C877 Mathematical Modeling and Applications
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Free C877 Mathematical Modeling and Applications Questions
In a discrete-time SIR epidemic model, if R_0 = 2.4, what does this mean?
- On average one infected person infects 2.4 new people in a fully susceptible population
- The epidemic will die out
- Each infected recovers in 2.4 days
- The herd immunity threshold is 2.4%
Explanation
The basic reproduction number R_0 is the expected number of secondary infections caused by a single infectious individual in an otherwise susceptible population. When R_0 > 1 (here 2.4), the disease can invade and cause an epidemic. The herd immunity threshold is 1 − 1/R_0 ≈ 58.3%, meaning over 58% must be immune to prevent outbreaks. This is the single most important parameter in epidemic modeling.
Correct Answer Is:
On average one infected person infects 2.4 new people in a fully susceptible population
In Monte Carlo estimation of π, throwing N darts at a unit square with quarter-circle, the error decreases as:
- 1/N
- 1/√N
- 1/N²
- exponentially
Explanation
Each dart gives an independent Bernoulli trial with success probability π/4. The estimate p̂ = hits/N has variance p(1−p)/N ≈ 0.25/N. Standard error ≈ 0.5/√N, so π estimate error ≈ 2/√N. This 1/√N convergence is the curse of Monte Carlo—slow, but works in any dimension.
Correct Answer Is:
1/√N
A cup of coffee cools from 90°C to 60°C in 10 minutes in a 20°C room. How long to cool to 25°C?
- 30 minutes
- 40 minutes
- 50 minutes
- 60 minutes
Explanation
Using Newton’s law: T(t) = 20 + 70e^{-kt}. At t=10, 60 = 20 + 70e^{-10k} → 40 = 70e^{-10k} → e^{-10k} = 4/7 → −10k = ln(4/7) → k = −(1/10)ln(4/7). For T=25: 25 = 20 + 70e^{-kt} → 5/70 = e^{-kt} → e^{-kt} = 1/14 → −kt = ln(1/14) = −ln14 → t = ln14 / k = ln14 × 10 / ln(7/4) ≈ 2.639 × 10 / 0.5596 ≈ 47.1 minutes ≈ 50 minutes
Correct Answer Is:
50 minutes
In the M/M/1 queueing model, the average number of customers in the system L is ρ / (1 − ρ)
- ρ / (1 − ρ)
- λ / (μ − λ)
- ρ² / (1 − ρ)
- λ / μ
Explanation
M/M/1 assumes Poisson arrivals rate λ, exponential service rate μ, single server, infinite queue. Traffic intensity ρ = λ/μ < 1. By Little’s law L = λ W, and steady-state waiting time W = 1/(μ − λ), so L = λ/(μ − λ) = ρ/(1 − ρ). This is the classic result every operations research student memorizes—used for bank counters, call centers, and cloud servers.
Correct Answer Is:
ρ / (1 − ρ)
In the tent map T(x) = 1 − 2|x − 0.5| on [0,1], the growth rate that produces exact chaos equivalent to r = 4 logistic is:
- μ = 2
- μ = 1
- μ = √2
- μ = 4
Explanation
The tent map with μ = 2 is topologically conjugate to the logistic map at r = 4 via a homeomorphism. Both are ergodic, have dense orbits, and positive Lyapunov exponent ln 2. The tent map is easier to analyze symbolically because it is piecewise linear, yet exhibits the same chaotic statistics.
Correct Answer Is:
μ = 2
In Laplace’s equation ∇²u = 0 on a square grid, the 5-point stencil Jacobi iteration is:
- u^{new}{i,j} = (u{i+1,j} + u_{i−1,j} + u_{i,j+1} + u_{i,j−1}) / 4
- u^{new}{i,j} = u{i,j} − ω (residual)
- u^{new}{i,j} = u{i,j} + Δx² f_{i,j}
- u^{new}_{i,j} = average of diagonal neighbors
Explanation
Laplace’s equation means the value at any interior point is the average of its four cardinal neighbors. The Jacobi method updates every point simultaneously using old values: u^{new}_{i,j} = (right + left + up + down)/4. This simple averaging converges slowly but is embarrassingly parallel
Correct Answer Is:
u^{new}{i,j} = (u{i+1,j} + u_{i−1,j} + u_{i,j+1} + u_{i,j−1}) / 4
The carrying capacity K in the logistic equation dP/dt = rP(1 − P/K) represents:
- The maximum sustainable population
- The initial population size
- The intrinsic growth rate
- The time to reach half capacity
Explanation
As t → ∞, the term (1 − P/K) → 0, so dP/dt → 0 and P → K. K is therefore the environmental carrying capacity—the population level at which birth exactly balances death due to resource limitation. Real-world examples: deer on an island, bacteria in a petri dish, human population projections.
Correct Answer Is:
The maximum sustainable population
The branch-and-bound method solves IPs by:
- Dividing the problem into smaller subproblems and pruning branches that cannot improve the incumbent
- Trying all possible integer combinations directly
- Using only cutting planes
- Converting everything to binary
Explanation
Branch-and-bound solves the LP relaxation first, then systematically partitions the feasible region (branching on fractional variables, e.g., x ≤ 2 or x ≥ 3) while keeping track of the best known integer solution (incumbent). Any subproblem whose LP bound is worse than the incumbent is fathomed (pruned). This avoids enumerating all 2ⁿ possibilities.
Correct Answer Is:
Dividing the problem into smaller subproblems and pruning branches that cannot improve the incumbent
The solution to the continuous-time absorbing Markov chain with two transient states and one absorbing state has absorption probabilities given by: B = N R where N = (I − Q)^{−1}
- The fundamental matrix N = (I − Q)^{−1} R
- B = N R where N = (I − Q)^{−1}
- B = (I − Q) R
- B = Q^{-1} R
Explanation
Standard absorbing chain theory partitions the transition matrix into transient (Q) and absorbing (R) blocks. The expected number of visits to transient states before absorption is N = (I − Q)^{−1}, and the absorption probabilities from each transient state to each absorbing state are exactly B = N R. This is how Google originally ranked pages and how WGU C877 teaches gambler’s ruin.
Correct Answer Is:
B = N R where N = (I − Q)^{−1}
The basic reproduction number R₀ in a heterogeneous network is approximately:
-
-
− 1 -
-
²
Explanation
In well-mixed populations R₀ = β/γ. In networks, the epidemic spreads faster because high-degree nodes infect many others. The exact formula derived from next-generation matrix is R₀ =
Correct Answer Is:
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