<p><a href="http://www.juliathomas.net/PalgraveArticleFeb2007.pdf" rel="nofollow">The New Palgrave Dictionary of Economics Article on Adjustment Costs</a> gives a nice overview of the use of adjustment costs in economic models. </p>

<h2>What is the purpose of scaling investment costs with K here?</h2>

<p>This function form implies that adjustment costs of size X% of capital $I/k$ cost the same fraction of capital ($c(I/k)$) regardless of the amount of capital($k$) involved. It is a constant returns to scale form of adjustment cost. </p>

<h2>It seems that this gives the property that marginal cost of investment decreases in existing capital stock. </h2>

<p>I think you have the marginal cost of adjustment wrong here.</p>

<p>$$ \frac{\partial}{\partial K} K \cdot c\left(\frac{I}{K}\right)$$
$$ = c(\frac{I}{K})  + K \cdot c'(\frac{I}{K})$$</p>

<p>The first term is decreasing in $K$ for fixed $I$ but the second need not be. </p>

CC BY-SA 3.0
2015-03-13T12:13:05.147Z
2015-03-13T12:13:05.147Z
BKay
<p>The "adjustment cost" function is modeled as a mark-up, so it needs a base to be expressed in comparable units.  </p>

<p>A frequent modeling approach is to make Total Adjustment Cost a function of gross investment and write</p>

<p>$${\rm Adjustment \; Cost\; of \;Investment} = I \cdot \phi(I/K)$$</p>

<p>The rationale here is that Adjustment Costs are "Costs of Change", and so the modeling should be all about that. Therefore  </p>

<p><strong>a)</strong> The larger the change compared to where we are now ($I/K$), the larger the extra <em>unit</em> cost, so we have the $\phi()$ function as non-decreasing in $(I/K)$ and convex,
and<br>
b) This additional unit cost is measured per "Unit of Change" (Investment).</p>

<p>In this approach we find that</p>

<p>$$\frac {\partial}{\partial I}\big[ I\cdot \phi(I/K)\big] = \phi(I/K) + (I/K)\cdot \phi'(I/K) &gt;0$$</p>

<p>and</p>

<p>$$\frac {\partial}{\partial K}\big[ I\cdot \phi(I/K)\big] = -(I/K)^2\cdot \phi'(I/K) &lt; 0$$</p>

<p>The message:<br>
a) Total adjustments costs are everywhere increasing in the <em>level</em> of gross investment<br>
b) But they are everywhere <em>decreasing</em> in the level of Capital: the bigger you are, the better, in terms of adjustment costs, for the same intended Change (because in relative terms, the change is smaller). This is intuitive at least from the point of view that abrupt large changes in the size of a company may have very large side-costs, which is in general verified by real world experience.</p>

<hr>

<p>The formulation</p>

<p>$${\rm Adjustment \; Cost\; of \;Investment} = K\cdot c(I/K)$$</p>

<p>uses the same rationale as regards the unit cost, but it postulates that this extra unit cost is per Unit of Current State (Capital) rather than per Unit of Change (Investment). So the function $c()$ has a different interpretation, and it is not conceptually comparable with $\phi()$ -the latter is "unit adjustment cost per unit of Investment" while the former "unit adjustment cost per unit of installed Capital" (so if one attempted to empirically estimate both functions, one would not arrive at the same function/parameters). </p>

<p>Here we have</p>

<p>$$\frac {\partial}{\partial I}\big[ K\cdot c(I/K)\big] = c'(I/K) &gt; 0 $$</p>

<p>which gives the same message as the previous formulation as regards Total Adjustment Costs and the level of investment.  </p>

<p>But
$$\frac {\partial}{\partial K}\big[ K\cdot c(I/K)\big] = c(I/K) - (I/K)\cdot c'(I/K) &lt;&gt; 0 $$</p>

<p>and it would be positive (negative) depending on whether the <em>elasticity</em> of the unit adjustment cost function with respect to $(I/K)$ is lower (higher) than unity.</p>

