Line balancing & EOQ

7 – 1
Line Balancing..Line Balancing..
Purpose is to minimize the number
of people and/or machines on an
assembly line that is required to
produce a given number of units
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

7 – 2Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Line Balancing ExampleLine Balancing Example
EXAMPLE
Green Grass’s plant manager just received marketing’s latest
forecasts of fertilizer spreader sales for the next year. She
wants its production line to be designed to make 2,400
spreaders per week. The plant will operate 40 hours per week.
a. What should be the line’s cycle time or throughput rate per
hour be?
Throughput rate/hr = 2400 / 40 = 60 spreaders/hr
Cycle Time = 1/Throughput rate= 1/60 = 1 minute = 60 seconds

Line balancing Example continued:Line balancing Example continued:
Assume that in order to produce the new fertilizer spreader on
the assembly line requires doing the following steps in the order
specified:
b.What is the total number of stations or machines required?
TM (total machines) = total production time / cycle time = 244/60 = 4.067 or 5
Work
Element
Description
Time
(sec)
Immediate
Predecessor(s)
A Bolt leg frame to hopper 40 None
B Insert impeller shaft 30 A
C Attach axle 50 A
D Attach agitator 40 B
E Attach drive wheel 6 B
F Attach free wheel 25 C
G Mount lower post 15 C
H Attach controls 20 D, E
I Mount nameplate 18 F, G
Total 244

Draw a Precedence DiagramDraw a Precedence Diagram
SOLUTION
The figure shows the complete diagram. We begin with work
element A, which has no immediate predecessors. Next, we
add elements B and C, for which element A is the only
immediate predecessor. After entering time standards and
arrows showing precedence, we add elements D and E,
and so on. The diagram simplifies
interpretation. Work element F,
for example, can be done
anywhere on the line after
element C is completed.
However, element I must
await completion of
elements F and G.
D
40
I
18
H
20
F
25
G
15
C
50
E
6
B
30
A
40
Precedence Diagram for
Assembling the Big Broadcaster

Allocating work or activities toAllocating work or activities to
stations or machinesstations or machines
 The goal is to cluster the work elements into
workstations so that
1. The number of workstations required is minimized
2. The precedence and cycle-time requirements are not
violated
 The work content for each station is equal (or
nearly so, but less than) the cycle time for the line
 Trial-and-error can be used but commercial
software packages are also available

Finding a SolutionFinding a Solution
The minimum number of workstations is 5 and the
cycle time is 60 seconds, so Figure 5 represents
an optimal solution to the problem
Firtilizer Precedence Diagram Solution
D
40
I
18
H
20
F
25
C
50
E
6
B
30
A
40
G
15

Calculating Line EfficiencyCalculating Line Efficiency
c. Now calculate the efficiency measures of a five-station
solution:
Efficiency = (100) =
Σt
nc
244
5(60)
= 81.3%
Idle time = nc – Σt = 5(60) – 244 = 56 seconds
Balance delay (%) = 100 – Efficiency = 100% - 81.3% = 18.7%

A Line ProcessA Line Process
The desired output rate is matched to the
staffing or production plan
Line Cycle Time is the maximum time
allowed for work at each station is
c =
1
r
where
c = cycle time in hours
r = desired output rate

The theoretical minimum number of
stations is
TM =
Σt
c
where
Σt = total time required to
assemble each unit

Idle time, efficiency, and balance delay
Idle time = nc – Σt
where
n = number of stations
Efficiency (%) = (100)
Σt
nc
Balance delay (%) = 100 – Efficiency

Solved Problem 2Solved Problem 2
A company is setting up an assembly line to produce 192 units
per 8-hour shift. The following table identifies the work
elements, times, and immediate predecessors:
Work Element Time (sec) Immediate Predecessor(s)
A
40
None
B
80
A
C
30
D, E, F
D
25
B
E
20
B
F
15
B
G
120
A
H
145
G

a. What is the desired cycle time (in seconds)?
b. What is the theoretical minimum number of stations?
c. Use trial and error to work out a solution, and show your
solution on a precedence diagram.
d. What are the efficiency and balance delay of the solution
found?
SOLUTION
a. Substituting in the cycle-time formula, we get
c = =
1
r
8 hours
192 units
(3,600 sec/hr) = 150 sec/unit

b. The sum of the work-element times is 720 seconds, so
TM =
Σt
c
= = 4.8 or 5 stations
720 sec/unit
150 sec/unit-station
which may not be achievable.

