Computational Methods

Maximize $Z=3x+2y$ subject to $x+y\le4,\ x+2y\le6,\ x,y\ge0$.

Standard form. Add slack variables $s_1,s_2\ge0$: $x+y+s_1=4,\qquad x+2y+s_2=6,\qquad Z-3x-2y=0.$

Initial tableau

The most negative entry is $-3$, so $x$ enters. The minimum ratio is $4$, so $s_1$ leaves; the pivot is $1$.

Iteration 1: $R_2\to R_2-R_1$, $\ Z\to Z+3R_1$.

All entries in the $Z$-row are non-negative, so the solution is optimal.

$x=4,\quad y=0,\quad \boxed{Z_{\max}=12}$

Check: the corner points $(0,0),(4,0),(2,2),(0,3)$ give $Z=0,12,10,6$.

The question text is incomplete; the standard form of this problem is solved below.

Maximize $Z=3x+5y$ subject to $x\le4,\ 2y\le12,\ 3x+2y\le18,\ x,y\ge0$.

Step 1: Standard form. Introduce slack variables $s_1,s_2,s_3\ge0$:

$x+s_1=4,\quad 2y+s_2=12,\quad 3x+2y+s_3=18,\quad Z-3x-5y=0$

Step 2: Initial tableau.

The most negative entry is $-5$, so $y$ enters; the minimum ratio is $6$, so $s_2$ leaves. The pivot is $2$.

Step 3: Iteration 1. $R_2\to R_2/2$, $R_3\to R_3-2R_2$, $Z\to Z+5R_2$.

Now $x$ enters; the minimum ratio is $2$, so $s_3$ leaves. The pivot is $3$.

Step 4: Iteration 2. $R_3\to R_3/3$, $R_1\to R_1-R_3$, $Z\to Z+3R_3$.

All $Z$-row entries are non-negative, so the solution is optimal:

$x=2,\quad y=6,\quad Z_{\max}=3(2)+5(6)=36.$

Check: $x=2\le4$, $2y=12\le12$, $3x+2y=18\le18$.

Maximize $Z = 3x_1 + 2x_2$ subject to $x_1 + x_2 \le 4$, $x_1 - x_2 \le 2$, $x_1, x_2 \ge 0$.

Standard form. Add slack variables: $x_1 + x_2 + s_1 = 4$, $x_1 - x_2 + s_2 = 2$.

Initial table

$x_1$ enters (largest positive $C_j - Z_j = 3$). Ratios: $4/1 = 4$, $2/1 = 2$, so $s_2$ leaves; pivot $= 1$.

Iteration 1: $R_1 \to R_1 - R_2$.

$x_2$ enters. Only row 1 has a positive entry: $2/2 = 1$, so $s_1$ leaves; pivot $= 2$.

Iteration 2: $R_1 \to R_1/2$, $R_2 \to R_2 + R_1$.

All $C_j - Z_j \le 0$, so the solution is optimal:

$x_1 = 3,\quad x_2 = 1,\quad Z_{\max} = 3(3) + 2(1) = 11.$

Maximize $Z=7x+5y$ subject to $x+2y\le6,\ 4x+3y\le12,\ x,y\ge0$.

Standard form: add slack variables $s_1,s_2\ge0$: $x+2y+s_1=6,\qquad 4x+3y+s_2=12,\qquad Z-7x-5y=0.$

Initial tableau

The most negative entry in the $Z$-row is $-7$, so $x$ enters. The minimum ratio is in row $s_2$, so $s_2$ leaves; the pivot is 4.

Iteration 1: $R_2\to R_2/4$; $R_1\to R_1-R_2$; $R_Z\to R_Z+7R_2$.

All entries in the $Z$-row are non-negative, so the solution is optimal.

Result: $x=3,\ y=0$, maximum $Z=7(3)+5(0)=21$.

Check at the other corners: $(0,3)$ gives $15$, $\left(\frac65,\frac{12}5\right)$ gives $20.4$, $(0,0)$ gives $0$; all are below $21$.

(a) $3x = 6$

Adding the two equations eliminates $y$: $(x+y)+(2x-y)=5+1$, so $3x=6$.

Maximize $Z = 3x + 4y$ subject to $x + y \le 4$, $x + 2y \le 6$, $x, y \ge 0$.

Standard form. Add slack variables $s_1, s_2$: $x + y + s_1 = 4$, $x + 2y + s_2 = 6$, $Z - 3x - 4y = 0$.

