01:640:354 – Linear Optimization – Practice questions for
Final Exam
Instructor: Izar Alonso Lorenzo
1. Tommy’s mother is planning his 10th birthday party and will serve a variety of soft drinks,
which will be chosen from the list below.
Cola Rootbeer Cherry Lemon Orange Grape Gingerale
Drink
Price per bottle (cents) 69
59
62
62
65
55
65
From past experience it has been determined that at least 12 bottles of soft drinks are needed.
Also, at least 2 bottles of ginger ale, at least 2 bottles of cola, and no more than 3 bottles
of fruit-flavored soft drinks are needed. How many bottles of each type should be bought to
minimize the total cost?
Set up a linear programming model of the situation described above. You do NOT need to
solve this problem.
1
01:640:354
Practice questions for Final Exam
2. Consider the following pure integer programming problem:
Maximize z = 3x1 + 11x2
subject to the constraints
2x1 +6x2 ≤ 4
−x1 +2x2 ≤ 1
x1 ≥ 0, x2 ≥ 0, x1 , x2 integral.
After solving it using the simplex method, we got the following tableau:
cB
3 x1
11 x2
3 11
0
0
x 1 x2
x3
x4
xB
1 0
1/5 −3/5 1/5
0 1 1/10 1/5 3/5
0 0 17/10 2/5 36/5
(a) Write the cutting plane constraint. (Hint: use the basic variable x2 )
(b) Solve the pure integer programming problem.
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01:640:354
Practice questions for Final Exam
The final exam is cumulative, so there will be questions about the materials from Midterm 1
and 2. Make sure to revise the following resources (see Files/Revision):
• 354 M1 practicequestions.pdf
• 354 1 M1 Revision.pdf (section 1) or 354 3 M1 Revision.pdf (section 3)
• 354 M2 practicequestions.pdf
• 354 1 M2 Revision.pdf (section 1) or 354 3 M2 Revision.pdf (section 3)
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01:640:354
Practice questions for Final Exam
Answer key
1. Let xi , i = 1, 2, . . . , 7 be the number of bottles of cola, root beer, cherry, lemon, orange, grape,
and ginger ale, respectively. Mathematical model:
Minimize z = 69x1 + 59x2 + 62x3 + 62x4 + 65x5 + 55x6 + 65x7
subject to
7
X
xi ≥ 12,
i=1
x7 ≥ 2,
x1 ≥ 2,
x3 + x4 + x5 + x6 ≤ 3,
xi ≥ 0, integer, i = 1, 2, . . . , 7.
2. (a) We compute f1 = 1/5 and f2 = 3/5, and then choose x2 as the variable to generate the
cutting plane (3/5 > 1/5), and get:
−
1
1
3
x3 − x4 + u 2 = − .
10
5
5
(b) We add the new constraint to the old optimal tableau:
cB
3 x1
11 x2
3 11
x1 x2
1 0
0 1
0
x3
1/5
1/10
0
0
x4
u2
−3/5 0
1/5
0
xB
1/5
3/5
0
0
0
−1/10
-1/5
1
−3/5
0
0
17/10
2/5
0
36/5
u2
We iterate the dual simplex method. We choose u2 as departing variable, the entering variable must
be x4 , ratios: −17, −2. We get the following tableau.
3 11 0
0 0
cB x1 x2 x3 x4 u2 xB
3 x1 1 0 1/2 0 −3 2
11 x2 0 1
0
0 1
0
0 x4 0 0 1/2 1 −5 3
0 0 3/2 0 2
6
This is optimal and feasible, so we have found an optimal solution of the relate linear programming
problem. Because this solution is also integral, we have found a solution of the integer programming
problem. It is x1 = 2, x2 = 0, z = 6.
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