# Group-1 Sec-A OMDM (1)

```OPTIMISATION MODEL IN DECISION MAKING
PROJECT
(2021)
TOPIC
UNBALANCED TRANSPORTATION PROBLEM
WITH MINIMIZATION
SUBMITTED TO:MR.ALOK KUMAR
\
SUBMITTED BY:
GROUP 1:
ALBHVYA ACKSHATT 301007
AMAN NAHATA 301009
ANVEET JOSHI 301013
GAURAV SINGH RAWAT 301022
SAHIL 301046
Introduction
Transportation problem is a particular class of linear programming, which is associated with day to-day activities in our real life and mainly deals with logistics. It helps in solving problems on
distribution and transportation of resources from one place to another. The goods are transported
from a set of sources (e.g., factory) to a set of destinations (e.g., warehouse) to meet the specific
requirement. There is a type of linear programming problem that may be solved using a simplified
version of the simplex technique called transportation method. Because of its major application in
solving problems involving several product sources and several destinations of products, this type
of problem is frequently called the transportation problem. It gets its name from its application to
problems involving transporting products from several sources to several destinations. Although
the formation can be used to represent more general assignment and scheduling problems as well
as transportation and distribution problems.
The two common objectives of such problems are either
(1) minimize the cost of shipping m units to n destinations or
(2) maximize the profit of shipping m units to n destinations.
A transportation problem is said to be unbalanced if the supply and demand are not equal
Problem Statement
Guinness Ghana LTD is one of the top five worldwide brewery companies. Transportation cost
represents about 25% of the total production cost. The company has outsourced its transportation
to external logistics services Providers. Guinness Ghana Ltd has registered about 20 transporters
who operate with 97 trucks. Each of the plants at the various sites namely Achimota and Kaasi has
its own constraint with respect to plant and warehouse capacity. Thus, there is a limited capacity
at each plant. The total plant capacity for Kaasi (S1) and Achimota( S2) per day is 1736 and 2419
crates depending on the plant efficiency. This project is intended to minimize the total
transportation cost from two production sites namely Kaasi (Kumasi), and Achimota (Accra) to its
numerous key distributors geographically scattered all over Ghana. As each site has its limit that
supply and each customer a certain demand at a time. The required data includes: A list of all
products, sources, demand for each product by customer, the full truck transportation cost,. The
study concerned the supply of Malta Guinness from two production sites Kaasi and Achimota to
8 key distributors geographically scattered in the regions of Ghana. The study covered data
gathered on the periods July07-June08, and Sep 08-June09. The demand for each destination was
also known in advance in table 1.1. The data included the transport cost per truck from each source
to each destination. For example- the transportation cost from Kaasi to D1 is 90 and Kaasi to D2
is 88 and so on. All cost with total demand and supply is given in table 1.1
Unbalanced Transportation - Manual Method
Table 1.1
D1
D2
D3
D4
D5
D6
D7
D8
Supply
S1
90
88
82
69
30
424
30
13
1736
S2
228
37
176
73
114
173
73
38
2419
576
445
335
272
431
304
120
Demand 907
Total demand = 3390
Total supply= 4155
As total demand is not equal to total supply therefore it is a case of unbalanced transportation
problem. Thus to make it balanced we are introducing dummy column D9.
So the balanced table is-
D1
D2
D3
D4
D5
D6
D7
D8
D9
Supply
S1
90
88
82
69
30
424
30
13
0
1736
S2
228
37
176
73
114
173
73
38
0
2419
576
445
335
272
431
304
120
0
Demand 907
To get initial basic feasible solution we are using Vogel’s approximation method (VAM)
S1
D1
90
D2
88
D3
82
445
D4
69
D5
30
272
D6
424
D7
30
D8
13
D9
0
Supply
1736
S2
907
228
37
176
73
114
173
112
73
38
0
2419
907
576
576
445
335
335
272
431
431
192
304
120
120
765
765
Demand
138
138
-
51
51
51
51
51
-
94
94
94
-
4
4
4
4
4
4
84
84
84
84
-
258
-
43
43
43
43
43
43
25
25
25
25
25
25
13
13
13
13
37
37
37
37
0
0
0
0
0
0
13
13
37
38
Initial basic feasible solution =
90*907+ 37*576+82*445+73*335+30*272+173*431+30*112+73*192+38*120+0*765
= 2,68,546
Total no of allocation = 10
No of rows(m)= 2
No of column(n)=9
So total allocation= m+n-1= 10
To further improve the solution, we used MODI Metho
Step1The most number of allocation are in row 2 therefore we are taking value of U2=0
Step2Now calculating all Vj and Ui by using allocated cells i.e Cij values
Cij= Ui+Vj
C22= U2+V2
37=0+V2
Therefore= V2=37
Similarly, calculating all the values of Vj and Ui
U1=-43
V1= 115
V6= 173
V2= 37
V7= 73
V3= 107
V8= 38
V4= 73
V9= 0
V5=55
D2
88
S2
D1
90
907
228
Demand
907
576
S1
Vij
37
576
D3
82
445
176
D4
69
445
335
73
335
D5
30
272
114
D6
424
D8
13
D9
0
Supply Uij
1736
U1=-43
173
431
D7
30
112
73
192
38
120
0
765
2419
272
431
304
120
765
V1=133 V2=37 V3=125 V4=73 V5=73 V6=173 V7=73 V8=38 V9=0
U2=0
Calculating Dij values from unoccupied cellsDij=Cij-(Ui+Vj)
D12= +81
D14= +26
D16= +281
D18= +5
D21=+90
D23=+23
D25=+41
Since all Dij values are positive thus the optimality condition is satisfied and the solution is optimal
solution.
