ADI Schemes for the Heston PDE¶

A fully implicit discretization of the two-dimensional Heston PDE produces a large, sparse linear system with $N_x \times N_v$ unknowns at each time step. Solving this system directly has $\mathcal{O}(N_x^2 N_v^2)$ cost---prohibitively expensive for production grids. Alternating direction implicit (ADI) schemes split the 2D problem into sequences of 1D implicit solves, each requiring only a tridiagonal matrix inversion at $\mathcal{O}(N)$ cost. This section develops three ADI schemes for the Heston PDE: Douglas-Rachford, Craig-Sneyd, and Hundsdorfer-Verwer, each handling the mixed derivative term differently.

Learning Objectives

By the end of this section, you will be able to:

Explain why ADI splitting reduces the computational cost from $\mathcal{O}(N^4)$ to $\mathcal{O}(N^2)$
Implement the Douglas-Rachford, Craig-Sneyd, and Hundsdorfer-Verwer schemes
Analyze the stability properties and accuracy order of each scheme
Choose the appropriate scheme based on the strength of correlation $\rho$

Prerequisites

This section requires the 2D PDE formulation and its operator splitting $\mathcal{L} = \mathcal{L}_x + \mathcal{L}_v + \mathcal{L}_{xv}$.

The ADI Principle¶

Recall the semi-discrete Heston PDE in backward time $\tau = T - t$:

\[ \frac{\partial V}{\partial \tau} = (\mathcal{L}_x + \mathcal{L}_v + \mathcal{L}_{xv} - rI) V \]

where $\mathcal{L}_x$, $\mathcal{L}_v$ are the directional operators and $\mathcal{L}_{xv}$ is the mixed derivative operator. Denote the discrete matrix representations as $A_x$, $A_v$, and $A_{xv}$, and let $A_0 = A_x + A_v + A_{xv} - rI$ be the full spatial operator.

A fully implicit step would solve:

\[ (I - \Delta\tau \, A_0) V^{n+1} = V^n \]

The matrix $(I - \Delta\tau \, A_0)$ is of size $(N_x N_v) \times (N_x N_v)$ and is sparse but not tridiagonal. ADI schemes instead solve sequences of tridiagonal systems by treating each direction implicitly in turn.

Douglas-Rachford Scheme¶

The Douglas-Rachford (DR) scheme is the simplest ADI method. It treats the $x$-direction implicitly first, then the $v$-direction, while handling the mixed derivative explicitly.

Stage 0 (explicit predictor):

\[ Y_0 = V^n + \Delta\tau \, A_0 V^n \]

Stage 1 ($x$-direction implicit correction):

\[ (I - \theta \Delta\tau \, A_x) Y_1 = Y_0 - \theta \Delta\tau \, A_x V^n \]

Stage 2 ($v$-direction implicit correction):

\[ (I - \theta \Delta\tau \, A_v) V^{n+1} = Y_1 - \theta \Delta\tau \, A_v V^n \]

Here $\theta \in [0, 1]$ is a parameter controlling the implicitness. The choice $\theta = \frac{1}{2}$ gives second-order accuracy in time; $\theta = 1$ gives first-order accuracy but improved stability.

Each stage requires solving a tridiagonal system: Stage 1 involves $N_v$ independent tridiagonal systems of size $N_x$, and Stage 2 involves $N_x$ independent systems of size $N_v$. The total cost per time step is $\mathcal{O}(N_x N_v)$.

Mixed Derivative Treatment

In the Douglas-Rachford scheme, the mixed derivative $A_{xv}$ appears only in the explicit predictor $Y_0$. This preserves the tridiagonal structure of each implicit stage but limits the scheme's stability when $|\rho|$ is large.

Craig-Sneyd Scheme¶

The Craig-Sneyd (CS) scheme improves the treatment of the mixed derivative by adding an explicit correction step that partially compensates for the explicit handling.