<p>This seems to provide a higher degree of flexibility as regards the modeling of real world situations. It permits for example to account for the costs of "rigidity" that may go together with a company that is "large" in size, where due to them Total Adjustment Costs for given $I$ may be increasing after some level of Capital.  </p>

<p>An empirically inclined and extensive discussion on various issues surrounding Adjustment Costs (functional forms, micro/macro point of view, etc) can be found in 
<a href="http://www.jstor.org/stable/2729502" rel="nofollow">Hamermesh, D. S., &amp; Pfann, G. A. (1996). Adjustment costs in factor demand. Journal of Economic Literature, 1264-1292.</a></p>

CC BY-SA 3.0
2015-03-13T23:57:13.973Z
2015-03-14T00:09:25.543Z
2015-03-14T00:09:25.543Z
Alecos Papadopoulos
<p>Formulation like in your post ensure that adjustment costs are homogeneous of degree 1 in $K,I$. If the production function has also constant returns, we can typically rescale the whole firm's problem by $K$, which is very convenient:</p>

<ul>
<li><p>value function will be linear in capital, so marginal and average value of capital (Tobin's q) are equal and don't depend on firm's size.</p></li>
<li><p>if there's technological growth, such formulation allows for balanced growth path.</p></li>
<li><p>we may be able to drop one state variable, which is always nice if the model must be solved numerically.</p></li>
</ul>

CC BY-SA 3.0
2015-03-13T15:49:47.933Z
2015-03-13T15:49:47.933Z
ivansml
<p>Take a cont. diff, convex and increasing cost function $c(X)$. Say you start with a stock $K$, and want to (dis)invest $I$. Many adjustment cost functions (<a href="http://www3.nd.edu/~esims1/investment_notes.pdf" rel="nofollow">for example, the first example on page 2 here</a>) then are of the type</p>

<p>$$ K c\left(\frac{I}{K}\right)$$</p>

<p>What is the purpose of scaling investment costs with $K$ here? Marginal cost of investment is</p>

<p>$$ c'\left(\frac{I}{K}\right)$$</p>

<p>It seems that this gives the property that marginal cost of investment decreases in existing capital stock. Is this observed in the real world, or does it rather simplify computations? Are there (popular) alternative formulations of adjustment costs?</p>

CC BY-SA 3.0
2015-03-12T23:41:50.410Z
I suppose that the scaling is just so that it's easier to interpret $c$. This way, $c$ represents the a per-unit of $K$ cost.
jmbejara
<macroeconomics><academic-graduate><investment><adjustment-costs>
Scaling Adjustment Costs
FooBar
# Scaling Adjustment Costs

(Asked by: username_0 on 2015-03-12)

Take a cont. diff, convex and increasing cost function $c(X)$. Say you start with a stock $K$, and want to (dis)invest $I$. Many adjustment cost functions ([for example, the first example on page 2 here](http://www3.nd.edu/~esims1/investment_notes.pdf)) then are of the type

$$ K c\left(\frac{I}{K}\right)$$

What is the purpose of scaling investment costs with $K$ here? Marginal cost of investment is

$$ c'\left(\frac{I}{K}\right)$$

It seems that this gives the property that marginal cost of investment decreases in existing capital stock. Is this observed in the real world, or does it rather simplify computations? Are there (popular) alternative formulations of adjustment costs?

**Tags:** macroeconomics, academic-graduate, investment, adjustment-costs
> **Comments:**
> **username_1**: I suppose that the scaling is just so that it's easier to interpret $c$. This way, $c$ represents the a per-unit of $K$ cost.  

---

## 3 answers

### Answer by username_2 on 2015-03-13. Score: 4

[The New Palgrave Dictionary of Economics Article on Adjustment Costs](http://www.juliathomas.net/PalgraveArticleFeb2007.pdf) gives a nice overview of the use of adjustment costs in economic models.

## What is the purpose of scaling investment costs with K here?

This function form implies that adjustment costs of size X% of capital $I/k$ cost the same fraction of capital ($c(I/k)$) regardless of the amount of capital($k$) involved. It is a constant returns to scale form of adjustment cost.

