The Economics of Production
Make or Buy Decisions
Capacity Expansion
Learning-Curves
Break-even Analysis
Production Functions
Make or Buy?
Let c1 = unit purchase price
c2 = unit production cost (c2 < c1)
K = fixed cost of production
x = number of units required
Produce if K + c2 x  c1x
or
x  K / (c1 - c2 )
Make or Buy Example
• It costs the Maker Bi Company $20 a unit to
purchase a critical part used in the manufacture of
their primary product line – a thing-um-a-jig
• It is estimated that the part could be produced
internally at a unit cost of $16 after incurring a
fixed cost of $20,000 for the necessary equipment.
• What to do?
a thing-um-a-jig
What to do?
Let c1 = $20
c2 = $16
K = $200,000
x = number of units required
Produce if
x  K / (c1 - c2 ) = 20,000 /(20 – 16) = 5,000
Nonlinear Cost Function
Let c1 = unit purchase price
c2 = K + axb where K, a, b > 0
x = number of units required
If c1 = 20 and c2 = 20,000 + 100x.7 , then
c2
Cumulative Cost
90000
80000
70000
60000
50000
40000
30000
20000
10000
0
0
2000
4000
6000
x
8000
10000
12000
More on that Nonlinear Cost Function
x
1900
1925
1950
1975
2000
2025
2035
2050
2075
2100
2125
2150
2175
2200
2225
2250
2275
Make
$39,730
$39,911
$40,092
$40,272
$40,451
$40,630
$40,701
$40,808
$40,985
$41,162
$41,338
$41,513
$41,688
$41,862
$42,036
$42,209
$42,381
Buy
$38,000
$38,500
$39,000
$39,500
$40,000
$40,500
$40,700
$41,000
$41,500
$42,000
$42,500
$43,000
$43,500
$44,000
$44,500
$45,000
$45,500
Strategic Decisions
Capacity Expansion
• Capacity Growth Planning
– when to construct new facilities
– where to locate facilities
– how large to size a facility
• Economies of scale
– advantage of expanding existing facilities
– share plant, equipment, support personnel
– avoid duplication at separate locations
Capacity Expansion
competing objectives: maximize market share
maximize capacity utilization
number units
number units
demand
time
capacity leads demand
demand
time
capacity lags demand
We need a model
let D = annual increase in demand
x = time interval (in yrs) between capacity increases
r = annual discount rate, compounded continuously
f(y) = cost of expansion of capacity y
assume y = xD, then
cost = C(x) = f(xD) [1 + e-rx + (e-2rx ) + (e-3rx ) + …]
= f(xD) [1 + e-rx + (e-rx )2 + (e-rx )3 + …]
= f(xD) / [1 – e-rx]
a
assume f(y) =
kya
, then
k ( xD)
C ( x) 
 rx
1 e
find the x that minimizes C(x)
A Diversion - the Geometric Series

1
1  y  y  ...  y  ...   y 
1 y
n0
2
1 e  e
rx
n

 rx 2
n
 ...   e

 rx n

 ...    e
n0
You see? It
does converge.

 rx n
1

1  e  rx
Discounting – another diversion
Consider the time value of money
$1.00 today is worth more than a $1.00 next year
How much more is it worth?
If r = annual interest rate, then it is worth (1+r) $1.00
After two years, it is worth (1+r)2 $1.00 (compounded)
Compounded quarterly for 1 yr = 1  r / 4 
4
Compounded continuously for one year = lim 1  r / n   e r
n
n 
t
1  r / n    ert

After t years = lim
n 

n
More diversionary discounting
A stream of costs: C1, C2, …, Cn incurred at times
t1, t2,…, tn has a present value of:
n
C e
i 1
 rti
i
For an infinite planning horizon
where x is the time between expansions:

