R solve.QP tracking error minimization constraints inconsistent - r
I am struggling with Solve.QP to get a solution to minimize tracking error. I have a benchmark consisting of 6 assets (asset_a to asset_f). For my portfolio I have upper and lower bounds (I cannot have a position in asset_f). The cov matrix is also given. I want to get the portfolio weights for the 6 assets that minimizes tracking error vs the benchmark (with position in asset_f equal to zero).
benchmark:
asset_a: 0.3
asset_b: 0.3
asset_c: 0.1
asset_d: 0.1
asset_e: 0.1
asset_f: 0.1
lowerbounds:
asset_a: 0.166
asset_b: 0.133
asset_c: 0.037
asset_d: 0.035
asset_e: 0.039
asset_f: 0
upperbounds:
asset_a: 1
asset_b: 1
asset_c: 1
asset_d: 1
asset_e: 1
asset_f: 0
benchmark weights and bounds:
test.benchmark_weights = c(0.3, 0.3, 0.1, 0.1, 0.1, 0.1)
test.lowerbound = c(0.166, 0.133, 0.037, 0.035, 0.039,0)
test.upperbound = c(1, 1, 1, 1, 1, 0)
cov matrix (test.Dmat):
test.dmat = matrix(c(0.0119127162, 0.010862842, 0.010266683, 0.0009550136, 0.008242322, 0.00964462, 0.0108628421, 0.010603072, 0.009872992, 0.0011019412, 0.007422522, 0.0092528873, 0.0102666826, 0.009872992, 0.010487808, 0.0012107665, 0.006489204, 0.0096216627, 0.0009550136, 0.001101941, 0.001210766, 0.0115527788, 0.001181745, 0.0008387247, 0.0082423222, 0.007422522, 0.006489204, 0.0011817453, 0.012920482, 0.005973886, 0.00964462, 0.009252887, 0.009621663, 0.0008387247, 0.005973886, 0.0089904809), nrow=6, ncol=6)
dvec (test.dvec):
test.dvec = matrix(c(0, 0, 0, 0, 0, 0), nrow=6, ncol=1)
Amat constraints matrix (test.Amat):
test.amat = matrix(c(1,1,1,1,1,1, 1,1,1,1,1,0, -1,0,0,0,0,0, 0,-1,0,0,0,0, 0,0,-1,0,0,0, 0,0,0,-1,0,0, 0,0,0,0,-1,0, 0,0,0,0,0,-1, 1,0,0,0,0,0, 0,1,0,0,0,0, 0,0,1,0,0,0, 0,0,0,1,0,0, 0,0,0,0,1,0, 0,0,0,0,0,0, -1,0,0,0,0,0, 0,-1,0,0,0,0, 0,0,-1,0,0,0, 0,0,0,-1,0,0, 0,0,0,0,-1,0, 0,0,0,0,0,0), nrow=6, ncol=20)
bvec (test.bvec)
test.bvec =cbind(0, 1, t(test.benchmark_weights), t(test.lowerbound), -t(test.upperbound)) %>% as.matrix()
then running the solver
solve.QP(as.matrix(test.Dmat), test.dvec, test.Amat, test.bvec)
gives me
constraints are inconsistent, no solution!
