Yesterday, commenter SRK emailed me with a query about a 2023 Further (= General) Mathematics question, from the 2023 NHT Exam 1. SRK has concerns about the question, and about this type of question in general (which, SRK noted, had also appeared in a commercial trial exam).
Treat this post as a proto-WitCH. I haven’t worked through or even really pondered the question, and even if it is a WitCH perhaps it is not a major concern. NHT exams are notoriously clunky(er) and, although SRK has always been solid with his comments, anyone is capable of jumping at shadows.
The question is below. It is perhaps preferable to have readers come to the question fresh, so I have not indicated SRK’s concerns or provided the exam report‘s answer and solution (W, I). But SRK and anyone is welcome to chime in whenever and however they believe will help the discussion.
The question is MCQ 8 from Module 2 – Networks and Decision Mathematics – of the 2023 Exam 1, and the exam report is here (W, I).
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UPDATE (01/11/25)
The exam report indicates that the answer is D (four immediate predecessors), with the following explanation:
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UPDATE (03/11/25)
Thanks very much to SRK for flagging the question and thanks to everyone for their comments. It’s not my home turf, but it has been very interesting, and there are a few things to say.
The first thing to say is that SRK was not jumping at shadows, and the exam question is stuffed. As noted in James‘s and SRK‘s comments, the project is not uniquely determined as presumed-implied by the exam report: any of C or D or E could be correct, depending upon the project. This is illustrated in the following network picture of the project(s) (which hopefully I haven’t screwed up).
Here, the red dummy activities are optional, and including none or some or all of them leads to the various answers to the exam question. (In the report’s solution, all the optional activities are included, and there is no need for the dummies or the dummy junctions.)
The second thing to say is that the exam question, even if it hadn’t been stuffed, is a bad GM exam question, for a couple reasons. To begin with, it is not a natural scenario, to know the ESTs and LSTs for a project but to not know the IPs. Then, as a pretty automatic consequence of this unnaturalness, the method for solving such a project is far from clear. The examiners have seemingly suggested the Nike method: just do it, try to construct a network and cross your fingers. Of course it “worked” in this case, because the examiners began with the network and then composed the exam question. But the Nike method will not always work. Typically, there will be no consistent project or, as in the exam scenario, multiple consistent projects.
The final point is that, even if the scenario makes for a bad exam question, purely as a mathematical question it is interesting, a non-obvious inverse problem. To this end, SRK in his comment outlines a very nice approach (which James enacted in an ad hoc manner). In brief:
*) The EST of any activity X is there for a reason: some other activity Y must be blocking X from starting any earlier.
*) Similarly, some activity Z is blocking X from beginning any later than the given LST.
*) That information together determines that certain activities are IPs and/or have other activities as IPs.
*) Other activities, for which there is a positive “float time”, may or may not be IPs of certain other activities.
It still seems somewhat messy, but it’s a very nice approach to an exam question that should not exist.
In my opinion a good question. Can’t be simply “CAS-crunched” and actually necessitates conceptual understanding as you have to create a diagram based off the ESTs and LSTS; challenging and well written.
Only had a quick run through the question but is the problem that there is more than one correct answer? I’m guessing the answer they want is 3 but 4 and 5 also seem possible.
James, yes, that is my concern. In fact the answer they want is 4. I found that 3 is also possible. Curious how you might have gotten 5.
The problem with the question is clear now that I’ve seen the suggested solution. There’s not enough information in the question to conclude that A is a predecessor to G or either C or E are predecessors to I. They can be but it’s not necessary. If you don’t consider them you get an answer of 3, if you do consider all of them (as VCAA did) you get 4. You get 5 if you consider only one of C or E to be an immediate predecessor if I.
So one must reason like examination setter…
Funny, 2 years after the exam some students will learn.
I think such a question will repeat in the form:
“Construct a possible flow chart corresponding to info given, then state the number of activities that have exactly 2 predecessors”.
At least you can then pretend the question can have more than 1 solution.
Would this whole thing happen if a competent uni level mathematician wrote / vetted it?
