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IGNOU MMT-009 Solved Assignment | 1st January, 2025 to 31st December, 2025 | MATHEMATICAL MODELLING | MSc MACS Sample Solution

MMT-009 Solved Assignment 2025

MATHEMATICAL MODELLING

a) Companies located on the banks of a river, dumping their chemicals waste into river, causing high levels of pollution. Local authorities passed new legislation with very high fines if the pollution in the river exceeds certain specified concentration limits. State, giving reasons, the type of modeling you will use to find a policy for discharging the waste to ensure that the concentration level never exceeds the specified limits. Also state four essentials and two non-essentials for the problem.

b) Consider the following data

x	2	3	4	5	6
y	8.3	16.5	30.2	65.2	125.6

Use a best fit line to estimate the value of y when

x = 4.5

a) Five securities have the following expected returns
$A = 20 %, B = 15 %, C = 25 %, D = 22 %, E = 18 %$ . Calculate the expected returns for a portfolio consisting of all five securities under the following conditions
i) The portfolio weights are of equal percentage in each
ii) The portfolio weights are $32 %$ in A and remaining are equally divided among other four securities.

b) Let

P = (w_{1}, w_{2})

be a portfolio of two securities. If variance of P is minimum then find the value of

w_{1}

and

w_{2}

in the following situations.
i)

ρ_{12} = - 1

ii)

σ_{1} = σ_{2}

iii)

ρ_{12} = - 0.5, σ_{1} = 1.5

and

σ_{2} = 2.5

a) Assume that the return distribution on the two securities $X$ and $Y$ be as given below:

	Market 1	Market 2	Market 3
Probability	0.2	0.5	0.3
Security X	$- 20 %$	$18 %$	$40 %$
Security Y	$- 10 %$	$20 %$	$15 %$

which security is more risky in the Markowitz sense. Also find the correlation coefficient of securities X and Y.

b) In a species of animals a constant fraction of the population

α = 6.2

are born each breeding season and a constant fraction

β = 4.5

die. Formulate a difference equation for the population and find out the number of individuals after fifteen seasons given that the initial number is 987 . Find the closed form solution of the formulated difference equation. If the growth rate of the population is represented by r then interpret the solution obtained when i)

r > 0

and ii)

r < 0

A model for insect populations leads to the difference equations

N_{k + 1} = \frac{λ N_{k}}{1 + {aN}_{k}}

where

λ

and a are positive constants.

i) Write the equation in the form

N_{k + 1} = N_{k} + R (N_{k}) N_{k}

and hence identify the growth rate.

ii) Plot the graph of

R (N_{k})

as a function of

N_{k}

iii) Express the intrinsic growth rate r and the carrying capacity K , for this model, in terms of the parameters, a and

λ

iv) Find the steady-state solution of this model and analyse the solution.

Do the stability analysis of the following competing species system of equations with diffusion and advection

\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}, 0 \leq x \leq L \end{aligned}

where

V_{1}

and

V_{2}

are advection velocities in x direction of the two populations with densities

N_{1}

and

N_{2}

respectively.

a_{1}

is the growth rate,

b_{1}

is the predation rate,

d_{1}

is the death rate,

C_{1}

is the conversion rate.

D_{1}

and

D_{2}

are diffusion coefficients. The initial and boundary conditions are:

\begin{aligned} N_{i} (x, 0) = f_{i} (x) > 0, 0 \leq x \leq L, i = 1, 2 \dots \\ N_{i} = {\overset{―}{N}}_{i} at x = 0 and x = L \forall t, i = 1, 2 \dots \end{aligned}

where

{\overset{―}{N}}_{i}

are the equilibrium solutions of the given system of equations. Interpret the solution obtained and also write the limitations of the model.

The population dynamics of a species is governed by the discrete model

x_{n + 1} = x_{n} \exp [r (1 - \frac{x_{n}}{K})],

where r and k are positive constants. Determine the steady states and discuss the stability of the model. Find the value of

r

at which first bifurcation occurs. Describe qualitatively the behaviors of the population for

r = 2 + ε

, where

0 < ε ≪ 1

. Since a species becomes extinct if

x_{n} \leq 1

for any

n > 1

, show using iterations, that irrespective of the size of

r > 1

the species could become extinct if the carrying capacity

k < r \exp [1 + e^{r - 1} - 2 r]

Do the stability analysis of the following model formulated to study the effect of toxicant on one competing species where the environment toxicant concentration is being taken to change w.r.t. time.