Seems like there is something wrong with your Amat and bvec, i.e. you need not have to pass in both sum of weights on first 5 assets equal to 1 and sum of 6 assets equal 1 and also benchmark weights are not constraints but the bounds are:
library(quadprog)
N = 6L
test.dvec = rep(0, N)
test.amat = cbind(
rep(1, N),
diag(1, N),
diag(-1, N))
test.bvec = c(1, test.lowerbound, -test.upperbound)
res = solve.QP(test.dmat, test.dvec, test.amat, test.bvec, meq=1L)
round(res$solution, 2)
#[1] 0.17 0.13 0.10 0.44 0.17 0.00
Related
Principal Component Analysis in R by hand
The questions is about Principal Component Analysis, partly done by hand. Disclaimer: My background is not in maths and I am using R for the first time. Given are the following five data points in R^3. Where xi1-3 are variables and x1 - x5 are observations. | x1 x2 x3 x4 x5 ---------------------- xi1 | -2 -2 0 2 2 xi2 | -2 2 0 -2 2 xi3 | -4 0 0 0 4 Three principal component vectors after the principal component analysis has been performed are given, and look like this: Phi1 = (0.41, 0.41, 0.82)T Phi2 = (-0.71, 0.71, 0.00)T Phi3 = (0.58, 0.58, -0.58)T The questions are as follows 1) Calculate the principal component scores zi1, zi2 and zi3 for each of the 5 data points. 2) Calculate the proportion of the variance explained by each principal component. So far I have answered question 1 with the following code, where Z represents the scores: A = matrix( c(-2, -2, 0, 2, 2, -2, 2, 0, -2, 2, -4, 0, 0, 0, 4), nrow = 3, ncol = 5, byrow = TRUE ) Phi = matrix ( c(0.41, -0.71, 0.58,0.41, 0.71, 0.58, 0.82, 0.00, -0.58), nrow = 3, ncol = 3, byrow = FALSE ) Z = Phi%*%A Now I am stuck with question 2, I am given the formula: But I am not sure how I can recreate the formula with an R command, can anyone help me?
#Here is the numerator: (Phi%*%A)^2%>%rowSums() [1] 48.4128 16.1312 0.0000 #Here is the denominator: sum(A^2) [1] 64 #So the answer is: (Phi%*%A)^2%>%rowSums()/sum(A^2) [1] 0.75645 0.25205 0.00000 we can verify with prcomp+summary: summary(prcomp(t(A))) Importance of components: PC1 PC2 PC3 Standard deviation 3.464 2.00 0 Proportion of Variance 0.750 0.25 0 Cumulative Proportion 0.750 1.00 1 This is roughly the same since your $\Phi$ is rounded to two decimals.
Quadratic optimization - portfolio maximization problems
In portfolio analysis, given the expectation, we aim to find the weight of each asset to minimize the variance
here is the code
install.packages("quadprog")
library(quadprog)
#Denoting annualized risk as an vector sigma
sigma <- c(0.56, 7.77, 13.48, 16.64)
#Formulazing the correlation matrix proposed by question
m <- diag(0.5, nrow = 4, ncol = 4)
m[upper.tri(m)] <- c(-0.07, -0.095, 0.959, -0.095, 0.936, 0.997)
corr <- m + t(m)
sig <- corr * outer(sigma, sigma)
#Defining the mean
mu = matrix(c(1.73, 6.65, 9.11, 10.30), nrow = 4)
m0 = 8
Amat <- t(matrix(c(1, 1, 1, 1,
c(mu),
1, 0, 0, 0,
0, 1, 0, 0,
0, 0, 1, 0,
0, 0, 0, 1), 6, 4, byrow = TRUE))
bvec <- c(1, m0, 0, 0, 0, 0)
qp <- solve.QP(sig, rep(0, nrow(sig)), Amat, bvec, meq = 2)
qp
x = matrix(qp$solution)
x
(t(x) %*% sig %*% x)^0.5
I understand the formulation of mu and covariance matrix and know the usage of the quadprog plot
However, I don‘t understand why Amat and bvec are defined in this way, why the are 6 by 4 matrix.
