IGNOU MMT-009 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMT-009 Assignment Question Paper 2024

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mmt-009-solved-assignment-2024-4b2dc12d-141c-4498-9d86-2c4f18822cb2

MMT-009 Solved Assignment 2024
1. a) A company manufacturing soft drinks is thinking of expanding its plant capacity so as to meet future demand. The monthly sale for the past 5 years are available. State, giving reasons, the type of modelling you will use to obtain good estimates for future demand so as to help the company make the right decisions. Also state four essentials and four non-essentials for the problem.
b) Which one of the following portfolios cannot lie on the efficient frontier as described by Markowitz?
 Portfolio Expected return Standard deviation W $9\mathrm{%}$$9\mathrm{%}$9%9 \%$9\mathrm{%}$ $21\mathrm{%}$$21\mathrm{%}$21%21 \%$21\mathrm{%}$ X $5\mathrm{%}$$5\mathrm{%}$5%5 \%$5\mathrm{%}$ $7\mathrm{%}$$7\mathrm{%}$7%7 \%$7\mathrm{%}$ Y $15\mathrm{%}$$15\mathrm{%}$15%15 \%$15\mathrm{%}$ $36\mathrm{%}$$36\mathrm{%}$36%36 \%$36\mathrm{%}$ Z $12\mathrm{%}$$12\mathrm{%}$12%12 \%$12\mathrm{%}$ $15\mathrm{%}$$15\mathrm{%}$15%15 \%$15\mathrm{%}$
Portfolio Expected return Standard deviation W 9% 21% X 5% 7% Y 15% 36% Z 12% 15%| Portfolio | Expected return | Standard deviation | | :—: | :—: | :—: | | W | $9 \%$ | $21 \%$ | | X | $5 \%$ | $7 \%$ | | Y | $15 \%$ | $36 \%$ | | Z | $12 \%$ | $15 \%$ |
1. a) Let $\mathrm{G}\left(\mathrm{t}\right)$$\mathrm{G}\left(\mathrm{t}\right)$G(t)\mathrm{G}(\mathrm{t})$\mathrm{G}\left(\mathrm{t}\right)$ be the amount of the glucose in the bloodstream of a patient at time $t$$t$tt$t$. Assume that the glucose is infused into the bloodstream at a constant rate of $\mathrm{k}\mathrm{g}/\mathrm{m}\mathrm{i}\mathrm{n}$$\mathrm{k}\mathrm{g}/\mathrm{m}\mathrm{i}\mathrm{n}$kg//min\mathrm{kg} / \mathrm{min}$\mathrm{k}\mathrm{g}/\mathrm{m}\mathrm{i}\mathrm{n}$. At the same time, the glucose is converted and removed from the bloodstream at a rate proportional to the amount of the glucose present. If at $\mathrm{t}=0,\mathrm{G}=\mathrm{G}\left(0\right)$$\mathrm{t}=0,\mathrm{G}=\mathrm{G}\left(0\right)$t=0,G=G(0)\mathrm{t}=0, \mathrm{G}=\mathrm{G}(0)$\mathrm{t}=0,\mathrm{G}=\mathrm{G}\left(0\right)$ then
i) formulate the model.
ii) find $g\left(t\right)$$g\left(t\right)$g(t)g(t)$g\left(t\right)$ at any time $t$$t$tt$t$.
iii) discuss the long term behavior of the model.
b) Tumour is developing from the organ of a human body with concentration $3.2×{10}^{9}$$3.2×{10}^{9}$3.2 xx10^(9)3.2 \times 10^9$3.2×{10}^{9}$ with growth and decay control parameters 9.2 and 2.7 respectively. In how many days the size of the tumor will be twice?
2. a) Return distributions of the two securities are given below:
 Return Probabilities $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ ${\mathrm{p}}_{\mathrm{x}\mathrm{j}}={\mathrm{p}}_{\mathrm{y}\mathrm{j}}={\mathrm{p}}_{\mathrm{j}}$${\mathrm{p}}_{\mathrm{x}\mathrm{j}}={\mathrm{p}}_{\mathrm{y}\mathrm{j}}={\mathrm{p}}_{\mathrm{j}}$p_(xj)=p_(yj)=p_(j)\mathrm{p}_{\mathrm{xj}}=\mathrm{p}_{\mathrm{yj}}=\mathrm{p}_{\mathrm{j}}${\mathrm{p}}_{\mathrm{x}\mathrm{j}}={\mathrm{p}}_{\mathrm{y}\mathrm{j}}={\mathrm{p}}_{\mathrm{j}}$ 0.16 0.14 0.33 0.12 0.08 0.25 0.08 0.05 0.17 0.11 0.09 0.25
Return Probabilities X Y p_(xj)=p_(yj)=p_(j) 0.16 0.14 0.33 0.12 0.08 0.25 0.08 0.05 0.17 0.11 0.09 0.25| Return | | Probabilities | | :—: | :—: | :—: | | $\mathrm{X}$ | $\mathrm{Y}$ | $\mathrm{p}_{\mathrm{xj}}=\mathrm{p}_{\mathrm{yj}}=\mathrm{p}_{\mathrm{j}}$ | | 0.16 | 0.14 | 0.33 | | 0.12 | 0.08 | 0.25 | | 0.08 | 0.05 | 0.17 | | 0.11 | 0.09 | 0.25 |
Find which security is more risky in the Markowitz sense. Also find the correlation coefficient of securities $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ and $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$.