Stage 0 (explicit predictor):

\[ Y_0 = V^n + \Delta\tau \, A_0 V^n \]

Stage 1 ($x$-direction implicit):

\[ (I - \theta \Delta\tau \, A_x) Y_1 = Y_0 - \theta \Delta\tau \, A_x V^n \]

Stage 2 ($v$-direction implicit):

\[ (I - \theta \Delta\tau \, A_v) \tilde{Y} = Y_1 - \theta \Delta\tau \, A_v V^n \]

Stage 3 (mixed derivative correction):

\[ \hat{Y}_0 = Y_0 + \frac{1}{2} \Delta\tau (A_{xv} \tilde{Y} - A_{xv} V^n) \]

Stage 4 ($x$-direction implicit, repeated):

\[ (I - \theta \Delta\tau \, A_x) \hat{Y}_1 = \hat{Y}_0 - \theta \Delta\tau \, A_x V^n \]

Stage 5 ($v$-direction implicit, repeated):

\[ (I - \theta \Delta\tau \, A_v) V^{n+1} = \hat{Y}_1 - \theta \Delta\tau \, A_v V^n \]

The additional Stages 3--5 re-process the solution with an updated mixed derivative contribution. With $\theta = \frac{1}{2}$, the Craig-Sneyd scheme achieves second-order accuracy and better stability than Douglas-Rachford for $|\rho|$ up to approximately 0.8.

Hundsdorfer-Verwer Scheme¶

The Hundsdorfer-Verwer (HV) scheme provides the best stability properties among the three, at the cost of one additional implicit sweep per direction.

Stage 0 (explicit predictor):

\[ Y_0 = V^n + \Delta\tau \, A_0 V^n \]

Stage 1 ($x$-direction implicit):

\[ (I - \theta \Delta\tau \, A_x) Y_1 = Y_0 - \theta \Delta\tau \, A_x V^n \]

Stage 2 ($v$-direction implicit):

\[ (I - \theta \Delta\tau \, A_v) \tilde{Y} = Y_1 - \theta \Delta\tau \, A_v V^n \]

Stage 3 (corrector predictor):

\[ \hat{Y}_0 = Y_0 + \frac{1}{2}\Delta\tau \left(A_0 \tilde{Y} - A_0 V^n\right) \]

Stage 4 ($x$-direction implicit, repeated):

\[ (I - \theta \Delta\tau \, A_x) \hat{Y}_1 = \hat{Y}_0 - \theta \Delta\tau \, A_x \tilde{Y} \]

Stage 5 ($v$-direction implicit, repeated):

\[ (I - \theta \Delta\tau \, A_v) V^{n+1} = \hat{Y}_1 - \theta \Delta\tau \, A_v \tilde{Y} \]

The key difference from Craig-Sneyd is in Stage 3: the full operator $A_0$ (including both directional and mixed derivative operators) is used in the correction, not just $A_{xv}$. This provides unconditional stability for all values of $\rho$ with $\theta = \frac{1}{2}$.

Stability and Accuracy Comparison¶

Property	Douglas-Rachford	Craig-Sneyd	Hundsdorfer-Verwer
Time accuracy ($\theta = \frac{1}{2}$)	$\mathcal{O}(\Delta\tau^2)$	$\mathcal{O}(\Delta\tau^2)$	$\mathcal{O}(\Delta\tau^2)$
Time accuracy ($\theta = 1$)	$\mathcal{O}(\Delta\tau)$	$\mathcal{O}(\Delta\tau)$	$\mathcal{O}(\Delta\tau)$
Stability for $	\rho	\leq 0.5$	Unconditional
Stability for $	\rho	\approx 0.7$	Conditional
Stability for $	\rho	\to 1$	Unstable
Tridiagonal solves per step	2	4	4
Cost per step (relative)	1.0	2.0	2.0

Correlation and Stability

For equity markets where $\rho \in [-0.9, -0.5]$ is typical, the Douglas-Rachford scheme may exhibit oscillations near $v = 0$ or large $S$. The Hundsdorfer-Verwer scheme is recommended as the default for Heston implementations because it maintains unconditional stability across all parameter regimes.

Implementation Structure¶

Each ADI time step follows this pattern:

Form the right-hand side using the explicit parts of $A_0$
Sweep in $x$-direction: for each fixed $v_j$, solve a tridiagonal system of size $N_x$
Sweep in $v$-direction: for each fixed $x_i$, solve a tridiagonal system of size $N_v$
For CS and HV: repeat steps 1--3 with the corrected right-hand side

The tridiagonal systems are solved using the Thomas algorithm (LU decomposition for tridiagonal matrices) in $\mathcal{O}(N)$ operations. The matrix entries depend on $v$ through the diffusion and drift coefficients but are constant along each $v$-line, allowing the LU factorization to be cached.

