Lab Handout 2
AGEC 352
January 31, 2012
Problem Setup for Linear Programming and Technical Analysis of Constraints
Part I:
Letterman’s Furniture Company produces two outputs, tables and chairs using two inputs
labor and lumber. The gross margin (revenue above variable cost) for producing a table is $
80.00 per table and the gross margin for chairs is $ 45.00 per chair. Letterman’s wants to
maximize the gross margin from chair and table production combined given the lumber they
have in current stock. They also want to limit each of their 15 employees to 30 hours of
work for the week. Historical records for the company show that producing a table requires
on average 20 board feet (b.f.) of lumber and 15 hours of labor. The records indicate that
producing a chair requires less labor and lumber, but that chair production is more labor
intensive. On average, a chair requires 5 b.f. of lumber and 10 hours of labor. The company
has 400 b.f. of lumber available to use in production. Our task is to take the above
information and recommend to Letterman’s Furniture Company a plan of production that
indicates the numbers of tables and chairs to produce that meets their objective.
This handout outlines each step in the process of setting up an LP problem. The
following text demonstrates via a set of tables a way of outlining the problem. We will use
the outline in writing the algebraic problem as well as a reference for any abbreviations we
might use for variable names.
Objective Variable and Goal
Objective:
Total Gross Margin GM
Goal:
Maximize
Activities and Coefficients of Objective Equation (GM per unit of output)
Chairs produced
C
@ $45 GM per chair
Tables produced
T
@ $80 GM per table
Resources
Lumber:
400 b.f. in stock
Labor:
450 hours available for the week
Activity Requirements of Resources (Units of Resource per unit of Activity)
Labor:
10 hours per chair
15 hours per table
Lumber:
5 b.f. per chair
20 b.f. per table
Non-negativity of Activities
C >= 0
T >= 0
Standard Algebraic Form of the LP Problem
max GM  45C  80T
subject to :
Labor : 10C  15T  450
Lumber : 5C  20T  400
Non  neg . : C  0; T  0
The key part of the algebraic form is recognizing that C, T, and GM are unknowns
and can take on any values allowed by the constraints. Thus, it is appropriate to think of the
equations and inequalities of an LP as an accounting system which you would use to find
information about a prospective C and T combination such as C = 2 and T =1 is 1) feasible,
2) worth a total GM of $170, 3) uses 35 hours of labor leaving 415 for other uses, and 4)
uses 30 board feet of lumber leaving 370 for other uses. It is only when we add the
statement max at the beginning that we see that we are looking for a specific combination in
this accounting system, the one that produces the largest value for GM.
We are going to use a spreadsheet to analyze the Tables and Chairs problem for a
best solution by identifying the corner points in a manner consistent with lecture of January
30. Download and open the spreadsheet Lab_2_Part_1.xlsx now for use in the remainder of
part I.
Labor
Lumber
Per Chair Per Table Total Available
10
15
450
5
20
400
Most Limiting Factor Labor
?
Max Chairs Max Tables
Lumber
?
The top of the spreadsheet is given above and we will use this section of the spreadsheet to
identify the most limiting factor for each output. Under the columns Max Chairs and Max
Tables we want to calculate the maximum number of each product that we could produce
considering only one constraint at a time. For example, under Max Chairs for Labor, we
would have 450/10 = 45. Using a cell formula in F4 this would be entered as =D4/B4.
Complete the table and identify the most limiting factor for each output. At the bottom of
this section of the spreadsheet you should enter the corner points in the table provided.
Recall that the most limiting factor is the one that has the lowest maximum output for a
given product. Also, we can use the most limiting factor analysis to identify two corner
points that intersect with non-negativity constraints that define the axes of the feasible space.
You should find that labor is the most limiting factor for chair production and the maximum
value is 45. This means we use up all the labor in chair production and make zero tables so
we have a corner point at (C,T) = (45, 0). Enter this into the spreadsheet section labeled
corner point in the first row. Calculate the other corner point from the most limiting factor
analysis and enter it into row 2 of that section.
90
80
70
60
50
40
30
20
10
0
0
5
10
15
20
Labor
25
30
35
Lumber
The graph above shows the intersection of the labor and lumber constraints for this
problem. Since they cross on the boundary of the feasible space, we need to calculate the
location of this point to complete our analysis. Rows 13 through 18 are set up to do this in
the Excel spreadsheet you have for this week.
Variable Level
Labor
Lumber
Chairs
0
LHS
Tables
0
RHS
Difference
0
450
-450
0
400
-400
-850 <-Total Difference
The cells of the spreadsheet are shown above and represent the equation system for the
labor and lumber constraint with chairs and tables as unknowns. In the first row, we have
zeroes to the right of the row heading ‘Variable Level.’ These cells are the current values of
the variables C and T and will be adjusted to find a solution for the corner point. Since Excel
doesn’t allow us to program equations we use formulas in cells that hold the result of an
equation. Considering the labor constraint which in equation form looks like
10C + 15T = 450
we can use the spreadsheet to calculate for any combination of C and T the left-hand-side
(LHS) of this equation. In the second column of the spreadsheet this is done for you using
the sumproduct formula. Open that formula and you will see
=SUMPRODUCT(B4:C4,B14:C14).