Initial table

Most negative entry in the $Z$-row is $-4$, so $y$ enters. Ratios: $4/1 = 4$, $6/2 = 3$. Minimum is 3, so $s_2$ leaves; pivot $= 2$.

Iteration 1: $R_2 \to R_2/2$, $R_1 \to R_1 - R_2$, $Z \to Z + 4R_2$.

$x$ enters; ratio $1/(1/2) = 2$ (row 1) and $3/(1/2) = 6$, so $s_1$ leaves; pivot $= 1/2$.

Iteration 2: $R_1 \to 2R_1$, $R_2 \to R_2 - \tfrac12 R_1$, $Z \to Z + R_1$.

All $Z$-row entries are non-negative, so the solution is optimal.

$x = 2,\quad y = 2,\quad Z_{\max} = 3(2) + 4(2) = 14.$

Check: $2+2 = 4$ and $2+4 = 6$, so both constraints are satisfied with equality.

(c) $7y = 7$

Multiply the second equation by 2: $2x + 8y = 12$. Subtract the first equation $2x + y = 5$:

$(2x + 8y) - (2x + y) = 12 - 5 \;\Rightarrow\; 7y = 7.$

(b) Infinitely many solutions

The last row of the reduced augmented matrix is a zero row in the coefficient part with the augmented entry also 0, i.e. it gives $0 = 0$. So

$\text{rank}(A) = \text{rank}([A|B]) = 2 < 3 \ (\text{number of unknowns})$

Since the ranks are equal but less than the number of unknowns, the system is consistent with a free variable, hence it has infinitely many solutions.

(Note: if the last row had read $0\ \ 0\ \ 0\ |\ k$ with $k \neq 0$, then $\text{rank}(A) \neq \text{rank}([A|B])$ and the answer would be (c) no solution.)

(b) 2

The pivot column is the one with the most negative entry in the $z$-row; here that is the $x$-column (entries 4 and 2). The pivot row is chosen by the minimum ratio test: $\frac{240}{4}=60$ and $\frac{100}{2}=50$, so the second row is the pivot row. The pivot element at the intersection of the $x$-column and the second row is 2.

For a projectile launched with speed $u$ at angle $\theta$ to the horizontal, the maximum height is

$H = \frac{u^2 \sin^2\theta}{2g}.$

Solving for the speed of projection,

$u = \frac{\sqrt{2gH}}{\sin\theta}.$

If the body is projected vertically upward ($\theta = 90^\circ$), this reduces to $u = \sqrt{2gH}$. Substitute the given values of $H$ and $\theta$ (with $g = 9.8\ \text{m/s}^2$) into this formula and choose the matching option.

Gauss–Seidel Method

System to be solved:

$20x + y - 2z = 17$ $3x + 20y - z = -18$ $2x - 3y + 20z = 25$

Step 1: Check diagonal dominance. In each equation the diagonal coefficient is larger in magnitude than the sum of the other two coefficients ($20 > 1+2$, $20 > 3+1$, $20 > 2+3$), so the Gauss–Seidel iteration will converge.

Step 2: Rearrange each equation for its diagonal unknown.

$x = \frac{17 - y + 2z}{20}, \qquad y = \frac{-18 - 3x + z}{20}, \qquad z = \frac{25 - 2x + 3y}{20}$

Step 3: Start with the initial approximation $x^{(0)} = 0,\; y^{(0)} = 0,\; z^{(0)} = 0$ and use each newly computed value immediately in the next equation.

Iteration 1

$x^{(1)} = \frac{17 - 0 + 0}{20} = 0.8500$ $y^{(1)} = \frac{-18 - 3(0.8500) + 0}{20} = \frac{-20.5500}{20} = -1.0275$ $z^{(1)} = \frac{25 - 2(0.8500) + 3(-1.0275)}{20} = \frac{20.2175}{20} = 1.0109$

Iteration 2

$x^{(2)} = \frac{17 - (-1.0275) + 2(1.0109)}{20} = \frac{20.0493}{20} = 1.0025$ $y^{(2)} = \frac{-18 - 3(1.0025) + 1.0109}{20} = \frac{-19.9966}{20} = -0.9998$ $z^{(2)} = \frac{25 - 2(1.0025) + 3(-0.9998)}{20} = \frac{19.9956}{20} = 0.9998$

