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Dear Sir,
Reference to the 2nd article written in the Student accountant talking about throughput accounting , and particularly the example (2) of the three machine X,Y and Z.
My question is how did the writer calculate the “Additional sales” for each of the machines X=100, Z&Y =300 and Z,Y & X=500 ?.
Appreciated your quick repsonseHello,
Could you please let me know which SA this is in as I can only find Decision Trees in recent SA articles….
Many Thanks
Hi Lou1,
Actually It was the edition published immediately before the recent one which you mentioned above, as I remember it was published in March 2013.
I hope you will get it .since, there is a great chance to be in the June exam.
Go over it and let me know if you understand the calculation of Example(2)
Regards
At the moment they can only produce 500 units in total (because Z is the bottleneck).
If they upgrade Z, then although Z would be capable of producing a lot more, Y becomes the bottleneck because Y can only deal with 600 units which means that only 600 units in total can be produced.
However this is an extra 100 units and so would generate more money.
If they carried on an also upgraded Y, then Y could produce a lot more, but X now becomes the bottleneck and because it can only handle 800 units, they can only produce 800 units in total (but this is 300 more than the present 500 units).
Finally, if they also upgrade X, then now although they could produce 1050 in total (Z being the one that again now limits production), there is in fact no bottleneck at all because the maximum demand is 100 units and they can now produce them all.
At the moment they can only produce and sell 500 units, and so upgrading all three means that they can now produce and sell 1000 units.
Dear John,
Many thanks for your quick response, actually this was my first participation in this helpful website, I really appreciate that.
Now I got the idea, I think it is tricky, really the word ‘In total” reveal the mystery.Thanks a lot.
Regards,
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