7 activites that MUST be happening on day 5 doesn't happen, it will mean that it will happen on another day, delaying the project.(Original post by Extricated)
Suppose there are 10 days :
DAY 1 : 3 activities must be happening
Day 2 : 5 activities must be happening
Day 3 : 3 activities
Day 4 : 3 activities
Day 5 : 7 activities
Day 6 : 2 activities
Day 7 : 1 activities
Day 8 : 4 activities
Day 9 : 2 activities
Day 10 : 1 activities
So you look at day 5 and see that 7 activities must be happening so there must be 7 workers?
Does this make sense? It is quite hard to explain without an actual question  but I'm happy to explain if there is a link to a past question you have seen and don't understand
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(Original post by CharrrrxD)
Yeah because say one of the 7 activites that MUST be happening on day 5 doesn't happen, it will mean that it will happen on another day, delaying the project.
Does this make sense? It is quite hard to explain without an actual question  but I'm happy to explain if there is a link to a past question you have seen and don't understand 
(Original post by Arsey)
Revision thread for the D1 exam.
I will upload the Jan 12 model answers on the day following the examination (Saturday morning!)
In the next few posts I will upload some past papers, mark schemes etc
IF YOU START FROM A GO TO C AND THEN RETURN TO A i.e route starts off with ACA does this count as traversing AC twice? or do you have to do AC then go round the network and do something like ACBDEGFCA 
how do you finish the paper on time

(Original post by Extricated)
Arsey, in the route inspection algorithm, suppose you have to repeat AC, BD, GF. AND YOUR START AND FINISH IS AT A.
IF YOU START FROM A GO TO C AND THEN RETURN TO A i.e route starts off with ACA does this count as traversing AC twice? or do you have to do AC then go round the network and do something like ACBDEGFCA
whichever you have to do twice, join them up as an extra path on your diagram and then use a highlighter to trace your path.
another tip is that if you have the order of each node, halve it and that is how many times it should appear in your path (excluding the starting node) 
(Original post by Arsey)
yes if you go ACA then you will be doing it twice.
whichever you have to do twice, join them up as an extra path on your diagram and then use a highlighter to trace your path.
another tip is that if you have the order of each node, that is how many times it should appear in your path (excluding the starting node) 
(Original post by Arsey)
yes if you go ACA then you will be doing it twice.
whichever you have to do twice, join them up as an extra path on your diagram and then use a highlighter to trace your path.
another tip is that if you have the order of each node, that is how many times it should appear in your path (excluding the starting node)
Thanks if you or anyone can explain this!
Your threads are awesome btw!
Thanks! 
(Original post by CharrrrxD)
Sorry Arsey could you explain the bit about the order is the amount of times it should appear in a path  Im just a bit confused because how I understand is if, for example B has order 2 and is attached to for example, C and D, on the path this is represented by CBD or DBC and therefore B only appears once?
Thanks if you or anyone can explain this!
Your threads are awesome btw!
Thanks!
so, a node with an order of 4 must appear twice in your path. 
(Original post by Extricated)
Nice, thanks . D1 is not too bad except for path analysis i think.....any tips for that? I really hate scheduling diagrams..
Once you have completed the forward scan, highlight all critical paths, you can then automatically fill in the late time of any node on a critical path, that will probably only leave 2 or 3 late times to fill in on the backward scan.
It is very important to realise that for any activity you should on consider the early time at the start and the late time at the end. 
Arsey, when there are two possible paths in critical path analysis and they say to state the critical activities, do you just state all of them or just state one of the possible paths which the critical activities would make if that makes sense? Wasn't quite sure what to do when i came across it in the jan/june 2011 papers

Can do everything but scheduling, hope it doesn't come up. practice is the key for this one hopefully get all the review exercises then past papers done before the exam. 12 days to go