\frac{d N_{1}}{dt} = r_{1} N_{1} - α_{1} N_{1} N_{2} - d_{1} C_{0} N_{1}

\begin{aligned} \frac{dN}{2} \\ dt \end{aligned} r_{2} N_{2} - α_{2} N_{1} N_{2} .

along with the initial conditions.

N_{1} (0) = N_{10}, N_{2} (0) = N_{20}, C_{0} (0) = 0, P (0) = P_{0} > 0

Here,

\begin{aligned} N_{1} (t) & = Density of prey population \\ N_{2} (t) & = Density of predator population \\ C_{0} (t) & = Concentration of the toxicant in the individual of the prey population \\ P & = Constant environmental toxicant concentration. \end{aligned}

α_{1}, α_{2}

are the predation rates,

r_{1}, r_{2}

are the growth rates or birth rates,

d_{1}

is the death rate due to

C_{0}, m_{1}

is the depuration rate,

Q, h, k, g

are positive rate constants.

The owner of a readymade garments store sells two types of shirts: Zee-shirts and Button-down shirts. He makes a profit of Rs. 5 and Rs. 10 per shirt on Zee-shirts and Button-down shirt, respectively. He has two tailors, A and B at his disposal to stitch the shirts. Tailors A and B can devote at the most 7 hours and 15 hours per day, respectively. Both these shirts are to be stitched by both the tailors. Tailors A and B spend 2 hours and 5 hours, respectively in stitching one Zee-shirts, and 4 hours and 3 hours, respectively in stitching a Button-down shirt. How many shirts of both types should be stitched in order to maximize daily profit?

a) Formulate and solve this problem as an LP problem.

b) If the optimal solution is not integer-valued, use Gomory technique to derive the optimal integer solution.

a) A company has three factories that supply to three markets. The transportation costs from each factory to each market are given in the table. Capacities of the factories and market requirements are shown. Find the minimum transportation cost.

	$M_{1}$	$M_{2}$	$M_{3}$	$a_{i}$
$F_{1}$	2	1	3	20
$F_{2}$	1	2	3	30
$F_{3}$	2	1	2	10
$b_{j}$	10	10	20	$40 / 60$

b) For a multi-channel queuing system with

λ = 12 /

hours,

μ = 5 /

hours,

c = 3

p_{0} = 0.056

, calculate

i) The average time a customer is in the system

ii) The average number of customers in the system

iii) Whether any time would be saved for customers if the three-channel system with the service rate of 5 per hour is replaced by a single-channel system with an average service rate of 15 per hour?

a) 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?

b) 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.

Answer:

Part A

Question:-1(a)

Companies located on the banks of a river, dumping their chemicals waste into river, causing high levels of pollution. Local authorities passed new legislation with very high fines if the pollution in the river exceeds certain specified concentration limits. State, giving reasons, the type of modeling you will use to find a policy for discharging the waste to ensure that the concentration level never exceeds the specified limits. Also state four essentials and two non-essentials for the problem.

Answer:

To address the problem of companies dumping chemical waste into a river while ensuring compliance with new legislation that imposes high fines for exceeding specified pollution concentration limits, a dynamic modeling approach using a control theory-based model, specifically an optimal control model, is recommended. Below, I outline the reasoning for this choice, the type of modeling, and the essentials and non-essentials for the problem.

