C877 Mathematical Modeling and Applications
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Free C877 Mathematical Modeling and Applications Questions
In difference equations, the solution to y_{n+1} = 1.5 y_n with y_0 = 100 is:
- y_n = 100 × 1.5^n
- y_n = 100 × (1.5^n − 1)
- y_n = 100 / 1.5^n
- y_n = 150 n
Explanation
This is a linear nonhomogeneous recurrence with no constant term, so geometric solution y_n = A r^n. Back-substitute: A r^{n+1} = 1.5 A r^n → r = 1.5. Initial condition y_0 = 100 → A = 100. Thus y_n = 100 × 1.5^n, which is discrete exponential growth, exactly analogous to continuous y(t) = 100 e^{rt} with r = ln(1.5).
Correct Answer Is:
y_n = 100 × 1.5^n
In nonlinear programming, the KKT conditions are necessary for optimality when:
- Constraint qualifications (e.g., LICQ) hold
- The problem is convex
- The objective is quadratic
- Only equality constraints exist
Explanation
Karush-Kuhn-Tucker (KKT) conditions generalize Lagrange multipliers to inequality constraints. They are necessary for local optimality if a constraint qualification holds (e.g., Linear Independence Constraint Qualification—gradients of active constraints are linearly independent). For convex problems, KKT are also sufficient. Every C877 student must check KKT for their final project.
Correct Answer Is:
Constraint qualifications (e.g., LICQ) hold
Crank-Nicolson scheme for the heat equation is:
- Unconditionally stable
- Second-order in time
- Averages explicit and implicit steps
- All of the above
Explanation
Crank-Nicolson is the average of forward and backward Euler at half-time points: (u^{n+1} − u^n)/Δt = (α/2) [∇² u^{n+1} + ∇² u^n] evaluated at i+1/2. This yields second-order accuracy in both space and time, is implicit, and unconditionally stable (amplification |g| ≤ 1 for all r), making it the industry standard for parabolic PDEs.
Correct Answer Is:
All of the above
In the Beverton-Holt stock-recruitment model, S_{t+1} = a R_t / (1 + b R_t), the carrying capacity for recruits is: a/b
- a/b
- 1/b
- A
- b
Explanation
As parental stock R_t → ∞, recruits S_{t+1} → a/b, so the maximum sustainable recruitment is a/b. This hyperbolic form appears in every C877 sustainable-yield question. Unlike Ricker, Beverton-Holt never overshoots and is always stable.
Correct Answer Is:
a/b
Newton’s law of cooling states that the rate of temperature change is proportional to the difference between object and ambient temperature. The DE is:
- dT/dt = k(T − T_a)
- dT/dt = −k(T − T_a)
- dT/dt = kT
- dT/dt = −kT
Explanation
Hot objects cool down and cold objects warm up toward ambient temperature T_a. The rate must be negative when T > T_a and positive when T < T_a, so the correct form is dT/dt = −k(T − T_a) where k > 0. The negative sign ensures temperature always moves toward T_a, leading to the solution T(t) = T_a + (T_0 − T_a)e^{-kt}.
Correct Answer Is:
dT/dt = −k(T − T_a)
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
- ∂V/∂t + (1/2)σ²S² ∂²V/∂S² + rS ∂V/∂S − rV = 0
- ∂V/∂t + (1/2)σ²S² ∂²V/∂S² + (r−q)S ∂V/∂S − rV = 0
- ∂V/∂t + σ S ∂V/∂S − rV = 0
- ∂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
In spatial epidemiology, the critical vaccination coverage p_c to achieve herd immunity in a random network is: p_c = 1 −
- p_c = 1 − 1/R₀
-
p_c = 1 −
/ -
p_c = 1 − 1/
- p_c = 1 − 1/R₀²
Explanation
In heterogeneous networks, the effective R₀ ≈
Correct Answer Is:
p_c = 1 −
In Euler’s method for solving dy/dt = f(t,y) with step size h, the error per step is proportional to:
- h
- h²
- H³
- exp(h)
Explanation
Euler’s method is a first-order technique: y_{n+1} = y_n + h f(t_n, y_n). The local truncation error comes from the Taylor expansion—the method ignores terms beyond the first derivative, so local error = O(h²). Global error accumulates to O(h) over fixed interval, making it crude but easy to implement in spreadsheets for C877 tasks.
Correct Answer Is:
h²
The coefficient of variation for an exponential distribution is: 1
- 1
- 0
- 0.5
- √2
Explanation
Exponential distribution has pdf λ e^{-λx}, mean 1/λ, variance 1/λ², standard deviation 1/λ. Coefficient of variation CV = σ/μ = 1. CV = 1 is the hallmark of memoryless processes—service times in M/M/1, time between customer arrivals, radioactive decay.
Correct Answer Is:
1
In the Ricker model for single-species discrete growth with overcompensation, N_{t+1} = N_t e^{r(1 − N_t/K)}, the critical r value where period-doubling begins is: r = 2
- r = 2
- r = 2.5
- r = 3
- r = 3.57
Explanation
The Ricker map is topologically conjugate to the logistic map via a change of variables. Bifurcations occur at exactly the same growth rates: stable fixed point for r < 2, period-2 cycle at r = 2, period-4 at r ≈ 2.526, and chaos beyond r ≈ 2.692.WGU C877 students see this every year when modeling fish stocks or insect outbreaks.
Correct Answer Is:
r = 2
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