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Operations Research (I)
2007/10/19
姓名：
學號：
班級：
1. Label the following statement as being true or false. If false, briefly state your
comments.
(20%)
a. If multiple optimal solutions exist in a 3-dimensional LP problem, At least 3 planes
must pass through a corner point to product a degenerate solution.
1. True
2. False
Explain:
If multiple optimal solutions exist in a 3-dimensional LP problem, At least 4 planes
must pass through a corner point to product a degenerate solution.
b. When an artificial problem is created by introducing artificial variables and using
the Big M method, if all artificial variables in current solution for the artificial
problem are equal to zero, then the current solution is an infeasible solution in the
real problem.
1. True
2. False
Explain:
If all artificial variables in current solution for the artificial problem are equal to
zero, then the current solution is a feasible solution in the real problem.
IF current solution is an infeasible solution in the real problem, then at least one
artificial variable is larger than zero.
c. The simplex method’s rule for choosing the entering basic variable is used
because it always leads to the best adjacent BF solution (Largest Z).
1. True
2. False
Explain:
The simplex method’s rule for choosing the entering basic variable is used because it
always leads to a better adjacent BF solution. The rule cannot assure to lead to the
best adjacent BF solution.
d. If there is no leaving basic variable at some iteration, then the problem has no
feasible solution.
1. True
2. False
Explain:
If there is no leaving basic variable at some iteration, then the solution of this problem
1
2
is unbounded.
e. The most important factor that affects the computation time to solve an LP
problem by the simplex method is the number of decision variables.
1. True
2. False
Explain:
The most important factor that affects the computation time to solve an LP problem
by the simplex method is the number of functional constraints.
2. Consider the linear programming problem: Maximize cx subject to Ax≦ b, x≧ 0,
where c is a nonzero vector. Suppose that the point x0 is such that A x0 < b and x0>
0. Explain why x0 cannot be an optimal solution. (10%)
Ans.
Because x0 is an interior feasible solution, x0 cannot be an optimal solution.
If optimal solution exists, it must be an extreme point or lie on the boundary
(multiple optimal solutions).
3. A manufacturer produce three models, I, II, and III, of a certain product using raw
materials A and B. The following table gives the data for the problem.
Requirements per unit
Raw material
A
B
Minimum demand
Profit per unit($)
I
2
4
200
30
availability
II
III
3
2
5
7
200
20
4000
6000
150
50
The labor time per unit of model I is twice that of model II and three times that of
model III. The entire labor force of the factory can produce the equivalent of 1500
units of model I.
Formulate a linear programming model for this problem to maximize the profit.
(15%)
Ans.
Let x1 : number of model I produced
x2 : number of model II produced
x3 : number of model III produced
2
3
Max Z  30 x1  20 x2  50 x3
s.t.
2x1  3 x2  5 x3  4000
4x1  2 x2  7 x3  6000
1
1
x2  x3  1500
2
3
x1  200
x1 
x2  200
x3  150
3
4
4. Consider the following problem:
Real Problem
Maximize Z  3x1  2 x2
Subject to
x1  x2  1
x1  x2  2
x1 , x2  0
and
Augmented form
Maximize
Z  3x1  2 x2
Subject to
and
x1  x2  x3
1
x1  x2
 x4  2
x1, x2 , x3 , x4  0
(a) Solve this problem graphically.(5%)
(b) Develop a table giving each of the CPF solutions and the corresponding defining
equations, BF solution, nonbasic variables. (15%)
(c) Adding the artificial variable x5 , using the two-phase method, construct the
complete first simplex tableau for phase 1. (5 %)
(d) work through phase 1 step by step.(3 %)
(e) work through phase 2 step by step to solve the problem. (5%)
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5
(a)
x* = (2, 0), Z* = 6
(b)
CPF
Defining
equations
BF
solution
Non-basic
variable
(1.5,0.5)
x1 + x2 = 2
x1 - x2 = 1
(1.5,0.5,0,0)
x3, x4
(2,0)
x1 + x2 = 2
x2 = 0
(2, 0, 1, 0)
x2, x4
(1,0)
x2 = 0
x1 - x2 = 1
(1, 0, 0, 1)
x2, x3,
5
6
(c)(d)
Minimize
x5
x1  x2  x3 
+ x5 = 1
x1  x2
+ x4
=2
Subject to
x1, x2 , x3, x4 , x5  0
and
Basic
Coefficient of:
Iteration Variable Eq.
