Lecture 1: Basics of Math and Economics

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Lecture 6:
Algorithm Approach to LP Soln
AGEC 352
Fall 2012 – Sep 12
R. Keeney
Linear Programming

Corner Point Identification
◦ Solution must occur at a corner point
◦ Solve for all corners and find the best solution

What if there were many (thousands) of
corner points?
◦ Want a way to intelligently identify candidate
corner points and check when we have found
the best

Simplex Algorithm does this…
Assigned Reading

5 page handout posted on the class
website
◦ Spreadsheet that goes with the handout

Lecture today will point out the most
important items from that handout
Problem Setup
Let:
C = corn production (measured in acres)
B = soybean production (measured in acres)
The decision maker has the following limited resources:
320 acres of land
20,000 dollars in cash
19,200 bushels of storage
The decision maker wants to maximize profits and
estimates the following per acre net returns:
C = $60 per acre
B = $90 per acre
Problem Setup (cont)
The two crops the decision maker produces use limited
resources at the following per acre rates:
Resource
Land
Cash
Storage
Corn
1
50
100
Soybeans
1
100
40
Algebraic Form of Problem
max   60 C  90 B
s .t .
land : C  B  320
cash : 50 C  100 B  20 , 000
storage : 100 C  40 B  19 , 200
non  neg : C  0 ; B  0
Problem Setup in Simplex

Note the correspondence between
algebraic form and rows/columns
Initial Tableau
C
land
cash
stor
obj
B
1
50
100
-60
s1
1
100
40
-90
s2
1
0
0
0
s3
0
1
0
0
P
0
0
1
0
RHS
0
320
0 20000
0 19200
1
0
Simplex

Procedure: Perform some algebra that is
consistent with equation manipulation
◦ Multiply by a constant
◦ Add/subtract a value from both sides of an
equation

Goal: Each activity column to have one cell
with a 1 and the rest of its cells with 0

Result: A solution to the LP can be read
from the manipulated tableau
Simplex Steps
The simplex conversion steps are as follows:
1) Identify the pivot column: the column with the most
negative element in the objective row.
2) Identify the pivot cell in that column: the cell with
the smallest RHS/column value.
3) Convert the pivot cell to a value of 1 by dividing the
entire row by the coefficient in the pivot cell.
4) Convert all other elements of the pivot column to 0
by adding a multiple of the pivot row to that row.
Step 1

B has the most negative ‘obj’ coefficient
◦ Most profitable activity
Initial Tableau
C
land
cash
stor
obj
B
1
50
100
-60
s1
1
100
40
-90
s2
1
0
0
0
s3
0
1
0
0
P
0
0
1
0
RHS
0
320
0 20000
0 19200
1
0
Step 2
ID pivot cell: Divide RHS by elements in B
column
 Most limiting resource identifcation

Initial Tableau
C
land
cash
stor
obj
B
1
50
100
-60
s1
1
100
40
-90
s2
1
0
0
0
s3
0
1
0
0
P
0
0
1
0
RHS
0 320
0 20000
0 19200
1
0
320/1
20K/100
19200/40
Step 3

Convert pivot cell to value of 1 (*1/100)
Initial Tableau
C
land
cash
stor
obj
B
1
50
100
-60
C
new cash row
s1
1
100
40
-90
B
0.5
s2
1
0
0
0
s1
1
s3
0
1
0
0
s2
0
P
0
0
1
0
s3
0.01
RHS
0
320
0 20000
0 19200
1
0
P
0
RHS
0
200
Step 4
Convert other elements of pivot col to 0,
by multiplying the new cash row and
adding to the other rows
 Multiplying factor for land row

◦ -1 (-1*1 + 1 = 0)

Multiplying factor for stor row
◦ -40 (-40*1 + 40 = 0)
Example
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