b) Let $P=\left({w}_{1},{w}_{2}\right)$$P=\left({w}_{1},{w}_{2}\right)$P=(w_(1),w_(2))P=\left(w_1, w_2\right)$P=\left({w}_{1},{w}_{2}\right)$ be a portfolio of two securities $X$$X$XX$X$ and $Y$$Y$YY$Y$. Find the values of ${w}_{1}$${w}_{1}$w_(1)w_1${w}_{1}$ and ${w}_{2}$${w}_{2}$w_(2)w_2${w}_{2}$ in the following situations:
i) ${\rho }_{\mathrm{x}\mathrm{y}}=-1$${\rho }_{\mathrm{x}\mathrm{y}}=-1$rho_(xy)=-1\rho_{\mathrm{xy}}=-1${\rho }_{\mathrm{x}\mathrm{y}}=-1$ and $\mathrm{P}$$\mathrm{P}$P\mathrm{P}$\mathrm{P}$ is risk free.
ii) $\phantom{\rule{1em}{0ex}}{\sigma }_{x}={\sigma }_{y}$$\phantom{\rule{1em}{0ex}}{\sigma }_{x}={\sigma }_{y}$quadsigma _(x)=sigma _(y)\quad \sigma_x=\sigma_y$\phantom{\rule{1em}{0ex}}{\sigma }_{x}={\sigma }_{y}$ and variance $P$$P$PP$P$ is minimum.
iii) Variance $P$$P$PP$P$ is minimum and ${\rho }_{xy}=-0.5,{\sigma }_{x}=2$${\rho }_{xy}=-0.5,{\sigma }_{x}=2$rho_(xy)=-0.5,sigma _(x)=2\rho_{x y}=-0.5, \sigma_x=2${\rho }_{xy}=-0.5,{\sigma }_{x}=2$ and ${\sigma }_{y}=3$${\sigma }_{y}=3$sigma _(y)=3\sigma_y=3${\sigma }_{y}=3$.
4. a) Companies considering the purchase of a computer must first assess their future needs in order to determine the proper equipment. A computer scientist collected data from seven similar company sites so that computer hardware requirements for inventory management could be developed. The data collected is as follows:
 Customer Orders (in thousands)
Customer Orders (in thousands)| Customer Orders | | :—: | | (in thousands) |
Add-delete items (in thousands)| Add-delete items | | :—: | | (in thousands) |
 CPU time (in hours)
CPU time (in hours)| CPU time | | :—: | | (in hours) |
123.5 2.108 141.5
146.1 9.213 168.9
133.9 1.905 154.8
128.5 0.815 146.5
151.5 1.061 172.8
136.2 8.603 160.1
92.0 1.125 108.5
“Customer Orders (in thousands)” “Add-delete items (in thousands)” “CPU time (in hours)” 123.5 2.108 141.5 146.1 9.213 168.9 133.9 1.905 154.8 128.5 0.815 146.5 151.5 1.061 172.8 136.2 8.603 160.1 92.0 1.125 108.5| Customer Orders <br> (in thousands) | Add-delete items <br> (in thousands) | CPU time <br> (in hours) | | :—: | :—: | :—: | | 123.5 | 2.108 | 141.5 | | 146.1 | 9.213 | 168.9 | | 133.9 | 1.905 | 154.8 | | 128.5 | 0.815 | 146.5 | | 151.5 | 1.061 | 172.8 | | 136.2 | 8.603 | 160.1 | | 92.0 | 1.125 | 108.5 |
i) Find a linear regression equation that best fit the data.
ii) Estimate the error variance for the regression model obtained in i) above.
(2)
b) The population consisting of all married couples is collected. The data showing the age of 12 married couples is as follows:
 Husband’s age (years)
Husband’s age (years)| Husband’s age | | :—: | | (years) |
 Wife’s age (years)
Wife’s age (years)| Wife’s age | | :—: | | (years) |
 Husband’s age (years)
Husband’s age (years)| Husband’s age | | :—: | | (years) |
 Wife’s age (years)
Wife’s age (years)| Wife’s age | | :—: | | (years) |
30 27 51 50
29 20 48 46
36 34 37 36
72 67 50 42
37 35 51 46
36 37 36 35
“Husband’s age (years)” “Wife’s age (years)” “Husband’s age (years)” “Wife’s age (years)” 30 27 51 50 29 20 48 46 36 34 37 36 72 67 50 42 37 35 51 46 36 37 36 35| Husband’s age <br> (years) | Wife’s age <br> (years) | Husband’s age <br> (years) | Wife’s age <br> (years) | | :—: | :—: | :—: | :—: | | 30 | 27 | 51 | 50 | | 29 | 20 | 48 | 46 | | 36 | 34 | 37 | 36 | | 72 | 67 | 50 | 42 | | 37 | 35 | 51 | 46 | | 36 | 37 | 36 | 35 |
i) Draw a scatter plot of the data
ii) Write two important characteristics of the data that emerge from the scatter plot.
iii) Fit a linear regression model to the data and interpret the result in terms of the comparative change in the age of husband and wife.
iv) Calculate the standard error of regression and the coefficient of determination for the data.