Iteration 3

$x^{(3)} = \frac{17 - (-0.9998) + 2(0.9998)}{20} = \frac{19.9994}{20} = 1.0000$ $y^{(3)} = \frac{-18 - 3(1.0000) + 0.9998}{20} = \frac{-20.0002}{20} = -1.0000$ $z^{(3)} = \frac{25 - 2(1.0000) + 3(-1.0000)}{20} = \frac{20.0000}{20} = 1.0000$

Summary of iterations

Verification

$20(1) + (-1) - 2(1) = 17 \checkmark$ $3(1) + 20(-1) - 1 = -18 \checkmark$ $2(1) - 3(-1) + 20(1) = 25 \checkmark$

Answer: $x = 1,\; y = -1,\; z = 1$ (correct to four decimal places after three iterations).

(a) pivot element

In Gauss elimination, the diagonal elements $a_{ij}$ with $i=j$ (i.e. $a_{11}, a_{22}, a_{33}, \dots$) are used to eliminate the entries below them, so they are called pivot elements.

Let the output vector be
$X=\begin{bmatrix}x_A\\x_B\end{bmatrix}$ and the final demand vector be
$D=\begin{bmatrix}d_A\\d_B\end{bmatrix}.$

1. Technical coefficient matrix

For industry A: $a_{AA}=\frac{500}{1000}=0.5,\qquad a_{AB}=\frac{350}{800}=0.4375$

For industry B: $a_{BA}=\frac{320}{1000}=0.32,\qquad a_{BB}=\frac{360}{800}=0.45$

So,

$$A=\begin{bmatrix} 0.5 & 0.4375\\ 0.32 & 0.45 \end{bmatrix}$$

2. New final demand

Original final demand: $d_A=150,\qquad d_B=120$

A increases by 10%: $d_A'=150\left(1+\frac{10}{100}\right)=165$

B decreases by 10%: $d_B'=120\left(1-\frac{10}{100}\right)=108$

Thus, $D'=\begin{bmatrix}165\\108\end{bmatrix}$

3. Leontief input-output equation

$(I-A)X=D'$

Now,

$$I-A=\begin{bmatrix} 1-0.5 & -0.4375\\ -0.32 & 1-0.45 \end{bmatrix} = \begin{bmatrix} 0.5 & -0.4375\\ -0.32 & 0.55 \end{bmatrix}$$

Determinant: $|I-A|=(0.5)(0.55)-(-0.4375)(-0.32)$ $=0.275-0.14=0.135$

Inverse:

$$(I-A)^{-1}=\frac{1}{0.135} \begin{bmatrix} 0.55 & 0.4375\\ 0.32 & 0.5 \end{bmatrix}$$

4. Output vector

$X=(I-A)^{-1}D'$

$$X=\frac{1}{0.135} \begin{bmatrix} 0.55 & 0.4375\\ 0.32 & 0.5 \end{bmatrix} \begin{bmatrix} 165\\ 108 \end{bmatrix}$$ $$X=\frac{1}{0.135} \begin{bmatrix} 0.55(165)+0.4375(108)\\ 0.32(165)+0.5(108) \end{bmatrix}$$ $$X=\frac{1}{0.135} \begin{bmatrix} 90.75+47.25\\ 52.8+54 \end{bmatrix}$$ $$X=\frac{1}{0.135} \begin{bmatrix} 138\\ 106.8 \end{bmatrix}$$ $$X= \begin{bmatrix} 1022.22\\ 791.11 \end{bmatrix}$$

Therefore, the required output vector is

$\boxed{X=\begin{bmatrix}1022.22\\791.11\end{bmatrix}}$

So, the total output of industry A is approximately Rs 1022.22 million, and the total output of industry B is approximately Rs 791.11 million.

$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$

This is the Newton–Raphson iterative formula for finding a root of $f(x)=0$, where $f'(x_n)\neq 0$.

Let $f(x) = x^3 - x - 1$. Then $f(1) = -1 < 0$ and $f(2) = 5 > 0$, so a root lies in $[1, 2]$.

The root lies between 1.3125 and 1.3281, so $x \approx 1.32$ (the exact root is about $1.3247$).

(b) eqn (i) + 3 eqn (ii)

Write (i) $3x+4y=18$ and (ii) $-x+3y=7$. Adding 3 times (ii) to (i) gives $(3x-3x)+(4y+9y)=18+21$, i.e. $13y=39$, which removes $x$.

