- This topic has 5 replies, 2 voices, and was last updated 11 months ago by
.
- Posts
The following observations of output and cost have been made:
Output Cost
Units $
8,000 39,400
20,000 68,000
It is known that at output levels above 15,000 units, variable cost per unit drops by $1 per unit for all
subsequent units produced.
What is the variable cost for each unit of output above 15,000 units?Hello professor, how to do this sum?
The following shows the total overhead costs for given levels of a company’s total output.
Cost Output
$ Units
4,000 1,000
7,000 2,000
10,000 3,000
9,500 4,000
A step up in fixed costs of $500 occurs at an output level of 3,500 units.
What would be the variable overhead cost per unit (to the nearest $0.01) using the high-low technique?Also i am quite confused with this sum too.
please help sir!!For the second question the difference in cost between the highest and lowest levels of output is $5,500.
$500 of this is due to the increase in the fixed costs, so only the remaining $5,000 is due to the variable cost.
So the variable cost per unit is 5,000 / 3,000 = $1.67 per unit
For the first question, if the variable cost per unit for 8,000 units was the same as for 20,000 units, then the total cost for 8,000 units would be 39,400 – (8,000 x $1) = $31,400.
Using the normal high low approach on costs of 31,400 and 68,000 will then give you the variable cost per unit at the higher level.
sir in the second question there’s a step up in fixed COST so sir, why do we subtract it from output? very confused sir in the high low method.
I did not subtract it from output!!
The total cost at each level of output is the total of the variable cost and the fixed cost.
If there was no step-up in the fixed cost then the difference in the two costs would be just the total variable cost of the extra units.
Since there is a step-up of $500 then this is the reason for $500 of the difference in the two costs so only the rest of the difference is the total variable cost of the extra units.
- You must be logged in to reply to this topic.