1. Consider a discrete model given by
${\mathrm{N}}_{\mathrm{t}+1}=\frac{{\mathrm{r}\mathrm{N}}_{\mathrm{t}}}{1+{\mathrm{b}\mathrm{N}}_{\mathrm{t}-1}^{2}}=\mathrm{f}\left({\mathrm{N}}_{\mathrm{t}}\right),\mathrm{r}>1.$${\mathrm{N}}_{\mathrm{t}+1}=\frac{{\mathrm{r}\mathrm{N}}_{\mathrm{t}}}{1+{\mathrm{b}\mathrm{N}}_{\mathrm{t}-1}^{2}}=\mathrm{f}\left({\mathrm{N}}_{\mathrm{t}}\right),\mathrm{r}>1.$N_(t+1)=(rN_(t))/(1+bN_(t-1)^(2))=f(N_(t)),r > 1.\mathrm{N}_{\mathrm{t}+1}=\frac{\mathrm{rN}_{\mathrm{t}}}{1+\mathrm{bN}_{\mathrm{t}-1}^2}=\mathrm{f}\left(\mathrm{N}_{\mathrm{t}}\right), \mathrm{r}>1 .${\mathrm{N}}_{\mathrm{t}+1}=\frac{{\mathrm{r}\mathrm{N}}_{\mathrm{t}}}{1+{\mathrm{b}\mathrm{N}}_{\mathrm{t}-1}^{2}}=\mathrm{f}\left({\mathrm{N}}_{\mathrm{t}}\right),\mathrm{r}>1.$
Investigate the linear stability about the positive steady state ${\mathrm{N}}^{\ast }$${\mathrm{N}}^{\ast }$N^(**)\mathrm{N}^*${\mathrm{N}}^{\ast }$ by setting ${\mathrm{N}}_{\mathrm{t}}={\mathrm{N}}^{\ast }+{\mathrm{n}}_{\mathrm{t}}$${\mathrm{N}}_{\mathrm{t}}={\mathrm{N}}^{\ast }+{\mathrm{n}}_{\mathrm{t}}$N_(t)=N^(**)+n_(t)\mathrm{N}_{\mathrm{t}}=\mathrm{N}^*+\mathrm{n}_{\mathrm{t}}${\mathrm{N}}_{\mathrm{t}}={\mathrm{N}}^{\ast }+{\mathrm{n}}_{\mathrm{t}}$. Show that ${\mathrm{n}}_{\mathrm{t}}$${\mathrm{n}}_{\mathrm{t}}$n_(t)\mathrm{n}_{\mathrm{t}}${\mathrm{n}}_{\mathrm{t}}$ satisfies the equation
${\mathrm{n}}_{\mathrm{t}+1}-{\mathrm{n}}_{\mathrm{t}}+2\left(\mathrm{r}-1\right){\mathrm{r}}^{-1}{\mathrm{n}}_{\mathrm{t}-1}=0.$${\mathrm{n}}_{\mathrm{t}+1}-{\mathrm{n}}_{\mathrm{t}}+2\left(\mathrm{r}-1\right){\mathrm{r}}^{-1}{\mathrm{n}}_{\mathrm{t}-1}=0.$n_(t+1)-n_(t)+2(r-1)r^(-1)n_(t-1)=0.\mathrm{n}_{\mathrm{t}+1}-\mathrm{n}_{\mathrm{t}}+2(\mathrm{r}-1) \mathrm{r}^{-1} \mathrm{n}_{\mathrm{t}-1}=0 .${\mathrm{n}}_{\mathrm{t}+1}-{\mathrm{n}}_{\mathrm{t}}+2\left(\mathrm{r}-1\right){\mathrm{r}}^{-1}{\mathrm{n}}_{\mathrm{t}-1}=0.$
Hence show that $\mathrm{r}=2$$\mathrm{r}=2$r=2\mathrm{r}=2$\mathrm{r}=2$ is a bifurcation value and that as $\mathrm{r}\to 2$$\mathrm{r}\to 2$rrarr2\mathrm{r} \rightarrow 2$\mathrm{r}\to 2$ the steady state bifurcates to a periodic solution of period 6.
6. a) The population dynamics of a species is governed by the discrete model
${x}_{n+1}={x}_{n}\mathrm{exp}\left[r\left(1-\frac{{x}_{n}}{K}\right)\right],$${x}_{n+1}={x}_{n}\mathrm{exp}\left[r\left(1-\frac{{x}_{n}}{K}\right)\right],$x_(n+1)=x_(n)exp[r(1-(x_(n))/(K))],x_{n+1}=x_n \exp \left[r\left(1-\frac{x_n}{K}\right)\right],${x}_{n+1}={x}_{n}\mathrm{exp}\left[r\left(1-\frac{{x}_{n}}{K}\right)\right],$
where $\mathrm{r}$$\mathrm{r}$r\mathrm{r}$\mathrm{r}$ and $\mathrm{k}$$\mathrm{k}$k\mathrm{k}$\mathrm{k}$ are positive constants. Determine the steady states and discuss the stability of the model. Find the value of $\mathrm{r}$$\mathrm{r}$r\mathrm{r}$\mathrm{r}$ at which first bifurcation occurs. Describe qualitatively the behaviours of the population for $\mathrm{r}=2+\epsilon$$\mathrm{r}=2+\epsilon$r=2+epsi\mathrm{r}=2+\varepsilon$\mathrm{r}=2+\epsilon$, where $0<\epsilon <<1$$0<\epsilon <<1$0 < epsi<<10<\varepsilon<<1$0<\epsilon <<1$. Since a species becomes extinct if ${x}_{n}\le 1$${x}_{n}\le 1$x_(n) <= 1x_n \leq 1${x}_{n}\le 1$ for any $n>1$$n>1$n > 1n>1$n>1$, show using iterations, that irrespective of the size of $r>1$$r>1$r > 1r>1$r>1$ the species could become extinct if the carrying capacity $\mathrm{k}<\mathrm{r}\mathrm{exp}\left[1+{\mathrm{e}}^{\mathrm{r}-1}-2\mathrm{r}\right]$$\mathrm{k}<\mathrm{r}\mathrm{exp}\left[1+{\mathrm{e}}^{\mathrm{r}-1}-2\mathrm{r}\right]$k < rexp[1+e^(r-1)-2r]\mathrm{k}<\mathrm{r} \exp \left[1+\mathrm{e}^{\mathrm{r}-1}-2 \mathrm{r}\right]$\mathrm{k}<\mathrm{r}\mathrm{exp}\left[1+{\mathrm{e}}^{\mathrm{r}-1}-2\mathrm{r}\right]$.
