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Homework 7 Due: October 14, 2016
- (10 points) Use the Big M method and the simplex algorithm to solve the following problem.
State the optimal solution and optimal value of the objective function.
minimize 2x 1 + x 2
x 1 + 2x 2 ≤ 4
x 1 + x 2 = 3
x 1 , x 2 ≥ 0
Solution: First we introduce a slack variable (x 3 ) to the first constraint to put the formu-
lation in standard form. Then we add an artificial variable a 1 to the second constraint, and
add M a 1 to the objective function. This gives us the following Big-M formulation:
minimize 2x 1 + x 2 + M a 1
subject to x 1 + 2x 2 + x 3 = 4
x 1 + x 2 + a 1 = 3
x 1 , x 2 , x 3 , a 1 ≥ 0
The initial tableau (after adjusting row 0 to make a 1 basic) and optimal tableau are shown
below. Since this is a minimization problem, we will pivot on positive values in row 0, and
stop when all values in row zero are non-positive. Recall that you may also multiply row 0
by -1, and pivot on negative values.
Initial tableau:
x 1 x 2 x 3 a 1 RHS
M -2 M -1 0 0 3 M
1 2 1 0 4
1 1 0 1 3
After two iterations you obtain a feasible solution (where a 1 = 0)
x 1 x 2 x 3 a 1 RHS
0 0 -1 3-M 5
0 1 1 -1 1
1 0 -1 2 2
Based on the optimal tableau, we see that the optimal solution (x 1 , x 2 ) = (2, 1).
- (10 points) Consider the following standard form LP:
maximize 4x 1 + 2x 2 + x 3
subject to x 1 + 3x 2 + x 3 = 10
2 x 1 − 4 x 2 + x 4 = 5
x 1 , x 2 , x 3 , x 4 ≥ 0
Homework 7 Due: October 14, 2016
(a) Use the Simplex algorithm to find the optimal solution. Hint: no artificial variables are
needed.
Solution: After adding row 1 to the objective function row, we obtain a feasible basis
with x 3 and x 4 as basic variables.
B.V. x 1 x 2 x 3 x 4 RHS
z -3 1 0 0 10
x 3 1 3 1 0 10
x 4 2 -4 0 1 5
After two iterations of the Simplex method we end up with the following optimal
tableau:
B.V. x 1 x 2 x 3 x 4 RHS
z 0 0 1 1 25
x 2 0 1
1 5
1 10
3 2 x 1 1 0
2 5
3 10
11 2
(b) Suppose the objective coefficient of x 3 is changed from 1 to 1 + δ. For what range of δ does
the current optimal solution remain optimal?
Solution: First we observe that x 3 is non-basic, so changing its cost only affects the
reduced cost of variable x 3. If the coefficient of x 3 becomes 1 + δ, then the coefficient
of x 3 in row 0 of the optimal tableau becomes 1 − δ, meaning we must have 1 − δ ≥ 0
for the optimal basis to not change, or equivalently, δ ≤ 1.
(c) Suppose the objective coefficient of x 2 is changed from 2 to 2 + δ. For what range of δ does
the current optimal solution remain optimal?
Solution: We see that x 2 is basic. If the coefficient of x 2 becomes 2 + δ, then the
coefficient of x 2 in the objective value row becomes 0 − δ. The corresponding tableau
is then:
B.V. x 1 x 2 x 3 x 4 RHS
z 0 0 − δ 1 1 25
x 2 0 1
1 5 −^
1 10
3 2 x 1 1 0
2 5
3 10
11 2
To correct the objective row of the tableau we need to make the reduced cost of the
basic variable x 2 zero again. We do this by adding δ times the first row (corresponding
to x 2 ).
B.V. x 1 x 2 x 3 x 4 RHS
z 0 0 1 +
1 5
δ 1 −
1 10
δ 25 +
3 2
δ
x 2 0 1
1 5 −^
1 10
3 2 x 1 1 0
2 5
3 10
11 2
We observe that for each unit of δ the objective function value goes up by
3 2
δ. Next,
Homework 7 Due: October 14, 2016
B.V. x 1 x 2 x 3 e 1 s 2 a 1 a 3 RHS
z 0 −M −
1 2
1 2
M −
5 2
M M +
1 2
0 0 −M + 2
a 1 0
1 2
1 2
x 1 1
1 2
1 2
1 2
a 3 0
1 2
1 2
1 2
We choose x 2 as the entering variable and a 1 as the leaving variable. This gives us the
following tableau:
B.V. x 1 x 2 x 3 e 1 s 2 a 1 a 3 RHS
z 0 0 −
1 2 M^ −^
5 2 −M^ −^1 0 2 M^ + 1^0 −M^ + 2
x 2 0 1 0 -2 − 1 2 0 0
x 1 1 0 −
1 2
a 3 0 0
1 2 1 0 -1^1
We choose e 1 as the entering variable and a 3 as the leaving variable. This gives us the
following tableau:
B.V. x 1 x 2 x 3 e 2 s 2 a 1 a 3 RHS
z 0 0 -2 0 0 M M + 1 3
x 2 0 1 1 0 − 1 0 2 2
x 1 1 0 -1 0 1 0 -1 1
e 1 0 0
1 2 1 0 -1^1
Next, we take x 3 to be the entering variable and x 2 to be the leaving variable. This
gives the tableau:
B.V. x 1 x 2 x 3 e 2 s 2 a 1 a 3 RHS
z 0 2 0 0 − 2 M M + 5 7
x 3 0 1 1 0 − 1 0 2 2
x 1 1 1 0 0 0 0 1 3
e 1 0 −
1 2
1 2
We choose s 2 as entering variable and e 1 as the leaving variable. This gives the following
tableau:
B.V. x 1 x 2 x 3 e 2 s 2 a 1 a 3 RHS
z 0 0 0 4 0 M − 4 M + 5 7
x 3 0 0 1 2 0 -2 2 2
x 1 1 1 0 0 0 0 1 3
s 2 0 − 1 0 2 1 -2 0 0
This tableau and the corresponding solution (x 1 , x 2 , x 3 ) = (3, 0 , 2) is optimal, since all
reduced cost are positive.