Type of Modeling: Optimal Control Model (Dynamic Modeling)

Reasons for Choosing Optimal Control Modeling:

Dynamic Nature of the System: Pollution levels in the river change over time due to continuous waste discharge, river flow, natural degradation of pollutants, and seasonal variations (e.g., rainfall affecting dilution). A dynamic model captures these time-dependent processes effectively.
Regulatory Compliance: The goal is to ensure that the pollution concentration never exceeds the specified limits. An optimal control model allows for designing a policy that dynamically adjusts the rate of waste discharge based on real-time or predicted river conditions to stay within legal limits.
Cost Optimization: Companies aim to minimize the cost of waste management (e.g., treatment, storage, or reduced production) while avoiding fines. Optimal control models can balance the trade-off between operational costs and compliance by optimizing the discharge policy.
Feedback Mechanism: The model can incorporate feedback from monitoring systems (e.g., sensors measuring pollutant concentrations) to adjust discharge rates in real time, ensuring robustness against unexpected changes in river conditions.
Predictive Capability: By integrating hydrological and chemical models of the river, the optimal control model can predict future concentration levels based on current discharge rates and environmental factors, enabling proactive decision-making.

How the Model Works:

State Variables: Represent the pollutant concentration in the river at different points and times.
Control Variables: Represent the rate of waste discharge from each company.
Objective Function: Minimize costs (e.g., waste treatment, production adjustments) while ensuring the pollutant concentration remains below the legal limit.
Constraints: The pollutant concentration must not exceed the specified limits, and discharge rates must be non-negative and within operational limits.
Dynamics: A differential equation models the evolution of pollutant concentration based on discharge rates, river flow, dilution, and degradation processes.

The model can be solved using techniques like Model Predictive Control (MPC), which iteratively optimizes the discharge policy over a finite time horizon, updating based on new measurements.

Four Essentials for the Problem

Pollutant Concentration Limits: The specific concentration thresholds set by the legislation are critical, as the policy must ensure these are never exceeded to avoid fines.
River Flow and Hydrological Data: Accurate data on river flow rates, velocity, and seasonal variations are essential to model dilution and transport of pollutants.
Waste Discharge Characteristics: Information on the type, volume, and chemical composition of the waste discharged by each company is necessary to quantify its impact on river pollution.
Monitoring and Feedback Systems: Real-time or frequent measurements of pollutant concentrations in the river are essential for adjusting discharge rates dynamically and verifying compliance.

Two Non-Essentials for the Problem

Historical Pollution Data: While useful for understanding long-term trends, historical data is not strictly necessary for developing a forward-looking discharge policy, as the model relies on current and predicted conditions.
Detailed Company Financials: While cost optimization is part of the objective, detailed financial data (e.g., profit margins, revenue) is not essential. A simplified cost function (e.g., treatment costs, production losses) is sufficient.

Question:-1(b)

Consider the following data

x	2	3	4	5	6
y	8.3	16.5	30.2	65.2	125.6

Use a best fit line to estimate the value of y when

x = 4.5

Answer:

To estimate the value of

y

when

x = 4.5

using a best fit line for the given data, we need to determine the equation of the line that best fits the points

(2, 8.3)

(3, 16.5)

(4, 30.2)

(5, 65.2)

, and

(6, 125.6)

. A best fit line is typically a linear regression line of the form

y = m x + b

, where

m

is the slope and

b

is the y-intercept. However, the data suggests a non-linear relationship (as

y

increases rapidly with

x

), so we’ll first try a linear fit and then consider if a non-linear transformation (e.g., exponential) is more appropriate based on the fit.

Step 1: Linear Regression

We assume a linear model

y = m x + b

and use the least squares method to find

m

and

b

. The formulas for the slope

m

and intercept

b

are:

m = \frac{n \sum (x_{i} y_{i}) - \sum x_{i} \sum y_{i}}{n \sum x_{i}^{2} - (\sum x_{i})^{2}}, b = \frac{\sum y_{i} - m \sum x_{i}}{n}

where

n

is the number of data points, and the sums are over all data points.