b) Do the stability analysis of the following model formulated to study the effect of toxicant on prey-predator population and interpret the solution.
$\begin{array}{rl}& \frac{{\mathrm{d}\mathrm{N}}_{1}}{\mathrm{d}\mathrm{t}}={\mathrm{r}}_{0}{\text{}\mathrm{N}}_{1}-{\mathrm{r}}_{1}{\mathrm{C}}_{0}{\text{}\mathrm{N}}_{1}-{\mathrm{b}\mathrm{N}}_{1}{\text{}\mathrm{N}}_{2}\\ & \frac{\mathrm{d}\mathrm{N}}{\mathrm{d}\mathrm{t}}=-{\mathrm{d}}_{0}{\text{}\mathrm{N}}_{2}-{\mathrm{d}}_{1}{\text{}\mathrm{V}}_{0}{\text{}\mathrm{N}}_{2}+{\beta }_{0}{\mathrm{b}\mathrm{N}}_{1}{\text{}\mathrm{N}}_{2}\\ & \frac{\mathrm{d}\mathrm{C}0}{\mathrm{d}\mathrm{t}}={\mathrm{k}}_{1}\mathrm{P}-{\mathrm{g}}_{1}{\mathrm{C}}_{0}-{\mathrm{m}}_{1}{\mathrm{C}}_{0}\\ & \frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{t}}={\mathrm{k}}_{2}\mathrm{P}-{\mathrm{g}}_{2}{\text{}\mathrm{V}}_{0}-{\mathrm{m}}_{2}{\text{}\mathrm{V}}_{0}\\ & \frac{\mathrm{d}\mathrm{P}}{\mathrm{d}\mathrm{t}}=\mathrm{Q}-\mathrm{h}\mathrm{P}-\mathrm{k}\mathrm{P}\left({\mathrm{N}}_{1}+{\mathrm{N}}_{2}\right)+{\mathrm{g}\mathrm{C}}_{0}{\text{}\mathrm{N}}_{1}+\ell {\mathrm{V}}_{0}{\text{}\mathrm{N}}_{2}.\end{array}$$\begin{array}{rl}& \frac{{\mathrm{d}\mathrm{N}}_{1}}{\mathrm{d}\mathrm{t}}={\mathrm{r}}_{0}{\text{}\mathrm{N}}_{1}-{\mathrm{r}}_{1}{\mathrm{C}}_{0}{\text{}\mathrm{N}}_{1}-{\mathrm{b}\mathrm{N}}_{1}{\text{}\mathrm{N}}_{2}\\ & \frac{\mathrm{d}\mathrm{N}}{\mathrm{d}\mathrm{t}}=-{\mathrm{d}}_{0}{\text{}\mathrm{N}}_{2}-{\mathrm{d}}_{1}{\text{}\mathrm{V}}_{0}{\text{}\mathrm{N}}_{2}+{\beta }_{0}{\mathrm{b}\mathrm{N}}_{1}{\text{}\mathrm{N}}_{2}\\ & \frac{\mathrm{d}\mathrm{C}0}{\mathrm{d}\mathrm{t}}={\mathrm{k}}_{1}\mathrm{P}-{\mathrm{g}}_{1}{\mathrm{C}}_{0}-{\mathrm{m}}_{1}{\mathrm{C}}_{0}\\ & \frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{t}}={\mathrm{k}}_{2}\mathrm{P}-{\mathrm{g}}_{2}{\text{}\mathrm{V}}_{0}-{\mathrm{m}}_{2}{\text{}\mathrm{V}}_{0}\\ & \frac{\mathrm{d}\mathrm{P}}{\mathrm{d}\mathrm{t}}=\mathrm{Q}-\mathrm{h}\mathrm{P}-\mathrm{k}\mathrm{P}\left({\mathrm{N}}_{1}+{\mathrm{N}}_{2}\right)+{\mathrm{g}\mathrm{C}}_{0}{\text{}\mathrm{N}}_{1}+\ell {\mathrm{V}}_{0}{\text{}\mathrm{N}}_{2}.\end{array}${:[(dN_(1))/(dt)=r_(0)N_(1)-r_(1)C_(0)N_(1)-bN_(1)N_(2)],[(dN)/(dt)=-d_(0)N_(2)-d_(1)V_(0)N_(2)+beta_(0)bN_(1)N_(2)],[(dC0)/(dt)=k_(1)P-g_(1)C_(0)-m_(1)C_(0)],[(dV)/(dt)=k_(2)P-g_(2)V_(0)-m_(2)V_(0)],[(dP)/(dt)=Q-hP-kP(N_(1)+N_(2))+gC_(0)N_(1)+ℓV_(0)N_(2).]:}\begin{aligned} & \frac{\mathrm{dN}_1}{\mathrm{dt}}=\mathrm{r}_0 \mathrm{~N}_1-\mathrm{r}_1 \mathrm{C}_0 \mathrm{~N}_1-\mathrm{bN}_1 \mathrm{~N}_2 \\ & \frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{d}_0 \mathrm{~N}_2-\mathrm{d}_1 \mathrm{~V}_0 \mathrm{~N}_2+\beta_0 \mathrm{bN}_1 \mathrm{~N}_2 \\ & \frac{\mathrm{dC} 0}{\mathrm{dt}}=\mathrm{k}_1 \mathrm{P}-\mathrm{g}_1 \mathrm{C}_0-\mathrm{m}_1 \mathrm{C}_0 \\ & \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{k}_2 \mathrm{P}-\mathrm{g}_2 \mathrm{~V}_0-\mathrm{m}_2 \mathrm{~V}_0 \\ & \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{Q}-\mathrm{hP}-\mathrm{kP}\left(\mathrm{N}_1+\mathrm{N}_2\right)+\mathrm{gC}_0 \mathrm{~N}_1+\ell \mathrm{V}_0 \mathrm{~N}_2 . \end{aligned}$\begin{array}{rl}& \frac{{\mathrm{d}\mathrm{N}}_{1}}{\mathrm{d}\mathrm{t}}={\mathrm{r}}_{0}{\text{}\mathrm{N}}_{1}-{\mathrm{r}}_{1}{\mathrm{C}}_{0}{\text{}\mathrm{N}}_{1}-{\mathrm{b}\mathrm{N}}_{1}{\text{}\mathrm{N}}_{2}\\ & \frac{\mathrm{d}\mathrm{N}}{\mathrm{d}\mathrm{t}}=-{\mathrm{d}}_{0}{\text{}\mathrm{N}}_{2}-{\mathrm{d}}_{1}{\text{}\mathrm{V}}_{0}{\text{}\mathrm{N}}_{2}+{\beta }_{0}{\mathrm{b}\mathrm{N}}_{1}{\text{}\mathrm{N}}_{2}\\ & \frac{\mathrm{d}\mathrm{C}0}{\mathrm{d}\mathrm{t}}={\mathrm{k}}_{1}\mathrm{P}-{\mathrm{g}}_{1}{\mathrm{C}}_{0}-{\mathrm{m}}_{1}{\mathrm{C}}_{0}\\ & \frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{t}}={\mathrm{k}}_{2}\mathrm{P}-{\mathrm{g}}_{2}{\text{}\mathrm{V}}_{0}-{\mathrm{m}}_{2}{\text{}\mathrm{V}}_{0}\\ & \frac{\mathrm{d}\mathrm{P}}{\mathrm{d}\mathrm{t}}=\mathrm{Q}-\mathrm{h}\mathrm{P}-\mathrm{k}\mathrm{P}\left({\mathrm{N}}_{1}+{\mathrm{N}}_{2}\right)+{\mathrm{g}\mathrm{C}}_{0}{\text{}\mathrm{N}}_{1}+\ell {\mathrm{V}}_{0}{\text{}\mathrm{N}}_{2}.\end{array}$
Where all the variables and constants are same as defined in the system (32)-(35) except for the following
$\mathrm{Q}=$$\mathrm{Q}=$Q=\mathrm{Q}=$\mathrm{Q}=$ constant input rate
$\mathbf{h}=$$\mathbf{h}=$h=\mathbf{h}=$\mathbf{h}=$ decay rate
$\mathrm{P}\left(\mathrm{t}\right)=$$\mathrm{P}\left(\mathrm{t}\right)=$P(t)=\mathrm{P}(\mathrm{t})=$\mathrm{P}\left(\mathrm{t}\right)=$ environmental toxicant concentration
$\mathrm{k}=$$\mathrm{k}=$k=\mathrm{k}=$\mathrm{k}=$ ingestion rate of toxicant by the populations
$\mathrm{g},\ell =$$\mathrm{g},\ell =$g,ℓ=\mathrm{g}, \ell=$\mathrm{g},\ell =$ return rate of toxicant in the environment after the death of the populations, assuming that toxicant is non-degradable
$\mathrm{Q}>0,\text{}\mathrm{h},\mathrm{k},\mathrm{g},\ell$$\mathrm{Q}>0,\text{}\mathrm{h},\mathrm{k},\mathrm{g},\ell$Q > 0,h,k,g,ℓ\mathrm{Q}>0, \mathrm{~h}, \mathrm{k}, \mathrm{g}, \ell$\mathrm{Q}>0,\text{}\mathrm{h},\mathrm{k},\mathrm{g},\ell$ are positive constants.
