Matching Supply with Demand:
An Introduction to Operations Management
Gérard Cachon
ChristianTerwiesch
All slides in this file are copyrighted by Gerard Cachon and Christian
Terwiesch. Any instructor that adopts Matching Supply with
Demand: An Introduction to Operations Management as a required
text for their course is free to use and modify these slides as desired.
All others must obtain explicit written permission from the authors to
use these slides.
Slide ‹#›
Batching
Slide ‹#›
Buying Custom Shirts: The Demand Size
Custom shirts ordered online
Large variety of styles
Basically infinitely many sizes
Four weeks lead time
Slide ‹#›
Custom Tailored Shirts: The Supply Side
Cutting Department
Draw a shirt pattern on a large paper. Fabric is overlaid in layers according to the number of orders.
Then, the large patterned paper is laid on top of the fabric and the fabric is cut according to the pattern.
Sewing Department
Sewing Section – Cut pieces of fabric are sewn together and inspected
Assembly Section - Responsible for assembling shirts and measuring the size.
Finishing Department
Responsible for ironing shirts before folding, packaging and delivery to customers.
Source: http://hosting.thailand.com/MWT00255/process1.htm
Slide ‹#›
Process Analysis with Batching
• Capacity calculation for the resource with set-up changes:
Batch Size
Capacity given Batch Size=
Set-up time + Batch-size*Time per unit
• Capacity increases with batch size:
0.5
Capacity
1/p
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
650
610
570
530
490
450
410
370
330
290
250
210
170
130
90
50
10
0
Batch Size
• Example: Cutting Machine for shirts
20 minute cutting time (irrespective of the number of shirts)
4 minute/unit preparation
time
Slide ‹#›
The Downside of Large Batches
Production with large batches
Cycle
Inventory
Production with small batches
Cycle
Inventory
Produce Sedan
Produce Station wagon
Beginning of
Month
End of
Month
Beginning of
Month
• Large batch sizes lead to more inventory in the process
• This needs to be balanced with the need for capacity
• Implication: look at where in the process the set-up occurs
If set-up occurs at non-bottleneck => decrease the batch size
If set-up occurs at the bottleneck => increase the batch size
Slide ‹#›
End of
Month
Example Calculations
What is the capacity of the cutting machine with a batch size of 15?
What is the capacity of the overall process?
How would you set the batch size?
Cutting
Set-up time: 20 minutes
Activity time: 4 min/unit
Resources:
1 Cutting machine
Section 1
40 min/unit
8 workers
Slide ‹#›
Section 2
30 min/unit
5 workers
Finishing
3 min/unit
1 worker
How to Set the Batch Size – An Intuitive Example
- one cart every 10 seconds
- 2 sec boarding time per passenger
- 2 sec exit time per passenger
- 2 minutes to go down the elevator
Capacity given Batch Size=
1/10 [units/sec]
Batch Size
=
=
Batch Size
120sec + Batch-size*4sec
Batch Size
120sec + Batch-size*4sec
20 units
Slide ‹#›
External
Internal
The 6-stage SMED approach
Stage
Before/after shutdown During shutdown
1 Measure total changeover
time
2 Determine internal and
external activities
3 Move external activities to before
or after the shutdown
4 Improve the internal
activities
5 Improve the external
activities
6 Standardize procedures
Reduce set-up so that you can change models as often as needed
=> Mixed model production (Heijunka)
Slide ‹#›
Source: McKinsey Ops Training Material
Process Analysis with Batching: Summary
• Batching is common in low volume / high variety operations
• Capacity calculation changes:
Capacity given Batch Size=
Batch Size
Set-up time + Batch-size*Time per unit
• This reflects economies of scale (similar to fix cost and variable cost)
• You improve the process by:
Setting the batch size:
(a) If set-up occurs at the bottleneck => Increase the batch size
(b) If set-up occurs at a non-bottleneck => Reduce the batch size
(c) Find the right batch size by solving equation
Reducing set-up times:
(a) SMED method separates between internal and external set-ups
(b) Do external set-ups off-line, i.e., while the process is still running
=> enables mixed model production (Heijunka)
• Set-up time reduction is also powerful in other settings, such as OR’s or airplanes
Slide ‹#›
Setup Costs
EOQ Formula
Quantity Discounts
Converting Setup Times to Setup Costs
Slide ‹#›
Ordering handle caps for the Xootr