c. The precedence diagram is shown in Figure 7.6. Each row in
the following table shows work elements assigned to each
of the five workstations in the proposed solution.
J
115
C
30
D
25
E
20
F
15
I
130
H
145
B
80
G
120
A
40
Figure 7.6 – Precedence Diagram
Work
Element
Immediate
Predecessor(s)
A None
B A
C D, E, F
D B
E B
F B
G A
H G
I H
J C, I

Station Candidate(s) Choice
Work-Element
Time (sec)
Cumulative
Time (sec)
Idle Time
(c= 150 sec)
S1
S2
S3
S4
S5
J
115
C
30
D
25
E
20
F
15 I
130
H
145
B
80
G
120
A
40

J
115
C
30
D
25
E
20
F
15 I
130
H
145
B
80
G
120
A
40
A A 40 40 110
B B 80 120 30
D, E, F D 25 145 5
E, F, G G 120 120 30
E, F E 20 140 10
F, H H 145 145 5
F, I I 130 130 20
F F 15 145 5
C C 30 30 120
J J 115 145 5
Station Candidate(s) Choice
Work-Element
Time (sec)
Cumulative
Time (sec)
Idle Time
(c= 150 sec)
S1
S2
S3
S4
S5

d. Calculating the efficiency, we get
Thus, the balance delay is only 4 percent (100–96).
Σt
nc =
720 sec/unit
5(150 sec/unit)
= 96%

In classIn class - Example- Example
A plant manager needs a design for an assembly line to
assembly a new product that is being introduced. The time
requirements and
immediate
predecessors for the
work elements are
as follows:
Work Element Time (sec)
Immediate
Predecessor
A 12 ―
B 60 A
C 36 ―
D 24 ―
E 38 C, D
F 72 B, E
G 14 ―
H 72 ―
I 35 G, H
J 60 I
K 12 F, J
Total = 435

K
Draw a precedence diagram, complete I, F, J, and K
Work
Element Time (sec) Immediate
Predecessor
A 12 ―
B 60 A
C 36 ―
D 24 ―
E 38 C, D
F 72 B, E
G 14 ―
H 72 ―
I 35 G, H
J 60 I
K 12 F, J
Total = 43
5
F
J
B
E
I
A
C
G
H
D

If the desired output rate is 30 units per hour, what are the
cycle time and theoretical minimum?
c = =
1
r
1
30
(3600) = 120 sec/unit
TM =
Σt
c
= = 3.6 or 4 stations
435
120

Suppose that we are fortunate enough to find a solution with
just four stations. What is the idle time per unit, efficiency, and
the balance delay for this solution?
Idle time = nc – Σt
Σt
nc
Balance delay (%) = 100 – Efficiency
= 4(120) – 435 = 45 seconds
= 100 – 90.6 = 9.4%
= (100) = 90.6%
435
480

Station
Work
Elements
Assigned Cumulative Time
Idle Time
(c = 120)
1
2
3
4
5
Using trial and error, one possible solution is shown below.

Using trial and error, one possible solution is shown below.
H, C, A 120 0
B, D, G 98 22
E, F 110 10
I, J, K 107 13
A fifth station is not needed
Station
Work
Elements
Assigned Cumulative Time
Idle Time
(c = 120)
1
2
3
4
5

Managerial ConsiderationsManagerial Considerations
Pacing is the movement of product from
one station to the next
Behavioral factors such as absenteeism,
turnover, and grievances can increase after
installing production lines
The number of models produced
complicates scheduling and necessitates
good communication
Cycle times are dependent on the desired
output rate

7 – 25
InventoryInventory
Management & theManagement & the
Economic OrderEconomic Order
Quantity (EOQ)Quantity (EOQ)

7 – 26
Lecture todayLecture today
Why is inventory so bad?
Why hold inventory?
Where to hold inventory?
What are types of inventory to keep?
What are the inventory costs?
How much inventory to keep?
When to order & how much to order?
What do I need to know to make those
decisions?

