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Edurev123
6. Transportation Problems
6.1 Determine an optimal transportation programme so that transportation cost of
340 tons of a certain type of material from three factories ?? ?? ,?? ?? ,?? ?? to five
warehouses ?? ?? ,?? ?? ,?? ?? ,?? ?? ,?? ?? is minimized. The five warehouses must receive 40
tons, 50 tons, 70 tons, 90 tons and 90 tons respectively. The availability of the
material at ?? ?? ,?? ?? ,?? ?? is 100 tons, 120 tons, 120 tons respectively. The
transportation costs per ton from factories to warehouses are given in the table
below :
?? ?? ?? ?? ?? ?? ?? ?? ?? ??
?? ?? 4 1 2 6 9
?? ?? 6 4 3 5 7
?? ?? 5 2 6 4 8
Use Vogel's approximation method to obtain the initial basic feasible solution.
(2010 : 30 Marks)
Solution:
Given : ? Supply =340 tons
Demand =40+50+70+90+90=340 tons
? Problem is balanced. Finding initial basic feasible solution by Vogel's approximation
method:
Page 2
Edurev123
6. Transportation Problems
6.1 Determine an optimal transportation programme so that transportation cost of
340 tons of a certain type of material from three factories ?? ?? ,?? ?? ,?? ?? to five
warehouses ?? ?? ,?? ?? ,?? ?? ,?? ?? ,?? ?? is minimized. The five warehouses must receive 40
tons, 50 tons, 70 tons, 90 tons and 90 tons respectively. The availability of the
material at ?? ?? ,?? ?? ,?? ?? is 100 tons, 120 tons, 120 tons respectively. The
transportation costs per ton from factories to warehouses are given in the table
below :
?? ?? ?? ?? ?? ?? ?? ?? ?? ??
?? ?? 4 1 2 6 9
?? ?? 6 4 3 5 7
?? ?? 5 2 6 4 8
Use Vogel's approximation method to obtain the initial basic feasible solution.
(2010 : 30 Marks)
Solution:
Given : ? Supply =340 tons
Demand =40+50+70+90+90=340 tons
? Problem is balanced. Finding initial basic feasible solution by Vogel's approximation
method:
50 at ?? 3
?? 2
;70 at ?? 1
?? 3
;30 at ?? 1
?? 1
;70 at ?? 3
?? 4
;10 at ?? 2
?? 1
;20 at ?? 2
?? 4
;90 at ?? 2
?? 5
So, initial basic feasible solution is :
No. of rows =?? =3
No. of columns =?? =5
??????????????????????????????????????????????????? +?? -1=3+5-1=7= No. of solutions
So, given solution is initial basic feasible solution.
Cost Optimization : For this, we will use u-v method.
iteration 1 :
?? 1
?? 2
?? 3
?? 4
???
5
30 (0) 70 (-3) (-4) -2
10 (-1) (1) 20 90 0
(0) 50 (-3) 70 (-2) -1
6 3 4 5 7
Now,
?? 2
+?? 7
-6=0??? 1
=6
?? 2
+?? 4
-5=0??? 4
=5
?? 2
+?? 5
-7=0??? 5
=7
?? 4
+?? 3
-4=0?5+?? 3
-4=0??? 3
=-1
?? 3
+?? 2
-2=0?-1+?? 2
-2=0??? 2
=3
?? 1
+?? 1
-4=0?6+?? 1
-?? =0??? 1
=-2
?? 1
+?? 3
-2=0?-2+?? 3
-2=0??? 3
=4
Calculate ?
????
for every blank cell where
?
????
=?? ?? +?? ?? -?? ????
Page 3
Edurev123
6. Transportation Problems
6.1 Determine an optimal transportation programme so that transportation cost of
340 tons of a certain type of material from three factories ?? ?? ,?? ?? ,?? ?? to five
warehouses ?? ?? ,?? ?? ,?? ?? ,?? ?? ,?? ?? is minimized. The five warehouses must receive 40
tons, 50 tons, 70 tons, 90 tons and 90 tons respectively. The availability of the
material at ?? ?? ,?? ?? ,?? ?? is 100 tons, 120 tons, 120 tons respectively. The
transportation costs per ton from factories to warehouses are given in the table
below :
?? ?? ?? ?? ?? ?? ?? ?? ?? ??
?? ?? 4 1 2 6 9
?? ?? 6 4 3 5 7
?? ?? 5 2 6 4 8
Use Vogel's approximation method to obtain the initial basic feasible solution.
(2010 : 30 Marks)
Solution:
Given : ? Supply =340 tons
Demand =40+50+70+90+90=340 tons
? Problem is balanced. Finding initial basic feasible solution by Vogel's approximation
method:
50 at ?? 3
?? 2
;70 at ?? 1
?? 3
;30 at ?? 1
?? 1
;70 at ?? 3
?? 4
;10 at ?? 2
?? 1
;20 at ?? 2
?? 4
;90 at ?? 2
?? 5
So, initial basic feasible solution is :
No. of rows =?? =3
No. of columns =?? =5
??????????????????????????????????????????????????? +?? -1=3+5-1=7= No. of solutions
So, given solution is initial basic feasible solution.
