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Et

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Active 12 years ago
  • Topics: 2
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Viewing 2 posts - 1 through 2 (of 2 total)
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  • May 28, 2012 at 12:09 am #97444
    f09c3499849bc5718199f6b704f1eece2757d9c968e650928051cec876af5ab9 80Et
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    ACCA has now removed the Exam questions and answers prior to 2007. The only past exam papers I could see on the site is the ones after 2007.That might be why the link is not active any more.

    June 4, 2011 at 11:14 am #82932
    f09c3499849bc5718199f6b704f1eece2757d9c968e650928051cec876af5ab9 80Et
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    I found the following note by alkemist useful
    alkemist https://opentuition.com/groups/f5-performance-management/forum/topic/iso-contribution-line/#post-11320’>said:
    You need to know ISO contribution lines. It isn’t complicated and can be exmained in paper F5. That being said……

    ISO contribution line is the line given by the solution of a set maximum profit. The objective is to find the gradient of the line for which profit is maximised and hence will give the optimal solution for profit. I am sure there are other postings there (I think I did one as well) but I will give a synopsis here.After finding the boundaries giving the feasibility area of a graph, one has to find the optimal solution.Let us assume product X costs $100 and product Y costs $200. Set a sales amount you wish to make…1. sales of $100002. sales of $50000In case 1., $10000 = 100X (when no Y is sold) and $10000 = 200Y (when no X is sold). The solution is thus when Y=0, X=100 and when X=0, Y=50. Plot the two points on the graph and draw a straight line between them. The equation of the line is 100X + 200Y = 10000 (gradient = -100/200 = -0.5)In case 2., $50000 = 100X (when no Y is sold) and $50000 = 200Y (when no X is sold). The solution is thus when Y=0, X=500 and when X=0, Y=250. Plot the two points on the graph and draw a straight line between them. The equation of the line is 100X + 200Y = 50000 (gradient = -100/200 = -0.5) In both cases, the gradients are the same so the line are called iso-profit lines. This gradient represents the gradient of the line on which the optimal point is found……so no matter what profit figure you chose, by sinply moving the line parallel up the graph drawn, the optimal point will be found as the last point in the feasibility area which falls on the line being drawn

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