Capacity Management with Setups:
Production/Inventory Models
What are Set-Ups (Video27)?
When multiple products (types) are processed through a
bottleneck recourse, there is usually a setup involved.
If the video clip does not play properly, please watch this through the link
which I provide in the description of the page. https://www.youtube.com/watch?v=fHI8LN2F0hk
What are Set-Ups (Video27)?
When multiple products (types) are processed through a
bottleneck recourse, there is usually a setup involved.
A
B
C
7 min/unit
20 min/unit
8 min/unit
5 min/setup
A
B
C
7 min/unit
20 min/unit
8 min/unit
A
B
C
7 min/unit
20 min/unit
8 min/unit
5 min/setup
A
B
C
7 min/unit
20 min/unit
8 min/unit
5 min/setup
A
B
C
7 min/unit
20 min/unit
8 min/unit
20 min/unit → 25 min/unit
What are Set-Ups?
The term Set-up – Comes from manufacturing where
machines needed to be prepared (set-up) to make a
product.
Set-Ups & Capacity
Set-ups tend to require resources
One product: Available time: 8hr/day.
Two products: At least 1 changeover → 1 setup
Available time: 8hr/day. Each setup time: 30mins.
Results in a reduction of available resource time
Could lead to a decrease in process capacity!
How managers often minimize the impact of
set-ups on capacity?
Increase available time for a resource – but this leads to
additional out-of pocket costs.
A potential solution which does not require an increase in
additional out-of-pocket cost is to process products in
batches
Batch
A collection of flow units.
Batches are produced in succession.
Once the production of one batch is completed, the production of
the next batch begins
All batches contain the same number and type of flow unit.
Computing Cycle Times
Processing a fixed amount of work
Example: Producing 50 cars. On average, production takes
5 hours per car. It takes 50 hours to set up the production
line.
Cycle Time =
Set-up Time + (Batch size) x (Time per unit)
Batch size
=
50 + (50) x (5)
50
=
300
50
=6
Computing Cycle Times
Processing a fixed amount of work
Example: Producing 100 cars. On average, production
takes 5 hours per car. It takes 50 hours to set up the
production line.
Cycle Time =
Set-up Time + (Batch size) x (Time per unit)
Batch size
=
50 + (100) x (5)
100
=
550
100
= 5.5
The Impact of Set-ups on Capacity
Production cycle
Batch of 12
Production cycle
Batch of 60
Batch of 120
Batch of 300
60
120
180
240
300
Produce steer supports (1 box corresponds to 12 units = 12 scooters)
Set-up from Ribs to Steer support
Produce ribs (1 box corresponds to 24 units = 12 scooters)
Set-up from steer support to ribs
Time [minutes]
Larger Batch Leads to a Lot of Inventory
Production with large batches
Cycle
Inventory
Production with small batches
Cycle
Inventory
Produce Sedan
Produce Station wagon
Beginning of
Month
End of
Month
Example: 2 car models. Average demand is 400 per day.
Beginning of
Month
End of
Month
Capacity Management with Set-Ups
Set-up times
dominate
Compute
Capacity as
function of
batch size
Compute cycle
time (CT) of the
rest of the
process
Compute
set-up costs
& inventory
costs
Use EOQ
model or one
of its variants
Solve for
batch size:
Cap(B)=1/CT
Analyze
Set-up times
& Set-up
Costs
Set-up costs
dominate
Reduce the need
for batches
• Set-up time reduction, SMED
• Process lay-out
•10-minute rule
• Externalize internal set-ups
Figure 7.11.: Summary of batching
Batching Under Set-Up Costs
Capacity Management with Set-Ups – Focused on Costs
Set-up times
dominate
Compute
Capacity as
function of
batch size
Compute cycle
time (CT) of the
rest of the
process
Compute
set-up costs &
inventory costs
Use EOQ
model or one
of its variants
Solve for
batch size:
Cap(B)=1/CT
Analyze
Set-up times &
Set-up Costs
Set-up costs
dominate
Reduce the need
for batches
• Set-up time reduction, SMED
• Process lay-out
•10-minute rule
• Externalize internal set-ups
Figure 7.11.: Summary of batching
Trade off between Holding cost and Set-up cost.
