Consistent framing
Notes for chapter 8
1
Accounting and decisions
• In accounting decision problems are often modified
Fixed costs are not included
– Often only incremental costs are considered
– Costs are approximated
•
Profit maximization is often assumed
– Cost minimization
• Uncertainty is often not considered
– Risk and risk premiums
– Variance and covariance among projects
• Time preferences are also disregarded
– No discounting
2
More accounting
• Product classification
– Primary product
– Secondary products
– Scrap
• Cost allocations
• Transfer pricing
• ABC costing
3
Rational Behavior
• Intelligent, wise, and enlightened.
• Economic setting pursue
– Self-interest
– Wealth
• How do we describe rationality
• How do we model rationality
4
Rational Behavior
• Consistency
– Complete and transitive ranking
• Two statements equivalent.
– Ranking complete and transitive.
– Exists a function on A, ω(a),
• a`, a ́ ∈ A, ω(a`) ≥ ω(a ́)
• Only when a` is ranked as good as a ́.
• (The set A is finite)
• Smoothness
5
A generic decision problem
We want to maximize a generic function
subject to some constraints of feasibility.
This could take the form as:
max  (a)
subject to a  A
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Decision with two variables
7
Irrelevance of Increasing
Transformations
graph 1:
1 (a)  10a  a 2  20
graph 2 :
2 (a)  1 ( a)  20  10a  a 2
graph 3 :
3 (a)  1  [1 ( a)]2  1  [10a  a 2  20]2
graph 4 :
4 (a)  ln[2 ( a)]  ln[1 ( a)  20]
8
Irrelevance of Increasing
Transformations
9
Irrelevance of Increasing
Transformations
10
Irrelevance of Increasing
Transformations
• Definition 19 Function T is an increasing
transformation of function ω(a) if ω (a) > ω
(â) if and only if T[ω(a)] > T[ω(â)] for every
a and â in the domain of the original
function
• The solution to a decision problem is
unaffected by an increasing transformation
of the objective function.
11
Shadow prices
max(40 x  42 y )  (30 x  30 y )  max10 x  12 y
x  y  400
x  2 y  500
(300,100)
12
Local search - Shadow prices
max  ( x, y )  10 x  12 y
x  0, y  0
subject to
x  y   8
x  2 y  12
Optimal choice is x = 4, y = 4
13
Component searches are possible
Interactions
max  ( x, y )  10 x  12 y
x  0, y  0
subject to
x y 8
x  2 y  12
First constraint
y  8 x
Second constraint
y  0.5(12-x)
14
Component searches are possible
This reduces to:
y  g ( x)  min{8  x;.5(12  x)}
Then we get:
w( x, g ( x))  ˆ ( x)  10 x  12  min{8  x;.5(12  x)}
10 x  6(12  x)  72  4 x, if 0  x  4

10 x  12(8  x)  96  2 x, if 4  x  8
15
Component searches are possible
Interactions
16
Component searches are possible
C (q; P)  min  ( z1 , z2 )  5 z1  20 z2
z1  0, z2  0
subject to q  z1 z2
z1  15
q2
z2 
z1
q2
C (q; P)  min ˆ ( z1 )  5 z1  20
z1  0
z1
subject to z1  15
17
Component searches are possible
max  ( x, y)
xX , yY
max{max ( x, y)}
xX
yY
max{max ( x, y)}  max ( x, g ( x))  max ˆ ( x)
xX
yY
xX
xX
18
Component search
• When faced with an optimization problem
of several variables we do component
search when we solve the problem
sequentially, by first optimizing with
respect to one variable then the next etc.
19
y  g ( x)  min{400  x;.5(500  x)}
  10 x  12 g ( x)  10 x  12 min{400  x;.5(500  x)} 
 3000  4 x 0  x  300

4200  2 x 300  x  400
0  x  400
20
21
22
23
Consistent Framing
• 3 principles:
– Irrelevance of Increasing Transformation
– Local searches are possible
– Component searches are Possible
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27
28
29
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Application of framing principles
and cost functions
max 90q 1 + 152q 2 - 20 z 1 - 10 z 2 - 15 z 3
s .t .
q1 + q 2 £ 8
q 1 + 2q 2 £ 12
z 1 ³ q 1 + 2q 2
z 2 ³ 3q 1 + 4q 2
z 3 ³ 2q 1 + 4 q 2
Optimal solution: q 1* = 4;q 2* = 4; z 1* = 12; z 2* = 28; z 3* = 24
Dual variables : l1 = 8,l2 = 2,l3 = -20,l4 = -10,l5 = -15.
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New framing!