1. Do the stability analysis of the following competing species system of equations with diffusion and advection
$\begin{array}{rl}& \frac{\mathrm{\partial }{N}_{1}}{\mathrm{\partial }t}={a}_{1}{N}_{1}-{b}_{1}{N}_{1}{N}_{2}+{D}_{1}\frac{{\mathrm{\partial }}^{2}{N}_{1}}{\mathrm{\partial }{x}^{2}}-{v}_{1}\frac{\mathrm{\partial }{N}_{1}}{\mathrm{\partial }x}\\ & \frac{\mathrm{\partial }{N}_{2}}{\mathrm{\partial }t}=-{d}_{1}{N}_{2}+{C}_{1}{N}_{1}{N}_{2}+{D}_{2}\frac{{\mathrm{\partial }}^{2}{N}_{2}}{\mathrm{\partial }{x}^{2}}-{V}_{2}\frac{\mathrm{\partial }{N}_{2}}{\mathrm{\partial }x},\phantom{\rule{1em}{0ex}}0\le x\le L\end{array}$$\begin{array}{rl}& \frac{\mathrm{\partial }{N}_{1}}{\mathrm{\partial }t}={a}_{1}{N}_{1}-{b}_{1}{N}_{1}{N}_{2}+{D}_{1}\frac{{\mathrm{\partial }}^{2}{N}_{1}}{\mathrm{\partial }{x}^{2}}-{v}_{1}\frac{\mathrm{\partial }{N}_{1}}{\mathrm{\partial }x}\\ & \frac{\mathrm{\partial }{N}_{2}}{\mathrm{\partial }t}=-{d}_{1}{N}_{2}+{C}_{1}{N}_{1}{N}_{2}+{D}_{2}\frac{{\mathrm{\partial }}^{2}{N}_{2}}{\mathrm{\partial }{x}^{2}}-{V}_{2}\frac{\mathrm{\partial }{N}_{2}}{\mathrm{\partial }x},\phantom{\rule{1em}{0ex}}0\le x\le L\end{array}${:[(delN_(1))/(del t)=a_(1)N_(1)-b_(1)N_(1)N_(2)+D_(1)(del^(2)N_(1))/(delx^(2))-v_(1)(delN_(1))/(del x)],[(delN_(2))/(del t)=-d_(1)N_(2)+C_(1)N_(1)N_(2)+D_(2)(del^(2)N_(2))/(delx^(2))-V_(2)(delN_(2))/(del x)”,”quad0 <= x <= L]:}\begin{aligned} & \frac{\partial N_1}{\partial t}=a_1 N_1-b_1 N_1 N_2+D_1 \frac{\partial^2 N_1}{\partial x^2}-v_1 \frac{\partial N_1}{\partial x} \\ & \frac{\partial N_2}{\partial t}=-d_1 N_2+C_1 N_1 N_2+D_2 \frac{\partial^2 N_2}{\partial x^2}-V_2 \frac{\partial N_2}{\partial x}, \quad 0 \leq x \leq L \end{aligned}$\begin{array}{rl}& \frac{\mathrm{\partial }{N}_{1}}{\mathrm{\partial }t}={a}_{1}{N}_{1}-{b}_{1}{N}_{1}{N}_{2}+{D}_{1}\frac{{\mathrm{\partial }}^{2}{N}_{1}}{\mathrm{\partial }{x}^{2}}-{v}_{1}\frac{\mathrm{\partial }{N}_{1}}{\mathrm{\partial }x}\\ & \frac{\mathrm{\partial }{N}_{2}}{\mathrm{\partial }t}=-{d}_{1}{N}_{2}+{C}_{1}{N}_{1}{N}_{2}+{D}_{2}\frac{{\mathrm{\partial }}^{2}{N}_{2}}{\mathrm{\partial }{x}^{2}}-{V}_{2}\frac{\mathrm{\partial }{N}_{2}}{\mathrm{\partial }x},\phantom{\rule{1em}{0ex}}0\le x\le L\end{array}$
where ${V}_{1}$${V}_{1}$V_(1)V_1${V}_{1}$ and ${V}_{2}$${V}_{2}$V_(2)V_2${V}_{2}$ are advection velocities in $x$$x$xx$x$ direction of the two populations with densities ${N}_{1}$${N}_{1}$N_(1)N_1${N}_{1}$ and ${\mathrm{N}}_{2}$${\mathrm{N}}_{2}$N_(2)\mathrm{N}_2${\mathrm{N}}_{2}$ respectively. ${\mathrm{a}}_{1}$${\mathrm{a}}_{1}$a_(1)\mathrm{a}_1${\mathrm{a}}_{1}$ is the growth rate, ${\mathrm{b}}_{1}$${\mathrm{b}}_{1}$b_(1)\mathrm{b}_1${\mathrm{b}}_{1}$ is the predation rate, ${\mathrm{d}}_{1}$${\mathrm{d}}_{1}$d_(1)\mathrm{d}_1${\mathrm{d}}_{1}$ is the death rate, ${\mathrm{C}}_{1}$${\mathrm{C}}_{1}$C_(1)\mathrm{C}_1${\mathrm{C}}_{1}$ is the conversion rate. ${\mathrm{D}}_{1}$${\mathrm{D}}_{1}$D_(1)\mathrm{D}_1${\mathrm{D}}_{1}$ and ${\mathrm{D}}_{2}$${\mathrm{D}}_{2}$D_(2)\mathrm{D}_2${\mathrm{D}}_{2}$ are diffusion coefficients. The initial and boundary conditions are:
$\begin{array}{rl}& {\mathrm{N}}_{\mathrm{i}}\left(\mathrm{x},0\right)={\mathrm{f}}_{\mathrm{i}}\left(\mathrm{x}\right)>0,0\le \mathrm{x}\le \mathrm{L},\mathrm{i}=1,2.\\ & {\mathrm{N}}_{\mathrm{i}}={\overline{\mathrm{N}}}_{\mathrm{i}}\text{at}\mathrm{x}=0\text{and}\mathrm{x}=\mathrm{L}\mathrm{\forall }\mathrm{t},\mathrm{i}=1,2.\end{array}$$\begin{array}{rl}& {\mathrm{N}}_{\mathrm{i}}\left(\mathrm{x},0\right)={\mathrm{f}}_{\mathrm{i}}\left(\mathrm{x}\right)>0,0\le \mathrm{x}\le \mathrm{L},\mathrm{i}=1,2.\\ & {\mathrm{N}}_{\mathrm{i}}={\overline{\mathrm{N}}}_{\mathrm{i}}\text{at}\mathrm{x}=0\text{and}\mathrm{x}=\mathrm{L}\mathrm{\forall }\mathrm{t},\mathrm{i}=1,2.\end{array}${:[N_(i)(x”,”0)=f_(i)(x) > 0″,”0 <= x <= L”,”i=1″,”2.],[N_(i)= bar(N)_(i)” at “x=0” and “x=LAAt”,”i=1″,”2.]:}\begin{aligned} & \mathrm{N}_{\mathrm{i}}(\mathrm{x}, 0)=\mathrm{f}_{\mathrm{i}}(\mathrm{x})>0,0 \leq \mathrm{x} \leq \mathrm{L}, \mathrm{i}=1,2 . \\ & \mathrm{N}_{\mathrm{i}}=\overline{\mathrm{N}}_{\mathrm{i}} \text { at } \mathrm{x}=0 \text { and } \mathrm{x}=\mathrm{L} \forall \mathrm{t}, \mathrm{i}=1,2 . \end{aligned}$\begin{array}{rl}& {\mathrm{N}}_{\mathrm{i}}\left(\mathrm{x},0\right)={\mathrm{f}}_{\mathrm{i}}\left(\mathrm{x}\right)>0,0\le \mathrm{x}\le \mathrm{L},\mathrm{i}=1,2.\\ & {\mathrm{N}}_{\mathrm{i}}={\overline{\mathrm{N}}}_{\mathrm{i}}\text{at}\mathrm{x}=0\text{and}\mathrm{x}=\mathrm{L}\mathrm{\forall }\mathrm{t},\mathrm{i}=1,2.\end{array}$
where ${\overline{\mathrm{N}}}_{\mathrm{i}}$${\overline{\mathrm{N}}}_{\mathrm{i}}$bar(N)_(i)\overline{\mathrm{N}}_{\mathrm{i}}${\overline{\mathrm{N}}}_{\mathrm{i}}$ are the equilibrium solutions of the given system of equations.
Interpret the solution obtained and also write the limitations of the model.
1. a) Maximize $\mathrm{Z}=3{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+3{\mathrm{x}}_{3}$$\mathrm{Z}=3{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+3{\mathrm{x}}_{3}$Z=3x_(1)+x_(2)+3x_(3)\mathrm{Z}=3 \mathrm{x}_1+\mathrm{x}_2+3 \mathrm{x}_3$\mathrm{Z}=3{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+3{\mathrm{x}}_{3}$, subject to the constraints $-{x}_{1}+2{x}_{2}+{x}_{3}\le 4,4{x}_{2}-3{x}_{3}\le 2,{x}_{1}-3{x}_{2}+2{x}_{3}\le 3$$-{x}_{1}+2{x}_{2}+{x}_{3}\le 4,4{x}_{2}-3{x}_{3}\le 2,{x}_{1}-3{x}_{2}+2{x}_{3}\le 3$-x_(1)+2x_(2)+x_(3) <= 4,4x_(2)-3x_(3) <= 2,x_(1)-3x_(2)+2x_(3) <= 3-x_1+2 x_2+x_3 \leq 4,4 x_2-3 x_3 \leq 2, x_1-3 x_2+2 x_3 \leq 3$-{x}_{1}+2{x}_{2}+{x}_{3}\le 4,4{x}_{2}-3{x}_{3}\le 2,{x}_{1}-3{x}_{2}+2{x}_{3}\le 3$ and $\left({x}_{1},{x}_{2},{x}_{3}\right)\ge 0$$\left({x}_{1},{x}_{2},{x}_{3}\right)\ge 0$(x_(1),x_(2),x_(3)) >= 0\left(x_1, x_2, x_3\right) \geq 0$\left({x}_{1},{x}_{2},{x}_{3}\right)\ge 0$ and are integers.
b) Ships arrive at a port at the rate of one in every 4 hours with exponential distribution of interarrival times. The time a ship occupies a berth for unloading has exponential distribution with an average of 10 hours. If the average delay of ships waiting for berths is to be kept below 14 hours, how many berths should be provided at the port?