Data:
 $0.85 = cost to Nova Cruz to purchase
each handle cap from its supplier in
Taiwan.
 $300 = customs fee for each shipment,
independent of the amount ordered.
 700 = demand for handle caps per week.
Note, each “handle cap” is actually a
pair, so one is needed per Xootr.
 40% = Nova Cruz’s annual inventory
holding cost.


Question:
 How many handle caps should they order
each time they order from their supplier?
Slide ‹#›
The inventory “saw-tooth” pattern
Inventory
Q
R
Time
Q/R
Shipment
arrives




Shipment
arrives
Assume we can time when the shipments arrive so that they arrive when we
have zero inventory.
Q = Quantity in each order (what we need to choose)
R = Flow Rate of demand (700 per week)
Q / R = Time between shipments
Slide ‹#›
Costs

Purchase costs:
 $0.85 per unit x 700 per week = $595 per week
 Q cannot influence our weekly purchase cost!

h = Inventory holding cost per unit time:
 40% annual holding cost, so …
 0.4 x $0.85 = $0.34 = cost to hold a unit for one year…
 h = $0.34 / 52 = $0.006538 = cost to hold a unit for one week.
 Average inventory = Q / 2
 Average inventory cost per unit time = h x Q / 2

K = Setup cost:
 This is the cost per order and it is independent of the amount ordered.
 K = $300
 Q / R = time between orders, so …
 Setup cost per unit time = K / (Q / R)
Slide ‹#›
Objective and solution
Objective:
 Choose Q to minimize the
average (setup and holding)
cost per unit time, C(Q):
KR 1
C(Q) 
 hQ
Q
2

Solution:
 Order the Economic Order
Quantity (EOQ) = Q*
Q* 
120
Costs for Xootr handle caps
100
C(Q) = cost per week

80
60
40
C(Q)
Setup
costs
Holding
costs
20
2 K  R
h
0
2000
2  300 700

 8015
0.006538
4000
6000
8000 10000 12000 14000
Order Quantity
Slide ‹#›
EOQ and economies of scale

Setup and inventory holding costs per unit:
 
C Q*
2 K  h

R
R

Per unit costs decrease in the demand volume for an item:
Flow Rate, R
(units / week)
300
500
700
900
1100
EOQ, Q *
5247
6774
8015
9088
10047
Per-unit ordering and
inventory cost, C(Q*) / R
($ / unit)
0.11
0.09
0.07
0.07
0.06
Slide ‹#›
Ordering and Inventory
Costs as a % of Total
Procurement Costs
13.5%
10.4%
8.8%
7.8%
7.0%
Two different shopping experiences
Slide ‹#›
Costco vs. Walmart
Costco
70,977
10.5%
1.8%
12.6
4,000
17.7
518,080
138.6
Sales ($m)
Gross Margin/Sales
Net Income/Sales
Inventory turns
Number of SKUs per store
Sales per SKU ($m)
Sales per employee ($)
Sales per store ($m)

Walmart
374,526
23.5%
3.4%
8.1
60,000
6.2
178,346
52.8
Costco must be hyper efficient with restocking shelves because it
has a much smaller margin.
 Consequently, Costco has much less variety, and much higher
volume per stock keeping unit (SKU)
Slide ‹#›
Quantity discount for handle caps