Tridiagonal System Structure

For the $x$-sweep at variance level $v_j$, the system has the form:

\[ -\alpha_i V_{i-1,j}^{n+1} + (1 + \beta_i) V_{i,j}^{n+1} - \gamma_i V_{i+1,j}^{n+1} = \text{RHS}_i \]

where the coefficients are:

\[ \alpha_i = \theta \Delta\tau \left[\frac{v_j}{h_i^-(h_i^+ + h_i^-)} + \frac{r - q - v_j/2}{h_i^+ + h_i^-}\cdot\frac{h_i^+}{h_i^-}\right] \]

and $\beta_i, \gamma_i$ are analogous. These are assembled once per time step and reused for the repeated sweeps in CS and HV.

Worked Example¶

Consider the standard Heston parameters with $N_x = 100$, $N_v = 50$, $N_t = 100$:

Grid	Total unknowns	DR time (relative)	HV time (relative)	HV error vs Fourier
$50 \times 25$	1,250	1.0	2.0	$\$0.08$
$100 \times 50$	5,000	4.0	8.0	$\$0.02$
$200 \times 100$	20,000	16.0	32.0	$\$0.005$

The quadratic scaling in grid size is clearly visible. For the $100 \times 50$ grid with the HV scheme and $\theta = \frac{1}{2}$, the pricing error for an ATM European call is within $\$0.02$ of the Fourier reference, with a computation time of a few seconds on a modern workstation.

Summary¶

ADI schemes transform the two-dimensional Heston PDE into sequences of one-dimensional tridiagonal solves, reducing the per-step cost from $\mathcal{O}(N_x^2 N_v^2)$ to $\mathcal{O}(N_x N_v)$. The Douglas-Rachford scheme is simplest but can be unstable for large $|\rho|$. The Craig-Sneyd scheme improves mixed derivative treatment. The Hundsdorfer-Verwer scheme provides unconditional stability for all correlation values and is the recommended default. All three achieve second-order temporal accuracy with $\theta = \frac{1}{2}$. The ADI framework extends naturally to American options via PSOR, where the early exercise constraint is applied after each time step.

Exercises¶

Exercise 1. A fully implicit Crank-Nicolson scheme on a grid with $N_x = 200$ and $N_v = 100$ produces a sparse system of size $20{,}000 \times 20{,}000$. If solving this system with a banded solver costs $\mathcal{O}(N_x^2 N_v)$ per time step, compare the total cost for $N_t = 200$ steps with the ADI cost of $\mathcal{O}(N_x N_v \cdot N_t)$. Compute the speedup factor.

Solution to Exercise 1

Banded Crank-Nicolson cost. The grid has $N_x = 200$, $N_v = 100$, so there are $N = N_x N_v = 20{,}000$ unknowns. A banded solver with bandwidth $\mathcal{O}(N_x)$ costs $\mathcal{O}(N_x^2 N_v)$ per time step. For $N_t = 200$ steps:

\[ C_{\text{CN}} = N_t \cdot N_x^2 \cdot N_v = 200 \times 200^2 \times 100 = 8 \times 10^8 \]

ADI cost. Each ADI time step costs $\mathcal{O}(N_x N_v)$ (with a small constant multiplier for the number of tridiagonal sweeps). For $N_t = 200$ steps:

\[ C_{\text{ADI}} = N_t \cdot N_x \cdot N_v = 200 \times 200 \times 100 = 4 \times 10^6 \]

Speedup factor.

\[ \text{Speedup} = \frac{C_{\text{CN}}}{C_{\text{ADI}}} = \frac{8 \times 10^8}{4 \times 10^6} = 200 \]

The speedup equals $N_x = 200$, which is not a coincidence. The banded solver has an extra factor of $N_x$ in its cost (from the bandwidth), so the ADI speedup scales linearly with $N_x$. For a $400 \times 200$ grid, the speedup would be approximately 400.

Exercise 2. The Douglas-Rachford scheme treats the mixed derivative explicitly. For $\rho = -0.9$, explain why the CFL-type stability constraint becomes binding. If $\Delta t = T/N_t$ with $T = 1$ and $N_t = 100$, is this time step small enough for stability on a typical Heston grid?