Data:

$x = [2, 3, 4, 5, 6]$
$y = [8.3, 16.5, 30.2, 65.2, 125.6]$
$n = 5$

Calculate the sums:

$\sum x_{i} = 2 + 3 + 4 + 5 + 6 = 20$
$\sum y_{i} = 8.3 + 16.5 + 30.2 + 65.2 + ක = 245.8$
$\sum x_{i}^{2} = 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} = 4 + 9 + 16 + 25 + 36 = 90$
$\sum x_{i} y_{i} = 2 \cdot 8.3 + 3 \cdot 16.5 + 4 \cdot 30.2 + 5 \cdot 65.2 + 6 \cdot 125.6 = 16.6 + 49.5 + 120.8 + 326 + 753.6 = 1266.5$

Compute the slope $m$ :

m = \frac{5 \cdot 1266.5 - 20 \cdot 245.8}{5 \cdot 90 - 20^{2}} = \frac{6332.5 - 4916}{450 - 400} = \frac{1416.5}{50} = 28.33

Compute the intercept $b$ :

b = \frac{245.8 - 28.33 \cdot 20}{5} = \frac{245.8 - 566.6}{5} = \frac{- 320.8}{5} = - 64.16

So, the linear equation is:

y = 28.33 x - 64.16

Estimate $y$ at $x = 4.5$ :

y = 28.33 \cdot 4.5 - 64.16 = 127.485 - 64.16 = 63.325

Thus, for a linear fit,

y \approx 63.33

when

x = 4.5

Step 2: Assessing the Fit

To evaluate the linear model, we compare predicted

y

-values to actual values:

At $x = 2$ : $y = 28.33 \cdot 2 - 64.16 = 56.66 - 64.16 = - 7.5$ (actual: 8.3)
At $x = 4$ : $y = 28.33 \cdot 4 - 64.16 = 113.32 - 64.16 = 49.16$ (actual: 30.2)
At $x = 6$ : $y = 28.33 \cdot 6 - 64.16 = 169.98 - 64.16 = 105.82$ (actual: 125.6)

The predictions show significant errors, especially at lower

x

-values, and the negative prediction at

x = 2

is unrealistic for a positive quantity like concentration. The rapid increase in

y

(from 8.3 to 125.6) suggests an exponential or polynomial relationship rather than linear.

Step 3: Exponential Model

Given the non-linear growth, an exponential model

y = a e^{b x}

may fit better. To linearize, take the natural logarithm of

y

\ln (y) = \ln (a) + b x

Let

Y = \ln (y)

A = \ln (a)

, so

Y = A + b x

. Perform linear regression on

x

and

Y

Compute $Y = \ln (y)$ :

$x = 2$ , $y = 8.3$ , $Y = \ln (8.3) \approx 2.1163$
$x = 3$ , $y = 16.5$ , $Y = \ln (16.5) \approx 2.8034$
$x = 4$ , $y = 30.2$ , $Y = \ln (30.2) \approx 3.4078$
$x = 5$ , $y = 65.2$ , $Y = \ln (65.2) \approx 4.1772$
$x = 6$ , $y = 125.6$ , $Y = \ln (125.6) \approx 4.8332$

Sums:

$\sum Y_{i} = 2.1163 + 2.8034 + 3.4078 + 4.1772 + 4.8332 = 17.3379$
$\sum x_{i} Y_{i} = 2 \cdot 2.1163 + 3 \cdot 2.8034 + 4 \cdot 3.4078 + 5 \cdot 4.1772 + 6 \cdot 4.8332 = 4.2326 + 8.4102 + 13.6312 + 20.886 + 28.9992 = 76.1592$
Use $\sum x_{i} = 20$ , $\sum x_{i}^{2} = 90$ , $n = 5$ .

Slope $b$ :

b = \frac{5 \cdot 76.1592 - 20 \cdot 17.3379}{5 \cdot 90 - 20^{2}} = \frac{380.796 - 346.758}{450 - 400} = \frac{34.038}{50} = 0.68076

Intercept $A$ :

A = \frac{17.3379 - 0.68076 \cdot 20}{5} = \frac{17.3379 - 13.6152}{5} = \frac{3.7227}{5} = 0.74454

So,

\ln (a) = 0.74454

, hence

a = e^{0.74454} \approx 2.106

The exponential model is:

y = 2.106 e^{0.68076 x}

Estimate $y$ at $x = 4.5$ :

y = 2.106 e^{0.68076 \cdot 4.5} = 2.106 e^{3.06342}

e^{3.06342} \approx 21.385, y \approx 2.106 \cdot 21.385 \approx 45.037

Thus, for the exponential fit,

y \approx 45.04

when

x = 4.5

Step 4: Compare Fits

The exponential model’s predictions are closer to actual values:

At $x = 4$ : $y = 2.106 e^{0.68076 \cdot 4} \approx 2.106 e^{2.72304} \approx 30.43$ (actual: 30.2)
At $x = 5$ : $y = 2.106 e^{0.68076 \cdot 5} \approx 63.57$ (actual: 65.2)

The exponential model fits the data better, as the residuals are smaller and it captures the rapid growth.

