BSc CSIT (TU) Science B.Sc. Mathematics 407 & 408 (Model) Question Paper 2079 Nepal

(a) With a constant per-year growth rate of $5\%$, the population follows geometric growth $P_t = P_0(1+r)^t$ with $P_0 = 200$, $r = 0.05$, $t = 5$.

So after 5 years the population is approximately 255 elephants.

(b) The model is $\Delta P = 1.3\,P\left(1 - \dfrac{P}{10}\right)$.

• $\Delta P > 0$ (population grows) when $P\left(1 - \frac{P}{10}\right) > 0$, i.e. for $0 < P < 10$.
• $\Delta P < 0$ (population declines) when $P > 10$ (or $P < 0$, which is not biologically meaningful).
• $\Delta P = 0$ at $P = 0$ and $P = 10$, the equilibria.

Biological significance: $K = 10$ is the carrying capacity. Below it the environment supports growth; above it resources are insufficient and the population shrinks back toward $K$. Thus the population self-regulates toward the stable equilibrium $P = 10$.

(a) Equilibria satisfy $P^* = P^* + 0.7\,P^*\left(1 - \frac{P^*}{10}\right)$, i.e. $0.7\,P^*\left(1 - \frac{P^*}{10}\right) = 0$, giving $P^* = 0$ and $P^* = 10$.

Let $f(P) = P + 0.7P\left(1 - \frac{P}{10}\right) = 1.7P - 0.07P^2$. Then $f'(P) = 1.7 - 0.14P$.

• At $P^* = 0$: $f'(0) = 1.7$, $|f'| > 1$ → unstable.
• At $P^* = 10$: $f'(10) = 1.7 - 1.4 = 0.3$, $|f'| = 0.3 < 1$ → stable.

So the population converges to the carrying capacity $P^* = 10$.

(b) Separate variables:

Using partial fractions $\dfrac{1}{N(1 - N/K)} = \dfrac{1}{N} + \dfrac{1/K}{1 - N/K}$, integrating gives

Applying $N(0) = N_0$ yields the logistic solution

As $t \to \infty$, $N(t) \to K$.

Formulation. The stage transitions over one time step are:

(a) Stepping forward,

When the cohort moves through all three stages, the net per-generation reproduction is

Since $1.1388 > 1$, the adult population grows about $13.88\%$ each generation.

(b) Writing the state vector $x_t = \begin{bmatrix} E_t \\ L_t \\ A_t \end{bmatrix}$, the model is $x_{t+1} = P\,x_t$ with projection matrix

The dominant eigenvalue of $P$ gives the long-term growth rate of the population.

Eigenvalues. Solve $\det(P - \lambda I) = 0$:

For a column-stochastic-type matrix the trace is $0.9925 + 0.9875 = 1.98$ and the determinant is

The characteristic equation is $\lambda^2 - 1.98\lambda + 0.9800125 = 0$, giving

Thus $\lambda_1 = 1$ and $\lambda_2 = 0.98$.

Eigenvectors.

For $\lambda_1 = 1$: $(P - I)v = 0 \Rightarrow -0.0075\,v_1 + 0.0125\,v_2 = 0 \Rightarrow v_1 = \frac{0.0125}{0.0075}v_2 = \frac{5}{3}v_2$. Taking $v_2 = 3$, $v_1 = 5$:

For $\lambda_2 = 0.98$: $(P - 0.98 I)v = 0 \Rightarrow 0.0125\,v_1 + 0.0125\,v_2 = 0 \Rightarrow v_1 = -v_2$:

Derivation. The prey $P$ grows logistically $rP_t(1 - P_t/K)$ but is reduced by predation proportional to encounters $sP_tQ_t$. Predators $Q$ die at rate $u$ ($-uQ_t$) but reproduce in proportion to prey consumed $vP_tQ_t$. This gives the coupled system stated.