Solution to Exercise 2

Why the mixed derivative causes instability. In the Douglas-Rachford scheme, the mixed derivative operator $A_{xv}$ appears only in the explicit predictor stage $Y_0 = V^n + \Delta\tau A_0 V^n$. For the explicit treatment to be stable, the time step must satisfy a CFL-type condition relating $\Delta\tau$ to the magnitude of $A_{xv}$.

The mixed derivative coefficient is $\rho \xi v$. At a typical variance level $v \approx 0.04$ and with $\rho = -0.9$, $\xi = 0.3$:

\[ |\rho \xi v| = 0.9 \times 0.3 \times 0.04 = 0.0108 \]

The explicit stability condition for the mixed derivative on a grid with spacings $\Delta x$ and $\Delta v$ is approximately:

\[ \Delta\tau \lesssim \frac{\Delta x \, \Delta v}{2 |\rho \xi v|} \]

For a typical grid with $\Delta x \approx 0.02$ and $\Delta v \approx 0.004$:

\[ \Delta\tau \lesssim \frac{0.02 \times 0.004}{2 \times 0.0108} \approx 0.0037 \]

With $T = 1$ and $N_t = 100$, the time step is $\Delta t = 0.01$, which exceeds this stability bound. The Douglas-Rachford scheme is therefore not stable with these parameters. One would need $N_t \gtrsim 270$ to satisfy the CFL constraint, or alternatively, one should use the Craig-Sneyd or Hundsdorfer-Verwer scheme, which handle the mixed derivative more robustly.

The instability manifests as oscillations, particularly near $v = 0$ (where the CIR drift is strongest) and for deep in-the-money or out-of-the-money options where $\partial^2 V / \partial x \partial v$ is largest.

Exercise 3. The Craig-Sneyd scheme adds a correction step that updates the cross-term. Write the four stages explicitly for the Heston PDE operators $\mathcal{A}_0$, $\mathcal{A}_1$, $\mathcal{A}_2$ with $\theta = 1/2$. Explain why Stage 4 (cross-term correction) improves the local truncation error from $\mathcal{O}(\Delta t)$ to $\mathcal{O}(\Delta t^2)$.

Solution to Exercise 3

Craig-Sneyd stages with Heston operators. Let $\mathcal{A}_0 = A_x + A_v + A_{xv} - rI$ be the full operator, and set $\theta = 1/2$. The six stages are:

Stage 0 (explicit predictor):

\[ Y_0 = V^n + \Delta\tau \, \mathcal{A}_0 V^n \]

Stage 1 ($x$-implicit):

\[ \left(I - \tfrac{1}{2}\Delta\tau \, A_x\right) Y_1 = Y_0 - \tfrac{1}{2}\Delta\tau \, A_x V^n \]

Stage 2 ($v$-implicit):

\[ \left(I - \tfrac{1}{2}\Delta\tau \, A_v\right) \tilde{Y} = Y_1 - \tfrac{1}{2}\Delta\tau \, A_v V^n \]

Stage 3 (cross-term correction):

\[ \hat{Y}_0 = Y_0 + \tfrac{1}{2}\Delta\tau \left(A_{xv} \tilde{Y} - A_{xv} V^n\right) \]

Stage 4 ($x$-implicit, repeated):

\[ \left(I - \tfrac{1}{2}\Delta\tau \, A_x\right) \hat{Y}_1 = \hat{Y}_0 - \tfrac{1}{2}\Delta\tau \, A_x V^n \]

Stage 5 ($v$-implicit, repeated):

\[ \left(I - \tfrac{1}{2}\Delta\tau \, A_v\right) V^{n+1} = \hat{Y}_1 - \tfrac{1}{2}\Delta\tau \, A_v V^n \]

Why Stage 3 improves accuracy. Without the correction (i.e., in the Douglas-Rachford scheme), the cross-derivative is evaluated only at $V^n$. The local truncation error from the mixed derivative is:

\[ \Delta\tau \, A_{xv} V^n = \Delta\tau \, A_{xv} V^{n+1} - \Delta\tau^2 \, A_{xv} \dot{V} + \mathcal{O}(\Delta\tau^3) \]

The $\mathcal{O}(\Delta\tau^2)$ term is a first-order temporal error in the mixed derivative treatment.

