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Solutions to review problems for Midterm
First: Midterm #1 covers Chapter 1 and 2. In particular, this means that it does not explicitly cover linear algebra. Also, I promise there will not be any proofs.
1 Consider the following scenario. Model it as a linear programming problem. Be sure to state explicitly what each of your decision variables x 1 , x 2 ,... represent. Do not attempt to solve the LPP. A cattle rancher uses three types of cattle feed: type 1, type 2, and type 3. Type 1 costs $1. per pound, type 2 costs $3.50 per pound, and type 3 costs $2.00 per pound. The rancher wants to meet the following minimum daily requirements for each animal. Each day, each animal should have at least 120 mg of vitamin A, 180 mg of vitamin B, and 100 mg of vitamin C. The following chart shows the number of mg per pound of each vitamin in the three types of feed:
Vitamin Type 1 Type 2 Type 3 A 8 2 20 B 9 11 5 C 1 10 20
Because of protein content, however, an animal cannot eat more than 15 pounds of type 1, 10 pounds of type 2, and 5 pounds of type 3 cattle feed per day. How many pounds of each type of cattle feed should the rancher purchase per day in order to minimize the cost, while still meeting the nutritional minimum daily requirements?
Answer: Letting x 1 = amount of Type 1 feed, x 2 = amount of Type 2 feed, x 3 = amount of Type 3 feed,
the problem can be written as
Minimize z = 1. 50 x 1 + 3. 50 x 2 + 2. 00 x 3 subject to 8 x 1 +2x 2 +20x 3 ≥ 120 , 9 x 1 +11x 2 +5x 3 ≥ 180 , x 1 +10x 2 +20x 3 ≥ 100 , x 1 ≤ 15 , x 2 ≤ 10 , x 3 ≤ 5 , x 1 , x 2 , x 3 ≥ 0.
Note that this is our answer, the problem specifically told us not to try to solve the problem.
2 Convert the following LPP into (a) standard form, and (b) canonical form:
minimize z = x 1 − 4 x 2 + 5x 3 subject to x 1 +x 3 ≤ 5 x 2 +3x 3 ≥ 7 2 x 1 +9x 2 = 11
x 1 , x 2 ≥ 0 x 3 unconstrained
Answer: To convert a problem to standard form, we generally have to do three things:
- Make the problem a maximization problem,
- Make every constraint except for constraints of the form xj ≥ 0 involve a ≤,
- Make every variable constrained to be non-negative.
To do that in this example we have to do quite a bit of work. The final answer is:
maximize z = −x 1 + 4x 2 − 5 x 3 subject to x 1 +x+ 3 −x− 3 ≤ 5 −x 2 − 3 x 3 +3x− 3 ≤ − 7 2 x 1 +9x 2 ≤ 11 − 2 x 1 − 9 x 2 ≤ − 11
x 1 , x 2 , x+ 3 , x− 3 ≥ 0
The answer for canonical form is:
maximize z = −x 1 + 4x 2 − 5 x 3 subject to x 1 +x+ 3 −x− 3 +u = 5 −x 2 − 3 x 3 +3x− 3 +v = − 7 2 x 1 +9x 2 = 11
x 1 , x 2 , x+ 3 , x− 3 , u, v ≥ 0
4 Find an optimal solution to the following LPP using the simplex method.
maximize z = x 1 + 3x 2 + 5x 3 subject to 2 x 1 − 5 x 2 +x 3 ≤ 3 x 1 +4x 2 ≤ 5
x 1 , x 2 , x 3 ≥ 0
Answer: First we have to add slack variables to both inequalities. Then our initial tableau will be
x 1 x 2 x 3 u 1 u 2 u 1 2 − 5 1 1 0 3 u 2 1 4 0 0 1 5 − 1 − 3 − 5 0 0 0
First iteration: Entering x 3 , departing u 1 :
x 1 x 2 x 3 u 1 u 2 x 3 2 − 5 1 1 0 3 u 2 1 4 0 0 1 5 9 − 28 0 5 0 15
Second iteration: Entering x 2 and departing u 2 :
x 1 x 2 x 3 u 1 u 2 x 3 13 / 4 0 1 1 5 / 4 37 / 4 x 2 1 / 4 1 0 0 1 / 4 5 / 4 16 0 0 5 7 50
Therefore the optimal solution is x 1 = 0, x 2 = 5/ 4 , x 3 = 37/ 4 , x 4 = 0, and z = 50 is the best we can do.
5 Find an optimal solution to the following LPP using the two-phase simplex method.
maximize z = −x 1 − 2 x 2 subject to x 1 +2x 2 −x 3 = 3 3 x 1 +4x 2 −x 4 = 10
x 1 , x 2 , x 3 , x 4 ≥ 0
Answer: We have to add artificial variables y 1 and y 2 to the two constraints, so our auxiliary problem is: maximize z′^ = −y 1 − y 2 subject to x 1 +2x 2 −x 3 +y 1 = 3 3 x 1 +4x 2 −x 4 +y 2 = 10
x 1 , x 2 , x 3 , x 4 , y 1 , y 2 ≥ 0
This corresponds to the tableau
x 1 x 2 x 3 x 4 y 1 y 2 y 1 1 2 − 1 0 1 0 3 y 2 3 4 0 − 1 0 1 10 0 0 0 0 1 1 0
But before we get started, we have to get 0s in the objective row in the y 1 and y 2 columns. (As we discussed in class, our book does this a little differently, but I think this is easier.) So, we subtract the first and second rows from the third row to obtain the following initial tableau:
x 1 x 2 x 3 x 4 y 1 y 2 y 1 1 2 − 1 0 1 0 3 y 2 3 4 0 − 1 0 1 10 − 4 − 6 1 1 0 0 − 13
Our entering variable will be x 2 , and the θ-ratios are 3 and 10/3, respectively, so the departing variable is y 1. After pivoting, this gives the following tableau:
x 1 x 2 x 3 x 4 y 1 y 2 x 2 1 / 2 1 − 1 / 2 0 1 / 2 0 3 / 2 y 2 1 0 − 2 − 1 − 2 1 4 − 1 0 − 2 1 3 0 − 4
So our next entering variable will be x 3 , and since the θ-ratios are 3 and 4, respectively, the departing variable is y 2. After pivoting, we get
x 1 x 2 x 3 x 4 y 1 y 2 x 2 3 / 4 1 0 − 1 / 4 0 1 / 4 5 / 2 x 3 1 / 2 0 1 − 1 / 2 − 1 1 / 2 2 0 0 0 0 1 1 0
