Lecture 6:
Algorithm Approach to LP Soln
AGEC 352
Fall 2012 – Sep 12
R. Keeney
Lab Discussion Questions

Opportunity cost of specialization
Total Gross
Corner Point Chairs Tables
Margin
1
45
0
2025
2
0
20
1600
3
24
14
2200
GM/unit
45
80


2200 – 2025 = 175 = Op cost chair spec
2200/45 = ? <- GM/chair required for spec to be equally
profitable
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   60C  90 B
s.t.
land : C  B  320
cash : 50C  100 B  20,000
storage : 100C  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
C
B
s1
s2
s3
P
RHS
New cash
row
0.5
1
0
0.01
0
0
200
New cash
row *-1
-0.5
-1
0
-0.01
0
0
-200
Old land
row
1
1
1
0
0
0
320
New land
row
0.5
0
1
-0.01
0
0
120
Explanation of rows:
New Cash Row: Multiply all elements by 1/100
New Cash Row *-1: Multiply new cash row by -1
Old Land Row: From initial tableau
New Land Row: Add New cash row*-1 to old land row
Tableau after 1st iteration
C
land
cash
stor
obj
B
0.5
0.5
80
-15
s1
0
1
0
0
s2 s3
1 -0.01
0 0.01
0 -0.4
0 0.9
P
0
0
1
0
RHS
0 120
0 200
0 11200
1 18000
IF all negative values are eliminated from obj row, we are
done
If negative values remain, repeat the four Simplex steps
1) New pivot column is C, …(see handout)
Tableau after 2nd iteration of
Simplex
Final Tableau
C
land
cash
stor
obj
B
0
0
1
0
s1
0
1
0
0
s2 s3
P
1 -0.008 -0.00625
0 0.013 -0.00625
0 -0.005 0.0125
0 0.825 0.1875
RHS
0
50
0 130
0 140
1 20100
No negatives in obj row, so we are done.
Solution values:
Look under the activity column, find the one and read the
corresponding RHS value: C = 140, B = 130,
Slack Land = 50; P=Profit=20,100
Simplex
Programmed into Excel as one of the
solution options
 When we begin LP, we will make sure that
we are choosing the simplex method
 Time consuming by hand but we need to
understand how solutions are generated
by the computer
 Key points for exam:

◦ Steps, reading a solution
Next week

Solving LP in Excel
◦ Standard problem setup
◦ Solution elements
◦ Interpretation of plan and evaluation of how
sensitive the best plan is to changed
assumptions