Stage 3 replaces $A_{xv} V^n$ with $\frac{1}{2}(A_{xv} V^n + A_{xv} \tilde{Y})$. Since $\tilde{Y} = V^{n+1} + \mathcal{O}(\Delta\tau)$, this gives a trapezoidal average:

\[ \frac{1}{2}A_{xv}(V^n + \tilde{Y}) = A_{xv} V^{n+1/2} + \mathcal{O}(\Delta\tau^2) \]

The error in the mixed derivative drops from $\mathcal{O}(\Delta\tau)$ to $\mathcal{O}(\Delta\tau^2)$, making the overall scheme second-order in time.

Exercise 4. Compare the computational cost per time step of the three ADI schemes. If Thomas's algorithm costs $5N$ operations for a tridiagonal system of size $N$, compute the total operations for: (a) Douglas-Rachford ($N_v$ solves of size $N_x$ + $N_x$ solves of size $N_v$), (b) Craig-Sneyd (same plus one explicit cross-term evaluation), (c) Hundsdorfer-Verwer (double the implicit sweeps). Use $N_x = 200$, $N_v = 100$.

Solution to Exercise 4

Thomas algorithm cost. Each tridiagonal solve of size $N$ costs $5N$ operations (2 divisions, 2 multiplications, 1 subtraction in each of the forward elimination and back-substitution phases, approximately).

(a) Douglas-Rachford.

$x$-sweep: $N_v$ tridiagonal systems of size $N_x$, cost $= N_v \times 5N_x = 100 \times 5 \times 200 = 100{,}000$
$v$-sweep: $N_x$ tridiagonal systems of size $N_v$, cost $= N_x \times 5N_v = 200 \times 5 \times 100 = 100{,}000$
Explicit predictor $Y_0$: matrix-vector product with $A_0$, cost $\approx c \cdot N_x N_v$ where $c$ is the stencil width (about 7--9 nonzeros per row), so approximately $7 \times 20{,}000 = 140{,}000$

\[ C_{\text{DR}} \approx 100{,}000 + 100{,}000 + 140{,}000 = 340{,}000 \text{ operations} \]

(b) Craig-Sneyd.

Same as DR plus: one explicit cross-term evaluation ($A_{xv} \tilde{Y}$ costs $\approx 5 \times 20{,}000 = 100{,}000$ using the 4-point stencil) and one additional pair of implicit sweeps (same cost as the first pair: $200{,}000$)

\[ C_{\text{CS}} \approx 340{,}000 + 100{,}000 + 200{,}000 = 640{,}000 \text{ operations} \]

(c) Hundsdorfer-Verwer.

Same as DR plus: one explicit full-operator evaluation ($A_0 \tilde{Y}$, cost $\approx 140{,}000$) and one additional pair of implicit sweeps ($200{,}000$)

\[ C_{\text{HV}} \approx 340{,}000 + 140{,}000 + 200{,}000 = 680{,}000 \text{ operations} \]

Ratios: $C_{\text{CS}} / C_{\text{DR}} \approx 1.88$ and $C_{\text{HV}} / C_{\text{DR}} \approx 2.00$, confirming the factor-of-2 rule of thumb from the comparison table.

Exercise 5. Design a convergence study for the Craig-Sneyd scheme: compute the European call price on grids $(N_x, N_v, N_t) = (50, 25, 50)$, $(100, 50, 100)$, $(200, 100, 200)$ and compare with the COS reference. If the scheme is second-order, the error should decrease by a factor of 4 each time. Verify this by computing the convergence rate $p = \log_2(\text{error}_1 / \text{error}_2)$.

Solution to Exercise 5

Convergence study design. Let the three grid levels be indexed by $\ell = 1, 2, 3$:

Level $\ell$	$N_x$	$N_v$	$N_t$	$\Delta x$ factor	$\Delta v$ factor	$\Delta\tau$ factor
1	50	25	50	1	1	1
2	100	50	100	1/2	1/2	1/2
3	200	100	200	1/4	1/4	1/4

At each level, compute the European call price $V_\ell$ (e.g., at $S = K$, $v = v_0$) and the error $e_\ell = |V_\ell - V_{\text{ref}}|$ where $V_{\text{ref}}$ is the COS Fourier reference.

