IE 3265 Production &
Operations Management
Slide Series 2
Topics for discussion


Product Mix and Product Lifecycle – as they
affect the Capacity Planning Problem
The Make or Buy Decision



Its more than $ and ¢!
Break Even Analysis, how we filter in costs
Capacity Planning

When, where and How Much
Product Issues Related to Capacity
Planning

Typical Product Lifecycle help many
companies make planning decisions

Facility can be designed for Product Families
and the organization tries to match lifecycle
demands to keep capacity utilized
The Product Life-Cycle Curve
The Product/Process Matrix
Product Mix (Families) Typically Demand
Different Production Capacity Design

Is product Typically “One-Off”?




These systems have little standardization and require high
marketing investment per product
Typically ‘whatever can be made in house’ will be made ‘in
house’
Most designs are highly private and guarded as competitive
advantages
Multiple Products in Low Volume


Standard components are made in volume or purchased
Shops use a mixture of flow and fixed site manufacturing
layouts
Product Mix (Families) Typically Demand
Different Production Capacity Design

Few Major (discrete) Products in Higher Volume




Purchase most components (its worth standardizing nearly
all components)
Make what is highly specialized or provides a competitive
advantage
Make decisions are highly dependent of capacity issues
High Volume & Standardized “Commodity” Products


Flow processing – all feed products purchased
Manufacturing practices are carefully guarded ‘Trade
Secrets’
Make-Buy Decisions



A difficult problem address by the M-B matrix
Typically requires an analysis of the issues
related to People, Processes, and Capacity
Ultimately the problem is addressed
economically
Make – Buy Decision Process
Secondary Questions
1.
2.
3.
4.
1.
2.
3.
4.
Is the Item Available?
Will our Union Allow us
to buy?
Is outside Quality
Acceptable?
Are Reliable Sources
Available?
Is Manufacturing
Consistent with our
objectives?
Do we have Technical
Expertise?
Is L & MFG capacity
available?
Must we MFG to utilize
existing capacity?
Primary Question
Can Item be
Purchased?
Decision
NO
YES
Can Item be
Made?
YES
NO
NO = MAKE
(if yes continue
down)
NO = Buy
(if yes continue
down)
Make – Buy Decision Process
Secondary Questions
1.
2.
3.
4.
1.
2.
3.
What Alternatives are
available to MFG?
What is future demand?
What are MFG costs?
What are Reliability
issues that influence
purchase or MFG?
What other
opportunities are avail.
For Capital?
What are the future
investment implications
if item is MFG?
What are costs of
receiving external
Financing?
Primary Question
Is it cheaper
to make than
buy?
Decision
NO
YES
Is Capital
Available To
Make?
YES
NO
NO = Buy
(if yes continue
down)
NO = Buy
YES = MAKE
Break-even Curves for the Make or Buy Problem
Cost to Buy = c1x
Cost to make=K+c2x
K
Break-even quantity
Example M-B Analysis

Fixed Costs to Purchase consist of:

Vendor Service Costs:




Purchasing Agents Time
Quality/QA Testing Equipment
Overhead/Inventory Set Asides
Fixed Costs to Make (Manufacture)

Machine Overhead




Invested $’s
Machine Depreciation
Maintenance Costs
Order Related Costs (for materials purchase and storage
issues)
Example M-B Analysis

BUY Variable Costs:


Simply the purchase price
Make Variable Costs



Labor/Machine time
Material Consumed
Tooling Costs (consumed)
Example M-B Analysis


Make or Buy a Machined Component
Purchase:



Fixed Costs for Component: $4000 annually ($20000 over 5
years)
Purchase Price: $38.00 each
Make Using MFG Process A




Fixed Costs: $145,750 machine system
Variable cost of labor/overhead is 4 minutes @ $36.50/hr:
$2.43
Material Costs: $5.05/piece
Total Variable costs: $7.48/each
Example M-B Analysis

Make on MFG. Process B:




Fixed Cost of Machine System: $312,500
Variable Labor/overhead cost is 36sec @ 45.00/hr:
$0.45
Material Costs: $5.05
Formula for Breakeven:
Fa + VaX = Fb + VbX
X is Break even quantity
Fi is Fixed cost of Option i
Vi is Variable cost of Option i
Example M-B Analysis
Fa  Va X  Fb  Vb X
Break Even is X that satisfies this equation or:
F b  Fa
X
Va  Vb



