IE230 Operations Research I ::::: Homework #5
due Wednesday, 22 March 2006
1. Linear Programming Model Formulation: Brady Corporation produces cabinets. Each
week, they require 90,000 cu ft of processed lumber. They may obtain processed lumber in
two ways. First, they may purchase lumber from an outside supplier and then dry it at their
kiln. Second, they may chop down logs on their own land, cut them into lumber at their
sawmill, and finally dry the lumber at their kiln.
Brady can purchase grade 1 or grade 2 lumber. Grade 1 lumber costs $3 per cu ft (cubic
foot) and when dried yields 0.7 cu ft of useful lumber. Grade 2 lumber costs $7 per cu ft and
when dried yields 0.9 cu ft of useful lumber. It costs the company $3 per cu ft to chop down
a log. After being cut and dried, one cubic foot of log yields 0.8 cu ft of lumber. Brady
incurs costs of $4 per cu ft of lumber that is dried in their kiln.
It costs $2.50 per cu ft of logs sent through their sawmill. Each week, the sawmill can
process up to 35,000 cu ft of lumber. Each week, up to 40,000 cu ft of grade 1 lumber and up
to 60,000 cu ft of grade 2 lumber can be purchased.
Each week, 40 hours of time are available for drying lumber. the time it takes to dry 1
cubic foot of grade 1 lumber, grade 2 lumber, or logs is as follows:
grade 1: 2 seconds
grade 2: 0.8 seconds
log: 1.3 seconds
a. Formulate an LP to help Brady minimize the weekly cost of meeting the demand for
processed lumber. Be sure to state precisely the definitions of your decision variables, and
briefly (e.g., “sawmill capacity”) explain the purpose of each type of constraint.
Grade 1
Grade 2
Logs from their Capacity
own land
Cost
3$
7$
0
0
Useful lumber
0.7 cu. ft.
0.9 cu. ft
0.8 cu ft
90000 cu. Ft
Hours
2 sec
0.8 sec
1.3 sec
144000 sec
Chop down
0
0
3$
0
Dry
4$
4$
4$
0
Sawmill
0
0
2.5$
35000 cu. ft
Capacity
40000 cu ft
60000 cu ft
0
X1 – The quantity (cu.ft) of grade 1
X2 – The quantity (cu.ft) of grade 2
X3 – The quantity (cu.ft) of logs from their own land
Min z = 7*x1 + 11*x2 + 9.5*x3
0.7*x1 + 0.9*x2 + 0.8*x3 >= 90000
2*x1 + 0.8*x2 + 1.3*x3 <= 144000
X1
<=40000
X2
<=60000
X3<=35000
xi >=0, (i = 1, 2, 3)
IE230 HW#5
2/17/2016
page 1 of 4
Rows=
6 Vars=
3 No. integer vars=
0 ( all are linear)
Nonzeros=
17 Constraint nonz=
9(
3 are +- 1) Density=0.708
Smallest and largest elements in abs value=
0.700000
144000.
No. < :
4 No. =:
0 No. > :
1, Obj=MIN, GUBs <=
3
Single cols=
0
Optimal solution found at step:
Objective value:
Variable
X1
X2
X3
Row
1
2
3
4
5
6
1
1033585.
Value
40000.00
55471.70
15094.34
Slack or Surplus
1033585.
0.0000000
0.0000000
0.0000000
4528.302
19905.66
Reduced Cost
0.0000000
0.0000000
0.0000000
Dual Price
1.000000
-12.64151
0.4716981
0.9056604
0.0000000
0.0000000
b. Solve the problem, using LINGO, LINDO or other LP solver.
c. State (in words) the optimal solution in such a way that the manager of this company
could understand it.
2. Consider the following LP:
M inimize z = 3x1 + x2
s.t. x1 - 2x2  2
x1 + x2  3
x1 0, x2 0
a. Convert the constraints (excluding the nonnegativity constraints) into equations, by
the use of "surplus" variables.
Minimize z = 3x1 +x2
s.t. x1 – 2x2 – e1 +a1 = 2
x1 + x2
– e2 +a2 = 3
x1>=0, x2 >=0, e1 >=0, e2>=0
b. How many basic variables will the set of two equality constraints have?
The set of two equality constraints have two basic variables: a1, a2.
c. How many basic solutions (both feasible and infeasible) will the two equality
constraint have, i.e., how many ways may these basic variables be chosen from the set
of four variables in the equations?
Because we have 2 constraints and 6 variables (2 basic and 4 non-basic) in this case
6
we will have   =15 basic solutions (both feasible and infeasible).
 2
IE230 HW#5
2/17/2016
page 2 of 4
d. Prepare the tableau to use the two-phase method to find an initial basic feasible
solution of the problem above, using artificial variables. (You need not solve it).
Minimize w = a1+ a2
s.t. x1– 2x2 – e1 + a1
=2
x1+ x2
– e2 + a2 = 3
-w
1
0
0
0
-z
0
1
0
0
X1
0
3
1
1
X2
0
1
-2
1
E1
0
0
-1
0
E2
0
0
0
-1
A1
1
0
1
0
A2
1
0
0
1
rhs
0
0
2
3
3. Revised Simplex Method We wish to solve the LP problem
M aximize z=10X1 + 6X2 + 4X3
subject to: X1 + X2 + X3  100
10X1 + 4X2 + 5X3  600
2X1 + 2X2 + 6X3  300
Xj  0, j=1,2,3
a. Complete the original tableau for solving this problem using the simplex method:
-z
1
0
0
0
X1
10
1
10
2
X2
6
1
4
2
X3
4
1
5
6
X4
0
1
0
0
X5
0
0
1
0
X6
0
0
0
1
RHS
0
100
600
300
Suppose that, after several iterations, we obtain the tableau below (in which some values have been
omitted):
-z
X1
X2
X3
X4
X5
X6
RHS
10/3
5/6
5/3
-1/6
0
200/3
1
0
1/6
1/6
0
2/3
0
1
100
2
b. What is the current basis for the constraint rows? B = { 2___ , ___ , ___ }
c. What is the “substitution rate” of X4 for X1? ___________
d. If X4 increases by 1 unit, X1 will ( circle one: increase or decrease) by ________ units.
 ___

B
e. What is the basis matrix? A  ___

 ___
IE230 HW#5
___
___
___
2/17/2016
___ 
___ 

___ 
page 3 of 4
_____ 
_____ _____ 
f. What is the inverse of the basis matrix?  A 

_____ _____ 
g. What are the values of the simplex multipliers for this basis?    _____, _____, _____ 
B 1
 _____
  _____

 _____
_____
h. What is the “relative profit” of X3, i.e., the number in the objective row of the X3 column in the
current tableau? _______
i. Compute all of the missing portions of the tableau above, using the basis inverse and the original
tableau.
IE230 HW#5
2/17/2016
page 4 of 4