LOGISTICS AND SUPPLY CHAIN MANAGEMENT PROGRAM
Course:
WAREHOUSE ENGINEERING MANAGEMENT
Preparation for Mid-term Test
Lecturer: Assoc. Prof. Dr. Ho Thi Thu Hoa
1
All groups:
Mid-term Test:
1. Date: 04/04/2023
2. Time: 15h15
3. Manner: Offline-Test; this is an open-book examination.
ü Hard copies of teaching materials, hand written notes, calculator and dictionary are allowed.
ü Laptop use, discussion and materials transfer are strictly prohibited.
4. Duration: 90 minutes
5. Note: You are required to provide formulas, explanation and analysis for the mid-term test
questions (the marks will be not given if you do not follow the guidelines)
Good luck with your exam!
2
Contents
1.Chapter 1: Introduction to Warehousing (LP
1)
2.Chapter 2: Inventory, Stock Analysis and
classifying products (LP 2&3)
3.Chapter 3: Warehouse Operations (LP 4-5-6)
4.Chapter 4: Warehouse networking (LP 7-8)
3
References
1.
Richards, G. (2014). Warehouse management: a complete guide to improving efficiency and
minimizing costs in the modern warehouse. Kogan Page Publishers.
2.
Richards, G. (2018). Warehouse management: a complete guide to improving efficiency and
minimizing costs in the modern warehouse. Kogan Page Publishers.
3.
Manzini, Ricardo. Warehousing in the Global Supply Chain. Springer: 2012
4.
Tompkins, J. A., White, J. A., Bozer, Y. A., & Tanchoco, J. M. A. (2010). Facilities planning. John Wiley
& Sons
5.
Alan Hrrison and et. (2014), Logistics management and strategy competing through the supply
chain (fifth edition), Pearson
6.
Martin Christopher(2011), Logistics & Supply Chain Management (4th Edition), Prentice Hall
7.
Arnold, Tony J. R., Chapman, S. N., Clive, L. M. Introduction to Materials Management, 7ed.
Pearson: 2016
Assoc. Prof. Dr Ho Thi Thu Hoa
4
Structure of Mid-term Test
1. Question 1: (20 points)
2. Question 2: (20 points)
3. Question 3: (30 points)
4. Question 4: (30 points)
5
Chapter 1: Introduction to Warehousing
1.
2.
3.
4.
5.
What is warehouse
The importance of warehouse
Types of warehouse operation
Warehouse strategy
Supply chain trends affecting warehouses
Assoc. Prof. Dr Ho Thi Thu Hoa
6
Chapter 2: Inventory, Stock Analysis and classifying products
1. Product classification
2. Demand management and forecasting
3. Inventory function and principles
4. Inventory costs and service
5. Inventory models (Replenishment methods)
6. ABC and Pareto analysis
7. Exercises
7
Simple Moving Average
n
MAn =
Di

i=1
n
where
n = number of periods in
the moving average
Di = demand in period i
Assoc. Prof. Dr Ho Thi Thu Hoa
3-month Simple Moving Average
ORDERS
MONTH
Jan
PER MONTH
120
Feb
90
Mar
100
Apr
75
May
110
June
50
July
75
Aug
130
Sept
110
Oct
Nov
90
-
MOVING
AVERAGE
–
–
–
103.3
88.3
95.0
78.3
78.3
85.0
105.0
110.0
3

MA3 =
Assoc. Prof. Dr Ho Thi Thu Hoa
=
i=1
Di
3
90 + 110 + 130
3
= 110 orders for Nov
Weighted Moving Average
• Adjusts moving average method to more closely
reflect data fluctuations
WMAn =
where
n
 W i Di
i=1
Wi = the weight for period i,
between 0 and 100
percent
 Wi = 1.00
Assoc. Prof. Dr Ho Thi Thu Hoa
12-10
Example
MONTH
August: 1
September: 2
October: 3
Total: 6
August WMA= 1/6 = 17%
Sep WMA = 2/6 = 33.3%
Oct WMA = 3/6 = 50%
August
September
October
November Forecast
WEIGHT
DATA
17%
33%
50%
130
110
90
WMA3 =
3
W i Di

i=1
= (0.50)(90) + (0.33)(110) + (0.17)(130)
= 103.4 orders
Assoc. Prof. Dr Ho Thi Thu Hoa
12-11
Exponential Smoothing
•
•
•
•
Averaging method
Weights most recent data more strongly
Reacts more to recent changes
Widely used, accurate method
Assoc. Prof. Dr Ho Thi Thu Hoa
12-12
Exponential Smoothing
Ft +1 = Dt + (1 - )Ft
where:
Ft +1 = forecast for next period
Dt =
actual demand for present period
Ft = previously determined forecast for
present period
=
weighting factor, smoothing constant
Assoc. Prof. Dr Ho Thi Thu Hoa
12-13
Effect of Smoothing Constant
0.0  1.0
If = 0.20, then Ft +1 = 0.20Dt + 0.80 Ft
If = 0, then Ft +1 = 0Dt + 1 Ft = Ft
Forecast does not reflect recent data
If = 1, then Ft +1 = 1Dt + 0 Ft =Dt
Forecast based only on most recent data
Assoc. Prof. Dr Ho Thi Thu Hoa
12-14
Exponential Smoothing (α=0.30)
PERIOD
MONTH
DEMAND
1
Jan
37
2
Feb
40
3
Mar
41
4
Apr
37
5
May
45
6
Jun
50
7
Jul
43
8
Aug
47
F2 = D1 + (1 - )F1
