Question

Maximize z = 3a + b + 2c
Subject to: 1. a + b + 3c <= 30
2. 2a + 2b + 5c<=  24
3. 4a + b + 2c<=  36
4. a,b,c>=  0
NB : a= Computers b= Network devices c= IP cameras
Z= Performance
-Numbers are costs.
The problem above consist of maximizing the performance of our computer network by reducing
the total cost.

Answer 1

Answer to Question #88001 - Math – Operations Research

Question:

Maximize $z = 3a + b + 2c$

Subject to 1. $a + b + 3c \leq 30$

2. $2a + 2b + 5c \leq 24$

3. $4a + b + 2c \leq 36$

4. $a,b,c \geq 0$

Solution:

In slack form:

Maximize $z = 3a + b + 2c$

Subject to: $d = 30 - a - b - 3c$

$e = 24 - 2a - 2b - 5c$

$f = 36 - 4a - b - 2c$

$a,b,c,d,e,f \geq 0$

The basic solution is $a = 0, b = 0, c = 0, d = 30, e = 24, f = 36$, i.e. (0, 0, 0, 30, 24, 36) $

$z = 0.$

1. Pivot using a.

When we substitute the basic solution into inequality (1), $a \leq 30$. From 2., $a \leq 24/2 = 12$. From 3., $a \leq 9$. Therefore, we can increase a only by 9. So we swap $a$ and $f$ by rewriting equation (3) as $a = 9 - b/4 - c/2 - f/4$ and substituting it into all other constraints.

Maximize $z = 3(9 - b/4 - c/2 - f/4) + b + 2c$

Subject to: $d = 30 - (9 - b/4 - c/2 - f/4) - b - 3c$

$e = 24 - 2(9 - b/4 - c/2 - f/4) - 2b - 5c$

$a = 9 - b/4 - c/2 - f/4$

$a,b,c,d,e,f \geq 0$

Maximize $z = 27 + b/4 + c/2 - 3f/4$

Subject to: 1. $a = 9 - b/4 - c/2 - f/4$

2. $d = 21 - 3b/4 - 5c/2 + f/4$

3. $e = 6 - 3b/2 - 4c + f/2$

4. $a,b,c,d,e,f \geq 0$

The basic solution is $b = 0, c = 0, f = 0, a = 9, d = 21, e = 6$ i.e. (9,0,0, 21, 6,0) $

$z = 27.$

2. Pivot using b or c. Pick c.

Constraint 1 limits c to 18, 2 limits c to 42/5, and 3 limits it to 3/2. Therefore, rewrite equation (3) and swap e and c.

Maximize $z = 111/4 + b/16 - e/8 - 11f/16$

Subject to:

1. $a = 33/4 - b/16 - e/8 - 5f/16$

2. $c = 3/2 - 3b/8 - e/4 + f/8$

3. $d = 69/4 + 3b/16 + 5e/8 - f/16$

4. $a, b, c, d, e, f \geq 0$

The basic solution is $b = 0, e = 0, f = 0, a = 33/4, c = 3/2, d = 69/4$ i.e. (33/4, 0, 3/2, 69/4, 0, 0). $z = 111/4$ .

3. Pivot using b.

The three constraints give bounds of 132, 4, and infinity (because $d$ increases as $b$ increases). Therefore rewrite equation (2) and swap $b$ and $c$ .

Maximize $z = 28 - c / 6 - e / 6 - 2f / 3$

Subject to:

1. $a = 8 + c / 6 + e / 6 - f / 3$

2. $c = 4 - 8c / 3 - 2e / 3 + f / 3$

3. $d = 18 - c / 2 + e / 2$

4. $a, b, c, d, e, f \geq 0$

The basic solution is (8, 4, 0, 18, 0, 0).

$z = 28$

Since the objective function has no non-negative coefficients, this is the optimal solution.

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