Should Xootr purchase 10,000 units per order if this gets them a 5%
discount from the supplier?
KR 1
Use the equation for costs:
C(Q) 
 hQ
Q
R
K
Purchase cost per unit
h
Q
C(Q) (per week)
Purchase cost per week
Total cost per week

Original
EOQ
700
300
0.85
0.006538
8,015
52.40
595.00
647.40
EOQ with 5%
discount
700
300
0.8075
0.006212
8,223
51.08
565.25
616.33
2
5% discount
with large Q
700
300
0.8075
0.006212
10,000
52.06
565.25
617.31
5% discount with
very large Q
700
300
0.8075
0.006212
23,000
80.56
565.25
645.81
Xootr should be willing to buy even 23,000 units to get the 5% discount!
Slide ‹#›
Trade promotions and forward buying


Supplier gives retailer a temporary discount, called a trade promotion.
Trade promotions are typically in the 2-8% range.
Retailer purchases enough to satisfy demand until the next trade promotion.

Example: Campbell’s Chicken Noodle Soup over a one year period:

One retailer’s buy
Total shipments and consumption
7000
6000
S hipm e nts
Cases
Cases
5000
4000
3000
C ons um ption
2000
1000
Slide ‹#›
Nov
Oct
Sep
Aug
Jul
Jun
Apr
Mar
Feb
May
T im e (w e e ks )
Jan
Dec
0
Converting setup times to setup costs
Milling
Assembly
Milling Machine
Setup time
120
0
Activity time
2
3
Assembly process

Suppose the milling machine cost $9000 per month and Nova Cruz
operates 35 hours per week, 4.33 weeks per month.
 This translates into about $59 per hour = 9000 / (4.33 x 35)

Suppose 1 component set (a steering support and 2 ribs) costs $10:
 The annual holding cost rate is 40%.
 The holding cost per hour of a component set is about 0.002198
per hour = 0.4 x $10 / (52 x 35)
Slide ‹#›
Converting setup times to setup costs
Milling
Assembly
Milling Machine
Setup time
120
0
Activity time
2
3
Assembly process

If you apply EOQ:
 R = 20 per hour, K = 2 x 59 = 118, h = 0.002198
 Q* = 1465 = sqrt(2 x K x R / h)

Milling machine’s capacity with the EOQ batch size is 0.48 units/min
= 1465 / (120 + 2 x 1465)

But the milling machine only needs to operate with a batch size of
120 to match Assembly’s capacity of 1/3 units/min
Slide ‹#›
Converting setup times to setup costs
Milling
Assembly
Milling Machine
Setup time
120
0
Activity time
2
3
Assembly process

The EOQ batch size is much larger than necessary, which creates
more inventory than needed.

The EOQ doesn’t work in this setting because the setup cost is a
sunk cost – Nova Cruz incurs $9000 per month whether they
operate the milling machine or not.

With the setup cost sunk, the object should be to minimize inventory
without constraining the process flow, which is a batch size of 120.
Slide ‹#›
Buffer or Suffer
Blocking and Starving
Slide ‹#›
Orange juice production
Extraction
Setup time
Filtering
0
Bottling
30 min
0
Max flow rate (barrels/hr) 80
100
120
Capacity (barrels/hr)
88.9
120
80

No inventory is allowed between the tasks (i.e., no buffers)

After 4 hours of production Filtering must shut down for 30 mins before
production can resume (for another 4 hours followed, etc.)

The above capacities assume each step can work in isolation:
 e.g., Filtering capacity = 4 x 100 / (0.5 + 4) = 88.9 barrels/hr.
Slide ‹#›
Where’s the bottleneck?
Extraction
Setup time
Filtering
0
Bottling
30 min
0
Max flow rate (barrels/hr) 80
100
120
Capacity (barrels/hr)
88.9
120
80

It looks like Extraction is the bottleneck (because it has the lowest capacity)
and the maximum flow rate through the process is 80 barrels/hr.

