*** F5 June 2011 Exam was … Post your comments here and vote in Instant Poll ***

What a shocker this paper was. Haven’t seen anything like it my life! That throughput question had way too many figures and too much information it just confused me in the end. It will be interesting to find out whether the pass rate improves or drops for this session. I hope iv’e secured enough marks.

@platinumkp said:
in the payoff tables:

* if demand = production then its just the contribution times the number of bags
* if demand > production then there is no penalty and its just the contribution x the demanded units.
* if demand < production then its the contribution * demand - (unsold * 0.50). Thats how you get the payoff table. Then you can apply the mini max and the maxi max…. after that to get the expected values you just multiply each payoff table cell with the probability of that corresponding demand to get the value of x. Then sum all the productions to get the expected values… you then select the highest expected value out of the production choices. Thats what I got at least. I can’t remember the numbers but I do remember that the contribution from the 200,000 production level was $1m regardless of the level of demand.

Guys i did different when you have unsold if you are using the profit per unit (9-4) multiplied per the demand then you remove the COST OF PRODUCTION OF THE UNSOLD 4*unsold and then 0,5 per unsold (cost of disposal). HOPE I GOT IT RIGHT…

so demand equation is calculated …(735 – 720)/(1000-2000) = -0.015
720 = a -(0.015 * 2000)
720 = a -30
a = 720 + 30
a = 750
P = 750 – 0.015Q………hope I was correct……..

@winnaman123 said:
735 = a + b1000
750 = a+ b850
not sure this is right, value of a is the price when 0 demand, which was 750, as price increased by $15 when demand fell by 1000. Price was 735 at 1000 demand, so 750 is 0 demand. a is 750.

solve simultaneously: a= 838; b= -0.1
therefore p= 835 – 0.1x

Friends:
Demand equation was 50,000-66.66Q.

a negative of 1000demand for each $15 price range gives a gradient of -66.67.
Feeding that into the equation 1000=a-(66.67 x $735) means a is 50,000.

The EV in Q1 – I suspect we only had to get the probability of the 3 numbers given, ie (.45 x 350,000) + (.25 x 280,000) + (.35 x 200,000).

Looking at the blog, I think we are all missing the point. The maths is only 50% of the marks. The blurb about it that you can waffle on about is possibly easier to do, and just as important.

Jude

what variances were recomended to assess the performance of the restaurant manager?

@samygold said:
Guys i did different when you have unsold if you are using the profit per unit (9-4) multiplied per the demand then you remove the COST OF PRODUCTION OF THE UNSOLD 4*unsold and then 0,5 per unsold (cost of disposal). HOPE I GOT IT RIGHT…

I did similar calculations but removed not just 0.5 for disposal but also the profit element relating to unsold goods that would have to be disposed. e.g. if 350,000 units were produced but only 280,000 were sold then it would be 350,000 * 5 = 1, 750, 000 expected profit. and then i would calculate 350,000 – 280,000 = 70,000 unsold items multiplied by 5 profit + and 70,000 *0.5 disposal cost. Therefore :

total profit on 350, 000 production is 1,750,000
profit on unsold 70,000 @ 5 (350,000)
disposal cost on 70,000 @ 0.5 ( 35,000)
actual profit 1,365,000 – goes to payfoff table.

@eimearnid said:

lol

@winnaman123 said:
735 = a + b1000
750 = a + b 850
solve simultaneously: a = 835; b= -0.1

therefore p= 835 – 0.1x

lol

@jude28 said:
Friends:
Demand equation was 50,000-66.66Q.

a negative of 1000demand for each $15 price range gives a gradient of -66.67.
Feeding that into the equation 1000=a-(66.67 x $735) means a is 50,000.

The EV in Q1 – I suspect we only had to get the probability of the 3 numbers given, ie (.45 x 350,000) + (.25 x 280,000) + (.35 x 200,000).
Jude

LOL

I think ACCA really needs to get to grips with these papers. I mean every exam should be attempted by 10 randomly selected markers, and they should be given 2 hours rather than the 3 hr 15 mins given to students, and if all of them dont get at least 60%, then the exam should be re-written.

I know you’re not meant to complete these exams, but just out of interest, is there anyone here who actually completely every part of this exam today?

@kotesh said:
Just wanted to double check that you flexed the budget to the actual sales of meals of 1650?

thats how i did as well so average meals per day is 65 n then i charged rest accordingly

Can anyone else tell me what they did?

I don’t know why you are so angry at ann irons. I think it was a completely predictable paper. In the last two sessions linear programming and ABC were examined so I knew they won’t come up. Instead, I have expected this pay-off stuff or a decision tree and of course throughput accounting.

I studied two days for this paper (Saturday and Sunday) and I found it very straightforward and easy. I couldnt do the flexed budget because the information on the actual food + drink mix was hidden in the question at the bottom of the paper and I couldn’t find it… so -13 marks. But I did parts 🙂 and c). For the additional two variances I wrote revise the budget and calculate planning and operational variances.

Q1: I also needed 5-6 minutes of thinking, what to do with disposal but then I found the right wayy, as some of you already wrote above.
Q2 and Q4: easy, straightforward
Q5: I was afraid of TPA as I could solve no one of them when practising, but this one was quite straighforward and easy.

As I had only two days to learn, I didnt learn any theory, but I could solve the theoretical questions in the paper. (remembered BSC from university, market skimming and penetration were easy, also maximin and expected values).

if you have problems with F5… this is the easiest paper from all (I got exemption for F1-F4 so don’t know how they are), the other are much more diffucult and have a wider syllabus!

this paper was difficult but hope that we all can pass this exam ,as this was a complete transition from the usual.