CS 600 Solutions to Homework 3

3.1 - 4
(a) Max $Z = 120 x_{27} + 80 x_{20}$
Subject to:
$x_{27} \leq 40$
$x_{20} \leq 10$
$20 x_{27} + 10 x_{20} \leq 500$
$x_{20}, x_{27} \geq 0$

(b) Find the largest value of $c$ such that the line $c = 120 x_{27} + 80 x_{20}$ touches the feasible region by varying $c$.

(c) Introduce slack variables $x_{1}, x_{2}, x_{3}$. We then have
Max $Z = 120 x_{27} + 80 x_{20}$
Subject to:
$x_{27} + x_{1} = 40$
$x_{20} + x_{2} = 10$
$20 x_{27} + 10 x_{20} + x_{3} = 500$
$x_{i} \geq 0$

The initial dictionary is:
$x_{1} = 40 - x_{27}$
$x_{2} = 10 - x_{20}$
$x_{3} = 500 - 20 x_{27} - 10 x_{20}$
$Z = 120 x_{27} + 80 x_{20}$

Since all the coefficients in $Z$ are positive, $Z$ can be improved.
Select $x_{27}$ as the pivot since it has the largest coefficient in $Z$.
Then $x_{3}$ is the new non-basic variable sinvce it restricts $x_{27}$ most.
Hence the new dictionary is
$x_{27} = 25 - (1/2)x_{20} - (1/20) x_{3}$
$x_{2} = 10 - x_{20}$
$x_{1} = 40 - (25 - (1/2)x_{20} - (1/20) x_{3}$
$= 15 + (1/2)x_{20} + (1/20) x_{3}$
$Z = 3000 + 20 x_{20} - 6 x_{3}$

Since the coefficient of $x_{20}$ is positive, $Z$ can be improved.
Select $x_{20}$ as the pivot.
Then $x_{2}$ is the new non-basic variable sinvce it restricts $x_{20}$ most.
Hence the new dictionary is
$x_{20} = 10 - x_{2}$
$x_{27} = 25 - (1/2)(10 - x_{2}) - (1/20) x_{3}$
$= 20 + (1/2)x_{2} - (1/20) x_{3}$
$x_{1} = 15 + (1/2)(10 - x_{2}) + (1/20) x_{3}$
$= 20 - (1/2)x_{2} + (1/20) x_{3}$
$Z = 3200 - 20 x_{2} - 6 x_{3}$

Since all the coefficients in $Z$ are negative, $Z$ can not be improved any further. Hence the optimum value for $Z$ is 3200, $x_{20} = 10$ and $x_{27} = 20$.

3.4-9 (a)
Let $x_{ij}$ represent the weight (ton) of cargo $i$ to be placed in compartment $j$, where $i = 1, 2, 3 or 4$ and $j$ = f(front), c(center) or b(back).

Then the objective function to be maximized is
$Z = 320 \Sigma_{j} x_{1,j} + 400 \Sigma_{j} x_{2,j} + 360 \Sigma_{j} x_{3,j} + 290 \Sigma_{j} x_{4,j}$

The constraints are:
From the weights
$x_{1f} + x_{2f} + x_{3f} + x_{4f} \leq 12$
$x_{1c} + x_{2c} + x_{3c} + x_{4c} \leq 18$
$x_{1b} + x_{2b} + x_{3b} + x_{4b} \leq 10$
and
$x_{1f} + x_{1c} + x_{1b} \leq 20$
$x_{2f} + x_{2c} + x_{2b} \leq 16$
$x_{3f} + x_{3c} + x_{3b} \leq 25$
$x_{4f} + x_{4c} + x_{4b} \leq 13$
From the weight balance
$(x_{1f} + x_{2f} + x_{3f} + x_{4f})/S \leq 12/40$
$(x_{1c} + x_{2c} + x_{3c} + x_{4c})/S \leq 18/40$
$(x_{1b} + x_{2b} + x_{3b} + x_{4b})/S \leq 10/40$,
where $S = \Sigma_{j} x_{1,j} + \Sigma_{j} x_{2,j} + \Sigma_{j} x_{3,j} + \Sigma_{j} x_{4,j}$
From the space constraints
$500x_{1f} + 400x_{2f} + 360x_{3f} + 290x_{4f} \leq 7000$
$500x_{1c} + 400x_{2c} + 360x_{3c} + 290x_{4c} \leq 9000$
$500x_{1b} + 400x_{2b} + 360x_{3b} + 290x_{4b} \leq 5000$

Also all $x_{ij}$'s must be non-negative.