## It seems that this gives the property that marginal cost of investment decreases in existing capital stock.

I think you have the marginal cost of adjustment wrong here.

$$ \frac{\partial}{\partial K} K \cdot c\left(\frac{I}{K}\right)$$
$$ = c(\frac{I}{K}) + K \cdot c'(\frac{I}{K})$$

The first term is decreasing in $K$ for fixed $I$ but the second need not be.


---

### Answer by username_3 on 2015-03-13. Score: 2

The "adjustment cost" function is modeled as a mark-up, so it needs a base to be expressed in comparable units.

A frequent modeling approach is to make Total Adjustment Cost a function of gross investment and write

$${\rm Adjustment \; Cost\; of \;Investment} = I \cdot \phi(I/K)$$

The rationale here is that Adjustment Costs are "Costs of Change", and so the modeling should be all about that. Therefore

**a)** The larger the change compared to where we are now ($I/K$), the larger the extra *unit* cost, so we have the $\phi()$ function as non-decreasing in $(I/K)$ and convex,
and  
b) This additional unit cost is measured per "Unit of Change" (Investment).

In this approach we find that

$$\frac {\partial}{\partial I}\big[ I\cdot \phi(I/K)\big] = \phi(I/K) + (I/K)\cdot \phi'(I/K) >0$$

and

$$\frac {\partial}{\partial K}\big[ I\cdot \phi(I/K)\big] = -(I/K)^2\cdot \phi'(I/K) < 0$$

The message:  
a) Total adjustments costs are everywhere increasing in the *level* of gross investment  
b) But they are everywhere *decreasing* in the level of Capital: the bigger you are, the better, in terms of adjustment costs, for the same intended Change (because in relative terms, the change is smaller). This is intuitive at least from the point of view that abrupt large changes in the size of a company may have very large side-costs, which is in general verified by real world experience.

---

The formulation

$${\rm Adjustment \; Cost\; of \;Investment} = K\cdot c(I/K)$$

uses the same rationale as regards the unit cost, but it postulates that this extra unit cost is per Unit of Current State (Capital) rather than per Unit of Change (Investment). So the function $c()$ has a different interpretation, and it is not conceptually comparable with $\phi()$ -the latter is "unit adjustment cost per unit of Investment" while the former "unit adjustment cost per unit of installed Capital" (so if one attempted to empirically estimate both functions, one would not arrive at the same function/parameters).

Here we have

$$\frac {\partial}{\partial I}\big[ K\cdot c(I/K)\big] = c'(I/K) > 0 $$

which gives the same message as the previous formulation as regards Total Adjustment Costs and the level of investment.

But
$$\frac {\partial}{\partial K}\big[ K\cdot c(I/K)\big] = c(I/K) - (I/K)\cdot c'(I/K) <> 0 $$

and it would be positive (negative) depending on whether the *elasticity* of the unit adjustment cost function with respect to $(I/K)$ is lower (higher) than unity.

This seems to provide a higher degree of flexibility as regards the modeling of real world situations. It permits for example to account for the costs of "rigidity" that may go together with a company that is "large" in size, where due to them Total Adjustment Costs for given $I$ may be increasing after some level of Capital.

An empirically inclined and extensive discussion on various issues surrounding Adjustment Costs (functional forms, micro/macro point of view, etc) can be found in
[Hamermesh, D. S., & Pfann, G. A. (1996). Adjustment costs in factor demand. Journal of Economic Literature, 1264-1292.](http://www.jstor.org/stable/2729502)


---

### Answer by username_4 on 2015-03-13. Score: 2

Formulation like in your post ensure that adjustment costs are homogeneous of degree 1 in $K,I$. If the production function has also constant returns, we can typically rescale the whole firm's problem by $K$, which is very convenient:

* value function will be linear in capital, so marginal and average value of capital (Tobin's q) are equal and don't depend on firm's size.
* if there's technological growth, such formulation allows for balanced growth path.
* we may be able to drop one state variable, which is always nice if the model must be solved numerically.


---

