List no 1

LIST 1
EXERCISES of "LOGISTICS" COURSE
1. ABC Classification
There are 10 materials in a stock of warehouse of different quantity in [ton] and of different
economic value [zł/ton] (table 1). Make ABC classification.
Suggestion: Apply the following classification structure of ABC using cumulated quantity share and
cumulated value share of materials on inventory (stock): A class – 20%/80%, B class – 50%/95%, C
class – 100%/100%. To verify statistical concentration of distribution, the following ratio coefficients can
be calculated: V(A)=CVA/CQA, V(B)=(CVB-CVA)/(CQB-CQA), and V(C)=(CVC-CVB)/(CQC-CQB). According
to a recommendation: V(A)>3, 0.7≤V(B)≤3, V(C)<0.7.
Table 1.
Material
Inventory [ton]
Price [zł/ton]
Value [zł]
Order (Priority)
1
2
3
4
5
6
7
8
9
10
Table 2.
Material
8
10
5
5
10
15
12
20
5
10
 100
1.00
0.30
8.00
0.20
0.10
2.00
0.50
0.20
0.40
0.50
 100.00
Cumulated quantity share CQ
[%]
Cumulated value share CV
[%]
A, B, C class
and its value share
Solution:
Table 1.
Material
1
2
3
4
5
6
7
8
9
10
Inventory [ton]
Price [zł/ton]
Value [zł]
Order
8
10
5
5
10
15
12
20
5
10
 100
1.00
0.30
8.00
0.20
0.10
2.00
0.50
0.20
0.40
0.50
8.00=8*1.00
3.00
40.00
1.00
1.00
30.00
6.00
4.00
2.00
5.00
 100.00
3
7
1
9
10
2
4
6
8
5
1
Table 2.
Material
Cumulated quantity share [%]
Cumulated value share [%]
A, B, C class
and its value share
5=(5/100)*100%
20=5+15
28
40
50
70
80
85
90
100
40=(40.00/100.00)*100%
70=40+30
78
84
89
93
96
98
99
100
A
(70%)
3
6
1
7
10
8
2
9
4
5
B
(19%)
C
(11%)
2. ABC/XYZ Classification, JIT
There are 10 material items in a stock of warehouse (table 1) in different quantity and value
(monthly usage). Make ABC/XYZ classifications (table 2) and determine (select) materials for
pilot JIT (Just in Time) programme.
Suggestion: Apply the following classification structure of ABC using cumulated quantity share and
cumulated value share of materials on inventory (stock): A class – 20%/80%, B class – 50%/95%, C
class – 100%/100%. Apply the following classification structure of XYZ with usage deviation around
average usage: X class – deviation up to 20%, Y – usage deviation from 20% to 50%, Z – usage
deviation above 50%.
Table 1.
Table 2.
Material
M1
M2
M3
M4
M5
M6
M7
M8
M9
M10
Monthly
usage
[zł]
Usage
deviation
[zł]
Quantity
share
[%]
60
55
300
150
70
45
250
20
10
40
 1000
± 20
± 10
± 40
± 30
± 10
± 30
± 110
± 5
± 10
± 35
12.0
10.0
7.0
6.0
7.0
15.0
5.0
13.0
15.0
10.0
 100
Material
Value share
[%]
Cumulated
value share
[%]
Cumulated
quantity share
[%]
ABC/XYZ
Materials for JIT (AX): …………..
3. Changing number of stores (stocks)
In modern logistics there are tendencies to reduce intermediate elements of supply chains. As
an example, to reduce logistics cost by reducing number of stocks it can be applied to inventory
systems in distribution of products. While the transport cost is increasing the overall logistics
cost is decreasing. Calculate and explain effects of changing the number of stocks (table 1).
Suggestion: Decreasing the total inventory in logistics system by decreasing number of stocks is
described by the "square root principle" (Z2/Z1 = L2/L1):
Z2
L2