But Extraction must also shut down for the 30 minutes Filtering is idle
because there is no place to put its output!
Slide ‹#›
Process interruption - blocking
Extraction
Setup time
Filtering
0
Bottling
30 min
0
Max flow rate (barrels/hr) 80
100
120
Capacity (barrels/hr)
88.9
120
80

The process produces 4 x 80 = 320 barrels every 4.5 hours, so its capacity
is 320 / 4.5 = 71 barrels per hour

If inventory were allowed between Extraction and Filtering, the process
would produce 80 barrels per hour (Extraction would always be working).

Lesson: add inventory to the process so that you don’t “block” the
bottleneck.
Slide ‹#›
Process variability
Buffer or Suffer (again)
Slide ‹#›
A Ski Rental Process – no variability
50
Forms
& Pay
{5}
50
Boots
50
Skis &
Boards
{5}
{5}
{ } = capacity







5 customers arrive per minute to rent skis or snow boards.
Forms & Pay – They fill out forms and pay for their rental.
Boots – They are given boots and try them on to check the fit.
Skis & Boards – They are given their skis or snow boards.
All three tasks can process 5 customers per minute.
All three buffers can have at most 50 customers.
This is a well balanced process and the process capacity is 5 customers per
minute.
Slide ‹#›
A Ski Rental Process – restricted buffers
50
Forms
& Pay
6
{5}
Boots
{5}
6
Skis &
Boards
{5}
{ } = capacity

Buffers are restricted:
 There can be at most 6 customers waiting to get boots and 6 customers
waiting to get skis & boards.

Also, if a customer arrives and there is no room in front of Forms & Pay,
then the customer goes away (angry).

The capacity of this process is?
Slide ‹#›
Restricted buffers and some variability
50
Forms
& Pay
6
{4,5,6}
Boots
{4,5,6}
6
Skis &
Boards
{4,5,6}
{ } = capacity

Buffers are restricted and…

Output of each task is now uncertain:
 Each task can serve 4, 5 or 6 customers per minute, each equally likely.
 Hence, the average capacity remains 5 customers per minute.

A task can be starved (lack “inventory” to work on) or blocked (have no
place to put “inventory”).

The capacity of this process is?
Slide ‹#›
Restricted buffers and considerable variability
50
Forms
& Pay
6
{0,…,10}
Boots
6
{0,…,10}
Skis &
Boards
{0,…,10}
{ } = capacity

Buffers are restricted and…

Output of each task is now highly uncertain:
 Now each task can serve 0,1,2,…,10 customers, each equally likely.

Average capacity again remains 5 customers per minute.

The capacity of this process is?
Slide ‹#›
Ample buffers and considerable variability
50
Forms
& Pay
50
{0,…,10}
Boots
{0,…,10}
50
Skis &
Boards
{0,…,10}
{ } = capacity

Add back the buffers (up to 50 customers before each task)…

But as before, each task can serve 0,1,2,…,10 customers, each equally
likely.

As before, a task can be blocked or starved.

The capacity of this process is?
Slide ‹#›
Summary

Setup costs provide a motivation to batch – the EOQ formula gives the
optimal batch size.
 Be very cautious when converting a setup time to a setup cost.
 Very large orders can be justified by seemingly small price discounts.

Buffer or Suffer:
 Setup times may cause process interruptions in other resources – use
inventory to decouple their production.
 Variability can reduce capacity considerably without buffers.