Equilibria $(P^*, Q^*)$ satisfy $\Delta P = 0$ and $\Delta Q = 0$:

From the second equation: $Q^* = 0$ or $P^* = u/v$.

• Trivial: $(0,0)$.
• Prey-only: $Q^* = 0 \Rightarrow P^* = K$, giving $(K, 0)$.
• Coexistence: $P^* = u/v$, substitute into the first: $r\left(1 - \frac{u}{vK}\right) = sQ^* \Rightarrow Q^* = \frac{r}{s}\left(1 - \frac{u}{vK}\right)$, giving $\left(\dfrac{u}{v},\ \dfrac{r}{s}\left(1 - \dfrac{u}{vK}\right)\right)$.

OR (specific model). Let $f(P,Q) = P(1 + 1.3(1 - P)) - 0.5PQ = 2.3P - 1.3P^2 - 0.5PQ$ and $g(P,Q) = 0.3Q + 1.6PQ$.

(a) Jacobian

(b) Equilibria: $g=0 \Rightarrow Q=0$ or $P = (1-0.3)/1.6 = 0.4375$.

• At $(0,0)$: $J = \begin{bmatrix} 2.3 & 0 \\ 0 & 0.3 \end{bmatrix}$.
• Prey-only $f=0,\ Q=0 \Rightarrow 2.3 - 1.3P = 0 \Rightarrow P = 1.0$: at $(1,0)$, $J = \begin{bmatrix} 2.3 - 2.6 & -0.5 \\ 0 & 0.3 + 1.6 \end{bmatrix} = \begin{bmatrix} -0.3 & -0.5 \\ 0 & 1.9 \end{bmatrix}$.
• Coexistence $P = 0.4375$, $f=0 \Rightarrow 2.3 - 1.3(0.4375) - 0.5Q = 0 \Rightarrow Q = (2.3 - 0.56875)/0.5 = 3.4625$: at $(0.4375, 3.4625)$, $J = \begin{bmatrix} 2.3 - 2.6(0.4375) - 0.5(3.4625) & -0.5(0.4375) \\ 1.6(3.4625) & 0.3 + 1.6(0.4375) \end{bmatrix} = \begin{bmatrix} -0.86875 & -0.21875 \\ 5.54 & 1.0 \end{bmatrix}.$

The sequence is $A\,G\,G\,G\,A\,T\,A\,C\,A\,T\,G\,A\,C\,C\,C\,A\,T\,A\,C\,A$.

(a) First 5 bases: $AGGGA$ — counts $A=2,\ G=3,\ C=0,\ T=0$.

(b) First 10 bases: $AGGGATACAT$ — counts $A=4,\ G=3,\ C=1,\ T=2$.

(c) All 20 bases — counts $A=8,\ G=3,\ C=5,\ T=4$ (sum $= 20$).

The estimates stabilize as the sample (sequence length) increases, illustrating that larger samples give more reliable frequency estimates.

Let $T(n)$ be the number of unrooted bifurcating trees for $n$ taxa (defined for $n \ge 3$). Using $T(n) = (2n-5)!! = 1\cdot 3\cdot 5\cdots(2n-5)$:

Graph: plotting $T(n)$ against $n$ gives a super-exponentially increasing curve (each value multiplied by the next odd number). Because the growth is so steep, a $\log$-scale plot of $\ln T(n)$ vs $n$ is clearer and is approximately linear-to-convex, showing that the number of possible tree topologies explodes combinatorially — by $n = 10$ there are already over 2 million unrooted trees.

Genetics cross $DdWw \times ddWw$. Treat the two genes independently. Assume $D$ (tall) is dominant over $d$ (dwarf), and $W$ (round) dominant over $w$ (wrinkled); 'dwarf' = $dd$, 'round' = $W\_$.

• Height: $Dd \times dd \Rightarrow \tfrac12\,Dd : \tfrac12\,dd$. So $P(\text{dwarf}, dd) = \tfrac12$.
• Seed: $Ww \times Ww \Rightarrow \tfrac14\,WW : \tfrac12\,Ww : \tfrac14\,ww$. So $P(\text{round}, W\_) = \tfrac34$.