Local search (only consider equality):
z1 = q1 + 2q2
z2 = 3q1 + 4q2
z3 = 2q1 + 4q2
max 90q1 + 152q2 - 20(q1 + 2q2 ) - 10(3q1 + 4q2 ) - 15(2q1 + 4q2 )
max(90 - 80)q1 + (152 - 140)q2 = max10q1 + 12q2
s.t.
q1 + q2 £ 8
q1 + 2q2 £ 12
Optimal solution: q1* = 4;q2* = 4; l1 = 8; l2 = 2.
32
Yet another framing!
Using Component search this reduces to:
q2  g (q1 )  min{8  q1;.5(12  q1 )}
Then we get:
w( x, g (q1 ))  ˆ (q1 )  10q1  12  min{8  q1;.5(12  q1 )}
10q  6(12  q1 )  72  4q1 , if 0  q1  4
 1
10q1  12(8  q1 )  96  2q1 , if 4  q1  8
33
Cost function I
C (q1 , q2 )  20(q1  2q2 )  10(3q1  4q2 )  15(2q1  4q2 )  80q1  140q2
OV  50000  1.5DL
Product 1
Product 2
Direct labor
Direct material
Variable overhead
20
30
30
40
40
60
Variable product cost
80
140
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Cost function II
C (q1 )  (80  6)q1  72
0  q1  4
C (q1 )  (80  12)q1  96
4  q1  8
0  q1  4
Direct labor
Direct material
Variable overhead
Externality
Variable product cost
20
30
30
6
86
4  q1  8
20
30
30
12
92
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Cost function III
Frame
Explicit
choices
Implicit
choices
I
q1,q2,
z1,z2,z3
N/A
Marginal
cost of first
product
N/A
II
q1,q2
z1,z2,z3
80
III
q1
q2,z1,z2,z3
86 or 92
36
Changed parameters
max 90q1  152 149q2  20 z1  10 z2  15 z3
s.t.
q1  q2  8
q1  2q2  12
z1  q1  2q2
z2  3q1  4q2
z3  2q1  4q2
Optimal solution: q1*  8; q2*  0; z1*  8; z2*  24; z3*  16
Dual variables:1  10, 2  0, 3  20, 4  10, 5  15.
Cost function:C ( q1 , q2 )  80q1  140q2
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Cost function –Changed
parameters
Frame
Explicit
choices
Implicit
choices
I
q1,q2,
z1,z2,z3
N/A
Marginal
cost of first
product
N/A
II
q1,q2
z1,z2,z3
80
III
q1
q2,z1,z2,z3
80
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Frame I – Short Term
Fix : z1 = 12000;
max 90q1 + 152 149q2 - 10z2 - 15z3
s.t.
q1 + q2 £ 8
q1 + 2q2 £ 12
z2 ³ 3q1 + 4q2
z3 ³ 2q1 + 4q2
Optimal solution: q1* = 4;q2* = 4;(z1* = 12); z2* = 28; z3* = 24
Dual variables:l1 = 11, l2 = 10, l3 = N / A, l4 = -20, l5 = -15.
Cost function:C(q1 ,q2 ) = 60q1 + 100q2
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Frame II – Short Term
90q1  149q2  10 z2  15 z3 
90q1  149q2  10(3q1  4q2 )  15(2q1  4q2 ) 
(90  30  30)q1  (149  40  60) q2 
(90  60)q1  (149  100)q2
or
max 30q1  49q2
st.
q1  q2  8
q1  2q2  8
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Frame II – Short Term
•
•
•
•
Short run cost: C(q1,q2)=60q1 +100q2
No labor cost
New use of the accounting library
What we mean by cost – depends!
Product 1
Product 2
Direct labor
Direct material
Variable overhead
0
30
30
0
40
60
Variable product cost
60
100
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Cost terminology
• Cost and benefit
• F(z) = B(z) – C(z)
– Separation always possible
– Separation hardly unique
• Relevant cost
– Is simply the portion of the cost function that
varies with the options at hand
– Depends upon framing
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Our objective
• Are we maximizing profit?
• Are we maximizing wealth?
• Are we maximizing utility?
– Are these different?
• What happened to uncertainty?
• How do we cope with uncertainty?
• Is risk aversion part of our story?
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Back to accounting
• Cost function
– Which products are included?
– How is scrap accounted for?
• Cost allocation
– Are externalities accounted for?
• Transfer pricing
– Linear pricing – first order condition maintained?
• ABC costing
– Approximation of cost function?
44
Conclusions
•
•
•
•
•
Ease of analysis vs complete specification
Framing is this decision
Notion of cost follows frame
Cost allocation might be part of framing
Where did the problem set-up come from
– Out of the blue
– A handy and clever representation of the problem at
hand
• Professional judgment
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