Firm's Decisions - Cost Minimization

Intermediate Microeconomics (Econ 100A)

Kristian López Vargas

UCSC - Spring 2017

Cost Minimization Problem - intro idea

  • The firm operates in two kinds of markets:
    1. Inputs/factor markets (e.g. it buys labor and capital)
    2. Final product market
  • Let's focus on optimal decisions regarding the first kind of market.

  • We will assume for now the firms has a target prod level $ q_0 $. (i.e. an isoquant!)

  • And, it aims to achieve that level of production in the best (most efficient) way possible.

  • let the fun begin

Cost Minimization Problem

  • The only decision the firm controls at this point is how much of inputs it uses.

  • So the most efficient way in this context refers to what is the "right" combination of (L,K) so achieve $ q_0 $.

  • The right combination is the one that minimize the cost of producing the given target level of output $ q_0 $.

  • Suppose wages are denoted by $ w $ and rental price of capital is denoted by $ r $.

  • So the firm wants to:

minimize: $ cost = w L + r K, \quad $ subject to: $ f(L,K) = q_0 $.

Isocost

  • Isocost: Combinations of input usage that cost the same (say $C):

  • Example: This is isocost at a cost of $100:

    • $ w L + r K = 100 $
  • Example: This is isocost at a cost of $50 when wages = 20 and price of capital = 10:

    • $ 20 L + 10 K = 50 $
  • Draw draw draw more examples until you dream of isocosts a little.

Solving the cost-minimization problem

  • Conceptually: find best(lowest) isocost given the required isoquant.

  • Graphically...

    1. Draw the "target" output level (isoquant) $ q_0 $.

    2. Ask: what is the cheapest combination of L,K that makes $ q_0 $ possible.

Solving the cost-minimization problem

  • What if prices of factors change?

  • What if target output changes?

Conditional factor demand functions

  • Optimal choices of factors are called the conditional factor demand functions

    • That is: $ L^* = L(w,r,q_0) $ and $ K^* = K(w,r,q_0) $
  • Optimal cost is the cost function

    • That is: $ c(w,r,q_0) = w L(w,r,q_0) + r K(w,r,q_0) $
  • Notice: all this is in the "long run" because we are able to adjust all inputs.

Cost-minimization problem - The Math


$$ \underset{L,K}{\text{minimize:}} ~ w L + r K \\ \text{Subject to:} ~ f(L,K) = q_0 $$

Cost-minimization problem, Case 1: tangency.

  • If technology satisfies mainly convexity and monotonicity then (in most cases) tangency solution!

  • Tangency condition: slope of isoquant equals slope of isocost curve.

    • In equation: $ - \frac{w}{r} = - \frac{MP_L}{MP_K} $ (EQ. 1)
  • Constraint: $ q = f(L,K) $ (EQ. 2)

  • System of two equations (Eq1 and Eq2), and two unknowns ($ L $ and $ K $).

Example: Cobb-Douglass

If $ q = f(L,K) = L K $,


$$ L^{*}(w,r,q) = \big( \frac{ q r }{ w } \big)^{1/2} $$


$$ K^{*}(w,r,q) = \big( \frac{ q w }{ r } \big)^{1/2} $$


c(w, r, q)=wL + rK = 2(qrw)1/2

Home exercise: solve the more general case:

  • $ q = f(L,K) = A ~ L^{a} K^{b} $
  • Find $ ~ c(w,r,q) = ?????? $

Cost-minimization problem, Case 2: Corner Solution.

Mainly, but not exclusively:

  1. Perfect Substitutes:


q = f(L, K)=aL + bK

  • For example: $ q = f(L,K) = L + K $

Example: Linear technology (1)


q = f(L, K)=aL + bK

  • Compare RTS $ ( - \frac{ a }{ b }) $ Vs. slope of the isocosts $ ( - \frac{ w }{ r } ) $

  • Alternatively, compare: $ \frac{ a }{ w } $ Vs. $ \frac{ b }{ r } $

    1. If $ \frac{ a }{ w } > \frac{ b }{ r } $ firm should use labor only.

    2. If $ \frac{ a }{ w } < \frac{ b }{ r } $ firm should use capital only.

    3. If $ \frac{ a }{ w } = \frac{ b }{ r } $ any point along the isoquant, minimizes cost.

Example: Linear technology (2)

IF $ \frac{ a }{ w } > \frac{ b }{ r } $

The firm should use only Labor (corner solution).

  • Then: $ q = a L $ , so $ L^*(w,r,q) = \frac{ q }{ a } $

  • Then: $ K^*(w,r,q) = 0 $

  • Then: $ c(w,r,q) = w L^\star = \frac{ w }{ a } q $

Example: Linear technology (3)

IF $ \frac{ a }{ w } < \frac{ b }{ r } $

The firm should use only Capital (corner solution).

  • Then: $ q = b K $ , so $ K^*(w,r,q) = \frac{ q }{ b } $

  • Then: $ L^*(w,r,q) = 0 $

  • Then: $ c(w,r,q) = r K^\star = \frac{ r }{ b } q $

  • IF $ \frac{ a }{ w } = \frac{ b }{ r } $, any point along the isoquant, minimizes cost.

Example: Linear technology (4)

Putting all this together:


$$ c(w,r,q)= q \times \begin{cases} \frac{ w }{ a },& \text{if} ~ \frac{ w }{ a } \leq \frac{ r }{ b } \\ \frac{ r }{ b },& \text{if} ~ \frac{ r }{ b } < \frac{ w }{ a } \end{cases} $$

This can be written as:


$$ c(w,r,q) = q ~ \textrm{min} \{ \frac{ w }{ a } , \frac{ r }{ b } \} $$

Simpler example:

If $ q = f(L,K) = L + K $, then:


c(w, r, q)=q × min{w, r}

Cost-minimization problem, Case 3: Kink Solution

  1. Perfect Complements (kink solution):


$$ q = f(L,K) = \text{min} \{ \frac{L}{a} , \frac{K}{b} \} $$

  1. if $ q = f(L,K) = \textrm{min} \{ L ,K \} $, then:


c(w, r, q)=(w + r)q

Short-run conditional demand for labor, cost function

  • $ K = \bar{K} $

  • do the graph!

  • Short-run conditional demand of labor: $ L = L(w,r,q, \bar{K}) $

  • This demand is obtained from solving L from $ q = f( \bar{K}, L) $

  • If there are no other inputs it does not depend on prices of inputs.

Short-run - example

  • $ q = f(L, K) = K^{0.5} L^{0.5} $

  • $ K = \bar{K} $

  • $ L^{SR} = \frac{ q^2 }{ \bar{K} } $

  • $ c^{SR}(w,r,q, \bar{K}) = w \frac{ q^2 }{ \bar{K} } + r \bar{K} $

Returns to scale and the cost function

  • Let us define the average cost function:

  • $ AC(w,r,q) = \frac{ c(w,r,q) }{ q } $

  • IRS implies that AC is decreasing in $ q $. (e.g. if we want to double q, we can less than double costs).

  • CRS implies that AC is constant in $ q $. (e.g. if we want to double q, we need to double costs).

  • DRS implies that AC is increasing in $ q $. (e.g. if we want to double q, we need to more than double costs).

Types of costs: Fixed and quasi-fixed costs

  1. Fixed: costs that must be paid, regardless of output level.

  2. Quasi-fixed cost: costs that must be paid, only if output level > 0. (heating, lighting, etc.)

  3. Sunk cost: fixed costs that are not recoverable (painting your factory)