By independence,

OR (two-child family).

(a) Outcomes (eldest, youngest), each with probability $\tfrac14$: $\{BB,\ BG,\ GB,\ GG\}$.

(b) $P(\text{at least one female}) = 1 - P(BB) = 1 - \tfrac14 = \tfrac34$.

(c) $P(\text{youngest is female}) = P(\{BG, GG\}) = \tfrac12$.

(d) $P(\text{youngest F} \mid \text{at least one F}) = \dfrac{P(\text{youngest F})}{P(\text{at least one F})} = \dfrac{1/2}{3/4} = \dfrac{2}{3}.$

(e) $P(\text{at least one F} \mid \text{youngest F}) = \dfrac{P(\text{youngest F})}{P(\text{youngest F})} = 1$ (if the youngest is female there is certainly at least one female).

SIR model. Partition the population $N = S + I + R$ into Susceptible, Infected, Removed. With transmission rate $\beta$ and recovery rate $\gamma$:

Threshold. An epidemic grows only if $\dfrac{dI}{dt} > 0$ initially, i.e. $\beta S_0 - \gamma > 0 \Rightarrow S_0 > \dfrac{\gamma}{\beta}$. Defining the basic reproduction number

an epidemic occurs iff $R_0 > 1$; the threshold susceptible level is $S^* = \gamma/\beta$ (the relative removal rate $\rho = \gamma/\beta$).

OR — SI and SIS models.

SI model (no recovery), $N = S + I$:

Equilibria: $I^* = 0$ (disease-free, $S^* = N$) and $I^* = N$ (endemic, $S^* = 0$). With $I(0) > 0$ the disease grows logistically to $I^* = N$.

SIS model (infected return to susceptible at rate $\gamma$):

Set $=0$: $I^*\left[\beta(N - I^*) - \gamma\right] = 0$, so

The endemic equilibrium is positive (and stable) iff $R_0 = \beta N/\gamma > 1$.

Least-squares line for $(1,1),(0,3),(1,4)$. Let the line be $y = mx + b$. With $x = \{1,0,1\}$, $y = \{1,3,4\}$: $n = 3$, $\sum x = 2$, $\sum y = 8$, $\sum xy = (1)(1)+(0)(3)+(1)(4) = 5$, $\sum x^2 = 2$.

Best-fit line: $\boxed{y = -\tfrac12 x + 3}$. The plot shows the three points with this line passing through $(0,3)$ with slope $-\tfrac12$.

OR — exponential drug decay.

(a) Iterating $y_{t+1} = (1-r)y_t$ from $y_0$: $y_1 = (1-r)y_0$, $y_2 = (1-r)y_1 = (1-r)^2 y_0$, and by induction $y_t = y_0(1-r)^t$.

(b) With $k = \ln(1-r)$, we have $1 - r = e^{k}$, so $(1-r)^t = e^{kt}$. Hence $y_t = y_0 e^{kt} = a e^{kt}$ with $a = y_0$.

(c) Since $r$ is the fraction of drug removed per step, the amount removed cannot exceed the amount present ($r \le 1$) and must be positive for decay ($r > 0$), so $0 < r < 1$. Then $0 < 1 - r < 1$, and the logarithm of a number in $(0,1)$ is negative, so $k = \ln(1-r) < 0$ — confirming exponential decay.

(a) Market equilibrium is the state in which the quantity demanded equals the quantity supplied for each commodity, so there is no tendency for price to change ($Q_d = Q_s$).

A two-commodity market model with linear demand/supply:

Equilibrium requires $Q_{d1} = Q_{s1}$ and $Q_{d2} = Q_{s2}$, i.e. the excess-demand equations

solved simultaneously for $P_1^*, P_2^*$ (then $Q_1^*, Q_2^*$).