(b) Suppose the objective coefficient of x 2 is changed to 1 + δ. For what range of δ does the
current solution remain optimal?
Homework 7 Due: October 14, 2016
Solution: Since x 2 is non-basic, changing its reduced cost only affects the reduced
cost of x 2 itself. For a change of δ, the new reduced cost is 0 − δ. For the basis and the
solution to remain optimal, we require that 0 − δ ≥ 0. So the solution remains optimal
for δ ≤ 0.
(c) Suppose the objective coefficient of x 3 is changed to 2 + δ. For what range of δ does the
current solution remain optimal?
Solution: After changing the objective coefficient of x 3 , we obtain the following opti-
mal tableau:
B.V. x 1 x 2 x 3 e 2 s 2 a 1 a 3 RHS
z 0 0 0 − δ 4 0 M − 4 M + 5 7
x 3 0 0 1 2 0 -2 2 2
x 1 1 1 0 0 0 0 1 3
s 2 0 − 1 0 2 1 -2 0 0
To ensure that x 3 has zero reduced cost, we add the first row to the objective function
row to obtain:
B.V. x 1 x 2 x 3 e 2 s 2 a 1 a 3 RHS
z 0 0 0 4 + 2δ 0 M − 4 − 2 δ M + 5 + 2δ 7
x 3 0 0 1 2 0 -2 2 2
x 1 1 1 0 0 0 0 1 3
s 2 0 − 1 0 2 1 -2 0 0
To ensure that the reduced cost remain non-negative, we require the following:
4 + 2δ ≥ 0 ⇒ δ ≥ − 2
M − 4 − 2 δ ≥ 0 ⇒ δ ≤
1 2
M − 2 = ∞
M + 5 + 2δ ≥ 0 ⇒ δ ≥ −
1 2 M^ −^
5 2 =^ −∞
− 2 ≤ δ ≤ ∞
(d) Suppose the right hand side of the second constraint is changed to −4 + ∆. For what range
of ∆ does the current basis remain optimal? And what is the shadow price?
Solution: Note that we flipped the sign of the inequality, so we need analyze the case
that we change the right hand side to 4 − ∆. The new right hand side vector can be
split into the following:
− 5 M
4 − δ
3
− 5 M
Homework 7 Due: October 14, 2016
In, addition we observe from the column of a 3 that:
M
M + 5
Since we are only adding multiples of other rows to the objective function row, this is
equivalent to: 0
——
0
0
1
Hence, our basic variables are given by:
For the solution to remain optimal the basic variables have to remain non-negative.
Therefore, we know that ∆ ≥ −1. In addition, we see that the optimal objective value
increases by 5 units for each unit of ∆. Hence, the shadow price is 5.
- (10 points) Solve Problem 6 on page 288 from the textbook. You can skip part (e) and (f).
Solution:
(a) If the Type 1 candy bar profit is ≤ 6 then the current basis remains optimal. So if the
new profit was 7 cents we would need to resolve the problem.
(b) If the Type 2 candy bar profit is between 5 and 15 then the current basis remains
optimal. So if the new profit was 13 cents the current solution would remain optimal:
(x 1 , x 2 , x 3 ) = (0, 25 , 25), this corresponds to a profit of 450.
(c) If the amount of available sugar (b 1 ) is in the following range 100/ 3 ≤ b 1 ≤ 100 then
the current basis remains optimal.
(d) First note that 60 oz of sugar is within the range found in part c so you do not need to
resolve. In particular δ = 10. The new profit is 300+4δ = 340. The new RHS values are
25 + 1. 5 δ and 25 − 0. 5 δ, since δ = 10, these become 40 and 20: (x 1 , x 2 , x 3 ) = (0, 20 , 40).
For the last part of this question, 30 oz is outside the range so we cannot answer
without resolving.
- (10 points) Solve Problem 8 on page 289 from the textbook. You can skip part (c).
Homework 7 Due: October 14, 2016
Solution:
(a) Since x 1 is a basic variable, we add δ times the first row to the objective function row.
Since the reduced cost need to remain non-positive for a minimization problem, we
require that − 5 −
3 20
δ ≤ 0 and − 7
1 2
1 40
δ ≤ 0. This means that −
100 3
≤ δ ≤ 300. Thus, 50 3
≤ c 1 ≤ 350.
(b) By adding δ times the column of a 1 (without the M ) to the right hand side, we obtain
that the right hand side is equal to:
320 + 5δ
——
3 .6 +
3 20
δ
- 4 −
1 40
δ
To ensure that the solution remain feasible, the right hand side has to be non-negative.
This requires that − 24 ≤ δ ≤ 56. That is, 4 ≤ b 1 ≤ 84.
Observe that 40 is within the range, so the current basis remains optimal. Plugging in
δ = 12 gives us (x 1 , x 2 ) = (5. 4 , 1 .1) and an objective function value of 320+5×12 = 380.