Quality:
 Don’t feed the bottleneck poor quality product.
 Avoid rework through the bottleneck.
 Check quality at the source.
Slide ‹#›
Process Interruptions – Setup Times
Lecture 3
January 31, 2012
Slide ‹#›
Milling machine at Novacruz



A milling machine is needed to
make one steer support and two
ribs per Xootr.
Setup
time (min)
Processing
time (min)
Steer
support
60
1
Rib
60
0.5
The setup time is the time
needed to get ready for production
during which no production
actually occurs.
What is the capacity of the milling
machine?
The Xootr by Novacruz
Slide ‹#›
Capacity of a process with setups

Capacity can be defined like this
Num berof units produced
Capacity
Tim eto producethoseunits

For example, if a process produces 12 units every 4 minutes then
Capacity = 12 units / 4 minutes = 3 units / minute

How long does it take to produce a batch of items?
Time to produce a batch =
Setup time + Batch size x Processing time

So the capacity of a process with setup times that produces in batches is:
Capacity
Batch size
Setuptim e  Batch size  Processingtim e
Slide ‹#›
What are the batches for Novacruz?

Suppose Novacruz uses the following production cycle: 100 steer
supports, 200 ribs, 100 steer supports, 200 ribs, etc.
= Setup and produce 100 Steer Supports
= Setup and produce 200 Ribs
Time

What is the “batch”?
 A “batch” is defined by a set of flow units and an interval of time
such that these intervals of time are identical and repeat
themselves.
 It follows that …
successive intervals are of the same length
the same number of the flow unit are produced in each interval.


Slide ‹#›
Suppose “ribs” are the flow unit defining the batch
1st rib batch
2nd rib batch
3nd rib batch

Within each rib batch, the same number of ribs are produced

Each rib batch includes the same amount of time:
This implies that certain events, like the start of rib production,
occur at regular intervals


Two things occur during a rib batch interval:
production of ribs
activities that do not produce ribs – we call this “setup time” for
rib production


Slide ‹#›
Setup time in rib batches
1st rib batch

2nd rib batch
3nd rib batch
Within each rib batch interval, the “setup time” is all time in which
ribs are not produced:
Can include setting up to start rib production.
Can include idle time (in which workers/machines are doing
nothing).
Can include time producing other things.



Slide ‹#›
Analysis of rib batches
1st rib batch



2nd rib batch
Setup
time
(min)
Processing
time (min)
Steer
support
60
1
Rib
60
0.5
3nd rib batch
Batch size = 200 because 200 ribs are produced per rib batch
Setup time = time per rib batch in which ribs are not being produced
 = Rib setup time + Steer support setup time + Steer support
production time
 = 60 min + 60 min + 1 x 100
 = 220 min
With this schedule the milling machine can produce 0.625 ribs per min:
Capacity
Batch size
200

 0.625
Setuptim e  Batch size  Processingtim e 220 200 0.5
Slide ‹#›
Analysis of steer support batches
1st steer batch



2nd steer batch
Setup
time
(min)
Processing
time (min)
Steer
support
60
1
Rib
60
0.5
3nd steer batch
Batch size = 100 because 100 steer supports are produced per steer
support batch
Setup time = time per steer batch in which steers are not being produced:
 = Steer support setup time + Rib setup time + Rib production time
 = 60 min + 60 min + 0.5 x 200
 = 220 min
The milling machine can produce 0.3125 steer supports per min:
Capacity
Batch size
100

 0.3125
Setuptim e  Batch size  Processingtim e 220 1001
Slide ‹#›
A component set batch
1st comp. set




2nd comp. set
Setup
time
(min)
Processing
time (min)
Steer
support
60
1
Rib
60
0.5
3rd comp. set
Define a “batch” to be a “component set” (i.e., 1 steer support and 2 ribs)
 This makes sense because each Xootr needs 1 support and 2 ribs
Batch size = 100 component sets
Setup time = time not producing a component set
 = Rib setup time + Steer support setup time
 = 120 min
Processing time = time to produce one component set
 = time to produce 1 steer support + 2 ribs
 = 1 + 2 x 0.5
 = 2 min
Slide ‹#›
Component set batches – continued
1st comp. set
2nd comp. set
Setup
time
(min)
Processing
time (min)
Steer
support
60
1
Rib
60
0.5
3rd comp. set
Batch size
Setuptim e  Batch size  Processing tim e
100

120 100 2
 0.3125
Capacity

The milling machine can produce 0.3125 component sets per minute.

