---
title: "Sample Size Calculation With Fixed Follow-up"
author: "Kaifeng Lu"
date: "12/15/2021"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{Sample Size Calculation With Fixed Follow-up}
  %\VignetteEngine{knitr::rmarkdown}
  \usepackage[utf8]{inputenc}
---

```{r, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)
```

```{r setup}
library(lrstat)
```

This R Markdown document illustrates the sample size calculation for a fixed follow-up design, in which the treatment allocation is 3:1 and the hazard ratio is 0.3. This is a case for which neither the Schoenfeld method nor the Lakatos method provides an accurate sample size estimate, and simulation tools are needed to obtain a more accurate result. 

Consider a fixed design with the hazard rate of the control group being 0.95 per year, a hazard ratio of the experimental group to the control group being 0.3, a randomization ratio of 3:1, an enrollment rate of 5 patients per month, a 2-year drop-out rate of 10%, and a planned fixed follow-up of 26 weeks for each patient. The target power is 90%, and we are interested in the number of patients to enroll to achieve the target 90% power. 

Using the Schoenfeld formula, the required number of events is 39. This requires 191 patients enrolled over 38.2 months. Denote this design as design 1.  
```{r}
lrsamplesize(beta = 0.1, kMax = 1, criticalValues = 1.96, 
             allocationRatioPlanned = 3, accrualIntensity = 5, 
             lambda2 = 0.95/12, lambda1 = 0.3*0.95/12, 
             gamma1 = -log(1-0.1)/24, gamma2 = -log(1-0.1)/24, 
             accrualDuration = NA, followupTime = 26/4, 
             fixedFollowup = TRUE,
             typeOfComputation = "schoenfeld")
```


On the other hand, the output from the default lrsamplesize call implies that we only need 26 events with 127 subjects enrolled over 25.4 months, a dramatic difference from the Schoenfeld formula. Denote this design as design 2. 
```{r}
lrsamplesize(beta = 0.1, kMax = 1, criticalValues = 1.96, 
             allocationRatioPlanned = 3, accrualIntensity = 5, 
             lambda2 = 0.95/12, lambda1 = 0.3*0.95/12, 
             gamma1 = -log(1-0.1)/24, gamma2 = -log(1-0.1)/24, 
             accrualDuration = NA, followupTime = 26/4, 
             fixedFollowup = TRUE,
             typeOfComputation = "direct")
```

To check the accuracy of either solution, we run simulations using the lrsim function. 
```{r}
lrsim(kMax = 1, criticalValues = 1.96,  
      allocation1 = 3, allocation2 = 1,
      accrualIntensity = 5, 
      lambda2 = 0.95/12, lambda1 = 0.3*0.95/12, 
      gamma1 = -log(1-0.1)/24, gamma2 = -log(1-0.1)/24,
      accrualDuration = 38.2, followupTime = 6.5, 
      fixedFollowup = TRUE,  
      plannedEvents = 39, 
      maxNumberOfIterations = 10000, seed = 12345)

lrsim(kMax = 1, criticalValues = 1.96,  
      allocation1 = 3, allocation2 = 1,
      accrualIntensity = 5, 
      lambda2 = 0.95/12, lambda1 = 0.3*0.95/12, 
      gamma1 = -log(1-0.1)/24, gamma2 = -log(1-0.1)/24,
      accrualDuration = 25.4, followupTime = 6.5, 
      fixedFollowup = TRUE,  
      plannedEvents = 26, 
      maxNumberOfIterations = 10000, seed = 12345)
```
The simulated power is about 95% for design 1, and 83% for design 2. Neither is close to the target 90% power.  

We use the following formula to adjust the sample size to attain the target power, 
\[
D = D_0 \left( \frac{\Phi^{-1}(1-\alpha) + \Phi^{-1}(1-\beta)} {\Phi^{-1}(1-\alpha) + \Phi^{-1}(1-\beta_0)} \right)^2
\]
where $D_0$ and $\beta_0$ are the initial event number and the correponding type II error, and $D$ and $\beta$ are the required event number and the target type II error, respectively. For $\alpha=0.025$ and $\beta=0.1$, plugging in $(D_0=39, \beta_0=0.05)$ and $(D_0=26, \beta_0=0.17)$ would yield $D=32$ and $D=32$, respectively. For $D=32$, we need about 156 patients for an enrollment period of 31.2 months,  
\[
N = \frac{D}{ \frac{r}{1+r}\frac{\lambda_1}{\lambda_1+\gamma_1} (1 - \exp(-(\lambda_1+\gamma_1)T_f)) + \frac{1}{1+r}\frac{\lambda_2}{\lambda_2+\gamma_2} (1 - \exp(-(\lambda_2+\gamma_2)T_f)) }
\]
Simulation results confirmed the accuracy of this sample size estimate. 
```{r}
lrsim(kMax = 1, criticalValues = 1.96,  
      allocation1 = 3, allocation2 = 1,
      accrualIntensity = 5, 
      lambda2 = 0.95/12, lambda1 = 0.3*0.95/12, 
      gamma1 = -log(1-0.1)/24, gamma2 = -log(1-0.1)/24,
      accrualDuration = 31.2, followupTime = 6.5, 
      fixedFollowup = TRUE,  
      plannedEvents = 32, 
      maxNumberOfIterations = 10000, seed = 12345)
```