c) A library wants to improve its service facilities in terms of the waiting time of its borrowers. The library has two counters at present and borrowers arrive according to Poisson distribution with arrival rate 1 every 6 minutes and service time follows exponential distribution with a mean of 10 minutes. The library has relaxed its membership rules and a substantial increase in the number of borrowers is expected. Find the number of additional counters to be provided if the arrival rate is expected to be twice the present value and the average waiting time of the borrower must be limited to half the present value.
$$2\:cos\:\theta \:cos\:\phi =cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$

MMT-009 Sample Solution 2024

mmt-009-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

mmt-009-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

MMT-009 Solved Assignment 2024 SS
1. a) A company manufacturing soft drinks is thinking of expanding its plant capacity so as to meet future demand. The monthly sale for the past 5 years are available. State, giving reasons, the type of modelling you will use to obtain good estimates for future demand so as to help the company make the right decisions. Also state four essentials and four non-essentials for the problem.
For predicting future demand based on past sales data, a time series analysis model would be an appropriate choice. Time series analysis takes into account the temporal nature of the data, allowing for the identification of trends, seasonal patterns, and other temporal dynamics that are crucial for accurate forecasting.
Reasons for choosing time series analysis:
1. Trend Analysis: Time series models can capture underlying trends in the sales data, which is essential for long-term planning and forecasting.
2. Seasonality: Soft drink sales are likely to exhibit seasonal patterns (e.g., higher sales in summer). Time series models can account for this seasonality, providing more accurate estimates for future demand.
3. Forecasting: Time series models are specifically designed for forecasting future values based on past observations, making them well-suited for this application.
4. Historical Data: The company has 5 years of monthly sales data, providing a sufficient historical record for modeling and forecasting purposes.
Four essentials for the problem:
1. Historical Sales Data: Detailed monthly sales data for the past 5 years is essential for building and training the time series model.
2. Seasonal Factors: Understanding seasonal variations in soft drink sales (e.g., higher sales in summer) is crucial for accurate forecasting.
3. Trend Analysis: Identifying any long-term trends in the sales data (e.g., increasing or decreasing demand) is important for future planning.
4. Model Validation: Splitting the data into training and testing sets is essential for validating the accuracy of the time series model.
Four non-essentials for the problem:
1. Daily Sales Data: While potentially useful, daily data may not be necessary for monthly demand forecasting and could complicate the analysis.
2. Individual Product Sales: If the company is interested in overall plant capacity, detailed sales data for individual soft drink products may not be essential.
3. Competitor Sales Data: While competitor information can be useful for strategic planning, it may not be essential for forecasting the company’s own demand.
4. External Economic Indicators: Economic indicators like GDP or consumer spending indices may not be necessary for the specific task of forecasting soft drink demand, although they could provide additional context.
Overall, a time series analysis model that takes into account trends and seasonality, using the past 5 years of monthly sales data, would be a suitable approach for estimating future demand and assisting the company in making informed expansion decisions.
b) Which one of the following portfolios cannot lie on the efficient frontier as described by Markowitz?
 Portfolio Expected return Standard deviation W $9\mathrm{%}$$9\mathrm{%}$9%9 \%$9\mathrm{%}$ $21\mathrm{%}$$21\mathrm{%}$21%21 \%$21\mathrm{%}$ X $5\mathrm{%}$$5\mathrm{%}$5%5 \%$5\mathrm{%}$ $7\mathrm{%}$$7\mathrm{%}$7%7 \%$7\mathrm{%}$ Y $15\mathrm{%}$$15\mathrm{%}$15%15 \%$15\mathrm{%}$ $36\mathrm{%}$$36\mathrm{%}$36%36 \%$36\mathrm{%}$ Z $12\mathrm{%}$$12\mathrm{%}$12%12 \%$12\mathrm{%}$ $15\mathrm{%}$$15\mathrm{%}$15%15 \%$15\mathrm{%}$
Portfolio Expected return Standard deviation W 9% 21% X 5% 7% Y 15% 36% Z 12% 15%| Portfolio | Expected return | Standard deviation | | :—: | :—: | :—: | | W | $9 \%$ | $21 \%$ | | X | $5 \%$ | $7 \%$ | | Y | $15 \%$ | $36 \%$ | | Z | $12 \%$ | $15 \%$ |
The efficient frontier, as described by Markowitz, is a set of portfolios that offer the highest expected return for a given level of risk (standard deviation) or the lowest risk for a given level of expected return. Portfolios that lie below the efficient frontier are considered suboptimal because they provide a lower return for the same level of risk or a higher risk for the same level of return.
To determine which portfolio cannot lie on the efficient frontier, we can compare the portfolios’ risk-return profiles:
• Portfolio W: Expected return = 9%, Standard deviation = 21%
• Portfolio X: Expected return = 5%, Standard deviation = 7%
• Portfolio Y: Expected return = 15%, Standard deviation = 36%
• Portfolio Z: Expected return = 12%, Standard deviation = 15%
Comparing Portfolio W and Z, we see that Portfolio Z has a higher expected return (12% vs. 9%) with a lower standard deviation (15% vs. 21%). This means Portfolio W is dominated by Portfolio Z, as Z offers both higher return and lower risk.
Therefore, Portfolio W cannot lie on the efficient frontier.

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$$c^2=a^2+b^2-2ab\:Cos\left(C\right)$$

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