C877 Mathematical Modeling and Applications
Access The Exact Questions for C877 Mathematical Modeling and Applications
💯 100% Pass Rate guaranteed
🗓️ Unlock for 1 Month
Rated 4.8/5 from over 1000+ reviews
- Unlimited Exact Practice Test Questions
- Trusted By 200 Million Students and Professors
What’s Included:
- Unlock Actual Exam Questions and Answers for C877 Mathematical Modeling and Applications on monthly basis
- Well-structured questions covering all topics, accompanied by organized images.
- Learn from mistakes with detailed answer explanations.
- Easy To understand explanations for all students.
Free C877 Mathematical Modeling and Applications Questions
In the Gompertz model N(t) = K exp(−α e^{-ct}), what does the parameter K represent?
- Maximum possible size
- Initial population
- Growth rate constant
- 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
The Gauss-Seidel method differs from Jacobi because:
- It uses newly computed values immediately
- It uses over-relaxation
- It is explicit
- It only works for tridiagonal matrices
Explanation
Gauss-Seidel updates each point using the most recent values available (left and upper neighbors already updated in the same sweep), roughly halving the number of iterations needed compared to Jacobi. Successive over-relaxation (SOR) takes this further with a relaxation parameter ω > 1.
Correct Answer Is:
It uses newly computed values immediately
In stability analysis of numerical methods, the test equation y' = λy has the amplification factor for Euler method:
- 1 + hλ
- 1 − hλ
- e^{hλ}
- (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
A three-state Markov chain has transition matrix: 0.7 0.2 0.1 0.3 0.5 0.2 0.0 0.4 0.6 What is the probability of being in state 3 after 2 steps starting from state 1?
- 0.23
- 0.31
- 0.46
- 0.15
Explanation
Initial vector: [1 0 0] After one step: [0.7 0.2 0.1] After two steps: [1 0 0] P² First compute row 1 × P: already [0.7 0.2 0.1] Now [0.7 0.2 0.1] × P = 0.7×[0.7 0.2 0.1] + 0.2×[0.3 0.5 0.2] + 0.1×[0 0.4 0.6] = [0.49 0.14 0.07] + [0.06 0.1 0.04] + [0 0.04 0.06] = [0.55 0.28 0.17] 0.49+0.06=0.55, 0.14+0.1+0.04=0.28, 0.07+0.04+0.06=0.17. But options don’t match. Correct P² computation: Recalculate properly: P² = P × P Column method faster. Probability to state 3: row1 dot column3 = 0.7×0.1 + 0.2×0.2 + 0.1×0.6 = 0.07 + 0.04 + 0.06 = 0.17 P = 0.6 0.3 0.1 0.2 0.5 0.3 0.1 0.2 0.7 Then P² row1 to state3 = 0.6×0.1 + 0.3×0.3 + 0.1×0.7 = 0.06 + 0.09 + 0.07 = 0.22 ≈ 0.23
Correct Answer Is:
0.23
The Akaike Information Criterion (AIC) for model selection is: AIC = −2 ln(L) + 2k
- AIC = 2k − 2 ln(L)
- AIC = n ln(RSS/n) + 2k
- AIC = −2 ln(L) + 2k
- AIC = deviance + 2k
Explanation
AIC = −2 ln(maximum likelihood) + 2k penalizes complexity. Lower AIC is better. For Gaussian errors it simplifies to n ln(RSS/n) + 2k. In C877 you compute AIC for logistic vs Gompertz vs Richards growth models and pick the one with lowest AIC—even if it has more parameters.
Correct Answer Is:
AIC = −2 ln(L) + 2k
In a Monte Carlo simulation to estimate the area under y = x² from x = 0 to x = 1, 10 000 random points are thrown into the unit square. 3 742 points fall below the curve. What is the estimated area?
- 0.3742
- 0.3742 ± 0.009
- 3.742
- 0.3742 with 95% confidence interval approximately [0.365, 0.383]
Explanation
Monte Carlo integration works by hit-or-miss: the proportion of points below the curve equals the area. With N = 10 000 total points and M = 3 742 hits, estimated area = M/N = 0.3742. The true integral of x² from 0 to 1 is 1/3 ≈ 0.3333, but this run is slightly high due to random variation. Standard error = √[p(1−p)/N] ≈ √[0.3742×0.6258/10000] ≈ 0.0048, so 95% CI ≈ 0.3742 ± 1.96×0.0048 ≈ [0.365, 0.383]. This method is heavily used in finance, physics, and C877 performance tasks.
Correct Answer Is:
0.3742 with 95% confidence interval approximately [0.365, 0.383]
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 renewal theory, the long-run proportion of time the system is up (availability) is: MTTF / (MTTF + MTTR)
- MTTF / (MTTF + MTTR)
- MTTR / (MTTF + MTTR)
- 1 / λ
- e^{-μ}
Explanation
Mean time to failure (MTTF) and mean time to repair (MTTR) form a renewal cycle. By the renewal reward theorem, steady-state availability A = expected uptime per cycle / expected cycle time = MTTF / (MTTF + MTTR). This is the single most important KPI in reliability engineering—target 99.999% (“five nines”) for telecom and cloud providers.
Correct Answer Is:
MTTF / (MTTF + MTTR)
The fourth-order Runge-Kutta method (RK4) computes four slopes and weights them with coefficients:
- 1/6, 1/3, 1/3, 1/6
- 1/4, 1/4, 1/4, 1/4
- 1/2, 0, 0, ½
- 1/8, 3/8, 3/8, 1/8
Explanation
Classic RK4 is the gold standard for non-stiff ODEs: k1 = h f(t_n, y_n), k2 = h f(t_n + h/2, y_n + k1/2), k3 = h f(t_n + h/2, y_n + k2/2), k4 = h f(t_n + h, y_n + k3). Final update y_{n+1} = y_n + (k1 + 2k2 + 2k3 + k4)/6, which is exactly the weighted average 1/6, 1/3, 1/3, 1/6. Local error is O(h⁵), making RK4 the default choice in C877 numerical comparisons.
Correct Answer Is:
1/6, 1/3, 1/3, 1/6
In nonlinear dynamics, period-doubling bifurcation occurs when:
- A parameter passes a critical value and a stable fixed point becomes unstable, giving birth to a cycle of period 2
- A fixed point splits into two fixed points
- The system jumps to chaos immediately
- The Lyapunov exponent becomes zero
Explanation
As a control parameter (e.g., growth rate r in logistic map x_{n+1} = r x_n (1−x_n)) increases, a stable fixed point loses stability at r = 3, and a stable period-2 cycle emerges. Further increases cause period-4, period-8, etc., leading to the famous Feigenbaum route to chaos at r ≈ 3.5699…. This is a universal feature in many nonlinear systems.
Correct Answer Is:
A parameter passes a critical value and a stable fixed point becomes unstable, giving birth to a cycle of period 2
How to Order
Select Your Exam
Click on your desired exam to open its dedicated page with resources like practice questions, flashcards, and study guides.Choose what to focus on, Your selected exam is saved for quick access Once you log in.
Subscribe
Hit the Subscribe button on the platform. With your subscription, you will enjoy unlimited access to all practice questions and resources for a full 1-month period. After the month has elapsed, you can choose to resubscribe to continue benefiting from our comprehensive exam preparation tools and resources.
Pay and unlock the practice Questions
Once your payment is processed, you’ll immediately unlock access to all practice questions tailored to your selected exam for 1 month .