(b) Taking the model literally (combining like terms):

• Commodity 1: $Q_{d1} = 10 - 4P_1 + P_2$. Set $Q_{d1} = Q_{s1}$: $10 - 4P_1 + P_2 = -2 + 3P_1 \Rightarrow 7P_1 - P_2 = 12$. …(i)
• Commodity 2: $Q_{d2} = 15 - P_1 - P_2$. Set $Q_{d2} = Q_{s2} = -1$: $15 - P_1 - P_2 = -1 \Rightarrow P_1 + P_2 = 16$. …(ii)

Add (i) and (ii): $8P_1 = 28 \Rightarrow P_1^* = 3.5$. Then $P_2^* = 16 - 3.5 = 12.5$.

Equilibrium quantities: $Q_1^* = -2 + 3(3.5) = 8.5$; $Q_2^* = -1$.

(The model as printed yields $Q_{s2} = -1$, a negative quantity, suggesting a typo in the source; method shown is the standard solution procedure.)

(a) States $\{E, U\}$. An unemployed person finds a job with prob $0.7$ (stays unemployed $0.3$); an employed person loses the job with prob $0.1$ (stays employed $0.9$). With row vector convention $x_{t+1} = x_t P$:

(b) Start $x_0 = [0,\ 1200]$.

• $x_1 = x_0 P = [0(0.9)+1200(0.7),\ 0(0.1)+1200(0.3)] = [840,\ 360]$.
• $x_2 = x_1 P = [840(0.9)+360(0.7),\ 840(0.1)+360(0.3)] = [756+252,\ 84+108] = [1008,\ 192]$.
  • After 2 periods: 192 unemployed.
• Iterating toward the steady state, after 10 periods the distribution is essentially at steady state. Numerically $x_{10} \approx [1050,\ 150]$, so about 150 unemployed after 10 periods.

(c) Steady state $\pi = \pi P$ with $\pi_E + \pi_U = 1200$. From $\pi_U = 0.1\pi_E + 0.3\pi_U \Rightarrow 0.7\pi_U = 0.1\pi_E \Rightarrow \pi_E = 7\pi_U$. With $\pi_E + \pi_U = 1200$: $8\pi_U = 1200 \Rightarrow \pi_U = 150$, $\pi_E = 1050$.

Steady-state unemployment = 150 people (i.e. $12.5\%$).

(a) Marginal cost (MC) is the rate of change of total cost with respect to output, $MC = \dfrac{dC}{dQ}$. Average cost (AC) is total cost per unit of output, $AC = \dfrac{C}{Q}$.

For $C = Q^3 - 5Q^2 + 12Q + 75$, the constant $75$ is fixed cost, so the variable cost is

Its derivative is

which equals the marginal cost (since fixed cost has zero derivative). Economically it measures the additional cost of producing one more unit of output at the given level of $Q$.

(b) A partial derivative $\partial y/\partial x_i$ measures the rate of change of $y$ with respect to $x_i$ holding the other variables constant — the marginal effect of that variable.

Jacobian test for functional dependence.

Jacobian:

Since $|J| \equiv 0$ identically, the functions are functionally dependent. Indeed $y_2 = (3x_1^2 + x_2)^2 + 8(3x_1^2 + x_2) + 12 = y_1^2 + 8y_1 + 12$.

IS-LM model.

Assumptions: prices fixed (short run), two markets — goods (IS) and money (LM); income and interest rate ($Y, i$) adjust to clear both simultaneously.

IS curve (goods-market equilibrium $Y = C(Y) + I(i) + G$): differentiating, the slope $\dfrac{di}{dY}\Big|_{IS} = \dfrac{1 - C'(Y)}{I'(i)} < 0$ since $I'(i) < 0$ — downward sloping.

LM curve (money-market equilibrium $M/P = L(Y, i)$): slope $\dfrac{di}{dY}\Big|_{LM} = -\dfrac{L_Y}{L_i} > 0$ (since $L_Y > 0$, $L_i < 0$) — upward sloping.