We’ll use this definition of a batch in the remainder of our discussion.
Slide ‹#›
Capacity increases as the batch size increases
Capacity
Batch size
Setuptim e  Batch size  Processingtim e
0.50

As the batch size
increases, the setup
time is amortized over
more units, thereby
increasing the process’
capacity
The process capacity
can never be more
than 1 / Procesing time
(That is the capacity
with an infinite
batch size.)
Capacity of milling machine
0.45
0.40
Capacity (units/min)

0.35
0.30
0.25
0.20
0.15
0.10
0
50
100
150
Batch size
Slide ‹#›
200
250
300
Batch size and the location of the bottleneck
Milling Machine
Assembly process
Set-up time
120 minutes
-
Processing time
2 minutes/unit
3 minutes/unit
Capacity (B=12)
0.0833 units/minute
1/3 units/minute
Capacity (B=300)
0.4166 units/minute
1/3 units/minute

If the batch size is 12, the Milling Machine is the bottleneck (0.0833 units/min)

If the batch size is 300, Assembly is the bottleneck (1/3 units/min)

What is the smallest batch size such that the Milling Machine does not
constrain the process flow?
Slide ‹#›
Choosing the batch size to meet a target capacity

The target capacity for
milling is 1/3 units/min
because that is the capacity
of assembly:
 With a lower capacity,
milling becomes the
bottleneck.
 This presumes the
system is supply
constrained (otherwise
the demand rate would
be the target capacity).
What is the minimum batch
size that yields the target
capacity of 1/3 units/min?
0.50
Capacity of milling machine
0.45
0.40
Capacity (units/min)

0.35
0.30
0.25
0.20
0.15
0.10
0
50
100
150
Batch size
Slide ‹#›
200
250
300
Choosing the batch size to meet a target capacity

Start with the capacity equation:
Capacity

Batch size
Setuptim e  Batch size  Procesingtim e
Rearrange it to yield an equation for the batch size that yields the target
capacity:
Batch size 

Capacity Setuptim e
1 - Capacity Processingtim e
For our milling machine example:
1/3  120
Batch size 
 120
1 - 1/3  2

If you enter 0.33 as the
desired capacity you get
116.47 due to the rounding
error.
Hence with a batch size of 120 units or more, the milling machine’s
capacity will be at least 1/3 units / min, equal to assembly’s capacity.
Slide ‹#›
Batch size, idle time, utilization and inventory

Idle time is time not producing:
 Idle time has two parts: setup time and literal idle time (neither
producing, nor even setting up for production).

Utilization = Time producing / (Time producing + Idle Time)
 Increasing the batch size increases utilization of milling up to the point
at which it is no longer the bottleneck or the process becomes demand
constrained.

Inventory always increases as the batch size gets larger:
 Reducing batch size reduces inventory.
 Reducing inventory reduces flow time through the process (Little’s Law).
Slide ‹#›
“Small” batch – assembly is under utilized

Suppose the batch size is 80 component sets (and they are supply
constrained:
 Milling is the bottleneck – it needs 280 minutes to produce this batch.
 Assembly needs 80 x 3 = 240 minutes to produce this batch (they must
idle for 40 min with each batch, or else they will produce too much).
Assembly
Idle workers
240
60
80
40
60
80
Setup ribs Produce ribs Setup supports


Time
Assembly’s utilization = 240 / 280 = 85.7%
Milling’s utilization = 160 / 280 = 57%
Slide ‹#›
Produce supports
“Large” batch – assembly is fully utilized

Suppose the batch size is 200 component sets (and they are supply
constrained):
 Assembly is the bottleneck, it takes 3 x 200 = 600 min to produce 200
units.
Assembly
600
60
Setup ribs