Final optimal tableD2
88
S2
D1
90
907
228
Demand
907
576
S1
37
576
D3
82
445
176
D4
69
445
335
73
335
D5
30
272
114
D6
424
D8
13
D9
0
Supply
1736
173
431
D7
30
112
73
192
38
120
0
765
2419
272
431
304
120
765
The minimized cost is-
90*907+ 37*576+82*445+73*335+30*272+173*431+30*112+73*192+38*120+0*765
= 2,68,546
Solution in Excel with Sensitivity Report
Let 1 Y =plant site at ACH
2 Y =plant site at KAS
IJ X = the units shipped in crates from plant i to distribution centre j
i =1, 2, 3... 9. and j =1, 2, 3..., 9.
Objective function to Minimize cost of Transportation from Achitoma and Kassi plant
90x11 + 88x12 + 82x13 + 69x14 + 30x15 + 424x16 + 30x17 + 13x18 + 0x19
+ 228x21 + 37x22+ 176x23+ 73x24 + 114x25
+ 173x26+ 73x27 + 38x28 +0x29
Consider capacity constraint
x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 ≤ 1736
x21 + x22 + x23 + x24 + x25 + x26 + x27 + x28 + x29 ≤ 2419
Demand constraint
x11 + x21 =907
x12 + x22 =576
x13 + x23 =445
x14 + x24 =335
x15 + x25 =272
x16 + x26 =431
x17 + x27 =304
x18+ x28 =120
x19 + x29 =765
xij ≥ o for all i and j
Z(Objective
function)
268546
Achimota
Kaasi
Demand
D1
90
228
907
D2
88
37
576
D3
82
176
445
D4
69
73
335
D5
30
114
272
D6
424
173
431
D7
30
73
304
D8
13
38
120
D9
0
0
765
xij
Achitoma
Kaasi
907
0
0
576
445
0
0
335
272
0
0
431
112
192
0
120
0
765
907
=
907
576
=
576
445
=
445
335
=
335
272
=
272
431
=
431
304
=
304
120
=
120
765
=
765
Supply
1736
2419
1736
2419
=
=
1736
2419
D1
Achimota 90
Kaasi
228
Demand
907
D2
88
37
576
D3
82
176
445
D4
69
73
335
D5
30
114
272
D6
424
173
431
D7
30
73
304
D8
13
38
120
D9
0
0
765
Supply
1736
2419
x
i
j
907
0
445
0
272
0
112
0
0
0
576
0
335
0
431
192
120
765
=SU
M(C1
5:C1
6)
=
907
=SU
M(D1
5:D1
6)
=
576
=SU
M(E1
5:E1
6)
=
445
=SU
M(F1
5:F16
)
=
335
=SU
M(G1
5:G1
6)
=
272
=SU
M(H1
5:H1
6)
=
431
=SU
M(I1
5:I16
)
=
304
=SU
M(J1
5:J16
)
=
120
=SU
M(K1
5:K1
6)
=
765
Cell
\$C\$15
\$D\$15
\$E\$15
\$F\$15
\$G\$15
\$H\$15
\$I\$15
\$J\$15
\$K\$15
\$C\$16
\$D\$16
\$E\$16
\$F\$16
Final
Name Value
907
D2
0
D3
445
D4
0
D5
272
D6
0
D7
112
D8
0
D9
0
0
D2
576
D3
0
D4
335
Reduced
Cost
0
94
0
39
0
294
0
18
43
95
0
51
0
Objective
Coefficient
90
88
82
69
30
424
30
13
0
228
37
176
73
Allowable
Increase
95
1E+30
51
1E+30
41
1E+30
18
1E+30
1E+30
1E+30
94
1E+30
39
=SU
= 1
M(C1
7
5:K1
3
5)
6
=SU
= 2
M(C1
4
6:K1
1
6)
9
Allowable
Decrease
1E+30
94
1E+30
39
1E+30
294
41
18
43
95
1E+30
51
1E+30
\$G\$16
\$H\$16
\$I\$16
\$J\$16
\$K\$16
D5
D6
D7
D8
D9
0
431
192
120
765
41
0
0
0
0
114
173
73
38
0
TRANSPORTATION OUTPUT TABLE
PLANT
DISTRIBUTO FULLTRUCK
SITE(SOURC R(DESTINAT PER
E)
ION)
CASE(0000)
ACH
ACH
ACH
D1= FTA
907
D2= RICKY
0
D3=OBIBA
445
JK
ACH
0
ACH
D5=NAATO
272
ACH
D6=LESK
0
ACH
D7=DCEE
112
ACH
D8=JOEMA
0
ACH
D9=KBOA
0
KAS
FTA
0
KAS
RICKY
576
KAS
OBIBA JK
0
KAS
335
KAS
NAATO
0
KAS
LESK
431
KAS
DCEE
192
KAS
JOEMA
120
KAS
KBOA
765
The total transportation cost is ,: 268546
1E+30
294
41
18
43
41
1E+30
18
1E+30
1E+30
COST
PER TOTAL
FULL
COST(GH
TRUCK
000)
90
81630
88
0
82
36490
69
30
424
30
13
0
228
37
176
73
114
173
73
38
0
0
8160
0
3360
0
0
0
21312
0
24455
0
74563
14016
4560
0
Interpretation
