Transcript Capacity and Constraint Management

S7
Capacity and Constraint
Management
PowerPoint presentation to accompany
Heizer and Render
Operations Management, 10e
Principles of Operations Management, 8e
PowerPoint slides by Jeff Heyl
1
Outline
 Capacity
 Design and Effective Capacity
 Capacity and Strategy
 Capacity Considerations
 Managing Demand
 Demand and Capacity Management in the
Service Sector
 Bottleneck Analysis and Theory of
Constraints
 Process Times for Stations, Systems, and
Cycles
 Break-Even Analysis
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Learning Objectives
When you complete this supplement,
you should be able to:
1. Define capacity
2. Determine design capacity, effective
capacity, and utilization
3. Perform bottleneck analysis
4. Compute break-even analysis
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Capacity
 The throughput, or the number of
units a facility can hold, receive,
store, or produce in a period of time
 Determines
fixed costs
 Determines if
demand will
be satisfied
 Three time horizons
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Planning Over a Time
Horizon
Options for Adjusting Capacity
Long-range
planning
Add facilities
Add long lead time equipment
Intermediaterange
planning
Subcontract
Add equipment
Add shifts
Short-range
planning
*
Add personnel
Build or use inventory
*
Schedule jobs
Schedule personnel
Allocate machinery
Modify capacity
Use capacity
* Difficult to adjust capacity as limited options exist
Figure S7.1
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Design and Effective
Capacity
 Design capacity is the maximum
theoretical output of a system
 Normally expressed as a rate
 Effective capacity is the capacity a
firm expects to achieve given current
operating constraints
 Often lower than design capacity
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Utilization and Efficiency
Utilization is the percent of design capacity
achieved
Utilization = Actual output/Design capacity
Efficiency is the percent of effective capacity
achieved
Efficiency = Actual output/Effective capacity
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Efficiency = 148,000/175,000 = 84.6%
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Efficiency = 148,000/175,000 = 84.6%
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Efficiency = 84.6%
Efficiency of new line = 75%
Expected Output = (Effective Capacity)(Efficiency)
= (175,000)(.75) = 131,250 rolls
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Bakery Example
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Efficiency = 84.6%
Efficiency of new line = 75%
Expected Output = (Effective Capacity)(Efficiency)
= (175,000)(.75) = 131,250 rolls
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Managing Demand
 Demand exceeds capacity
 Curtail demand by raising prices,
scheduling longer lead time
 Long term solution is to increase capacity
 Capacity exceeds demand
 Stimulate market
 Product changes
 Adjusting to seasonal demands
 Produce products with complementary
demand patterns
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Complementary Demand
Patterns
Sales in units
4,000 –
Combining both
demand patterns
reduces the
variation
3,000 –
Snowmobile
motor sales
2,000 –
1,000 –
JFMAMJJASONDJFMAMJJASONDJ
Time (months)
Jet ski
engine
sales
Figure S7.3
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Demand and Capacity
Management in the
Service Sector
 Demand management
 Appointment, reservations, FCFS rule
 Capacity
management
 Full time,
temporary,
part-time
staff
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Break-Even Analysis
 Objective is to find the point in dollars and
units at which cost equals revenue
 Fixed costs are costs that continue even if no
units are produced

Depreciation, taxes, debt, mortgage payments
 Variable costs are costs that vary with the
volume of units produced

Labor, materials, portion of utilities
 Assumes - Costs and revenue are linear
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Break-Even Analysis
–
Total revenue line
900 –
800 –
Cost in dollars
700 –
Break-even point
Total cost = Total revenue
Total cost line
600 –
500 –
Variable cost
400 –
300 –
200 –
100 –
Fixed cost
|
|
|
|
|
|
|
|
|
|
|
–
0 100 200 300 400 500 600 700 800 900 1000 1100
|
Figure S7.5
Volume (units per period)
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Break-Even Analysis
BEPx = break-even point in
units
BEP$ = break-even point in
dollars
P = price per unit (after
all discounts)
x = number of units
produced
TR = total revenue = Px
F = fixed costs
V = variable cost per unit
TC = total costs = F + Vx
Break-even point occurs when
TR = TC
or
Px = F + Vx
F
BEPx =
P-V
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Break-Even Analysis
BEPx = break-even point in
units
BEP$ = break-even point in
dollars
P = price per unit (after
all discounts)
x = number of units
produced
TR = total revenue = Px
F = fixed costs
V = variable cost per unit
TC = total costs = F + Vx
BEP$ = BEPx P
F
=
P
P-V
F
=
(P - V)/P
F
=
1 - V/P
Profit = TR - TC
= Px - (F + Vx)
= Px - F - Vx
= (P - V)x - F
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Break-Even Example
Fixed costs = $10,000
Direct labor = $1.50/unit
Material = $.75/unit
Selling price = $4.00 per unit
$10,000
F
BEP$ =
=
1 - [(1.50 + .75)/(4.00)]
1 - (V/P)
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Break-Even Example
Fixed costs = $10,000
Direct labor = $1.50/unit
Material = $.75/unit
Selling price = $4.00 per unit
$10,000
F
BEP$ =
=
1 - [(1.50 + .75)/(4.00)]
1 - (V/P)
$10,000
=
= $22,857.14
.4375
$10,000
F
BEPx =
=
= 5,714
4.00 - (1.50 + .75)
P-V
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Break-Even Example
50,000 –
Revenue
Dollars
40,000 –
Break-even
point
30,000 –
Total
costs
20,000 –
Fixed costs
10,000 –
|
–
0
|
|
2,000
4,000
|
6,000
Units
|
|
8,000
10,000
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Break-Even Example
Multiproduct Case
BEP$ =
where
V
P
F
W
i
F
∑
Vi
1x (Wi)
Pi
= variable cost per unit
= price per unit
= fixed costs
= percent each product is of total dollar sales
= each product
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In-Class Problems from the
Lecture Guide Practice Problems
Problem 1:
The design capacity for engine repair in our company is 80
trucks/day. The effective capacity is 40 engines/day and the actual
output is 36 engines/day. Calculate the utilization and efficiency of the
operation. If the efficiency for next month is expected to be 82%, what
is the expected output?
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In-Class Problems from the
Lecture Guide Practice Problems
Problem 5:
Jack’s Grocery is manufacturing a “store brand” item that has a
variable cost of $0.75 per unit and a selling price of $1.25 per unit.
Fixed costs are $12,000. Current volume is 50,000 units. The Grocery
can substantially improve the product quality by adding a new piece
of equipment at an additional fixed cost of $5,000. Variable cost would
increase to $1.00, but their volume should increase to 70,000 units
due to the higher quality product. Should the company buy the new
equipment?
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In-Class Problems from the
Lecture Guide Practice Problems
Problem 6:
What are the break-even points ($ and units) for the two processes
considered in Problem S7.5?
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In-Class Problems from the
Lecture Guide Practice Problems
Problem 7:
Develop a break-even chart for Problem S7.5.
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