200
Produce ribs
Time
60
200
Setup
supports
Produce supports
Utilization for assembly = 600 / 600 = 100%
Utilization for milling is 400 / 600 = 67%.
Slide ‹#›
80
Idle
The “just right” batch size

Suppose now the batch size is 120 component sets (and they are supply
constrained):
 Assembly takes 3 x 120 = 360 min to produce 120 units.
360
Assembly
60
Setup ribs


120
60
Time
120
Produce ribs Setup supports
Produce supports
Utilization for assembly = 360 / 360 = 100%
Utilization for milling is 240 / 360 = 67%
 Milling’s utilization cannot be higher!
 If it were, milling’s actual production would exceed 67% x 1/2 units/min
= 1/3 units/min …
 … and then parts inventory would continue to grow without limit…
 … which is clearly not desirable.
Slide ‹#›
Inventory dynamics
Suppose the batch size is 200 component sets and the flow rate is 1/3
units/min.
140
Rib inventory
Steer inventory
120
Inventory (units)

100
80
60
40
20
0
200 260
Produce
ribs
Setup for
steer
supports
460 520
600
Produce
Idle
steer
supports Setup for
ribs
Slide ‹#›
Time
How much inventory?

Suppose the batch size is 200 units
(component sets) and the flow rate is 1/3
units/min.
140


Suppose production of ribs starts when its
inventory is zero.
We produce rib pairs at the rate of 1 per minute
for 200 minutes.
During 200 minutes, 200 x 1/3 = 67 rib pairs
are demanded.
120
Inventory (units)

100
Rib
inventory
Steer
inventory
80
60
40
20
0

So inventory at the end of production is 200 –
67 = 133.

Average inventory of rib pairs is then 133 / 2 =
66 units.
Slide ‹#›
Time
Inventory with the recommended batch size

Suppose the batch size is 120 units (components sets) and the flow rate
remains 1/3 units/min
140
Rib inventory
Inventory (units)
120
Steer inventory
100
80
60
40
20
0
Time



We produce rib pairs for 120 minutes during which inventory builds at the
rate of (1-1/3) = 2/3 units per minute.
Max inventory is thus 120 x 2/3 = 80 units. Average inventory is now 40
units.
If inventory were any lower, then the milling machine would become the
bottleneck.
Slide ‹#›
Inventory and product variety - two soups

Consider a process that makes two kinds of soup:
Demand (gal/hr)
Setup time (hr)
Production rate (gal/hr)




Chicken
100
0.5
300
Tomato
75
0.5
300
Define the flow unit to be 1 gallon of soup.
Assume we iterate between chicken and tomato production.
Define the batch to be the total quantity (in gallons) of chicken and tomato
soup.
What batch size minimizes inventory?
Slide ‹#›
Two soup analysis

Our target capacity (or flow rate) is 100 + 75 = 175 gal/hour

Each batch involves producing Chicken and Tomato soup, so the total setup
time is 2 x 0.5 = 1 hour per batch

Processing time = 1/300 hours/gal

So the recommended batch size is 420 gals
Capacity Setuptim e
175 (2  0.5)
Batch size 

 420
1 - Capacity Processingtim e 1  1751 / 300

Produce in proportion to demand:
 Chicken = (100 / 175) x 420 = 240 gals
 Tomato = (75 / 175 ) x 420 = 180 gals
Slide ‹#›
Three soups dividing demand

Now suppose one kind of soup is added but total demand stays the same:
Demand (gal/hr)
Setup time (hr)
Production rate (gal/hr)



Chicken
80
0.5
300
Tomato
65
0.5
300
Onion
30
0.5
300
Suppose we produce chicken, tomato, onion and then repeat
Define the batch to be the total quantity (in gallons) of chicken, tomato and
onion soup.
What batch size minimizes inventory?
Slide ‹#›
Three soups dividing demand - analysis

