C877 Mathematical Modeling and Applications
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Free C877 Mathematical Modeling and Applications Questions
In the SIR model, the epidemic threshold is:
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R₀ > 1
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R₀
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R₀ = 1
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β = γ
Explanation
R₀ = β/γ is the basic reproduction number. When R₀ > 1, a single infected individual sparks a major outbreak (exponential initial growth); R₀ = 1 is the knife-edge case (linear growth); R₀ < 1 means the disease dies out. This single inequality determines whether masks, vaccines, or lockdowns are needed—used daily by WHO and CDC.
Correct Answer Is:
R₀ > 1
In compartmental modeling, the differential equation for amount of drug in bloodstream with constant IV infusion rate R and elimination rate constant k is:
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dA/dt = R − kA
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dA/dt = R + kA
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dA/dt = −kA
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dA/dt = R/k
Explanation
Drug enters at constant rate R (mg/h) and leaves proportionally to current amount (−kA). Net rate of change is inflow minus outflow, giving the linear first-order DE dA/dt = R − kA. Steady-state occurs when dA/dt = 0 → A_ss = R/k, which is the target concentration in IV therapy.
Correct Answer Is:
dA/dt = R − kA
In the Gompertz model N(t) = K exp(−α e^{-ct}), what does the parameter K represent?
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Maximum possible size
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Initial population
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Growth rate constant
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Time to reach half capacity
Explanation
The Gompertz function N(t) = K exp(−α e^{-ct}) is written such that as t → ∞, e^{-ct} → 0, so N → K exp(0) = K. Therefore K is the upper asymptote or carrying capacity, exactly like in logistic models but approached more slowly at first and then faster. This makes Gompertz preferred for human mortality tables and tumor growth where early growth is slow.
Correct Answer Is:
Maximum possible size
In stability analysis of numerical methods, the test equation y' = λy has the amplification factor for Euler method:
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1 + hλ
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1 − hλ
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e^{hλ}
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(1 + hλ/2)/(1 − hλ/2)
Explanation
Applying forward Euler to y' = λy gives y_{n+1} = y_n + h λ y_n = (1 + hλ) y_n. The exact solution is y(t) = y_0 e^{λt}, so exact amplification over one step is e^{hλ}. Euler’s factor 1 + hλ is the first-order Taylor approximation of e^{hλ}, which is why Euler is only A-stable in the left half-plane when |1 + hλ| ≤ 1.
Correct Answer Is:
1 + hλ
In the Black-Scholes PDE for a European call option, the equation is: ∂V/∂t + (1/2)σ²S² ∂²V/∂S² + rS ∂V/∂S − rV = 0
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∂V/∂t + (1/2)σ²S² ∂²V/∂S² + rS ∂V/∂S − rV = 0
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∂V/∂t + (1/2)σ²S² ∂²V/∂S² + (r−q)S ∂V/∂S − rV = 0
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∂V/∂t + σ S ∂V/∂S − rV = 0
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∂V/∂t + rV = 0
Explanation
Derivation via Itô’s lemma on dS = μS dt + σS dW hedged with Δ shares yields the Black-Scholes PDE. For dividend yield q, the term becomes (r−q)S ∂V/∂S; for q=0 it simplifies to the classic form. Every finance student memorizes this PDE—solved backward from expiry payoff max(S−K,0).
Correct Answer Is:
∂V/∂t + (1/2)σ²S² ∂²V/∂S² + rS ∂V/∂S − rV = 0
Variance reduction in Monte Carlo can be achieved using:
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Antithetic variates
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Control variates
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Importance sampling
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All of the above
Explanation
All three are standard variance-reduction tricks: antithetic pairs negative-correlate samples, control variates subtract known-expectation random variables, importance sampling shifts density toward high-contribution regions. In this test performance tasks, students routinely cut error by factors of 10–100 using these methods.
Correct Answer Is:
All of the above
In a discrete logistic model P_{n+1} = P_n + r P_n (1 - P_n / K), what happens if r > 2.0 while keeping K fixed?
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The population oscillates with period 2
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The population goes extinct
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The population grows without bound
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The population stabilizes at K
Explanation
The discrete logistic map exhibits different behaviors based on the growth parameter r. For 0 < r ≤ 1, population goes to 0; 1 < r ≤ 3, it stabilizes at K; for 3 < r ≤ 3.57, period-doubling bifurcations lead to oscillations (starting with period 2 around r > 3); beyond ~3.57, chaos emerges. When r > 2 (specifically entering 3+), stable cycles appear instead of monotonic approach to equilibrium, illustrating sensitivity to parameters in discrete models, unlike continuous logistic which always stabilizes.
Correct Answer Is:
The population oscillates with period 2
In optimal control with Pontryagin’s maximum principle, the costate λ(t) satisfies:
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dλ/dt = −∂H/∂x where H is the Hamiltonian
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dλ/dt = ∂H/∂x
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dλ/dt = 0
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dλ/dt = λ r
Explanation
The Hamiltonian H = L(x,u) + λ f(x,u). Pontryagin requires the costate to evolve backward as dλ/dt = −∂H/∂x (adjoint equation). Terminal condition λ(T) = ∂Φ/∂x(T) for fixed endpoint, or λ(T) = 0 for free endpoint. This is how NASA solves minimum-fuel rocket trajectories and how C877 students solve cancer treatment schedules.
Correct Answer Is:
dλ/dt = −∂H/∂x where H is the Hamiltonian
In the 3-species Rock-Paper-Scissors model dx_i/dt = x_i (∑_{j≠i} x_j − x_k) where {i,j,k} cycle, the behavior is:
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Stable cycles around the center
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One species dominates
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Extinction of two species
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Chaos
Explanation
This is the replicator equation for RPS game. The only interior fixed point (1/3,1/3,1/3) is unstable, and trajectories spiral outward to the boundary in conservative systems, but with any mutation or spatial structure, the system sustains beautiful cyclic dominance forever—used to model coral reef invertebrates, E. coli strains, and lizard mating strategies.
Correct Answer Is:
Stable cycles around the center
The Fibonacci sequence is defined by F_{n} = F_{n-1} + F_{n-2}, F_1 = 1, F_2 = 1. The closed-form Binet formula is:
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F_n = (φ^n − (−φ)^{-n}) / √5
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F_n = φ^n / √5
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F_n = n(n+1)/2
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F_n = 2^n
Explanation
The characteristic equation r² − r − 1 = 0 has roots φ = (1+√5)/2 ≈ 1.618 (golden ratio) and ψ = (1−√5)/2 = −1/φ. General solution F_n = A φ^n + B ψ^n. Using F_1 = 1, F_2 = 1 yields A = 1/√5, B = −1/√5, giving Binet’s exact formula F_n = (φ^n − ψ^n)/√5, which amazingly gives integers despite irrational numbers.
Correct Answer Is:
F_n = (φ^n − (−φ)^{-n}) / √5
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