This is the spreadsheet equivalent of the LHS expression of the equation or of doing the
longer formula
= B4*B14 + C4*C14.
The SUMPRODUCT formula literally sums up a bunch of products. You can observe how
the SUMPRODUCT is working by changing the variable levels under Chairs and Tables
from zero to one. This should change the value in LHS for Labor to 25 indicating that if we
make one unit of each product we will use 10 hours on the chair and 15 hours on the table.
The column RHS (right-hand-side) is the upper limit on labor and lumber use and the
column difference calculates the difference between the LHS and RHS. Only when we have
zeroes in the column ‘Difference’ do we have an equation that holds. The last cell I have in
this section of the spreadsheet is a total difference which adds up the difference from the
labor and lumber comparisons of RHS and LHS.
We are going to use Solver to find the corner point in this problem so we first need to check
and see if Solver is installed.
Checking to see if Solver is installed
Click on the ‘Data’ tab at the top of the Excel screen. If Solver is installed it will appear
at the right of the screen as part of the ‘Analysis’ group of tools. If Solver does not appear
there, we will need to add it to your installation of MS Excel.
Installing Solver into Excel (only required if it is not present)
To install Solver complete the following steps:
1)
2)
3)
4)
Click on the Office button at the top left of the Excel program screen.
Click on Excel Options at the bottom of the Office button box that appears.
Choose Add-Ins in the left hand side of the Excel Options page
Highlight ‘Analysis ToolPak’ in the Add-ins list, then click the ‘Go…’ button at
the bottom of the page
5) In the Add-Ins choices that appear, click the box next to ‘Solver Add-in’ and
click OK
The correct setup for Solver should already be entered when you open solver. Make sure
it looks like the graphic on the next page and examine each item to understand what each
part means. In the ‘Set Objective’ area I have chosen D18 as the cell I want to minimize
(see the selection area directly below the objective window). As the ‘By Changing
Variables’ I have set B14 and C14. This means that I want to minimize the total
difference between the RHS and LHS of the two constraint equations by changing the
values of Chairs and Tables to produce. The area ‘Subject to the Constraints’ says that the
RHS and LHS differences need to be exactly equal to zero for both equations and the
check box right below that area indicates that C and T should be non-negative (i.e. >= 0).
After looking this section over you can click on ‘Solve’ and the values of C and T in
your spreadsheet should adjust to the solution values where the labor and lumber
constraints intersect. Just click OK on the box that appears afterward.
Enter the solution values for Table and Chairs as the 3rd corner point in the table at the
bottom of the spreadsheet.
At this point you should have all of the question marks filled in for rows 24 through 26
with 3 different corner points that need to be checked for the best solution.
Corner Point
GM/unit
Chairs
1
?
2
?
3
?
Tables
?
?
?
45
Total Gross Margin
80
The last column in this area is labeled Total Gross Margin and will be used to compare
the corner points in terms of what they are worth to the decision maker. The last row of
this section gives the gross margin per unit so we just need to enter formulas in the last
column for each corner point that calculates the total. The formula for the first corner
point (D24) should be =SUMPRODUCT(B27:C27,B24:C24). Enter a similar formula for
the other two corner points and compare the values in the Total Gross Margin to column
to find the best solution.
This completes part I of the assignment. You can now go to the discussion board for
this lab.
Part II
Springer’s Farm Problem
In part II we are going to be redoing the steps in part I for a different problem.
Refer back to part I for assistance in completing the questions 1-6.
Springer’s Farm has two output activities and three primary resources. Mr. Springer
needs to decide how much of each crop to produce given the limits of his resources in order
to maximize his farm’s total gross margin. The two outputs are corn and oats. His three
limiting resources are land, labor, and capital. The units of output for each cropping activity
is an acre and Mr. Springer’s historical records indicate that one acre of each of his crops
produce the following gross margins: $ 40 per acre of corn and $20 per acre of oats.
Obviously producing one acre of any of the crops requires the use of one acre of Springer’s
farm land. Springer’s farm is 1200 acres. Springer has 4800 hours of labor for use on his
small farm. Production of an acre of corn uses 6 hours of his labor time, while production of
oats will only use 2 hours of his labor time. Capital is measured in dollars and Mr. Springer
has a limit of $36,000 for his farm. An acre of corn production will use $36 of his capital and
an acre of oats will use $18 of his capital. Mr. Springer does not require that both outputs be
produced and is not concerned with rotational issues.
1)
2)
Write the algebraic form for Springer’s farm problem. You can set up a series of
tables that outline the pieces of the problem as was done in part I of this lab if you
wish but it is not required. It may help me see where you made errors if you include
it though.
What is the most limiting factor for corn? What is the most limiting factor for oats?
3)
Graph the feasible space for the Springer’s Farm problem using Excel. Label the
feasible space and copy and paste graph into your assignment.
4)
What are the corner points of the feasible space that need to be checked?
5)
What is the optimal production plan based on your analysis of the corner points?
What total GM is earned from this?
6)
In the optimal plan you found, which resources were completely used up and which
were not?