Z1
L1
where: Z1 – inventory prior the change of number of stocks, Z2 - inventory after the change of number of
stocks, L1 – number of stocks prior the change of number of stocks, L2 - number of stocks after the
2
change of number of stocks Therefore changing the number of stocks the following relative inventory
change can be observed:
Z1  Z2
L2
.
 1
Z1
L1
As an example, decreasing the number of stocks from L1=16 to L2=9, the inventory reduction is:
Z1  Z2
L2
9
3
 1
 1
 1   0. 25.
Z1
L1
16
4
It means that decreasing number of stocks from 16 to 9 (by 43.75%) a reduction of inventory by 25% is
observed.
Table 1.
1
L2 (after change)
3
4
2
5
6
1
2
L1
3
4
5
6
4. Equilibrium of market gravitation in distribution
Determine equilibrium of market gravitation in distribution of goods in 2 cities: Wrocław and
Jelenia Góra. Population of the cities is: Wrocław – 750.000, Jelenia Góra – 250.000, and the
distance between Wrocław and Jelenia Góra is: 125 km.
Suggestion: Demand from customers is straight proportional to customers’ population and opposite
proportional to a distance square between customers and supply location.
A
C
B
MA
MB
LCA
Z CA M A

Z CB M B
LCB
L 
  CB 
 LCA 
2
L AB
O AB 
1
MB
MA
ZCA – demand in A point by customers from C point, ZCB - demand in B point by customers from C point,
MA – population in A point, MB – population in B point, LCA – distance from C point to A point, LCB –
distance from C point to B point.
Equilibrium of demand (neutral area) as an influence of city A to city OAB: …
3
5. Locational break-even analysis
Solve the problem of automobile clutch factory location by break-even analysis for the following
data about manufacturing costs, product prices and expected production volumes (table 1).
What location is to be preferred when production volume is: a) less than 2000 units, b) greater
than 2000 units?
Table 1.
Factory location
Location A
Location B
Location C
Fixed cost in €
per year
30.000
60.000
110.000
Variable cost in €
per year/unit
75
45
25
Selling price in €
per unit
120
120
120
Expected volume
in units/year
2000
2000
2000
Suggestion: Determine the fixed and variable cost for each location, plot the costs for each location, with
costs on the vertical axis of the graph and annual volume on the horizontal axis, select the location that
has the lowest total cost for the expected production volume.
Total cost for location A =
Total cost for location B =
Total cost for location C =
Expected profit for selected location =
Location for production volume less than 2000 is …
Location for production volume greater than 2000 is …
6. Distribution centre location
Solve the problem of distribution centre location for the following data about suppliers and
customers’ (demand) locations (table 1).
Table 1.
Supplier/Customer
Supplier 1
Supplier 2
Customer 1
Customer 2
Customer 3
Customer 4
Location coordinates
(5, 2)
(4, 7)
(2, 4)
(7, 1)
(0, 0)
(3, 9)
Distributed quantity [units]
5000
3500
1900
1400
2700
2500
Data: (xi, yi) – supply points coordinates, (ui, vi) - customer points coordinates, pi – supply from i supplier,
qi – demand from i customer, ri – transport cost from i supplier to the centre, Ri – transport cost from
centre to j customer (ri=1, Ri=1). Distribution centre coordinates (x, y) are calculated by formulas:
Centre
Suppliers
r  p
i
x
(x,y)
i
i
j
r  p  R
i
i
i
Customers
(xi,yi)
 xi   R j  q j  u j
r  p
i
(uj, vj)
y
i
i
x=
y=
Centre coordinates (x,y): …
4
 yi   R j  q j  v j
r  p  R
i
(pi,ri)
qj
j
i
(qj,Rj)
j
j
i
j
j
qj
7. Distribution centre location
Determine location of a distribution centre for the following data (table 1). Data on supply points:
Białystok - 15000 [units], Suwałki - 10000 [units], Ostrów Mazowiecki - 5000 [units]. Data on
demand points: Suwałki - 5000 [units], Augustów - 2300 [units], Ełk - 1200 [units], Grajewo 1000 [units], Łomża - 3400 [units], Białystok - 12000 [units], Zambrów - 1000 [units], Bielsk
Podlaski - 2500 [units]. Assume that transportation cost for all deliveries is equal.
Table 1.
Supply/Demand
points
Białystok
Suwałki
Ostrów Mazowiecki
Suwałki
Augustów
Ełk
Grajewo
Łomża
Białystok
Zambrów
Bielsk Podlaski
Total
Centre coordinates
Delivery [items]
15000
10000
5000
5000
2300
1200
1000
3400
12000
1000
2500
58400
Coordinates
horizontal
280
840
60
840
700
670
580
290
280
180
60
5
vertical
530
430
100
430
450
230
280
150
530
220
560
Calculations:
(coordinate·delivery)
horizontal
vertical