Vetting is a thankless job. It is very difficult to see what you don’t see. And it’s NHT: whatever the offical line, I don’t know that anybody cares as much about. And, it’s General, which is a monster with the ninety-five different modules. But obviously once the mathematicians are watching over there is much less chance of such issues arising or remaining. Not that Smithy really gives a stuff.
I’ve updated the post with the exam report’s answer (D), and their explanation.
OK, I haven’t looked, but it sounds like SRK was not jumping at shadows, and the question is screwed? Is there any sense in which one legitimate answer is in some sense more natural?
It also sounds as if a question so framed would typically be screwed in the same manner?
The question is screwed. I guess 3 and 4 are, in a sense, more natural answers than 5, but all of them are valid.
I’m not sure if any question of this type would necessarily be screwed. It should be possible to create a table that leads to only one possible answer. (I think if you removed activities G and I (or A and I) from this question it would work). I could very well be wrong, but I also think it’s only determining the predecessors of activities NOT on the critical path that creates problems, so if the question were changed to “the number of activities *on the critical path* that have exactly two immediate predecessors…” then it should be fine for any such table.
Thanks, James. I’m not surprised that a question could be cooked to have just one solution. What I was asking was whether a generic question of this type would result in multiple solutions.
Once I get to look at this question, I’ll also look at your suggested alternative wording.
There can only be one critical path (fortunately), in any network drawn for these numbers, so the suggested improvement is a very good one. Now, did the original draft of the question say this…? (we will never know).
NVM about the alternative wording. Just did a quick test and it doesn’t work out. It seems that simply giving the ESTs and LSTs is just generally insufficient to properly determine all predecessors/dependencies.
So, unless a question of this type is very carefully cooked, it’ll be fundamentally screwed?
Certainly seems that way to me.
Ok, thanks, James, and everyone.
I’ll take a look at this, but can someone explain why VCAA regards D as *the* answer?
My best guess is that someone drew a network diagram, worked out all the EST and LSTs and then thought: let’s give them the table but without immediate predecessors and…
…no one saw an issue with that until it was too late.
The issue with the question is (ironically) the use of the word “THE” in the examiners’ report. If the report had said “A” network diagram, I would have some hope that alternative diagrams had been considered.
Is there any methodical way to determine any or all networks that fit with the provided information? I’m not familiar with this stuff but it’s not obvious how to best approach the problem.
Would the students have ever likely practised such a problem?
The answer to the second question is most likely no. A quick scan of the three textbooks I have on this show no similar examples and I can’t find anything remotely similar in 15 years of exam papers either.
As to the first question, I have two answers, take your pick:
1. This is General/Further – do you really care?
2. Even if there was a way, do you really think General/Further teachers/examiners are likely to spend the time to investigate?
Thanks, RF. Regarding 2, my question is, what did the examiners expect students to do? (Yeah, yeah …)
The only thing I can think of is that D is the answer provided that the network does not involve dummy activities. But I don’t see any good reason to think that dummy-less networks are more natural than dummy-containing networks. In my mind, the fundamental object here is the set of immediate-predecessor-relations, not the directed network we draw to represent that set.
Thanks, SRK. I agree, although your phrase “dummy-less networks” invites all manner of VCAA jokes.
There is no “network” here. What is provided is a list of activities, with their ESTs and LSTs. The question is then what can the immediate predecessors be that is consistent with the provided information. (A pretty artificial question, it seems to me.)
The network only comes later. It may be a good means to understand the information but it is not the information. And, sure, a dummy-less network may be easier to discover, but there is zero argument for such an activity being somehow better or more natural.
Since it is a multiple choice question, there seems to be no way of knowing how the students actually did the problem. Unless the students are specifically “trained” by their teacher in examinology — “when you see this, do that” (the trained seal approach) — they can come up with all sorts of cockamanny schemes to get the answer, some legit and some not. When you look at it from that point of view, the question seems to be a kind of lottery; some candidates are luckier than others.