7 – 27
Inventory ManagementInventory Management
Inventory management is the planning and
controlling of inventories in order to meet the
competitive priorities of the organization.
Inventory management requires information
about expected demands, amounts on hand and
amounts on order for every item stocked at all
locations.

7 – 28
Inventory BasicsInventory Basics
Inventory is created when the receipt of
materials, parts, or finished goods
exceeds their disbursement.
Inventory is depleted when their
disbursement exceeds their receipt.
An inventory manager’s job is to balance
the advantages and disadvantages of
both low and high inventories.

7 – 29
Inventory CostsInventory Costs
Cost of capital
Obsolescence
Storage
Insurance
Taxes
Security
Theft
Damage
Locating
Measurement
Management & Labor

7 – 30
Why hold Inventory?Why hold Inventory?
Customer Sales & Service: Avoid Retail
stock outs and thus customer goodwill
(Retailing)
Seasonal sales (Xmas trees)
Take advantage of quantity discounts
Balance process flow time
Uncertainty in supply and demand
Lead Time
Speculative inventory (wine, gold)

7 – 31
Inventory atInventory at WALWAL--MARTMART
 Making sure the shelves are stocked with tens of
thousands of items at their 5,379 stores in 10 countries is
no small matter for inventory managers at Wal-Mart.
 Knowing what is in stock, in what quantity, and where it is
being held, is critical to effective inventory management.
 With inventories in excess of $29 billion, Wal-Mart is
aware of the benefits from improved inventory
management.
 They know that effective inventory management must
include the entire supply chain.
 The firm is implementing radio frequency identification
(RFID) technology in its supply chain.

7 – 33
Macro Inventory DecisionsMacro Inventory Decisions
Where do we hold inventory?
◦ Manufacturers and suppliers
◦ warehouses and distribution centers
◦ retailers
Types of Inventory to keep?
◦ raw materials
◦ WIP
◦ finished goods

7 – 34
Micro Inventory DecisionsMicro Inventory Decisions
When to order items?
How much of each item to order?
How much safety stock to keep?
Objective: minimize overall cost of keeping
inventory!

7 – 35
Relevant Costs in an Inventory SystemRelevant Costs in an Inventory System
Procurement costs
 Ordering cost (administrative, inspection,
transportation etc.)
Holding costs
 Maintenance and Handling
 Taxes
 Obsolescence
Stock-outs costs
 Lost sales (Customer goodwill)
 Backorders

7 – 36
Relevant information to any inventoryRelevant information to any inventory
decisiondecision
Knowing how much demand there is
Knowing if this demand is fairly constant or
varies
Knowing what is in stock
Knowing where they exist in the supply chain
Knowing how long it will take to replenish
Knowing where it is going to be replenished from

7 – 37
Frequently used inventory termsFrequently used inventory terms
Inventory lot size
Replenishment Lead time
Stock out
Reorder Point
Safety stock

7 – 38
Thousands of items are held in inventory
by a typical organization, but only a
small % of them deserves management’s
closest attention and tightest control.
ABC analysis: The process of dividing
items into three classes, according to
their dollar usage, so that managers can
focus on items that have the highest
dollar value.
Knowing which Items are Critical

7 – 39
ABC AnalysisABC Analysis
1010 2020 3030 4040 5050 6060 7070 8080 9090 100100
Percentage of itemsPercentage of items
PercentageofdollarvaluePercentageofdollarvalue
100100 —
9090 —
8080 —
7070 —
6060 —
5050 —
4040 —
3030 —
2020 —
1010 —
00 —
Class C
Class A
Class B