Google image
Set-up Cost
Ordering cost or shipping cost…
Holding Cost
Notation we will use is h;
carrying cost of inventory.
expressed in terms of
$ per unit per unit time.
Fraction of unit cost, h = I x c,
where I is expressed as a percent
Example)
Annual carrying costs are $3.6/unit
Annual carrying costs are 12% of the purchase price (such as $30)
Inventory Carrying Charge
Category
Housing costs (building rent or
depreciation, operating costs, taxes,
insurance)
Material handling costs (equipment lease or
depreciation, power, operating cost)
Labor cost
Cost (and range)
as a Percent of
Inventory Value
6% (3 - 10%)
3% (1 - 3.5%)
3% (3 - 5%)
Investment costs (borrowing costs, taxes,
and insurance on inventory)
Pilferage, space, and obsolescence
11% (6 - 24%)
Overall carrying cost
26%
3% (2 - 5%)
Example (Noodle)
Demand: 1200 per year, Order size: 100,
Example (Noodle)
Demand: 1200 per year, Order size: 100,
Shipping cost: 10/order, Storage cost: 1 /unit/year
Number of orders per year:
Annual ordering cost
Average inventory level
Annual inventory
1200
= 12
100
1012 = 120
100
= 50
2
50 1
holding cost
Total annual cost
Length of order cycle
50 +120 = 170
1
12 months
year ×
= 1 month
12
1 year
Average Inventory = Order size / 2
100
100
90
90
80
80
70
70
60
60
50
50
40
40
30
30
20
20
10
10
0
0
1
2
3
…
Capacity Management with setup cost:
Economic Order Quantity Model
Economic Order Quantity Model (EOQ)
Assumptions (need to memorize this assumption):
There is a fixed setup cost S independent of the order quantity.
Demand is constant and not stochastic: D=annual demand
Lead time is zero (or constant) → Replenishment occurs instantaneously.
Full lot delivery
No backorder or shortage is allowed, No quantity discounts.
Related costs:
Ordering cost: S $/order
Inventory holding cost: H $/unit/year
Questions
When to order from the supplier
How much to order
Objective
Minimize total annual cost
The EOQ Model
Annual setup cost = S
D
Q
Q = Number of pieces per order
D = Annual demand in units for the inventory item
S = Setup or ordering cost for each order
H = Holding or carrying cost per unit per year
Annual setup cost =
=
Number of orders
placed per year
Annual demand
Number of units in each order
x
Set-up or order cost
Per order
Setup or order
cost per order
=
D
Q
(S)
Annual setup cost = S
The EOQ Model
Annual holding cost =
Q = Number of pieces per order
D = Annual demand in units for the inventory item
S = Setup or ordering cost for each order
H = Holding or carrying cost per unit per year
Annual holding cost =
=
Average Inventory
Q
2
(H)
x
Holding cost
per unit per year
D
Q
Q
H
2
Annual Cost
Cost Minimization Goal
TC =
Q
D
H+ S
2
Q
Ordering Costs
QO
Order Quantity (Q)
)
(optimal order quantity) = EOQ
TC curve reaches its minimum where the carrying & ordering costs are equal.
Example (Noodle)
Demand: 1200 per year, Order size: 100,
Shipping cost: 10/order, Storage cost: 1 /unit/year
Number of orders per year:
Annual ordering cost
Average inventory level
Annual inventory
1200
= 12
100
1012 = 120
100
= 50
2
50 1
holding cost
Total annual cost
Length of order cycle
50 +120 = 170
1
12 months
year ×
= 1 month
12
1 year
How much to buy and performance measures
If annual demand is D, ordering cost is S and holding
cost is H, then for any order quantity Q,
Number of orders per year:
Annual ordering cost
S
Average inventory level
Annual inventory
holding cost
Total annual cost
Length of order cycle
D
Q
D
Q
Q
2
Q
Q
H
2
Q
D
H + S
2
Q
Q
D
Q/2
D
Q
Q
Annual holding cost = H
2
Annual setup cost = S
The EOQ Model
Q = Number of pieces per order
D = Annual demand in units for the inventory item
S = Setup or ordering cost for each order
H = Holding or carrying cost per unit per year
Optimal order quantity is found
when annual setup cost equals annual holding cost
D
S =
Q
Solving for Q* Where Q* = Optimal number of pieces per order (EOQ)
2DS = Q2H
Q2 = 2DS/H
Q*= EOQ =
2DS/H
Q
H
2
The Performance of “approximate” EOQ
EOQ usually is an approximate quantity
How good is it in terms of minimizing cost?