Late to the party but I get 4 with no issues. I don’t see a huge issue with the question as you just build the network with the critical activities and then go backwards with the others by adding the duration and the LST and linking that end to it’s corresponding critical activity and its start using EST. Could someone show me how you’re getting another answer other than 4? Dummies are only used so immediate predecessors are accurately displayed, I would have only looked for dummies if I couldn’t get my first method to work.
Dunno, maybe I’ve done too many exams and I’m fluent in VCAA or brainwashed and can’t see the errors as well as others.
I agree with you, B: you’ve done too many exams and are too fluent in VCAA. The question is screwed.
Actually, I’ll amend that. IF there is more than one correct answer THEN the question is screwed. I have to learn this stuff enough to convince myself that SRK and co are correct.
I agree that no dummy activities ARE needed, but that does not (to my thinking anyway) mean they are automatically excluded either.
I want to think about this a lot more but I also know that doing so will likely result in my wanting to go outside and yell at a cloud for a few hours.
So… I may or may not reply again on this one.
Thanks everyone though, it has certainly been thought-provoking, which is not usually what General exams do for me.
As I replied to SRK, the requirement or not of dummy activities does not in any manner make a solution to the problem better or worse. The question is not about networks.
1. You are correct, as it is currently written it is quite possibly about nothing at all.
2. The question appears in the exam booklet under the sub-heading “Networks” so I think some assumptions are valid in this instance.
3. An assumption being valid does not make it correct.
I think if a dummy can be used to get another answer VCAA needs to review it and accept the alternatives especially because that is a more difficult approach. However, I would assume they will say ‘dummy links should only be used if required and it was not needed in this case’.
I always find the northern hemisphere papers a bit clunkier and bc the cohort is so small, I’d assume it doesn’t get the same amount of scrutiny when they stuff up.
Well, it’s a little late to “accept the alternatives”, but, yeah, something should be done.
But see my comment below. This is very much not my area, but I am not sure how to get answers other than D.
There is something I am not getting about this question. It could be my misunderstanding of “immediate predecessor” (which is not helped by Cambridge screwing up the definition), but I cannot see projects that give rise to answers C and E.
Could someone upload images of their relevant solutions?
As I mentioned in a previous comment we can’t say that A must be an immediate predecessor to G or that either C or E must be immediate predecessors to I. Attached is what a network diagram would look like if we assume they are not. This gives three activities that have exactly two immediate predecessors; F (A and D), J (G and H) and K (I and J).
Thank you, James (I presume). I’m slow at this but I’ll check it out.
My approach to answering this question did not involve drawing a network. Rather my reasoning was (i) any activity which doesn’t have EST = 0 has at least one immediate predecessor, and any activity with LST + Duration < 25 is an immediate predecessor of at least one other activity (ii) if an activity has EST = 8 (say), then it has an immediate predecessor with EST + Duration = 8 (it may have other immediate predecessors); similarly, if an activity has LST + Duration = 8 (say), then it is an immediate predecessor of an activity with LST = 8 (it may be an immediate predecessor of other activities as well)
Then I calculated:
A: EST = 0, so no immediate predecessors. LST + Duration = 8, so immediate predecessor of F.
B: EST = 0, so no immediate predecessors. LST + Duration = 6, so immediate predecessor of D.
C: EST = 0, so no immediate predecessors. LST + Duration = 14, so immediate predecessor of H.
D: EST = 6, has immediate predecessor B. LST + Duration = 8, so immediate predecessor of F.
E: EST = 6, has immediate predecessor B. LST + Duration = 14, so immediate predecessor of H.
F: EST = 8, has immediate predecessor D. LST + Duration = 14, so immediate predecessor of H.
G: EST = 8, has immediate predecessor D. LST + Duration = 18, so immediate predecessor of J.
H: EST = 14, has immediate predecessors E, F. LST + Duration = 18, so immediate predecessor of J.
I: EST = 14, has immediate predecessor F. LST + Duration = 20, so immediate predecessor of K.
J: EST = 18, has immediate predecessor G. LST + Duration = 20, so immediate predecessor of K.
I: EST = 20, has immediate predecessor J. LST + Duration = 25, so not an immediate predecessor of anything.