Trapezoidal rule: For $\int_a^b f(x)\,dx$ with $n$ equal sub-intervals of width $h=\dfrac{b-a}{n}$ and $y_i=f(a+ih)$, $\int_a^b f(x)\,dx\approx\frac h2\left[y_0+2(y_1+y_2+\cdots+y_{n-1})+y_n\right].$

Estimate of $\int_1^5x^4\,dx$, $n=4$: $h=\dfrac{5-1}{4}=1$.

$\int_1^5x^4\,dx\approx\frac12\left[1+625+2(16+81+256)\right]=\frac12[626+706]=666.$

(Exact value: $\dfrac{5^5-1}{5}=624.8$, so the trapezoidal estimate overshoots by $41.2$.)

OR

Simpson's 1/3 rule for $\int_0^2(3x+1)\,dx$: take $n=2$, $h=1$, with $y_0=f(0)=1,\ y_1=f(1)=4,\ y_2=f(2)=7$: $\int_0^2(3x+1)\,dx\approx\frac h3\left[y_0+4y_1+y_2\right]=\frac13[1+16+7]=8.$

Exact value: $\left[\dfrac{3x^2}{2}+x\right]_0^2=6+2=8$.

Error $=8-8=0$. Simpson's rule is exact for polynomials up to degree 3, so there is no error here.

(b) No solution

The two lines $x + y = 5$ and $x + y = 7$ are parallel (same slope $-1$, different intercepts), so they never intersect. Hence the system is inconsistent and has no solution.

(d) $y_{t+2} - 2y_{t+1} + y_t$

Since $\Delta y_t = y_{t+1} - y_t$, applying $\Delta$ again gives $\Delta^2 y_t = \Delta(y_{t+1} - y_t) = (y_{t+2} - y_{t+1}) - (y_{t+1} - y_t) = y_{t+2} - 2y_{t+1} + y_t.$

a) Solving by matrix method

The system is

$$x+y+z=4,\qquad 2x+y-3z=-9,\qquad 2x-y+z=-1.$$

Write it as (AX=B):

$$A=\begin{pmatrix} 1&1&1\\ 2&1&-3\\ 2&-1&1 \end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}4\\-9\\-1\end{pmatrix}.$$

Now,

$$\det A =1(1\cdot1-(-3)(-1))-1(2\cdot1-(-3)\cdot2)+1(2(-1)-1\cdot2) =-14\ne0.$$

So the system has a unique solution.

The cofactor matrix is

$$C=\begin{pmatrix} -2&-8&-4\\ -2&-1&3\\ -4&5&-1 \end{pmatrix}.$$

Hence,

$$\operatorname{adj}A=C^T= \begin{pmatrix} -2&-2&-4\\ -8&-1&5\\ -4&3&-1 \end{pmatrix}.$$

Therefore,

$$A^{-1}=\frac{\operatorname{adj}A}{\det A} = \begin{pmatrix} \frac17&\frac17&\frac27\\[2mm] \frac47&\frac1{14}&-\frac5{14}\\[2mm] \frac27&-\frac3{14}&\frac1{14} \end{pmatrix}.$$

Now,

$$X=A^{-1}B = \begin{pmatrix} \frac17&\frac17&\frac27\\[2mm] \frac47&\frac1{14}&-\frac5{14}\\[2mm] \frac27&-\frac3{14}&\frac1{14} \end{pmatrix} \begin{pmatrix}4\\-9\\-1\end{pmatrix} = \begin{pmatrix}-1\\2\\3\end{pmatrix}.$$

Hence,

$$\boxed{x=-1,\quad y=2,\quad z=3}.$$

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b) Using De Moivre’s theorem

$$2(\cos15^\circ+i\sin15^\circ)^6 =2(\cos(6\cdot15^\circ)+i\sin(6\cdot15^\circ))$$ $$=2(\cos90^\circ+i\sin90^\circ) =2(0+i) =2i.$$ $$\boxed{2(\cos15^\circ+i\sin15^\circ)^6=2i}$$

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c) Proof

We know the exponential series:

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

Taking (x=1),

$$1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots=e.$$

Taking (x=-1),

$$1-\frac1{1!}+\frac1{2!}-\frac1{3!}+\cdots=e^{-1}.$$

Therefore,

$$\left(1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots\right) \left(1-\frac1{1!}+\frac1{2!}-\frac1{3!}+\cdots\right) =e\cdot e^{-1}=e^0=1.$$ $$\boxed{1}$$

Hence proved.