The GGL problem was solved with the linear programming module and the transportation module
of The Management Scientist. The results from both the linear programming module and that of
the transportation module of The Management Scientist yielded the same values, in terms of the
optimal solution obtained. The excel solution shows the minimum total transportation cost is
268546. The value for the decision variables shows the optimal amount of drinks to be ship over
each route.
To minimize the transportation cost the management of Guinness Ghana Ltd should make the
following shipments:
Ship 907,000 cases of Malta Guinness from Plant ACH to Distributor FTA
Ship 445,000 cases of Malta Guinness from Plant ACH to distributor OBIBA JK
Ship 272,000 cases of Malta Guinness from Plant ACH to distributor NAATO
Ship 112,000 cases of Malta Guinness from Plant ACH to distributor DCEE
Ship 576,000 cases of Malta Guinness from Plant KAS to distributor RICKY
Ship 335,000 cases of Malta Guinness from Plant KAS to distributor KADOM
Ship 431,000 cases of Malta Guinness from Plant KAS to distributor LESK
Ship 192,000 cases of Malta Guinness from Plant KAS to distributor DCEE
Ship 120,000 cases of Malta Guinness from Plant KAS to distributor JOEMAN
Ship 765,000 cases of Malta Guinness from Plant KAS to distributor KBOA
Managerial Implication
Cost Minimization: The transportation problem involves the distribution of goods from the
manufacturing units to the warehouse, retailers, or the customers. The minimization of cost of
supplying through optimum solution will help the manager to reduce the expenditure and thus
increase the cash balance to the company which can be further used for the operations of the
company distribution.
This will also help to reduce the wastage and extra cost expenditure which is a liability to the
company
Flexible Practical Solution to the problem: The range found through sensitivity analysis will
help the manager to take flexible plausible solutions which are to be taken in a practical situation
thus avoiding any rigid solution which cannot be implemented on scale.
Conclusion
The minimization of cost to transport units from Achitoma plant and Kassi plant is done by finding
the optimal solution through both excel method and manual method of Transportation.
The transportation problem was formulated as a Linear Programming and solved with the standard
LP solvers such as the Management scientist module to obtain the optimal solution.
The sensitivity analysis was done to find the range where there will be no effect on the optimum
solution of the problem. The transportation cost is an important element of the total cost structure
Through the use of this Transportation Model the business (GGBL) can identify easily and
efficiently plan out its transportation, so that it not only minimizes the cost of transporting goods
and services but also create time utility by reaching the goods and services at the right place at
right time. This intend will enable them to meet the corporative objective such as education fund,
entertainment and other support they offered to people of Ghana.
```