Transcript Piraeus, Greece - Olive Yield: Pruning vs. non

Problem 7-26 Presentation
Piraeus, Greece - Olive Yield: Pruning vs. nonPruning Methods
Pruning Yield
More trees per acre
Increased output, smaller
olives
One barrel = 5 hours labor
on 1 acre
One barrel sells for $20
Non-pruning Yield
One barrel = 2 hours
labor on 2 acres
One barrel sells for $30
Grower will produce no more than 40 barrels of “pruned” olives.
Grower has 250 hours of labor available.
Grower has 150 acres available for growing.
Jan White - BUS340
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Use graphical Linear Programming to find:
• The maximum possible profit
• The best combination of barrels of pruned
and regular olives
• The number of acres that the olive grower
should devote to each growing process
Jan White - BUS340
2
Equations: Labor Hours and Acres
Available for Use
Labor Hours:
Pruned Olives Hours = 5P
5P + 2U <= 250
Unpruned Olives Hours = 2U
Acres Available:
Pruned Olives Acres = 1P
1P + 2U <= 150
Unpruned Olives Acres = 2U
Other Constraints:
P<= 40 barrels
Jan White - BUS340
3
Equations and Solutions
Labor:
When U = 0
When P = 0
5P + 2(0) = 250, so 5P = 250, then P = 50
5(0) + 2U = 250, so 2U = 250, then U = 125
Acres:
When U = 0
When P = 0
1P + 2(0) = 150, then P = 150
1(0) + 2U = 150, so 2U = 150, then U = 75
Use the points derived from the above equations to graph
the lines for labor hours and acres and to display the
“Feasible Region”
Jan White - BUS340
4
Graph of the Equations and “Feasible
Region”
Pruned Olives
“P” axis
To find the coordinates of the point
of maximum profit, the constraint
equations must be solved
simultaneously. The equations are:
5P + 2U <= 250, and
1P + 2U <= 150
(150, 0)
100
“Corner Point”, the possible
point of maximum profit within
the “Feasible Region” as outlined
by the constraint equations
(50,0)
Feasible
Region
(0,125)
10
10
(0, 75)
100
Jan White - BUS340
Unpruned Olives
“U” axis
(Not to Scale)
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Solving to find the solutions for maximum profit,
barrels of olives to be produced and amount of
acreage to be allocated for growing methods
To isolate one variable (P) for solution, multiply the second equation by -1, so
-1(1P + 2U = 150), which yields -P - 2U = -150. This equation is then added to the first.
-P - 2U = -150
5P + 2U = 250
which solves as 4P = 100, so P = 25. This solution for P is now
substituted back into the first equation to solve for U.
5(25) + 2U = 250, so 125 + 2U = 250, 2U = 125. U = 62.5
These solutions give you the coordinates for maximum profit, (25, 62.5) which is the
number of barrels of each type of olive to be grown (25 pruned olives and 62.5 unpruned).
Maximum profit (the Objective) can be solved (z = 20P + 30U) by substituting in the number
of barrels of each, so 20(25) + 30(62.5) = z. So, z = $500 + $1875 = $2375
The amount of acreage to be used for each growth method can be solved as
1(25) + 2(62.5) <= 150, so 25 + 125 <= 150. This is true, so 25 acres will be used for pruned
olives, and 125 acres for unpruned olives.
Jan White - BUS340
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