Equilibrium identities: $Y = C(Y) + I(i) + G_0$ and $\dfrac{M_0}{P} = L(Y, i)$, solved for $Y^*, i^*$. Comparative statics (e.g. $\partial Y^*/\partial G_0 > 0$, $\partial Y^*/\partial M_0 > 0$) follow by total differentiation and Cramer's rule.

OR — national-income model.

Differentials: $dy = f'(x)\,dx$ approximates the change in $y$ for a small change in $x$. Point elasticity: $\varepsilon = \dfrac{dy/y}{dx/x} = \dfrac{x}{y}\dfrac{dy}{dx}$, the proportional responsiveness of $y$ to $x$.

(a) $S' = dS/dY$ is the marginal propensity to save; $T' = dT/dY$ is the marginal tax rate; $I' = dI/dY$ is the marginal propensity to invest — each the change in that quantity per unit change in income.

(b) Let $F(Y, G_0) = S(Y) + T(Y) - I(Y) - G_0 = 0$. $F$ is continuous with continuous partials, and $\dfrac{\partial F}{\partial Y} = S' + T' - I' > 0$ (given $S' + T' > I'$), which is nonzero. So the implicit-function theorem applies and the equilibrium identity $S(Y^*) + T(Y^*) \equiv I(Y^*) + G_0$ defines $Y^* = Y^*(G_0)$.

(c) By the implicit-function rule,

Implication: an increase in government expenditure raises equilibrium national income (the government-expenditure multiplier); its size is the reciprocal of $(S' + T' - I')$ — larger leakages (saving, taxes) shrink the multiplier.

Wine storage problem. A merchant owns wine whose value grows with age as $V(t)$ (a known increasing, concave function). Storage is costless but money has a discount rate $r$. The present value of selling at time $t$ is

Maximize over $t$: First-order condition $\dfrac{dA}{dt} = 0$:

Sell when the proportional growth rate of value equals the discount rate. Second-order condition for a maximum: $\dfrac{d^2A}{dt^2} < 0$ at $t^*$, i.e. $V''(t) - rV'(t) < 0$, ensuring $A(t)$ is concave at the optimum.

OR — unconstrained extremum, $Z = -x_1^3 + 3x_1 x_3 + 2x_2 - x_2^2 - 3x_3^2$.

FOC (set first partials $= 0$):

From $x_3 = x_1^2$ and $x_1 = 2x_3$: $x_1 = 2x_1^2 \Rightarrow x_1(1 - 2x_1) = 0 \Rightarrow x_1 = 0$ or $x_1 = \tfrac12$.

• $x_1 = 0 \Rightarrow x_3 = 0$: critical point $(0, 1, 0)$.
• $x_1 = \tfrac12 \Rightarrow x_3 = \tfrac14$: critical point $(\tfrac12, 1, \tfrac14)$.

SOC (Hessian).

Leading principal minors for a maximum need signs $-,+,-$.

• At $(\tfrac12, 1, \tfrac14)$: $|H_1| = -6(\tfrac12) = -3 < 0$; $|H_2| = (-3)(-2) - 0 = 6 > 0$; $|H_3| = \det H$. With $-6x_1 = -3$: $\det = -3[(-2)(-6) - 0] - 0 + 3[0 - (-2)(3)] = -3(12) + 3(6) = -36 + 18 = -18 < 0$. Signs $-,+,-$ → local maximum. $Z = -(\tfrac12)^3 + 3(\tfrac12)(\tfrac14) + 2(1) - 1^2 - 3(\tfrac14)^2 = -\tfrac18 + \tfrac38 + 2 - 1 - \tfrac{3}{16} = 1\tfrac{1}{16} \approx 1.0625$.
• At $(0, 1, 0)$: $|H_1| = 0$, the determinantal test is inconclusive/fails the strict maximum signs; this point is a saddle (since the cubic term makes $Z$ unbounded along $x_1$).