 Ce
i 1
 irx
C

1  e  rx
Why can’t you
show us an
example?
The Example
• Chemical firm expanding at a cost ($M) of
f ( y)  .0107 y
.62
– where y is in tons per year.
• Demand is growing at the rate of D = 5,000 tons
per year and future costs are discounted at a rate of
r = 16 percent
.0107(5000 x).62
• Find x that minimizes C ( x) 
.16 x
1 e
Capacity Expansion Solution
C(x) - $M
15
14
13
12
11
10
9
8
0
2
4
6
8
years
.0107(5000 x).62
C ( x) 
.16 x
1 e
10
12
5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6
10.35653
10.35042
10.3456
10.34203
10.33964
10.33839
10.33823
10.33911
10.34098
10.34382
10.34757
alternately set C’(x) = 0
solve for x.
Learning Curves
Based upon the observation that unit labor hours or costs
decrease for each additional unit produced
Direct
labor hrs
per unit
Units produced
Why does this happen?
•
•
•
•
•
•
•
•
•
•
•
Employee learning
reduced set-up times
better routing and scheduling of material (WIP)
improved tool design
more efficient material handling equip. (MHE)
reduced lead-times
improved (simplified) product design
production smoothing
quality assurance
revised plant layout
increased machine utilization
Learning Curve
(experience curves)
Y(u) = labor hours to produce the uth unit
assume
Y(u) = au-b
a = hours to produce the first unit
b = rate at which production hours decline
labor coming
to work
Learning Curves
Assume hours to produce unit 2n is a fixed
percentage of the hours to produce unit n
Then for an 80 percent learning curve:
b
Y ( 2 n ) a ( 2 n)
b

 2  .80
b
Y ( n)
an
 ln .8
or b 
 .3219
ln 2
 ln(Pct/100)
b
ln 2
Observe the
simple
formula
Learning Curves
least-squares analysis
Unit
Number - x
Direct Labor
Hours -Y(x)
20
40
60
80
100
35.8
30.1
27.3
25.7
24.1
Fit Y(x) = ax-b
using Excel
Y(x) = 74x-.243
2-.243 = .845
or a 84.5% learning curve
Learning Curves
Cumulative Cost
Y (i )  ai
b
hours to produce ith unit
x
x
i 1
i 1
T ( x)   Y (i )   ai
T ( x)
V ( x) 
x
b
cumulative direct labor
hrs to produce x units
average unit hours to
produce x units
Learning Curves
Approximate Cumulative Cost
x
x
 b 1
ax
T ( x)   Y (i) di   a i di 
 b 1
0
0
b 1
b
b
ax
ax
V ( x) 

(b  1) x 1  b
Example
Y(x) = 74x-.243
2-.243 = .845 or a 84.5% learning curve
.2431
.757
74
x
74
x
T ( x)   74 i .243 di 

 97.75 x.757
 .243  1
.757
0
x
T (2000)  97.75(2000).757  30,833 hr.
V(2000)  30,8333/2000  15.42
X
100
200
300
400
500
600
700
800
900
1000
T(x)
3192.396
5395.062
7333.268
9117.509
10795.36
12393.02
13926.95
15408.34
16845.28
18243.86
V(x)
31.92396
26.97531
24.44423
22.79377
21.59072
20.65504
19.89564
19.26043
18.71698
18.24386
Break-Even Analysis
Let x = number of units produced and sold
x = S-1(unit selling price)
S(x) = unit selling price
F = fixed cost
g(x) = variable cost to produce x units
Sam Even
on a break
then break-even point occurs when revenue = cost; or
S(x) x = F + g(x)
and profit = revenue – cost or
P(x) = S(x)x – [F + g(x)]
Break-Even Analysis
$
Cost curve
Revenue
curve
profit
loss
loss
Diminishing
returns
F
Break
even pt
x
Max
profit
Break-Even Analysis
Demand Curve
S(x)
x
S(x) = d + e x + f x2 (quadratic)
d, e, and f are constants to be determined
Break-Even Analysis
Demand Curve
S(x)
Approximate as linear
S(x) = d + e x
x
S(x) = d + e x + f x2
d, e, and f are constants to be determined
Break-Even Analysis
Unit Cost
Let
M = direct material unit cost ($/unit)
L = direct labor rate ($/hour)
B = factory burden rate
Y(x) = direct labor hours to produce unit x
C(x) = cost to produce unit x
C(x) = M + L Y(x) + L B Y(x)
= M + (1+B) L Y(x)
= M + (1+B) L a x-b
Learning
curve effect
The Factory Burden Diversion
Manufacturing Costs
Direct costs
Direct
material
Direct
labor
Factory burden
Indirect
material
-Office &
janitorial
supplies
-Paint
Indirect
labor
-Supervision
-Engineering
-Maintenance
Indirect
expense
-Heating
-Lighting
-Depreciation
-Rent & Taxes
Factory Burden - example
Product
A
B
C
total
Category
Indirect material
Indirect labor
Indirect expenses
total
annual cost
$ 6,120
42,800
22,900
$71,820
annual production
100,000
140,000
80,000
labor hours
1000
1400
1600
4000
burden rate = 71,820 /30,000
= 2.394 per direct labor $
rate wages
$9/hr $9,000
$7/hr 9,800
$7/hr 11,200
$30,000
Administrative
Costs
Marketing
Costs
Development
Costs
General Overhead Costs
Manufacturing Costs
Profit
Selling
Price S(x)
Demands
Cumulative Cost
Unit cost:
C(x) = M + (1+B) L a
x-b
Learning curve
 b 1
ax
T ( x) 
 b 1
g(x) = M x + L (1+B) T(x)
= M x + L (1+B) [a x1-b / (1-b)]
total cost = F + M x + L (1+B) a x1-b / (1-b)
where F is a fixed cost to produce product x
Break-Even Analysis -Profit
Profit = P(x) = S(x) x - [F + g(x)]
letting S(x) = d + ex, e < 0
P(x) = (d + e x) x - F - M x - L (1+B) a x1-b /(1-b)
= d x + e x2 - F - M x - g x1-b
where g = L (1+B) a /(1-b)
More Break-Even Analysis
P(x) = (d - M) x + e x2 - g x1-b - F
break-even: set P(x) = 0 and solve for x
maximize profit: set
dP(x)
=0
dx
and solve for x
dP( x)
 d  M  2ex  (1  b) gx b  0
dx
d 2 P( x)
1 b 

