Iteratively Rebalancing Portfolio Weights After Transaction Costs

# Portfolio rebalancing to optimal weights including transaction costs and without cash component


# Portfolio rebalancing to optimal weights including transaction costs and without cash component












Consider a portfolio with 4 assets (A, B, C, D) with prices, quantities, current weights, and target weights as follows:

I want to rebalance the portfolio from the current weights to the target weights. There is a 0.2% transaction fee that is charged on the rebalance amount (i.e. the difference between the current and target weight) regardless of whether it is an increase or a decrease in that asset.

Assuming that the transaction fee is charged on the rebalanced amount of each asset, I would like to reach the target weights after taking into account the transaction fee. In other words, I want to achieve the Target Weight percentages after all fees have been paid from each asset.

While I know that funds usually have a cash component to cover these fees, I am interested in this scenario where there is no cash and the fee is deducted from each asset.

I am interested in calculating the amount to be rebalanced in each asset to achieve these target weights. What would be the general formula to do so?

## Answer by nbbo2 (score 2)

https://quant.stackexchange.com/a/60896

I do not think there is a closed form solution. I have applied a simple iterative method to your example problem. See below.

Let $N=4$ be the number of assets, indexed by $i$ ranging from 1 to $N$.

Let $x_i$ be the dollar allocations before rebalancing (in your example they are called "value (P*Q)").

Let $w_i$ be the desired post-rebalancing weights.

Let $\theta=0.002$ be the transactions cost per dollar transacted.

We want to find $y_i$, the dollar allocations after rebalancing.

The total transactions costs will be $T=\sum_i \theta |y_i-x_i|$

Unfortunately we cannot compute $T$ because it depends on the unknown $y_i$. If we had a good estimate of $T$ we could calculate the

Portfolio value after transaction costs $Y=\sum_i y_i=\sum_i x_i-T$

and then the dollar allocation after rebalancing $y_i=w_i Y$

The method I propose is: compute the allocation assuming no transactions cost, form an estimate of transaction costs, and find a refined allocation. From this allocation, find an improved estimate of transactions costs and so on.

Using you example, the convergence is very rapid, requiring just two or three iterations:

```
            Estimated                         ReEstimt'd        
 0costAlloc TransCost ValAftCost    NewAlloc  TransCost ValAftCost  NewAlloc
A    168000   64                   167890.2   64.21952              167890.1654
B    113400   73.2                 113325.9   73.34818              113325.8616
C     63000   46                    62958.8   45.91768               62958.8120
D     75600   91.2                  75550.6   91.10122               75550.5744
Tot. 420000  274.4    419725.6     419725.6  274.58659  419725.4134 419725.4134
```

If we start a third iteration the estimated transactions costs are 274.58671 which differs from the value above only in the 4th decimal place so I won't show the results here, the allocations are substantially the same.

I haven't proved that the results always converge. That's all I have for now.

## Answer by Attack68 (score 2)

https://quant.stackexchange.com/a/60899

Using some of @noob2 notation, if:

$x_i$: the initial dollar amount of asset i $t_i$: the transacted dollar amount of asset i $\theta$: the fee fraction (0.2%). $w_i$: the desired post balancing weights

Then you have the minimisation problem:

$$ \min_{t}{f(t)} = \sum_i \left ( \frac{x_i + t_i -\theta|t_i|}{\sum_j x_j +t_j -\theta|t_j|} - w_i\right )^2 $$

The minimum is zero (target weights achieved for each asset) and at this point we must have $\frac{\partial f}{\partial t_i} = 0$.

------- Contains errors: left for demonstration

$$ \frac{\partial f}{\partial t_i} = 2\frac{1-\theta sign(t_i)}{\sum_j x_j + t_j -\theta|t_j|} \left (1 - \frac{x_i + t_i -\theta |t_i|}{\sum_j x_j + t_j -\theta|t_j|} \right )$$

So multiplying through at with optimality conditions yields:

$$ x_i + t_i - \theta|t_i| = \sum_j x_j + t_j -\theta|t_j| \quad \forall i$$

Or in linear algebra notation, where $\mathbf{1}$ is a matrix of ones:

$$ \mathbf{(I - 1)}(\mathbf{t} - \theta |\mathbf{t}|) = (\mathbf{1-I}) \mathbf{x} $$

--------EDIT: Looking again, refreshed, the derivative above is wrong. The real derivative is more complicated. I leave this answer here if someone can complete but it might simply be too messy algebra to solve for a closed form optimisation solution in matrix notation

For

$X_i = x_i + t_i - \theta | t_i |$ (vector) $S = \sum_j x_j + t_j -\theta | t_j | $ (scalar)

Then:

$$ f(t) = \sum_i \left ( \frac{X_i}{S} - w_i \right )^2 $$

$$ \frac{\partial f}{\partial t_i} = 2 \left ( \frac{X_i}{S} - w_i \right ) \left( \frac{1 - \theta sign(t_i)}{S} \right ) + \sum_j 2 \left( \frac{X_j}{S} - w_j \right ) \left( \frac{-X_j (1-sign(t_i))}{S^2}\right )$$

When you set optimality condition and multiply through:

$$\left ( \frac{X_i}{S} - w_i \right ) + \sum_j \left( \frac{X_j}{S} - w_j \right ) \left( \frac{-X_j}{S}\right ) = 0$$ $$ \implies S X_i - S^2 w_i + \sum_j \left( X_j X_j - S w_j \right ) = 0 \quad \forall i$$

This looks suspiciously quadratic form-y and probably not going to yield a nice formula.

One other thing that I didn't include that might have improved matters is the constraint: $\sum_j t_j = 0$, ie. whatever you sell you buy in another asset. Would change the optimality conditions.