Cost Optimization : For this, we will use u-v method.
iteration 1 :
?? 1
?? 2
?? 3
?? 4
???
5
30 (0) 70 (-3) (-4) -2
10 (-1) (1) 20 90 0
(0) 50 (-3) 70 (-2) -1
6 3 4 5 7
Now,
?? 2
+?? 7
-6=0??? 1
=6
?? 2
+?? 4
-5=0??? 4
=5
?? 2
+?? 5
-7=0??? 5
=7
?? 4
+?? 3
-4=0?5+?? 3
-4=0??? 3
=-1
?? 3
+?? 2
-2=0?-1+?? 2
-2=0??? 2
=3
?? 1
+?? 1
-4=0?6+?? 1
-?? =0??? 1
=-2
?? 1
+?? 3
-2=0?-2+?? 3
-2=0??? 3
=4
Calculate ?
????
for every blank cell where
?
????
=?? ?? +?? ?? -?? ????
?? 2
?? 3
has positive ?
????
value. Adding ?? in ?? 2
?? 3
and subtracing ? in alternate cells as per
?? -?? method, we get
Iteration 2 :
Now,
?? 2
+?? 3
-3=0 ? ?? 3
=3
?? 2
+?? 4
-5=0 ? ?? 4
=5
?? 2
+?? 5
-7=0 ? ?? 5
=7
?? 3
+?? 1
-2=0 ? ?? 1
=-1
?? 4
+?? 3
-4=0 ? ?? 3
=-1
?? 1
+?? 1
-1=0 ? ?? 1
=5
?? 2
+?? 3
-2=0 ? ?? 2
=3
Calculating ?
????
for blank cells, we get positive value at ?? 1
?? 2
Adding ?? in this ceil as per ?? -?? method, we get
Page 4
Edurev123
6. Transportation Problems
6.1 Determine an optimal transportation programme so that transportation cost of
340 tons of a certain type of material from three factories ?? ?? ,?? ?? ,?? ?? to five
warehouses ?? ?? ,?? ?? ,?? ?? ,?? ?? ,?? ?? is minimized. The five warehouses must receive 40
tons, 50 tons, 70 tons, 90 tons and 90 tons respectively. The availability of the
material at ?? ?? ,?? ?? ,?? ?? is 100 tons, 120 tons, 120 tons respectively. The
transportation costs per ton from factories to warehouses are given in the table
below :
?? ?? ?? ?? ?? ?? ?? ?? ?? ??
?? ?? 4 1 2 6 9
?? ?? 6 4 3 5 7
?? ?? 5 2 6 4 8
Use Vogel's approximation method to obtain the initial basic feasible solution.
(2010 : 30 Marks)
Solution:
Given : ? Supply =340 tons
Demand =40+50+70+90+90=340 tons
? Problem is balanced. Finding initial basic feasible solution by Vogel's approximation
method:
50 at ?? 3
?? 2
;70 at ?? 1
?? 3
;30 at ?? 1
?? 1
;70 at ?? 3
?? 4
;10 at ?? 2
?? 1
;20 at ?? 2
?? 4
;90 at ?? 2
?? 5
So, initial basic feasible solution is :
No. of rows =?? =3
No. of columns =?? =5
??????????????????????????????????????????????????? +?? -1=3+5-1=7= No. of solutions
So, given solution is initial basic feasible solution.
Cost Optimization : For this, we will use u-v method.
iteration 1 :
?? 1
?? 2
?? 3
?? 4
???
5
30 (0) 70 (-3) (-4) -2
10 (-1) (1) 20 90 0
(0) 50 (-3) 70 (-2) -1
6 3 4 5 7
Now,
?? 2
+?? 7
-6=0??? 1
=6
?? 2
+?? 4
-5=0??? 4
=5
?? 2
+?? 5
-7=0??? 5
=7
?? 4
+?? 3
-4=0?5+?? 3
-4=0??? 3
=-1
?? 3
+?? 2
-2=0?-1+?? 2
-2=0??? 2
=3
?? 1
+?? 1
-4=0?6+?? 1
-?? =0??? 1
=-2
?? 1
+?? 3
-2=0?-2+?? 3
-2=0??? 3
=4
Calculate ?
????
for every blank cell where
?
????
=?? ?? +?? ?? -?? ????
?? 2
?? 3
has positive ?
????
value. Adding ?? in ?? 2
?? 3
and subtracing ? in alternate cells as per
?? -?? method, we get
Iteration 2 :
Now,
?? 2
+?? 3
-3=0 ? ?? 3
=3
?? 2
+?? 4
-5=0 ? ?? 4
=5
?? 2
+?? 5
-7=0 ? ?? 5
=7
?? 3
+?? 1
-2=0 ? ?? 1
=-1
?? 4
+?? 3
-4=0 ? ?? 3
=-1
?? 1
+?? 1
-1=0 ? ?? 1
=5
?? 2
+?? 3
-2=0 ? ?? 2
=3
Calculating ?