It is fairly robust
An EOQ Example
Determine optimal number of needles to order
D = 1,000 units
S = $10 per order
H = $.50 per unit per year
Q* =
2DS
H
Q* =
2(1,000)(10)
=
0.50
40,000 = 200 units
An EOQ Example
Determine optimal number of needles to order
D = 1,000 units
Q* = 200 units
S = $10 per order
H = $.50 per unit per year
Expected
number of = N =
orders
N=
Demand
Order quantity
1,000
200
=
D
Q*
= 5 orders per year
An EOQ Example
Determine optimal number of needles to order
D = 1,000 units
Q*= 200 units
S = $10 per order
N= 5 orders per year
H = $.50 per unit per year
Expected time
between
=T=
orders
T=
Number of working
days per year
N
250
5
= 50 days between orders
An EOQ Example
Determine optimal number of needles to order
D = 1,000 units
Q*
S = $10 per order N
= 200 units
= 5 orders per year
H = $.50 per unit per year
T
= 50 days
Total annual cost = Setup cost + Holding cost
TC =
TC =
D
Q
S +
H
Q
2
1,000
200
($10) +
($.50)
200
2
TC = (5)($10) + (100)($.50) = $50 + $50 = $100
Sample Quiz Example
Demand for a certain radial tires at a tire company is
800 units per month. Each tire costs the company $80.
Ordering costs are $75, and the annual carrying costs
are 20 percent of the purchase price. Assume that the
lead time is zero and that the tire company operates
288 days a year
D = 800 * 12 = 9600 /yr
S = $75 /order
H = $80 * 0.2 = $16 /unit · yr
Match!
D = 800 * 12 = 9600 /yr
S = $75 /order
H = $80 * 0.2 = $16 /unit · yr
Solution to SQ
1. How many tires should the manager order in each
lot?
EOQ =
2 DS
H
=
2(9600)75
= 300 units.
16
2. What is the company's average inventory of this tire?
EOQ 300
Averageinventory =
=
= 150 units.
2
2
Solution to SQ (Cont.)
3. How often will an order be placed (length of order
cycle)?
EOQ
D
300
=
= 0.03125 ( yr)  288 (days/yr) = 9 days.
9600
4. How many times per year will an order be placed?
D
EOQ
9600
=
= 32 / y r.
300
Solution to SQ (Cont.)
5. How much does the company spend annually on
ordering costs?
D
9600
S=
75 = $2,400
EOQ
300
6. How much does the company spend annually on
holding (carrying) costs?
EOQ
300
H=
16 = $2,400
2
2
Solution to SQ (Cont.)
7. What is the total annual cost if the EOQ quantity is
ordered?
D
EOQ
TC =
S+
H
EOQ
2
= $2,400 + $2,400 = $ 4,800
The ordering and carrying costs are equal at the EOQ
Capacity Management with setup cost:
Economic Production Quantity Model
Basic Economic Production Quantity Model (EPQ)
Assumptions:
1. Produces in batch and production rate is constant.
2. Only one product is involved.
3. Constant demand rate.
Demand is spread evenly throughout the year.
4. Constant lead time.
Lead time does not vary much for a long enough time.
5. Usage occurs continually, but production occurs periodically.
6. Capacity to produce a part exceeds the part's usage rate.
Economic Production Quantity (EPQ)
EPQ
Maximum
Inventory
Imax
Slope=
Production rate p
Slope=
Production rate p – usage rate
u
Slope = – usage rate u
Usage rate u is demand rate D in terms of the time measure same as
production rate p.
e.g. (a) D = 200 units / week, p = 50 units / day ➔ u = 200/5 = 40 units / day
(b) D = 5 units / hr, p = 8 units / hr ➔ u = 5 units / hr.
EPQ
p>0
We need to produce 100 units (=EPQ).
P=10
Product rate = 10 units / day.