7 – 40
 Economic Order Quantity (EOQ) is the
lot size that minimizes total annual
inventory holding and ordering costs.
 Assumptions of EOQ
1. The demand rate is constant and known with
certainty.
2. There are no constraints on lot size.
3. The only relevant costs are holding costs
and ordering/setup costs.
4. Decisions for items can be made
independently of other items.
5. Lead time is constant and known with
certainty.
Economic Order Quantity (EOQ)Economic Order Quantity (EOQ)

7 – 43
Inventory depletionInventory depletion
(demand rate)(demand rate)
ReceiveReceive
orderorder
1 cycle1 cycle
On-handinventory(units)On-handinventory(units)
TimeTime
QQ
AverageAverage
cyclecycle
inventoryinventory
QQ
——
22
Cycle-Inventory LevelsCycle-Inventory Levels

7 – 44
Annualcost(dollars)Annualcost(dollars)
Lot Size (Lot Size (QQ))
Total Annual Cycle-Inventory CostsTotal Annual Cycle-Inventory Costs
Holding cost = (Holding cost = (HH))
QQ
22
Ordering cost = (Ordering cost = (SS))
DD
QQ
Total cost = (Total cost = (HH) + () + (SS))
DD
QQ
QQ
22
Q = lot size; C = total annual cycle-inventory cost
H = holding cost per unit; D = annual demand
S = ordering or setup costs per lot

7 – 45
Costing out a Lot Sizing PolicyCosting out a Lot Sizing Policy
Bird feeder sales are 18 units per week, and the
supplier charges $60 per unit. The cost of placing an
order (S) with the supplier is $45.
Annual holding cost (H) is 25% of a feeder’s value,
based on operations 52 weeks per year.
Management chose a 390-unit lot size (Q) so that new
orders could be placed less frequently.
What is the annual cycle-inventory cost (C) of the
current policy of using a 390-unit lot size?
Museum of Natural History Gift Shop:

7 – 46
Costing out a Lot Sizing PolicyCosting out a Lot Sizing Policy
What is the annual cycle-inventory cost (C) of the
current policy of using a 390-unit lot size?
D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15
C = $2925 + $108 = $3033
C = (H) + (S) = (15) + (45)
Q
2
D
Q
936
390
390
2
Museum of Natural History Gift Shop:

7 – 47
30003000 —
20002000 —
10001000 —
00 —
| | | | | | | |
5050 100100 150150 200200 250250 300300 350350 400400
Lot Size (Q)
Annualcost(dollars)Annualcost(dollars)
Total costTotal cost
Holding costHolding cost
Ordering costOrdering cost
Current
cost
Current
Q
Lowest
cost
Best Q (EOQ)
Lot Sizing at the MuseumLot Sizing at the Museum
of Natural History Gift Shopof Natural History Gift Shop

7 – 48
Computing the EOQComputing the EOQ
C = (H) + (S)
Q
2
D
Q
EOQ =
2DS
H
D = annual demand
S = ordering or setup costs per lot
H = holding costs per unit
D = 936 units
H = $15
S = $45
EOQ =
2(936)45
15
= 74.94 or 75 units
C = (15) + (45)
75
2
936
75
C = $1,124.10
Bird Feeders:

7 – 49
Time Between OrdersTime Between Orders
Time between orders (TBO) is the average
elapsed time between receiving (or placing)
replenishment orders of Q units for a particular
lot size.
For the birdfeeder example, using an EOQ of 75
units.
TBOEOQ =
EOQ
D
TBOEOQ = = 75/936 = 0.080 year
EOQ
D
TBOEOQ = (75/936)(12) = 0.96 months
TBOEOQ = (75/936)(52) = 4.17 weeks
TBOEOQ = (75/936)(365) = 29.25 days

7 – 50
In Class ExampleIn Class Example

7 – 51

7 – 52
In Class Example (In Class Example (continued)continued)

7 – 53
continuedcontinued

7 – 54
Understanding the Effect of ChangesUnderstanding the Effect of Changes
What happens if there is a change in the
Demand Rate (D)?
What happens if the Setup Costs (S)
changes?
What happens if the holding Costs (H)
change?
What happens if there are errors in
estimating D, H, and S?

Line balancing & EOQ

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