Buy vs MFG1: BE is {(145750-20000)/(38-7.48)} = 4120
units
Buy vs MFG2: BE is {(312500-20000)/(38-5.5)} = 9000
units
MFG1 vs MFG2: BE is {(312500-145750)/(7.48-5.50)} =
68620 units
Break Even Analysis
4500000
4000000
Total Expenditure ($)
3500000
3000000
2500000
2000000
1500000
1000000
500000
0
0
20000
40000
60000
80000
Parts used each year
Purchase
MFG Proc 1
MFG Proc 2
100000
120000
Capacity Strategy
Fundamental issues:

Amount. When adding capacity, what is the optimal
amount to add?

Too little means that more capacity will have to be added
shortly afterwards.

Too much means that capital will be wasted.

Timing. What is the optimal time between adding new
capacity?

Type. Level of flexibility, automation, layout, process,
level of customization, outsourcing, etc.
Three Approaches to Capacity
Strategy

Policy A: Try not to run short. Here capacity must
lead demand, so on average there will be excess
capacity.

Policy B: Build to forecast. Capacity additions should
be timed so that the firm has excess capacity half
the time and is short half the time.

Policy C: Maximize capacity utilization. Capacity
additions lag demand, so that average demand is
never met.
Capacity Leading and Lagging
Demand
Determinants of Capacity Strategy


Highly competitive industries (commodities, large
number of suppliers, limited functional difference in
products, time sensitive customers) – here
shortages are very costly. Use Type A Policy.
Monopolistic environment where manufacturer has
power over the industry: Use Type C Policy.


(Intel, Lockheed/Martin).
Products that become obsolete quickly, such as
computer products. Want type C policy, but in
competitive industry, such as computers, you will be
gone if you cannot meet customer demand. Need
best of both worlds: Dell Computer. (tend toward A
with B in mind!)
Mathematical Model for Timing of
Capacity Additions
Let D = Annual Increase in Demand
x = Time interval between adding capacity
r = annual discount rate (compounded continuously)
f(y) = Cost of operating a plant of capacity y
Let C(x) be the total discounted cost of all capacity additions over
an infinite horizon if new plants are built every x units of time.
Then
 rx
2 rx
C ( x)  f ( xD )  e
f ( xD )  e
f ( xD ) 
 f ( xD )(1  e  rx  (e  rx ) 2  (e  rx )3 
f ( xD )

1  e  rx
Mathematical Model (continued)


xD is a desired future capacity
A typical form for the cost function f(y) is:
f ( y)  ky a

k is a constant of proportionality (Investment for Capacity), and a
measures the ratio of incremental to average cost of a unit of
plant capacity.


A typical value is a = 0.6
Note that since a < 1 we expect ‘economies of scale’ in plant
construction
f (2 y) k (2 y) a
a


2
( 1.516 for a=.6)
a
f ( y)
k ( y)
Economies of Scale



a has been found to be 0.5 – 0.7 for most
industries
Looking at the Example above (a=.6) we find
that to double the production capacity it takes
only 2a times the investment, an increase of
52% over the smaller size to double capacity
For a = .5 doubling capacity takes only a 41%
greater investment while for a = .7 doubling
capacity takes 62% more investment
Mathematical Model (continued)
Hence,

k ( xD)a
C ( x) 
1  e rx
It can be shown that this function is minimized at the
value of x that satisfies the equation:
rx
a
rx
e 1

This is a transcendental equation, with no algebraic
solution. However, using the graph (Fig. 1-14), one
can find the optimal value of x or any value of a: (0 <
a < 1) – thru function u = rx
The Function f (u )  a  u /(e 1)
u
To Use: Locate the value of a
on the y axis and the corresponding value
of u on the u axis.
where {u = rx}
or
x
u graph
rdecimal
Lets Try one







Cast Iron Production System
a = 0.55
k ($million/ton new capacity) = 0.0119
D is estimated to be 1000 ton/yr
Set r = 12% (.12) – typical MARR
Searching Fig 1-14 with a (.55) we find u is about
1.2
Solving for design:



X = 1.2/.12 = 10 years
Capacity required: Dx = 1000*10 = 10000
Investment: 0.0119*(10000).55 = $1.886 Million (every
10 years)
Issues in Plant Location








Size of the facility
Product lines
Process technology
Labor requirements
Utilities requirements
Environmental issues
International considerations
Tax Incentives