0
1
2
Right
x2
0
x3
0
x4
0
x5
side
1
0
Z
(0)
x1
-1 0
x5
0
0
1
1
-1
1
-1
0
0
1
x4
(1)
(2)
1
0
1
2
Z
(0)
-1 -1
1
1
0
0
-1
x5
(1)
-1
1
-1
0
1
(2)
1
1
0
x4
0
0
1
0
1
2
Z
(0)
-1
0
0
0
0
1
0
x1
(1)
0
1
-1
-1
0
1
1
x4
(2)
0
0
2
1
1
-1
1
Z
(e)
Basic
Iteration Variable Eq.
0
1
2
3
Coefficient of:
Z
(0)
x1 x2
-1 -3 -2
x1
(1)
0
1
-1
-1
0
1
x4
(2)
0
0
2
1
1
1
Z
(0)
-1
0
-5
-3
0
3
x1
(1)
0
1
-1
-1
0
1
x4
(2)
0
0
2
1
1
1
Z
(0)
-1
0
0
-0.5
2.5
5.5
x1
(1)
0
1
0
-0.5 0.5.
1.5
x2
(2)
0
0
1
0.5
0.5
0.5
Z
(0) -1
0
1
0
3
6
x1
(1)
0
1
1
0
1
2
x3
(2)
0
0
2
1
1
1
Z
x3
0
x4
0
Right
side
0
6
7
5.
Consider the following problem
Maximize
Z  x1  C 2 x 2
- x1  x 2  3
s.t.
x1  2 x 2  9
6
x1
x1  0, x 2  0
(a) Using graphical analysis to determine the optimal solution(s) for (x1, x2) for
the various possible values of C2 (-  C2  ) (10 %);
(b) Let C2=2, find all the optimal solutions (6 %);
(c) Use your result from part (b) to identify the shadow prices for the three
resources.(6 %)
(a)
-x1+x2=3
x1=6
(1,4)
(0,3)
(6,3/2)
x1+2x2=9
(0,0)
Let m  
(6,0)
1
be the slope of the objective function
C2
And c1=1
(1) C2<0
x1= (6,0 ) is the optimal solution.
(2) 0<C2<2, x3= (6,3/2) is the optimal solution..
(3) C2 =2 the optimal solutions are on the line segment connecting x2=(1,4) and
x3=(6,3/2) i.e. x*   * x2  (1   )x3 where  [0,1] .
(4) C2>2,
x2=(1,4) is the optimal solution.
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(b)
Maximize Z  x1  2 x2
-x1  x2  3
s.t.
x1  2x2  9
6
x1
x1  0,
Iteration 0
B.V.. Eq.No
x2  0
Z
x1
x2
x3
x4
x5
RHS
Z
(0)
1
-1
-2
0
0
0
0
x3
x4
x5
(1)
(2)
(3)
0
0
0
-1
1
1
1
2
0
1
0
0
0
1
0
0
0
1
3
9
6
Z
x1
x2
x3
x4
x5
RHS
Iteration 1
B.V.. Eq.No
Z
(0)
1
-3
0
2
0
0
6
x2
x4
(1)
(2)
0
0
-1
3
1
0
1
-2
0
1
0
0
3
3
x5
(3)
0
1
0
0
0
1
6
Z
x1
x2
x3
x4
x5
RHS
0
1
0
9
0
0
1
4
1
5
Iteration 2
B.V.. Eq.No
Z
(0)
1
0
0
x2
x1
x5
(1)
(2)
(3)
0
0
0
0
1
0
1
0
0
1/3 1/3
-2/3 1/3
2/3 -1/3
x3 is not a basic variable, but its coefficient in the row 0 is zero. It means that there are
multiple optimal solutions in this case.
B.V..
Eq.No
Z
x1
x2
x3
x4
x5
RHS
Z
(0)
1
0
0
0
1
0
9
x2
x1
x3
(1)
(2)
(3)
0
0
0
0
1
0
1
0
0
0
0
1
1/2 -1/2
0
1
-1/2 3/2
3/2
6
15/2
8
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Optimal solutions can be written as
( x1* , x2* )   *(1, 4)  (1   ) *(6,3 / 2) where   [0,1]
(c)
The shadow price of resource 1, resource 2, and resource 3 are 0, 1, and 0,
respectively.
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