Therefore,
1
D→u
we need 10 days to produce 100 units
without considering usages...
EPQ
At the same time, we spend 5 units / day = u
EPQ/p
= 10 days
Production rate – usage rate
= 10 units/day – 5 units/day = 5 units/day…
EPQ/u
= 20 days
2 DS
EPQ =
H
p
p −u
Economical Production/Optimal Run quantity
2 DS
EPQ =
H
p
p−u
Production cycle time
EPQ
=
u
Run time
(time of producing in a cycle)
I max
p−u
= EPQ
p
EPQ
=
p
I max
I average =
2
Economic Production Quantity (EPQ)
EPQ
Maximum
Inventory
Imax
I av erag e
Slope=
Production rate p
Production cycle time =
Run time =
EPQ
p
Slope= – usage rate u
EPQ
u
Slope= Production rate p – usage rate u
Economic Production Quantity (EPQ)
I max
Carrying cost =
H
2
D
Setup cost =
S
EPQ
Sample Quiz Example
A toy manufacturer uses 48,000 rubber wheels per year for its
popular dump truck series. The firm makes its own wheels, which
it can produce at a rate 800 per day. The toy trucks are
assembled uniformly over entire year. Carrying cost is $1 per
wheel a year. Setup costs for a production run of wheels is $45.
The firm operates 240 days per year.
D = 48,000 / yr
S = $45, H = $1 / yr
p = 800 / day
48,000
u=
= 200 /day
240
Solution to Example
1. What is the optimal size of a production run?
EPQ =
2 DS
H
p
2(48000)45
800
=
= 2,400 wheels
p−u
1
800 − 200
2. What is the length of each production run?
EPQ 2400
Run time =
=
= 3 days
p
800
3. What is cycle time for the optimal run size?
EPQ 2400
Production cycle time =
=
= 12 days
u
200
Solution to Example (Cont.)
4. What is the maximum inventory level? the average
inventory level?
EPQ
2400
I max =
( p − u) =
(800 − 200) = 1,800 wheels
p
800
I max 1800
I average =
=
= 900 wheels
2
2
5. What is the average annual cost for holding
inventory? for setting up production?
I max
1800
Carrying cost =
H=
 1 = $900
2
2
D
48,000
Setup cost =
S=
 45 = $900
EPQ
2400
TC = $1800
EPQ vs EOQ
p=∞
P=10
1
D→u
D
EOQ
EPQ
EOQ/D
EPQ =
=
2 DS
H
p
p −u
2 DS
= EOQ
H
As p = ∞
EPQ/p
= 10 days
EPQ/u
= 20 days
1
EPQ =
2 DS
H
p
p −u
Capacity Management with setup cost: Economies of Scale
Quantity Discount Models
▪ Reduced prices are often available when larger
quantities are purchased
Total cost = Setup cost + Holding cost + Product cost
Discount
Number
1
2
Discount Quantity
0 to 999
1,000 to 1,999
Discount (%)
no discount
4
Discount
Price (P)
$5.00
$4.80
3
2,000 and over
5
$4.75
Demand 2800
Order size = 700
Product cost = (700x5)x4
= 14,000
Order size = 1400
Product cost = (1400x4.80)x2 = 13,440
Order size = 2800
Product cost = (2800x4.75)x1 = 13,300
Quantity Discount Models
▪ Reduced prices are often available when larger
quantities are purchased
Total cost = Setup cost + Holding cost + Product cost
TC =
where P = unit price
D
Q
S+
H + PD
Q
2
EOQ =
2 DS
H
Quantity Discount Models
▪ Steps in analyzing a quantity discount
•
For each discount, calculate Q*
•
If Q* for a discount doesn't qualify, choose the smallest
possible order size to get the discount
•
Compute the total cost for each Q* or adjusted value from
Step 2
•
Select the Q* that gives the lowest total cost
Quantity Discount Example (H = $1 or H = 0.2P)
Calculate Q* for every discount
D = 5000, S = 49
Q* =
2DS
H
Total cost = Setup cost + Holding cost + Product cost
TC =
D S + Q H + PD
Q
2
Quantity Discount Example (H = $1)