????
for blank cells, we get positive value at ?? 1
?? 2
Adding ?? in this ceil as per ?? -?? method, we get
Iteration 3 :
Applying same method, we observe that none of the blank cell has ?
????
>0.
? This is the optimal solution, for minimum
Cost of transportation.
6.2 Find the initial basic feasible solution to the following transportation problem
by Vogel's approximation method. Also, find its optimal solution and the minimum
transportation cost :
(2014: 20 marks)
Solution:
Since the total supply and total demand being equal.
? The transportation problem is balanced.
Find the initial basic feasible solution:
Page 5
Edurev123
6. Transportation Problems
6.1 Determine an optimal transportation programme so that transportation cost of
340 tons of a certain type of material from three factories ?? ?? ,?? ?? ,?? ?? to five
warehouses ?? ?? ,?? ?? ,?? ?? ,?? ?? ,?? ?? is minimized. The five warehouses must receive 40
tons, 50 tons, 70 tons, 90 tons and 90 tons respectively. The availability of the
material at ?? ?? ,?? ?? ,?? ?? is 100 tons, 120 tons, 120 tons respectively. The
transportation costs per ton from factories to warehouses are given in the table
below :
?? ?? ?? ?? ?? ?? ?? ?? ?? ??
?? ?? 4 1 2 6 9
?? ?? 6 4 3 5 7
?? ?? 5 2 6 4 8
Use Vogel's approximation method to obtain the initial basic feasible solution.
(2010 : 30 Marks)
Solution:
Given : ? Supply =340 tons
Demand =40+50+70+90+90=340 tons
? Problem is balanced. Finding initial basic feasible solution by Vogel's approximation
method:
50 at ?? 3
?? 2
;70 at ?? 1
?? 3
;30 at ?? 1
?? 1
;70 at ?? 3
?? 4
;10 at ?? 2
?? 1
;20 at ?? 2
?? 4
;90 at ?? 2
?? 5
So, initial basic feasible solution is :
No. of rows =?? =3
No. of columns =?? =5
??????????????????????????????????????????????????? +?? -1=3+5-1=7= No. of solutions
So, given solution is initial basic feasible solution.
Cost Optimization : For this, we will use u-v method.
iteration 1 :
?? 1
?? 2
?? 3
?? 4
???
5
30 (0) 70 (-3) (-4) -2
10 (-1) (1) 20 90 0
(0) 50 (-3) 70 (-2) -1
6 3 4 5 7
Now,
?? 2
+?? 7
-6=0??? 1
=6
?? 2
+?? 4
-5=0??? 4
=5
?? 2
+?? 5
-7=0??? 5
=7
?? 4
+?? 3
-4=0?5+?? 3
-4=0??? 3
=-1
?? 3
+?? 2
-2=0?-1+?? 2
-2=0??? 2
=3
?? 1
+?? 1
-4=0?6+?? 1
-?? =0??? 1
=-2
?? 1
+?? 3
-2=0?-2+?? 3
-2=0??? 3
=4
Calculate ?
????
for every blank cell where
?
????
=?? ?? +?? ?? -?? ????
?? 2
?? 3
has positive ?
????
value. Adding ?? in ?? 2
?? 3
and subtracing ? in alternate cells as per
?? -?? method, we get
Iteration 2 :
Now,
?? 2
+?? 3
-3=0 ? ?? 3
=3
?? 2
+?? 4
-5=0 ? ?? 4
=5
?? 2
+?? 5
-7=0 ? ?? 5
=7
?? 3
+?? 1
-2=0 ? ?? 1
=-1
?? 4
+?? 3
-4=0 ? ?? 3
=-1
?? 1
+?? 1
-1=0 ? ?? 1
=5
?? 2
+?? 3
-2=0 ? ?? 2
=3
Calculating ?
????
for blank cells, we get positive value at ?? 1
?? 2
Adding ?? in this ceil as per ?? -?? method, we get
Iteration 3 :
Applying same method, we observe that none of the blank cell has ?
????
>0.
? This is the optimal solution, for minimum
Cost of transportation.
6.2 Find the initial basic feasible solution to the following transportation problem
by Vogel's approximation method. Also, find its optimal solution and the minimum
transportation cost :
(2014: 20 marks)
Solution:
Since the total supply and total demand being equal.
? The transportation problem is balanced.
Find the initial basic feasible solution:
Using Vogel's approximation method, the initial basic feasible solution is :
The difference between the smallest and next to the smallest cost in each row and each
column are first computed and displayed inside paranthesis against the respective rows
and columns.
The largest of these differences is (5) which is associated with the 2nd row.
Since ?? 23
=2, is the minimum cost, we allocate ?? 23
=min(15,16)=15 in the cell (2,3) .
This exhausts the demand of the 3rd column and therefore we cross it.
Proceeding in this way, the subsequent reduced transportation tables and differences for
the remaining rows and columns are as shown below :
? The initial basic feasible solution is
Now, find the values of ?? ?? and ?? ?? :
As the maximum number of basic cells exist in the first row.
? Let
?? 1
=0
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