= (0.30)(37) + (0.70)(37)
= 37
F3 = D2 + (1 - )F2
= (0.30)(40) + (0.70)(37)
= 37.9
F13 = D12 + (1 - )F12
= (0.30)(54) + (0.70)(50.84)
= 51.79
Assoc. Prof. Dr Ho Thi Thu Hoa
12-15
Exponential Smoothing
PERIOD
MONTH
DEMAND
1
2
3
4
5
6
7
8
9
10
11
12
13
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
Jan
37
40
41
37
45
50
43
47
56
52
55
54
–
FORECAST, Ft + 1
( = 0.3)
( = 0.5)
–
37.00
37.90
38.83
38.28
40.29
43.20
43.14
44.30
47.81
49.06
50.84
51.79
Assoc. Prof. Dr Ho Thi Thu Hoa
–
37.00
38.50
39.75
38.37
41.68
45.84
44.42
45.71
50.85
51.42
53.21
53.61
12-16
Example problem
A company manufactures a line of ten items. The usage and unit cost are
shown in the following table, along with the annual dollar usage. The latter
is obtained by multiplying the unit usage by the unit cost.
a. Calculate the annual dollar usage for each item.
b. List the items according to their annual dollar usage.
c. Calculate the cumulative annual dollar usage and the cumulative
percentage of items.
d. Group items into an A, B, C classification.
Assoc. Prof. Dr Ho Thi Thu Hoa
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Answer
Assoc. Prof. Dr Ho Thi Thu Hoa
18
Answer
In this table, look at second column, the annual usage are rearrange following decreasing order. After that, the
cummulative annual usage are calculated in third column.
Assoc. Prof. Dr Ho Thi Thu Hoa
19
Chapter 3: Warehouse Operations
1. Introduction (WHS-DC-Fulfillment Center – Crossdocking-Sorting centerCold WHS)
2. Materials handling equipment
3. Warehouse operations process
4. Receiving and put-away
5. Storage operations (general, cold- chilled, frozen)
6. Order picking operations
7. Replenishment – Returns and Despatch
8. Warehouse VAL (value-added services)
9. Exercises
20
Example
1. At the ABC manufacturing firm, the average daily withdrawal rate for
product A is 20 cartons/day, safety stock is 5 days, order lead time is 10
days, and the order quantity is 45 days. What are the maximum and
average quantities of unit loads to be stored?
Assoc. Prof. Dr Ho Thi Thu Hoa
22
Storage Space Planning (2 of 3)
Dedicated storage
Randomized storage
• The planned quantity of unit
• The planned quantity of unit loads =
loads = the sum of the openings
the number of openings required to
required for each SKU
store all SKUs.
Which would result in less storage space?
Suppose 6 products have inventory levels in pallet loads as the below table. Calculate the required
amount of space for (i) dedicated storage; (ii) randomized storage.
Period Product 1 Product 2 Product 3 Product 4 Product 5 Product 6
1
24
12
2
12
11
12
2
22
9
8
8
10
9
3
20
6
6
4
9
6
4
18
3
4
24
8
3
5
16
36
2
20
7
24
6
12
30
6
12
5
18
• Sum of individual maximum
inventory levels =
Max (Product 1+…+Product 6)
= (24+36+8+24+24+24) = 140
Dedicated storage: 140 pallet positions
Randomized storage: 105 pallet positions
Total:
1860
Average inventory
level = 1860/24 = 77.5
Assoc. Prof. Dr Ho Thi Thu Hoa
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Storage Space Planning (3 of 3)
• To maximize throughput when using dedicated storage, SKU should be assigned to storage
locations based on the ratio of their activity to the number of openings or slots assigned to
the SKU.
• Despite the greater throughput of dedicated storage, it is not used as often as it should be.
Why?
• The storage philosophy chosen for a specific SKU will not be strictly fixed or random
location storage, but hybrid location storage. Example?
Storage Layout Planning
• The objectives of a warehouse layout are:
- To use space efficiently;
- To allow for the most efficient material handling;
- To provide the most economical storage in relation to costs of equipment, use of space,
damage to material, handling labor, and operational safety;
- To provide maximum flexibility in order to meet changing storage and handling
requirements;
- To make the warehouse a model of good housekeeping.
• To accomplish the objectives, several storage area principles must be integrated,
including: popularity, similarity, size, characteristics, and space utilization
Popularity (1 of 3)
• Pareto’s law: 85% of the wealth of the world is held by 15% of the people  How to
interpret Pareto’s law into the popularity of materials stored?