Calculate Q* for every discount
D = 5000, S = 49, H = 1
Q* =
2DS
H
Total cost = Setup cost + Holding cost + Product cost
TC =
D S + Q H + PD
Q
2
Quantity Discount Example
Calculate Q* for every discount
Q* =
2DS
H
Q1* =
2(5,000)(49)
= 700 cars/order
(1)
Q2* =
2(5,000)(49)
= 700 cars/order
(1)
Q3* =
2(5,000)(49)
= 700 cars/order
(1)
Example
Calculate Q* for every discount
D = 5000, S = 49, H = 0.2P
Quantity Discount Example (H = 0.2 x p)
Calculate Q* for every discount
D = 5000, S = 49, H = 0.2P
Q* =
2DS
IP
Total cost = Setup cost + Holding cost + Product cost
TC =
D S + Q H + PD
Q
2
Quantity Discount Example
Calculate Q* for every discount
Q* =
2DS
IP
Q1* =
2(5,000)(49)
= 700 cars/order
(.2)(5.00)
Q2* =
2(5,000)(49)
= 714 cars/order
(.2)(4.80)
Q3* =
2(5,000)(49)
= 718 cars/order
(.2)(4.75)
Quantity Discount Models
Total cost $
Total cost
curve for
discount 1
1st price
break
0
2nd price
break
1,000
2,000
Order quantity
Quantity Discount Models
Total cost $
Total cost curve for discount 2
b
a
Q* for discount 2 is below the allowable range at point a
and must be adjusted upward to 1,000 units at point b
1st price
break
0
2nd price
break
1,000
2,000
Order quantity
Total cost $
Quantity Discount Models
Total cost curve for discount 3
1st price
break
0
2nd price
break
1,000
2,000
Order quantity
Quantity Discount
Calculate Q*
for every discount
Q1* =
2(5,000)(49)
= 700 cars/order
(.2)(5.00)
Q2* =
2(5,000)(49)
= 714 cars/order
(.2)(4.80)
1,000 — adjusted
Q3* =
2(5,000)(49)
= 718 cars/order
(.2)(4.75)
2,000 — adjusted
Quantity Discount
Order
Quantity
Annual
Product
Cost
Annual
Ordering
Cost
Annual
Holding
Cost
Discount
Number
Unit
Price
1
$5.00
700
$25,000
$350
$350
$25,700
2
$4.80
1,000
$24,000
$245
$480
$24,725
3
$4.75
2,000
$23.750
$122.50
$950
$24,822.50
Total
Table 12.3
Choose the price and quantity that gives the lowest total cost
Buy 1,000 units at $4.80 per unit
Extra Example
The maintenance department of a large hospital uses about 816
cases of liquid cleanser annually. Ordering costs are $12, carrying
costs are $4 per case a year, and the new price schedule indicates
that orders of less than 50 case will cost $20 per case, 50 to 79
cases will cost $18 per case, 80 to 99 cases will cost $17 per case,
and larger orders will cost $16 per case.
Q: Determine the optimal order quantity and the total cost.
Solution to Extra Example
D = 816 cases/yr
S = $12
H = $4/case  yr
Compute EOQ:
EOQ =
2 DS
=
H
Range
1 to 49
50 to 79
80 to 99
100 or more
Price
$20
$18
$17
$16
2(816)12
= 69.97  70 cases
4
Solution to Extra Example
D = 816 cases/yr
S = $12
H = $4/case  yr
Compute EOQ:
EOQ =
2 DS
=
H
Range
1 to 49
50 to 79
80 to 99
100 or more
Price
$20
$18
$17
$16
2(816)12
= 69.97  70 cases
4
The total cost to purchase at EOQ = 70 cases per order:
TC70 = Carrying cost + Order cost + Purchase cost
Q
D
= H+
S + PD
2
Qo
70
816
=
4+
12 + 18(816) = $14,968
2
70
Solution to Extra Example(Cont.)
3. Check next two lower cost ranges:
•
Range
1 to 49
50 to 79
80 to 99
100 or more
To buy at $17 per cases, order at least 80 cases:
TC 80 = (80 / 2)4 + (816 / 80)12 + 17(816) = $14154
•
To buy at $16 per cases, order at least 100 cases:
TC100 = (100 / 2)4 + (816 / 100)12 + 16(816) = $13354
Finally, 100 cases yields the lowest cost, therefore,
optimal order quantity is 100 cases.
