\section*{Problem 1}

\subsection*{Part A}

\(K_{eff}\) is the effective growth or decay of the neutron population based on
the prompt reactions. This block can be represented by basic physical quantities
by using the six factor formula:

\[ k_{eff} = \eta f p \epsilon P_{FNL} P_{TNL} \]

where:

\begin{itemize}
  \item \(\eta\) is the thermal neutron fission factor
  \item \(f\) is the thermal neutron absorption factor
  \item \(p\) is the resonance escape probability
  \item \(\epsilon\) is the fast fission neutron production factor
  \item \(P_{FNL}\) is the fast non-leakage probability
  \item \(P_{TNL}\) is the thermal non-leakage probability
\end{itemize}

\subsection*{Part B}

We start with a set of equations representing our neutron populations:

\[ N_{in} = N_{f} \]
\[ N_{f} = N_{p} + N_{decay} + S \Delta t \]
\[ N_{t} = N_{d} + N_{p} \]
\[ N_t = K_{eff}N_{in} \]

Now, how are these related? Well, the total number of neutrons is split between
the neutrons that are prompt neutrons and those that are destined to become
delayed neutrons. We can represent the fraction between the two as:

\[ N_t = (1-\beta)K_{eff} N_{in} + \sum_{i=1}^6 \lambda_i C_i \Delta t + S \Delta
t \]

where

\[ N_d = \sum_{i=1}^6 K_{eff} \beta_i N_{in} \]
and
\[ N_p = (1-\beta) K_{eff} N_{in} \]

Then we find the change in neutron population:

\[ N_f - N_i = (1-\beta) K_{eff} N_i + \sum_{i=1}^6 \lambda_i C_i \Delta t + S
\Delta t - N_i \]

and take the 'derivative':

\[ \frac{N_f - N_i}{\Delta t} = \frac{(1- \frac{1}{K_eff} - \beta) K_{eff} N_i}{\Delta t} + \sum_{i=1}^6 \lambda_i C_i + S
\]

then after a little more substitution found in Fundamental Kinetics Ideas:

\\[ \dot N(t) = \frac{(\rho - \beta) N(t)}{\Lambda} + \sum_{i=1}^6 \lambda_i C_i + S
\]

and not forgetting our precursors:

\[\dot C_i(t) = \frac{\beta_i}{\Lambda}N(t) - \lambda_i C_i\]


\subsection*{Part C} 

The prompt jump assumption assumes that the speed of the prompt cycle is so fast
that we can effectively ignore the dynamics of the prompt neutron effects. It is
equivalent to us only using the outside loop, and not considering additions from
\(N_p\).


\subsection*{Part D}

First, the number of prompt neutrons would increase dramatically. This would
happen over the first hundreds of milliseconds. Then, over a much longer time
horizon (seconds, minutes), the delayed number of neutrons would increase as the
precursors catch up to the reactivity increase.


\subsection*{Part E}

Power turning has to do with start-up rate and the rate at which the total
neutron population is changing. For this diagram, turning means that \(\dot
N_f\) changes sign.