• To maximize throughput, the most popular of materials should be stored such that travel
distance is minimized. As illustrated below, by storing popular materials in deep storage
areas, the travel distance to other materials will be less than if materials were stored in
shallow areas.
Assoc. Prof. Dr Ho Thi Thu Hoa
28
Popularity (2 of 3)
The impact of storage depth on travel distances
B1 B2 B3
Outsite
B4
wall
B1 B2
B5 B6 Aisle
B4
Reference
point
A5 A6
Outsite
wall
Aisle
B2
…
A4 A5 A6
B3
…
A1 A2 A3
Outsite
wall
B1
Reference
point
A6
Aisle
Reference
point
Storage depth
(a) Storage depth of 3 units
Distance from reference point
(b) Storage depth of 2 units
(c) Storage depth of 1 unit
Storage depth
3 units
2 units
1 unit
Distance to A6
2
2
2
Distance to A1
5
5
7
Distance to B6
4
5
8
Distance to B1
7
8
13
Average travel distance
4.5
5
7.5
Assoc. Prof. Dr Ho Thi Thu Hoa
30
Popularity (3 of 3)
Warehouse Receiving
entrance & & shipping
exit point
quantity
Same
Different
Different
Rules for stock location
-
Popular materials should be
positioned as close to this point
as possible (shown in picture)
Same
The most popular items should
be positioned along the most
direct route between the
entrance & departure points.
Different
The most popular items having
the smallest receiving/shipping
ratio should be positioned
close to the shipping point
along the most direct route
between the entrance &
departure points, while the
most popular ones with the
largest ratio close to the
receiving point.
Material storage by popularity
Example (Book 4 Facility Planning p.428)
The below table shows the most popular products in a warehouse. How
should these items be aligned along the main aisle?
Pr.
Quantity per receipt
Trips to receive
Average customer
order size
Trips to ship
A
40 pallets
40
1.0 pallet
40
B
100 pallets
100
0.4 pallet
250
C
800 cartons
200
2.0 cartons
400
D
30 pallets
30
0.7 pallet
43
E
10 pallets
10
0.1 pallet
100
F
200 cartons
67
3.0 cartons
67
G
1000 cartons
250
8.0 cartons
125
H
1000 cartons
250
4.0 cartons
250
Receive
Main aisle
Ship
Warehouse layout
Receiving/Shipping ratio = trips to receive/trips to ship
Assoc. Prof. Dr Ho Thi Thu Hoa
33
Assoc. Prof. Dr Ho Thi Thu Hoa
34
Chapter 4: Warehouse Networking
1. Introduction
2. Model classification
3. Rectilinear-distance facility location models
Euclidean-distance facility location models
Covering problems
4. Exercises
35
Example 1
Five warehouses currently have the following cooordinate locations: �1 =
1, 1 , �2 = 6, 2 , �3 = 2, 8 , �4 = 3, 6 , �5 = 8, 4 . The cost per unit
distance traveled between the new warehouse and each existing warehouse is
the same. The number of trips per month between the new warehouse and
existing warehouse 1, …, 5 are 10, 20, 25, 20, and 25, respectively.
a. Find the optimum location for a new warehouse.
b. If it is no possible to locate the new warehouse in the optimum location.
Instead, there are three feasible locations: �1 = 3, 5 , �2 = 4, 5 ,
and �3 = 2, 4 . Which is preferred?
Example 1 (1 of 3)
a. Solving for the optimum x-coordinate
�
Warehouse �
Coordinate ��
Weight ��
10
�=�
3
2
25
35 < 50
10+25
4
3
20
55 > 50
35+20
2
6
20
75
55+20
5
8
25
100
75+25
1
1
��
10
- Ordering the x-coordinates of the existing facilities give the sequence
1, 2, 3, 6, and 8, with the corresponding weights 10, 25, 20, 20, and 25.
- The sum of the weights is 100. The partial sum of the ordered sequence of weights
first equal or exceeds one-half the total (50) for � = 4 �∗ = �4 = 3.
Example 1 (2 of 3)
Warehouse �
Solving for the optimum y-coordinate: Likewise, �∗ = �5 = 4.
�
Coordinate ��
Weight ��
10
�=�
2
2
20
30 < 50
5
4
25
55 > 50
4
6
20
75
3
8
25
100
1
1
��
10
Assoc. Prof. Dr Ho Thi Thu Hoa
39
Example 1 (2 of 3)
b. Computing the value of � � for � = �� , � = 1, 2, 3 yeilds:
� 3, 5 = 60 + 120 + 100 + 20 + 150 = 450
� 4, 5 = 70 + 100 + 125 + 40 + 125 = 460
� 2, 4 = 40 + 120 + 100 + 60 + 150 = 470
 The best site is �1
Example 1 (3 of 3)
�3
�1
�4
X
�2
�5
Using Excel to solve the single-facility, rectilinear location problem
Good luck for Mid-term Test!
42