$mu0$ is the expectation we aim to have for the portfolio and it is fixed at value 8%
Attached is the question
As you are probably aware, the reason that Amat has four columns is that there are four assets that you are allocating over. It has six rows because there are six constraints in your problem: The allocations add up to 1 (100%) Expected return = 8% 'Money market' allocation >= 0 'Capital stable' allocation >= 0 'Balance' allocation >= 0 'Growth' allocation >= 0 Look at the numbers that define each constraint. They are why bvec is [1, 8, 0, 0, 0, 0]. Of these six, the first two are equality constraints, which is why meq is set to 2 (the other four are greater than or equal constraints). Edited to add: The way the constraints work is this: each column of Amat defines a constraint, which is then multiplied by the asset allocations, with the result equal to (or greater-than-or-equal-to) some target that is set in bvec. For example: The first column of Amat is [1, 1, 1, 1], and the first entry of bvec is 1. So the first constraint is: 1 * money_market + 1 * capital_stable + 1 * balance + 1 * growth = 1 This is a way of saying that the asset allocations add up to 1. The second constraint says that the expected returns add up to 8: 1.73 * money_market + 6.65 * capital_stable + 9.11 * balance + 10.32 * growth = 8 Now consider the third constraint, which says that the 'Money market' allocation is greater than or equal to zero. That's because the 3rd column of Amat is [1, 0, 0, 0] and the third entry of bvec is 0. So this constraint looks like: 1 * money_market + 0 * capital_stable + 0 * balance + 0 * growth >= 0 Simplifying, that's the same as: money_market >= 0
Divide rows into groups given the similarity between them
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You could do a cluster analysis and play around with different cutoffs h. cl <- hclust(dist(DF)) DF$group <- cutree(cl, h=.5) DF # x y z group # 1 1.00 0.00 0.0 1 # 2 0.85 0.00 0.0 1 # 3 0.90 0.10 0.0 1 # 4 0.00 0.90 0.9 2 # 5 0.00 1.00 0.9 2 # 6 0.90 0.90 0.0 3 # 7 0.95 0.97 0.9 4 A dendrogram helps to determine h. plot(cl) abline(h=.5, col=2)
VECM in R: Testing weak exogeneity and imposing restrictions
I estimated VECM and would like to make 4 separate tests of weak exogeneity for each variable. library(urca) library(vars) data(Canada) e prod rw U 1980 Q1 929.6105 405.3665 386.1361 7.53 1980 Q2 929.8040 404.6398 388.1358 7.70 1980 Q3 930.3184 403.8149 390.5401 7.47 1980 Q4 931.4277 404.2158 393.9638 7.27 1981 Q1 932.6620 405.0467 396.7647 7.37 1981 Q2 933.5509 404.4167 400.0217 7.13 ... jt = ca.jo(Canada, type = "trace", ecdet = "const", K = 2, spec = "transitory") t = cajorls(jt, r = 1) t$rlm$coefficients e.d prod.d rw.d U.d ect1 -0.005972228 0.004658649 -0.10607044 -0.02190508 e.dl1 0.812608320 -0.063226620 -0.36178542 -0.60482042 prod.dl1 0.208945048 0.275454380 -0.08418285 -0.09031236 rw.dl1 -0.045040603 0.094392696 -0.05462048 -0.01443323 U.dl1 0.218358784 -0.538972799 0.24391761 -0.16978208 t$beta ect1 e.l1 1.00000000 prod.l1 0.08536852 rw.l1 -0.14261822 U.l1 4.28476955 constant -967.81673980 I guess that my equations are: and I would like to test whether alpha_e, alpha_prod, alpha_rw, alpha_U (they marked red in the picture above) are zeros and impose necessary restrictions on my model. So, my question is: how can I do it? I guess that my estimated alphas are: e.d prod.d rw.d U.d ect1 -0.005972228 0.004658649 -0.10607044 -0.02190508 I guess that I should use alrtest function from urca library: alrtest(z = jt, A = A1, r = 1) and probably my A matrix for alpha_e should be like this: A1 = matrix(c(0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1), nrow = 4, ncol = 3, byrow = TRUE) The results of the test: jt1 = alrtest(z = jt, A = A1, r = 1) summary(jt1) The value of the likelihood ratio test statistic: 0.48 distributed as chi square with 1 df. The p-value of the test statistic is: 0.49 Eigenvectors, normalised to first column of the restricted VAR: [,1] RK.e.l1 1.0000 RK.prod.l1 0.1352 RK.rw.l1 -0.1937 RK.U.l1 3.9760 RK.constant -960.2126 Weights W of the restricted VAR: [,1] [1,] 0.0000 [2,] 0.0084 [3,] -0.1342 [4,] -0.0315 Which I guess means that I can't reject my hypothesis of weak exogeneity of alpha_e. And my new alphas here are: 0.0000, 0.0084, -0.1342, -0.0315. Now the question is how can I impose this restriction on my VECM model? If I do: t1 = cajorls(jt1, r = 1) t1$rlm$coefficients e.d prod.d rw.d U.d ect1 -0.005754775 0.007717881 -0.13282970 -0.02848404 e.dl1 0.830418381 -0.049601229 -0.30644063 -0.60236338 prod.dl1 0.207857861 0.272499006 -0.06742147 -0.08561076 rw.dl1 -0.037677197 0.102991919 -0.05986655 -0.02019326 U.dl1 0.231855899 -0.530897862 0.30720652 -0.16277775 t1$beta ect1 e.l1 1.0000000 prod.l1 0.1351633 rw.l1 -0.1936612 U.l1 3.9759842 constant -960.2126150 the new model don't have 0.0000, 0.0084, -0.1342, -0.0315 for alphas. It has -0.005754775 0.007717881 -0.13282970 -0.02848404 instead. How can I get reestimated model with alpha_e = 0? I want reestimated model with alpha_e = 0 because I would like to use it for predictions (vecm -> vec2var -> predict, but vec2var doesn't accept jt1 directly). And in general - are calculations which I made correct or not? Just for illustration, in EViews imposing restriction on alpha looks like this (not for this example):
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Generating random variables with specific correlation threshold value
I am generating random variables with specified range and dimension.I have made a following code for this.
generateRandom <- function(size,scale){
result<- round(runif(size,1,scale),1)
return(result)
}
flag=TRUE
x <- generateRandom(300,6)
y <- generateRandom(300,6)
while(flag){
corrXY <- cor(x,y)
if(corrXY>=0.2){
flag=FALSE
}
else{
x <- generateRandom(300,6)
y <- generateRandom(300,6)
}
}
I want following 6 variables with size 300 and scale of all is between 1 to 6 except for one variable which would have scale 1-7 with following correlation structure among them.
1 0.45 -0.35 0.46 0.25 0.3
1 0.25 0.29 0.5 -0.3
1 -0.3 0.1 0.4
1 0.4 0.6
1 -0.4
1
But when I try to increase threshold value my program gets very slow.Moreover,I want more than 7 variables of size 300 and between each pair of those variables I want some specific correlation threshold.How would I do it efficiently?
This answer is directly inspired from here and there. We would like to generate 300 samples of a 6-variate uniform distribution with correlation structure equal to Rhos <- matrix(0, 6, 6) Rhos[lower.tri(Rhos)] <- c(0.450, -0.35, 0.46, 0.25, 0.3, 0.25, 0.29, 0.5, -0.3, -0.3, 0.1, 0.4, 0.4, 0.6, -0.4) Rhos <- Rhos + t(Rhos) diag(Rhos) <- 1 We first generate from this correlation structure the correlation structure of the Gaussian copula: Copucov <- 2 * sin(Rhos * pi/6) This matrix is not positive definite, we use instead the nearest positive definite matrix: library(Matrix) Copucov <- cov2cor(nearPD(Copucov)$mat) This correlation structure can be used as one of the inputs of MASS::mvrnorm: G <- mvrnorm(n=300, mu=rep(0,6), Sigma=Copucov, empirical=TRUE) We then transform G into a multivariate uniform sample whose values range from 1 to 6, except for the last variable which ranges from 1 to 7: U <- matrix(NA, 300, 6) U[, 1:5] <- 5 * pnorm(G[, 1:5]) + 1 U[, 6] <- 6 * pnorm(G[, 6]) + 1 After rounding (and taking the nearest positive matrix to the copula's covariance matrix etc.), the correlation structure is not changed much: Ur <- round(U, 1) cor(Ur)