Final Answer

Given the context (likely pollution concentration, which grows non-linearly due to accumulation), the exponential model is more appropriate. The estimated value of

y

when

x = 4.5

is:

45.0

Question:-2(a)

Five securities have the following expected returns

A = 20 %, B = 15 %, C = 25 %, D = 22 %, E = 18 %

. Calculate the expected returns for a portfolio consisting of all five securities under the following conditions
i) The portfolio weights are of equal percentage in each
ii) The portfolio weights are

32 %

in A and remaining are equally divided among other four securities.

Answer:

To calculate the expected returns for a portfolio consisting of five securities with given expected returns

A = 20 %

B = 15 %

C = 25 %

D = 22 %

E = 18 %

, we use the formula for the expected return of a portfolio:

E (R_{p}) = \sum_{i = 1}^{n} w_{i} \cdot E (R_{i})

where

w_{i}

is the weight of security

i

, and

E (R_{i})

is the expected return of security

i

. The weights must sum to 1 (

\sum w_{i} = 1

). We compute the expected returns under two conditions.

i) Portfolio weights are of equal percentage in each

Each security has an equal weight. Since there are 5 securities, each weight is:

w_{i} = \frac{1}{5} = 0.2 = 20 %

Expected return:

E (R_{p}) = w_{A} \cdot E (R_{A}) + w_{B} \cdot E (R_{B}) + w_{C} \cdot E (R_{C}) + w_{D} \cdot E (R_{D}) + w_{E} \cdot E (R_{E})

E (R_{p}) = 0.2 \cdot 20 % + 0.2 \cdot 15 % + 0.2 \cdot 25 % + 0.2 \cdot 22 % + 0.2 \cdot 18 %

= 4 % + 3 % + 5 % + 4.4 % + 3.6 % = 20 %

Alternatively, since the weights are equal, the expected return is the average of the individual returns:

E (R_{p}) = \frac{20 % + 15 % + 25 % + 22 % + 18 %}{5} = \frac{100 %}{5} = 20 %

Answer for part i:

E (R_{p}) = 20 %

ii) Portfolio weights are 32% in A, and the remaining are equally divided among the other four securities

Weight of security $A$ : $w_{A} = 32 % = 0.32$ .
Remaining weight: $1 - 0.32 = 0.68$ (or 68%).
This remaining 68% is equally divided among securities $B$ , $C$ , $D$ , and $E$ . Thus, each of these four securities has a weight of:

w_{B} = w_{C} = w_{D} = w_{E} = \frac{0.68}{4} = 0.17 = 17 %

Verify the weights sum to 1:

0.32 + 0.17 + 0.17 + 0.17 + 0.17 = 0.32 + 0.68 = 1

Expected return:

E (R_{p}) = w_{A} \cdot E (R_{A}) + w_{B} \cdot E (R_{B}) + w_{C} \cdot E (R_{C}) + w_{D} \cdot E (R_{D}) + w_{E} \cdot E (R_{E})

E (R_{p}) = 0.32 \cdot 20 % + 0.17 \cdot 15 % + 0.17 \cdot 25 % + 0.17 \cdot 22 % + 0.17 \cdot 18 %

Calculate each term:

$0.32 \cdot 20 % = 6.4 %$
$0.17 \cdot 15 % = 2.55 %$
$0.17 \cdot 25 % = 4.25 %$
$0.17 \cdot 22 % = 3.74 %$
$0.17 \cdot 18 % = 3.06 %$

Sum:

E (R_{p}) = 6.4 % + 2.55 % + 4.25 % + 3.74 % + 3.06 % = 20 %

Answer for part ii:

E (R_{p}) = 20 %

Final Answer

i) 20 %, ii) 20 %

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