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From the indifference conditions, following equations are equal in QALY values.
\( 10 * u(T_2) = 6 * u(T_1) \)
\( 8 * u(D_2) = 4 * u(D_1) \)
\( 10 * u(D_1) = 8 * u(T_1) \)
$1 * u(T_1) = 1 $ QALY
By solving these the following are obtained.
\( u(T_2) = 6 / 10 * u(T_1) = 0.6 \)
\( u(D_1) = 8 / 10 * u(T_1) = 0.8 \)
\( u(D_2) = 4 / 8 * u(D_1) \) = 1 / 2 * 0.8 = 0.4
a. What is the quality-adjusted life expectancy for a patient undergoing dialysis?
The patient has a life expectancy of 10 years, and will be in state \( D_1 \) (0.5 chance) or in state \( D_2 \) (0.5 chance). By averaging out:
\( 10 * 0.5 * u(D_1) \) = 5 * 0.8 = 4
\( 10 * 0.5 * u(D_2) \) = 5 * 0.4 = 2
Therefore, 6 quality-adjusted life years is the answer.
b. What is the quality-adjusted life expectancy for a patient who survives a transplant operation and accepts the kidney?
The patient's life expectancy is 12.5 years. The \( T_1 \) state has a chance of 0.6, and carries utility of 1, whereas the \( T_2 \) state has a chance of 0.4, and carries utility of 0.4. Thus:
\( 12.5 * 0.6 * u(T_1) \) = 7.5 * 1 = 7.5
\( 12.5 * 0.4 * u(T_2) \) = 5 * 0.6 = 3
Therefore, 10.5 quality-adjusted life years is the answer.
c. What is the quality-adjusted life expectancy for a patient who survives a transplant operation?
After surviving the operation, the patient is faced with 0.4 risk of graft failure and 10 year life expectancy on dialysis. If the graft remains good, the paitent will live 12.5 years.
The QALYs from each possibility is:
Graft survival, \( T_1 \): 0.6 * 0.6 * 12.5 * 1 = 4.5
Graft survival, \( T_2 \): 0.6 * 0.4 * 12.5 * 0.6 = 1.8
Graft failure, \( D_1 \): 0.4 * 0.5 * 10 * 0.8 = 1.6
Graft failure, \( D_2 \): 0.4 * 0.5 * 10 * 0.4 = 0.8
The sum, 8.7 is the answer.
d. What is the quality-adjusted life expectancy for a patient given a transplant?
The surgical procedure carries a perioperative mortality risk of 0.2, thus only 80% survives to experience the chances presented in question c.
Thus,
8.7 * 0.8 = 6.96
is the answer.
e. Compare your answer for part (a) to that for part (d), and interpret the result.
The transplantation strategy confers of 6.96 QALYs compared to the continued dialysis strategy, which give only 6 QALYs. Thus, for this patient's preferences, the transplantation strategy is more favorable if both survival and utility are considered.
f. The most difficult value trade-off is the comparison between the quality of life with dialysis and the quality of life with a transplant. This trade-off is described by statement iii. Perform a threshold analysis indicating how the trade-off in statement iii would have to change in order to change the decision that gives the higher quality-adjusted life expectancy. (Assume that the trade-offs in i and ii still apply.)
As u(\( D_1 \)) remains as a variable, u(\( D_2 \)) = 0.5 * u(\( D_1 \)).
The QALY from the treatment strategy is 0.8 * the sum of the following:
Graft survival, \( T_1 \): 0.6 * 0.6 * 12.5 * 1 = 4.5
Graft survival, \( T_2 \): 0.6 * 0.4 * 12.5 * 0.6 = 1.8
Graft failure, \( D_1 \): 0.4 * 0.5 * 10 * \( u(D_1) \) = 2 * \( u(D_1) \)
Graft failure, \( D_2 \): 0.4 * 0.5 * 10 * 0.5 * \( u(D_1) \) = \( u(D_1) \).
Thus,
0.8 * (6.3 + 3 * \( u(D_1) \)) = 5.04 + 2.4 * \( u(D_1) \).
is the function for the QALY from the treatment strategy.
The QALY from the no treatment and continue dialysis strategy is the sum of these
\( 10 * 0.5 * u(D_1) \) = 5 * \( u(D_1) \)
\( 10 * 0.5 * u(D_2) \) = 5 * 0.5 * \( u(D_1) \) = 2.5 * \( u(D_1) \).
Thus, it is 7.5 * \( u(D_1) \) is the function for the QALY from the no treatment strategy.
These two functions should be equal at the cut off.
strategy.transplant <- function(x) 5.04 + 2.4 * x
strategy.dialysis <- function(x) 7.5 * x
f.qf <- function(x) strategy.transplant(x) - strategy.dialysis(x)
7.5 * x = 5.04 + 2.4 * x
x = 5.04 / 5.1 = 0.9882
\( u(D_1) \) = 0.9882 is the cut off, above which continuing dialysis is more favorable.
ggplot(data.frame(x = 0:1), aes(x)) +
stat_function(fun = strategy.transplant, aes(color = "transplant")) +
stat_function(fun = strategy.dialysis, aes(color = "dialysis")) +
scale_color_manual(values = c("transplant" = "green", "dialysis" = "red"),
name = "strategy", breaks = c("transplant","dialysis")) +
scale_x_continuous(limit = c(0.8,1.0))