Our target capacity (or flow rate) is 80 + 65 + 30 = 175 gal/hour

Each batch involves producing Chicken, Tomato and Onion soup, so the
total setup time is 3 x 0.5 = 1.5 hours per batch

Processing time = 1/300 hours/gal

The recommended batch size increases by 50% to 630 gals!
Batch size 

Capacity Setuptim e
175 (3  0.5)

 630
1 - Capacity Processingtim e 1  1751 / 300
Produce in proportion to demand:
 Chicken = (80 / 175) x 630 = 288 gals
 Tomato = (65 / 175 ) x 630 = 234 gals
 Onion = (30 / 175) x 630 = 108 gals
Slide ‹#›
Three soups expanding demand

Now suppose one kind of soup is added and demand for the others remains
the same:
Demand (gal/hr)
Setup time (hr)
Production rate (gal/hr)
Chicken
100
0.5
300
Tomato
75
0.5
300

A batch is still a set of chicken, tomato and onion.

What batch size minimizes inventory?
Slide ‹#›
Onion
50
0.5
300
Three soups expanding demand

Our desired capacity (or flow rate) is 100 + 75 + 50 = 225 gal/hour

Each batch involves producing Chicken, Tomato and Onion soup, so the
total setup time is 3 x 0.5 = 1.5 hours per batch

Processing time = 1/300 hours/gal

So the recommended batch size increases by 221% to 1350 gals!
Batch size 

Capacity Setuptim e
225 (3  0.5)

 1350
1 - Capacity Processingtim e 1  2251 / 300
Produce in proportion to demand:
 Chicken = (100 / 225) x 1350 = 600 gals
 Tomato = (75 / 225 ) x 1350 = 450 gals
 Onion = (50 / 225) x 1350 = 300 gals
Slide ‹#›
Henry Ford’s famous proclamation
Customers can have any color they
want, as long as it is black.
Slide ‹#›
Summary

If there are setup times, then capacity depends on the production
schedule:
 Capacity increases as the batch size gets larger
 Inventory increases as the batch size gets larger
 Utilization may increase as the batch size gets larger
 There is a tradeoff between capacity and inventory

Reducing setup times allows you to reduce batch sizes because the
recommended batch size is proportional to the setup time:
Batch size 

Capacity Setuptim e
1 - Capacity Processingtim e
Long setup times are not compatible with large product variety.
Slide ‹#›
Recitation Questions
(prepare the questions described on the following slides for
the R3 recitation)
Slide ‹#›
Kinga Doll Company:
A Little Background
Kinga Doll Company manufactures eight versions of its popular girl doll, Shari.
The company operates on a 40-hour work week. The eight versions differ in
doll skin, hair, and eye color, enabling most children to have a doll with a
similar appearance to them. It currently sells an average of 4,000 dolls (spread
equally among its eight versions) per week to boutique toy retailers.
Slide ‹#›
Kinga Doll Company:
A Little Background
In simplified terms, doll making at Kinga involves three basic operations:
molding the body and hair, painting the face, and dressing the doll. Changing
over between versions requires setup time at the molding and painting stations
due to the different colors of plastic pellets, hair, and eye color paint required.
The table below lists the setup times for a batch and the processing times for
each unit at each step. Unlimited space for buffer inventory exists between
these steps.
Setup time
Processing time
Molding
Painting
Dressing
15 min
30 min
none
0.25 min/unit
0.15 min/unit
0.30 min/unit
Slide ‹#›
Kinga Doll Company:
Analysis
Q1. What is the flow rate in units per hour
when, for each of the eight types, 500 dolls are
produced with each setup?
Q2. Given the production schedule described in
Q1, what is Kinga’s utilization (time producing
dolls / total time)? Note, “total time” includes
production time, setup time and literal idle time.
Q3. How many dolls should be produced with
each setup to minimize inventory without
decreasing the current flow rate?
Q4. Given the production schedule determined
in Q3, what is Kinga’s utilization?
Slide ‹#›