Ocr mei decision 1 maths exam -- 6th june 2013

I found the price increase different:

Maximise 40x+50y

The optimal solution before was (100, 500) to have a profit of €29,000.

I then used another point, and found the difference in profit to be €1,500.
But I figured that if you increased the prices of snowboards, then the profit of the original point also increased. Eventually they joined at €30,000 (difference of €1,000) so I said that it required snowboards to be increased in price by (at least) €10 (to change the optimal solution)

Could be interpreted either way I guess, but it sure isn't 6 euros

Fairly sure I put 35 (285 - 250)

My attempt at Q5, since it's the only one uploaded so far:



Therefore, the greatest number of extra snowboards it is worth Angelo accomodating is 36. (Thinking about this, I'm not sure. I put 35 in the exam, but 36 seems to make more sense now. Needs clarification!)

I guessed 35, so would love if that was the correct answer. (I just used my graph to find the points of intersection, between the 2 lines)

I mean, why would you round up to 36, you cannot rent 0.785 of a snowboard?

Well, with 35 snowboards the point of intersection of the two lines hasn't actually been reached - the limit is still the number of snowboards. By using 36 snowboards, the limit then becomes this point of intersection, so the profit is greater than when 35 is used. Anything greater than this is pointless though.

My hope would be that they would accept using your graph as the "working out" and if they see an answer between a range, it will be correct. But I don't know, I hated this exam, any shiny marks I can gain will be a good thing.

Eh, it won't be too many marks lost. Grade boundaries will be low. Lowest for quite a while I imagine.

My attempt at Q5, since it's the only one uploaded so far:

i)
Let $x$ be the number of snowbards
Let $y$ be the number of pairs of skis

$x \leq 250, y \leq 500$
$10y \geq 11x$
$x+y \leq 600$

The graph can be found here . There are three points of intersection, though one can be ignored since it will produce a lower profit anyway, one lies on $x=250$, one on the line $y=500$.

ii) Objective Function: $P=40x+50y$

P₁:$y=500, x+y=600 => x=100$
P₁$=40 \times 100+50 \times 500=29000$

P₂:$x=250, x+y=600 => y=350$
P₂$=40 \times 250+50 \times 350=27500$

Therefore, solution which will give max profit is 500 skis, 100 snowboards.

iii)

To change the optimal solution, P₂>P₁

Lets define the new profit as $s$

So,

$250s+350 \times 50>100s+50 \times 500$
$150s > 7500$

The minimum value for s is 50, therefore the profits will have to rise by 10 (since the profit was 40 before).

iv)

Taking a look at the graph, we can see that there is a point at which increasing the number of extra snowboards becomes pointless, because the point of intersection no longer lies on this line. The point at which this occurs is when the two lines

$10y=11x$
$x+y=600$

meet.

Solving,

$10x+10y=6000$
$21x=6000$
$x=285.714...$

Therefore, the greatest number of extra snowboards it is worth Angelo accomodating is 36. (Thinking about this, I'm not sure. I put 35 in the exam, but 36 seems to make more sense now. Needs clarification!)

hey, what did everyone get as the value of the last node on the dijkstras question? I think I got 52?

How I found it:

Graph Theory (1) last part - where it asks you to repeat the process for a different map. I wasn't quite sure if the second graph you had to draw was planar like the first bit - I didn't in the end and crossed arcs.

Algorithms (2) - This didn't seem TOO bad to me, long as you understood what it was asking before you ploughed on ahead trying to complete it. Basically, it was a bubble-sort algorithm. I think there were 6 numbers needed to be sorted? Either way, I remember every pass I had to do had 2 swaps (except the last one, no swaps), so it resulted in 15 comparisons and 8 swaps in total.

Networks (3) - The application of Dijkstra's algorithm was fairly standard and straightforward I thought, and so was the analysis of the shortest routes. I did like the part about how to adapt the algorithm, I wonder if others thought about converting the 'distance' weights to 'time to travel'. About how to adapt the node C, I tried to describe inserting a complete network in that point to connect each junction with each other at C, each with a weight of 20 (minutes, or 1/3 hour). No idea if this was a wise thing to do...!

Critical Path Analysis (4) - Oooh, i hate these. EVENTUALLY, I managed to draw an activty on arc diagram... Thankfully, the forward and backward pass seemed fairly simple, as was identifying critical activities. Thankfully no question on floats...! I found the earliest finish time to be 100 minutes. Oh, and I love cascade diagrams. That part also seemed average.

Linear Programming (5) - Skis (y) and snowboards (x). I never had to do this when I went to Austria, that's for sure. The relationships themselves weren't that bad, I remember x+y≤600 and 1.1y≥x, along with x≤350(?) and y≤500. The last half really got to me, however. I somehow found that it would have to be an increase over €10 to change the optimal point. I eventually got to two conflicting answers, and I plumped for what is now evident to be definitely wrong.

Simulation (6) - This was a bit tricky.. It seemed so simple at first, use a few rules to figure out a few times per person and do 10 simulations. The bits that required thought: Having a cumulative arrival time table, because the values you were simulating were the time intervals between people arriving; then the last but one part where you have to consider waiting times for the two events (shopping(?) bit and the paying bit), and having to add times on and carry them through; The last part seemed a bit empty to me, I only had one action to take in my simulation. Probably made a few slips here and there.


All in all, blegh. Thank god I can now stick to purity and mechanics. S2 was a nice relief to have after that.
Grade Boundaries? D1 usually tends to be quite low - I don't think this one will be different.

My attempt at Q5, since it's the only one uploaded so far:

i)
Let $x$ be the number of snowbards
Let $y$ be the number of pairs of skis

$x \leq 250, y \leq 500$
$10y \geq 11x$
$x+y \leq 600$

The graph can be found here . There are three points of intersection, though one can be ignored since it will produce a lower profit anyway, one lies on $x=250$, one on the line $y=500$.

ii) Objective Function: $P=40x+50y$

P₁:$y=500, x+y=600 => x=100$
P₁$=40 \times 100+50 \times 500=29000$

P₂:$x=250, x+y=600 => y=350$
P₂$=40 \times 250+50 \times 350=27500$

Therefore, solution which will give max profit is 500 skis, 100 snowboards.

iii)

To change the optimal solution, P₂>P₁

Lets define the new profit as $s$

So,

$250s+350 \times 50>100s+50 \times 500$
$150s > 7500$

The minimum value for s is 50, therefore the profits will have to rise by 10 (since the profit was 40 before).

iv)

Taking a look at the graph, we can see that there is a point at which increasing the number of extra snowboards becomes pointless, because the point of intersection no longer lies on this line. The point at which this occurs is when the two lines

$10y=11x$
$x+y=600$

meet.

Solving,

$10x+10y=6000$
$21x=6000$
$x=285.714...$

Therefore, the greatest number of extra snowboards it is worth Angelo accomodating is 36. (Thinking about this, I'm not sure. I put 35 in the exam, but 36 seems to make more sense now. Needs clarification!)

thank you, that is what I was looking for, I put it down as 6 euros but I then realised that changed the 29,000 euro point aswell, thought I was going to have to do something with limits or w/e, but I only noticed about 2 minutes from the end when I was going over it :/