Second-order convergence prediction. The Craig-Sneyd scheme with $\theta = 1/2$ is second-order in both space and time. Since all grid parameters are halved simultaneously:

\[ e_\ell = C \cdot h_\ell^2 + \mathcal{O}(h_\ell^3) \]

where $h_\ell$ represents the characteristic mesh size at level $\ell$. With halving:

\[ \frac{e_1}{e_2} \approx 4, \qquad \frac{e_2}{e_3} \approx 4 \]

Convergence rate computation. The numerical convergence rate between levels $\ell$ and $\ell+1$ is:

\[ p = \log_2\!\left(\frac{e_\ell}{e_{\ell+1}}\right) \]

For a second-order scheme, $p \approx 2$. In practice, one typically observes $p \in [1.8, 2.2]$ due to boundary effects, non-uniform grid stretching, and the explicit treatment of the mixed derivative. If $p$ is significantly less than 2, this may indicate that the mixed derivative treatment or the boundary conditions are limiting the convergence order.

Richardson extrapolation. With the three grid levels, one can also compute an extrapolated price:

\[ V_{\text{extrap}} = \frac{4 V_3 - V_2}{3} \]

which should be accurate to $\mathcal{O}(h_3^4)$ if the scheme is genuinely second-order.

Exercise 6. The Hundsdorfer-Verwer scheme performs two sets of implicit sweeps per time step. Describe the advantage: unconditional stability allows larger $\Delta t$. If the Craig-Sneyd scheme requires $N_t = 200$ steps for stability while Hundsdorfer-Verwer requires only $N_t = 100$ for the same accuracy, which scheme is faster overall? Account for the fact that Hundsdorfer-Verwer costs twice as much per step.

Solution to Exercise 6

Cost comparison. Let $C_1$ denote the cost of one tridiagonal sweep pair (one $x$-sweep and one $v$-sweep). Then:

Craig-Sneyd costs $2C_1$ per step (two pairs of sweeps plus explicit evaluations)
Hundsdorfer-Verwer costs $2C_1$ per step (same number of sweeps, slightly more expensive explicit evaluation, but approximately equal)

With the given time step requirements:

Craig-Sneyd total cost:

\[ C_{\text{CS}} = N_t^{\text{CS}} \times 2C_1 = 200 \times 2C_1 = 400 \, C_1 \]

Hundsdorfer-Verwer total cost:

\[ C_{\text{HV}} = N_t^{\text{HV}} \times 2C_1 = 100 \times 2C_1 = 200 \, C_1 \]

Hundsdorfer-Verwer is faster by a factor of 2 despite costing the same per step, because its unconditional stability allows a time step twice as large.

The key insight is that the per-step cost ratio between CS and HV is approximately 1:1 (both require four tridiagonal sweeps), but HV's unconditional stability means the number of time steps is determined purely by accuracy, not by stability. When $|\rho|$ is large, the CS scheme's conditional stability forces smaller $\Delta\tau$ than accuracy alone would require, creating a gap between the stability-limited and accuracy-limited step counts.

More generally, if the CS stability limit requires $N_t^{\text{CS}}$ steps while accuracy alone requires $N_t^{\text{acc}}$, then HV is faster whenever:

\[ \frac{N_t^{\text{CS}}}{N_t^{\text{acc}}} > 1 \]

For typical equity parameters with $\rho \in [-0.9, -0.5]$, this ratio ranges from 1.5 to 3, making HV the preferred scheme in practice.

Property	Douglas-Rachford	Craig-Sneyd	Hundsdorfer-Verwer
Time accuracy (\(\theta = \frac{1}{2}\))	\(\mathcal{O}(\Delta\tau^2)\)	\(\mathcal{O}(\Delta\tau^2)\)	\(\mathcal{O}(\Delta\tau^2)\)
Time accuracy (\(\theta = 1\))	\(\mathcal{O}(\Delta\tau)\)	\(\mathcal{O}(\Delta\tau)\)	\(\mathcal{O}(\Delta\tau)\)
Stability for $	\rho	\leq 0.5$	Unconditional
Stability for $	\rho	\approx 0.7$	Conditional
Stability for $	\rho	\to 1$	Unstable
Tridiagonal solves per step	2	4	4
Cost per step (relative)	1.0	2.0	2.0

Grid	Total unknowns	DR time (relative)	HV time (relative)	HV error vs Fourier
\(50 \times 25\)	1,250	1.0	2.0	\(\$0.08\)
\(100 \times 50\)	5,000	4.0	8.0	\(\$0.02\)
\(200 \times 100\)	20,000	16.0	32.0	\(\$0.005\)