TC1 00  TC 8 0  TC 7 0  Q * = 100 cases
Price
$20
$18
$17
$16
Example
Surge Electric uses 4,000 toggle switches a year. Switches are priced as
follows: 1 to 499, 90 cents each; 500 to 999, 85 cents each ; and 1000 or
more, 80 cent each. It costs approximately $30 to prepare an order and
receive it, and carrying costs are 40 percentage of purchasing cost per
unit on an annual basis. Determine the optimal order quantity and the
total annual cost.
Demand 4000 switches per year
Ordering cost: S = $30
Carrying cost: H = 0.4 p
Range
Price
H
1 to 499
$0.90
0.36
500 to 999
$0.85
0.34
1000 or more
$0.80
0.32
12-72
Example
D = 4000 /yr
S = $30
H = 0.40P
Range
1 to 499
500 to 999
1000 or more
Price
$0.90
$0.85
$0.80
1. First Compute EOQ with each price:
EOQ 0.90 =
2 DS
=
H 0.90
2(4000)30
= 816.49
0.36
EOQ 0.85 =
2 DS
=
H 0.85
2(4000)30
= 840.16
0.34
EOQ 0.80 =
2 DS
=
H 0.80
2(4000)30
= 866
0.32
H = 0.4P
0.4(0.90) = 0.36
0.4(0.85) = 0.34
0.4(0.80) = 0.32
Example
D = 4000 /yr
S = $30
H = 0.40P
Range
1 to 499
500 to 999
1000 or more
Price
$0.90
$0.85
$0.80
H = 0.4P
0.4(0.90) = 0.36
0.4(0.85) = 0.34
0.4(0.80) = 0.32
1. Total cost with each price:
EOQ 0.90 = 499
TC499
EOQ 0.85 = 840.16
499
4000
=
(0.36) +
(30) + 0.90(4000) = $3930.30
2
499
840
4000
TC840 =
(0.34) +
(30) + 0.85(4000) = $3686
2
840
EOQ 0.80 = 1000
1000
4000
TC1000 =
(0.32) +
(30) + 0.80(4000) = $3480
2
1000
Capacity Management with setup cost: Backorder
Planned Shortage Model
When an item is out of stock, customers may:
Go somewhere else (lost sales).
Place their order and wait (backordering).
In this model we consider the backordering case.
All the other EOQ assumptions are in place.
If backordered are considered
This EOQ modification is based on the assumption
that
in some cases the magnitude of penalty costs could
make it economical for the batching policy
Relevant Costs and Decisions
Relevant Costs
Setup Costs(S)
Holding Costs(H)
Penalty Costs(p)
Decisions:
The batch size (Q)
The number of units backordered (B)
The EOQ Model with Backorders
Inventory
Q-B
B
Shipment arrives
Order cycle = T
Time
EOQ Costs Structure with Backorders
Now Total Annual Relevant Costs G(Q,B) which are
a function of both batch size Q and the maximum
backordered quantity B are as follows (detail
omitted)
G(Q,B) = total setup costs + total holding costs + total backorder
costs
H
G(Q,B) = DS/Q + [(Q-B)2/2Q]H + [B2/2Q]p
You won’t need to remember this but you will need to know the
derived Q* and B*
The Results
Q* =
B* = Q*
2DS(H +p)
pH
H
H+p
High penalty cost…
Note: As p goes to infinity B* = 0
Q* =
B* = Q*
0
2DS(H +p)
pH
H
H+p
=
=
2DS 2DS
+
p
H
H
∞
=0
G(Q,B) = DS/Q + [(Q-B)2/2Q]H + [B2/2Q]p
High penalty cost…
Note: As p goes to infinity B* = 0
Q* =
B* = Q*
2DS(H +p)
pH
H
H+p
=
=
2DS 2DS
+
p
H
H
∞
=0
0
G(Q,B) = DS/Q + [(Q-0)2/2Q]H + [02/2Q]p
[Q/2]H
The EOQ Model with Backorders
Inventory
Q-B
B
Shipment arrives
Order cycle = T
Time
The EOQ Model with Backorders
Inventory
Q
Time
Shipment arrives
Order cycle = T