2
e

b
(1

b
)
gx
dx 2
for e < 0, a max point can exist
Break-Even Analysis - example
Data:
d = 100
e = - .01
F = $100,000
M = $4
B = .5
L = $20 / hr
a = 10
b = .60
2-.6
= 66%
P(x) = d x + e x2 - F - M x - g x-b+1
where g = (1+B) L a /(1-b)
P(x) = 100 x - .01 x2 – 100,000
- 4 x – (1+.5) (20) (10) x.4 / .4
= 96x -.01x2 –750 x.4 –100,000
The Math
P( x)  96x  .01x – 750 x
x  1382
2
P '( x)  96  .02 x  300 x
0.6
x*  4706
P ''( x)  .02  180 x
 .02 
x

 180 
1.6
0
1/1.6
 296.07
0.4
0
–100, 000  0
The Graph
P(x)
$150,000
$100,000
x = 1382
$50,000
$0
0
1000
2000
3000
4000
5000
-$50,000
-$100,000
-$150,000
x = 4706
6000
7000
Production Functions
A production function expresses the relationship between
an organization's inputs and its outputs. It indicates, in
mathematical or graphical form, what outputs can be obtained
from various amounts and combinations of factor inputs.
In its most general mathematical form, a production function is
expressed as:
Q = f(X1,X2,X3,...,Xn) where: Q = quantity of output and
X1,X2,X3,...,Xn = factor inputs (such as capital, labor, raw
materials, land, technology, or management)
Production Functions
There are several ways of specifying this function. One is as an
additive production function:
Q = a + bX1 + cX2 + dX3,...
where a,b,c, and d are parameters that are determined
empirically.
Another is as a Cobb-Douglas production function
Q = f(L,K,M) = A * (Lalpha) * (Kbeta) * (Mgamma)
where L = labor, K = capital, M = materials and supplies, and Q
= units of product.
Cobb-Douglas Production Function
Q = f(L,K,M) = A * (Lalpha) * (Kbeta) * (Mgamma)
Properties of the Cobb-Douglas production function:
Decreasing returns to scale: alpha + beta + gamma < 1
Increasing returns to scale: alpha + beta + gamma > 1
Let CL, CK, and CM = the unit cost of labor, capital, and material,
then
C(L,K,M) = CLL + CKK + CM M
is the total cost function
A Little Production Problem
An interesting problem: Given a monthly budget of $B, how
should the money be spent to obtain a specified output Q?
Find L, K, and M where L = dollars spent on labor, K =
dollars spent on facilities and equipment, and M = dollars
spent on material
Q  f ( L, K , M )  100 L.3 K .2 M .4
Subject to: L  K  M  $1, 000, 000
M  3L; K  2( L  M )
I know I can
work this one.
The Inevitable Example
A
100
budget
labor
capital
material
alpha
beta
gamma
0.3
0.2
0.4
L
K
M
$8,333 $66,667 $25,000
1
1
1
-3
1
-2
1
-2
Q
794,700 RHS
100,000
100,000
0
0
0
0
Stop the madness.
Optimize your production system!
profits
homework: turn-in breakeven problem
text: Chapter 1- 29, 30, 31, 32, 34, 35
36, 37 ,38, 43, 44