(Original post by ps3g4m3r)
Can do everything but scheduling, hope it doesn't come up. practice is the key for this one hopefully get all the review exercises then past papers done before the exam. 12 days to go 
(Original post by Arsey)
Definitions for the D1 Exam (and what they really mean!)
Graphs
A graph G consists of points (vertices or nodes) which are connected by lines (edges or arcs). A graph is just some points joined together with lines.
A subgraph of G is a graph, each of whose vertices belongs to G and each of whose edges belongs to G. A subgraph is just part of a bigger graph.
If a graph has a number associated with each edge (usually called its weight) then the graph is called a weighted graph or network. A weighted graph is one with values, eg distance, assigned to the edges.
The degree or valency of a vertex is the number of edges incident to it. A vertex is odd (even) if it has odd (even) degree. The degree of a vertex is how many edges go into it.
A path is a finite sequence of edges, such that the end vertex of one edge in the sequence is the start vertex of the next, and in which no vertex appears more than once. A path describes a ‘journey’ around a graph, where no vertex is visited more than once.
A walk is a path where you may visit vertices more than once.
A cycle (circuit) is a closed path, i.e. the end vertex of the last edge is the start vertex of the first edge.
Two vertices are connected if there is a path between them. A graph is connected if all its vertices are connected. Two vertices are connected if you can walk along edges from one to the othe, in a connected graph all vertices are connected.
If the edges of a graph have a direction associated with them they are known as directed edges and the graph is known as a digraph. A digraph will have arrows on the edges.
A tree is a connected graph with no cycles.
A spanning tree of a graph G is a subgraph which includes all the vertices of G and is also a tree.
A minimum spanning tree (MST) is a spanning tree such that the total length of its arcs is as small as possible. (MST is sometimes called a minimum connector.)
A graph in which each of the n vertices is connected to every other vertex is called a complete graph. In a complete graph every vertex is directly connected to every other edge.
Matchings
A bipartite graph consists of two sets of vertices X and Y. The edges only join vertices in X to vertices in Y, not vertices within a set. (If there are r vertices in X and s vertices in Y then this graph is Kr,s.) A bipartite graph has two sets of nodes. Nodes of one set can only be matched to nodes of the other set.
A matching is the pairing of some or all of the elements of one set, X, with elements of a second set, Y. If every member of X is paired with a member of Y the matching is said to be a complete matching. In a matching some of one set are joined to the other. In a complete matching they’re all joined.
An Alternating Path starts at an unconnected vertex in one set and ends at an unconnected vertex in the other set. Edges alternate between those ‘not in’ and ‘in’ the initial matching.

(Original post by CharrrrxD)
Sorry Arsey could you explain the bit about the order is the amount of times it should appear in a path  Im just a bit confused because how I understand is if, for example B has order 2 and is attached to for example, C and D, on the path this is represented by CBD or DBC and therefore B only appears once?
Thanks if you or anyone can explain this!
Your threads are awesome btw!
Thanks!(Original post by Arsey)
forgot the key words halve it
so, a node with an order of 4 must appear twice in your path. 
(Original post by CharrrrxD)
Sorry Arsey could you explain the bit about the order is the amount of times it should appear in a path  Im just a bit confused because how I understand is if, for example B has order 2 and is attached to for example, C and D, on the path this is represented by CBD or DBC and therefore B only appears once?
Thanks if you or anyone can explain this!
Your threads are awesome btw!
Thanks!
The ones you are confused about above start and finish at different vertices.
CBD would be CBDBC so it does appear twice. 
(Original post by Gibbo81)
It only works to halve the order if you are starting and finishing at the same vertex.
The ones you are confused about above start and finish at different vertices.
CBD would be CBDBC so it does appear twice.
I have a couple of questions regarding D1 if you could please give me some tips :
1) In Djikstra's algorithm, sometimes they actually add a value in and sometimes they just add the smallest value in the mark scheme...
for instance, if there are two paths to B, one of length 57 and one of length 46..sometimes they'll put 57 46 and sometimes it will be just 46. When do you know when to put each one in?
2) Any tips on drawing a network from a precedence table? I find it difficult.
3) How likely is scheduling to come up? Arsey said it hasn't come up since January 2007 and therefore is unlikely to come up, however I'd argue the opposite : If it hasn't come up since 07' then there may be a higher chance of it coming up!
Thanks 

(Original post by Extricated)
Hi Gibbo,
I have a couple of questions regarding D1 if you could please give me some tips :
1) In Djikstra's algorithm, sometimes they actually add a value in and sometimes they just add the smallest value in the mark scheme...
for instance, if there are two paths to B, one of length 57 and one of length 46..sometimes they'll put 57 46 and sometimes it will be just 46. When do you know when to put each one in?
2) Any tips on drawing a network from a precedence table? I find it difficult.
3) How likely is scheduling to come up? Arsey said it hasn't come up since January 2007 and therefore is unlikely to come up, however I'd argue the opposite : If it hasn't come up since 07' then there may be a higher chance of it coming up!
Thanks
2) practise... do a rough one in pencil first and then do a second one. There is no need to have straight lines, you can have overlapping lines as long as it is clear.
Look out for necessary dummies before you start
If you can see a letter appearing more than once in the depends on column and it is joined by different letters it will definitely need a dummy
i.e. if you have...
Activity  depends on
P  F
Q  F,H
Because you have F and you also have F and H a dummy is required.
3) his guess is no better than mine. 
(Original post by Arsey)
1) you have to consider the working values in the order of the nodes. It is only necessary to add a new working value if it is shorter than what is already there, you can include it if it is bigger but there is no need.
2) practise... do a rough one in pencil first and then do a second one. There is no need to have straight lines, you can have overlapping lines as long as it is clear.
Look out for necessary dummies before you start
If you can see a letter appearing more than once in the depends on column and it is joined by different letters it will definitely need a dummy
i.e. if you have...
Activity  depends on
P  F
Q  F,H
Because you have F and you also have F and H a dummy is required.
3) his guess is no better than mine.
Thanks 
I can never tell who's better arsey or gibbo... They should have a maths off
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