This gives the following table of immediate predecessors:
A: –
B: –
C: –
D: B
E: B
F: A, D
G: D
H: C, E, F
I: F
J: G, H
K: I, J.
I then quickly verified that this table of immediate predecessor relations gives the required ESTs and LSTs.
I might even be inclined to argue that this answer is more "natural" since I think it results in each activity having the minimal number of immediate predecessors that is consistent with the given ESTs and LSTs.
Edit: The network that would represent this is the one Anonymous (James, I assume?) posted in a comment above.
Thanks, SRK. This is very interesting.
This material is not natural to me, and the exam question is artificial. So i’m slow with the basic material, and very slow to understand a natural approach to the unnatural exam question. I’ll read carefully what you wrote and what “James” wrote, and reply and/or update the post.
Yeah, that was my comment and diagram earlier.
The question is unnatural and artificial so I wouldn’t say there is a natural approach to it. My approach was just some kind of haphazard mishmash of what both SRK and B did and I stopped when I realised the problem. I didn’t draw that diagram until well after the fact (and yes that diagram should represent the table SRK provided).
Thanks, James.
That is an interesting strategy SRK. Do you use a similar method for finding the critical Path, EST and LST when provided a precedence table which has immediate predecessors and activity durations?
Yes. I teach my students how to find those things without drawing a diagram. (Also how to determine if a dummy activity would be required, and where it would go).
Yes I also teach students how to find dummies from the table but I seem to lose them trying to get them calculate cps, est or lst just from the table and they prefer drawing the network. I’m sure I’m over complicating it.
I’ve updated the post with a summary of the comments, particularly those of James and SRK.
Some final thoughts.
Whilst the problems in this question all involve that fact that we can’t determine whether one non-critical activity is a predecessor to another non-critical activity, there could be others.
First, what if there’s more than one critical path? For example, in this question, change activity A to duration 8 and EST=LST=0. It is now a critical activity and there are two critical paths. G must now be preceded by at least one of A or D, but it’s impossible to know. Make G a critical activity too and there are even more possibilities.
More generally a problem arises from the different “paths” from start to finish. As long as an activity (X) on one path has a latest finish time less than or equal to the earliest starting time of an activity (Y) on a different path before those two paths meet, then it’s impossible to know of X is a predecessor of Y simply from a table of ESTs and LSTs. (I think the problem in this question might be a simple of example of this more general idea, but this isn’t my area either and I’ve already spent enough time thinking on this problem). As an extreme example, imagine a network with two entirely separate paths that meet only at the finish. Given only a table of their times (and assuming the numbers are nice enough that you can find those two separate paths), there would be no way of knowing if earlier activities on one path are predecessors of later activities on the other path.
Thanks again, James. Yes, it is clearly dumb luck or backwards engineering to have the EST/LST info give rise to a unique project. Still, it is an interesting question of how to determine if it is the case. I thought what you and SRK were doing was nice stuff.
All this discussion raises for me an important issue as to why this might not be a good question.
The candidates are writing the examination under a time constraint so there is a certain amount of stress. How certain can you be that a student who understands the situation will not make an honest slip or slight misapprehension that will lead to the wrong answer? How certain can you be that a student cannot arrive at the correct answer by specious reasoning?
How were the detractors chosen? Picked out of a hat, or designed to be diagnostic — if the student chose this detractor, it is likely that he solved the problem in this particular (correct or incorrect) way?
Ed, as an exam question, it is obviously a disaster, even ignoring the fact that it has multiple correct answers.
It is worth noting that the NHT (Northern Hemisphere Timetable) exams are prone to such eccentricities. The exams are only taken by, I think, a few hundred students, at campuses in China and the like. No one puts in the standard (but still inadequate) effort into writing and vetting the exams.
There were (I presume still are) private schools in Melbourne that have their “accelerated” students sit the General (or possibly Methods) NHT exams early, with the other accelerated students and then the Specialist exam together with the non-accelerating students.
An interesting approach when done school-wide.
Trying to find the data on this (even number of enrolments) is difficult.