Problem 5-30 Graphing; Incremental Analysis; Operating Leverage [LO5-2, LO5-4, LO5-5, LO5-6, LO5-8][The following information ap
1 answer:
Answer:
Break Even Point
In Units = 2,000 units
In value = $80,000
Explanation:
Break even Point =
When we use contribution per unit, we get the break even point in units sales.
When we use the contribution margin as a percentage of sales we get break even sales in value.
Contribution per unit = $20
Contribution margin in percentage = $20/$40 = 50%
Therefore, Break even Point in units =
Break even units = 2,000
Break Even Point in value =
Sales to be made in value at break even = $80,000
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Answer:
210
Explanation:
Let us consider that x is the number of soldiers produced each week and y is number of trains produced each week.
Also, weekly revenues and costs can be expressed in terms of the decision variables x and y.
Then,
Hence the profit which we want to maximize is given by,
Now the constraints are given as,
Finishing Constraint:
Each week, no more than 100 hours of finishing time may be used.
Carpentry Constraint:
Each week, no more than 80 hours of carpentry time may be used.
Demand Constraint:
Because of limited demand, at most 40 soldiers should be produced each week.
Combining the sign restrictions and with the objective function and constraints,and yield the following optimization model:
Such that,
First convert the given inequalities into equalities:
From equation (1):
If x=0 in equation (1) then (0,100)
If y=0 in equation (1) then (50,0)
From equation (2):
If x=0 in equation (2) then (0,80)
If y=0 in equation (2) then (80,0)
From equation (3):
Equation (3) is the line passing through the point x=40.
Therefore, the given LPP has a feasible solution first image
The optimum solution for the given LPP is obtained as follows in the second image
The optimal solution to this problem is,
And the optimum values are .
Let c be the contribution to profit by each train. We need to find the values of c for which the current, basis remain optimal. Currently c is 2, and each iso-profit line has the form
3x + 2y = constant
y = 3x/2 +constant/ 2
And so, each iso-profit line has a slope of .
From the graph we can see that if a change in c causes the isoprofit lines to be flatter than the carpentry constraint, then the optimal solution will change from the current optimal solution to a new optimal solution, If the profit for each train is c, the slope of each isoprofit line will be.
-3/c
Because the slope of the carpentry constraint is –1, the isoprofit lines will be flatter than the carpentry constraint.
If,
-3/c<-1
c >3
and the current basis will no longer be optimal. The new optimal solution will be point A of the graph.
If the is oprofit lines are steeper than the finishing constraint, then the optimal solution will change from point B to point C. The slope of the finishing constraint is –2.
If,
-3/c < -2 or
C < 1.5
Then the current basis is no longer optimal and point C,(40,20), will be optimal. Hence when the contribution to the profit for trains is between $1.50 and $3, the current basis remains optimal.
Again, consider the contribution to the profit for trains is $2.50, then the decision variables remain the same since the contribution to the profit for trains is between $1.50 and $3. And the optimal solution is given by,
z = 3× (20) + 2.5 × (60)
= 60 + 150
= 210
