1. Transportation Problems
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2. What is a Transportation Problem ?
There are ‘m’ plant locations(Origins) and ‘n’ distribution centers(Destinations). The production capacity of the ith plant is ai and the number of units required at the jth destination is bj. The transportation cost of one unit from the ith plant to jth destination is cij. The problem is to determine the number of units to be transported from ith plant to jth destination so that the total transportation cost is minimum.
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3. Dr.G.SureshKumar@KLUniversity
Model as LPP
Minimize Z =
subject to:
xi1 + xi xin = ai (i = 1, 2, …, m)
x1j + x2j xmj = bj (j = 1, 2, …, n) or
and xij ≥ 0.
xij = number of units shipped from ith origin to jth destination.
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4. Dr.G.SureshKumar@KLUniversity
Transportation Table
Distribution Centers (Destinations)
Availability 1 2 j n
c11
c12
c1j
c1n
c21
c22
c2j
c2n
ci1
ci2
cij
cin
cm1
cm2
cmj
cmn 1 a1 2 a2
Plants
(Origins) i ai am m
Requirement b1 b2 bj bn
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5. Dr.G.SureshKumar@KLUniversity
Problem#1
Solve the following transportation problem
Source
Destinations
Availability I 21 16 25 13 11 II 17 18 14 23
III 33 27 41 19
Requirement 6 10 12 15 43
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6. Dr.G.SureshKumar@KLUniversity
Solution
Total number of availabilities = =43 Total number of requirement = = 43 Total number of availabilities = Total number of requirement = 43 Therefore, given TPP is balanced or Homogeneous TPP. Now, finding initial basic solution by Vogel’ method.
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7. Vogel’s approximation Method
Source
Destinations
Avail.
Differences (penalties) 1
--- 21 16 25 13
11 (X) 3 -- 2 17 18 14 23
13 x
( 9)(3) 4 33 27 41 19
(13)12 9
Req.
6 x
10(7)x 12
(15) 4x 10 15 11 6 3 4 7 12
Penalties 10 18 15 9 27
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8. Dr.G.SureshKumar@KLUniversity
Solution
The number of allocations = m + n – 1 = 6. Now we can apply MODI method.
Determine ui, vj from
dij = ui + vj – cij = 0 for basic cells.
i.e u1 + v4 = 13, u2 + v1 =17, u2 + v2 = 18,
u2 + v4 = 23, u3 + v2 = 27, u3 + v3 = 18.
putting u2 = 0, we have u1 = -10, u3 = 9,
v1 =17, v2 = 18, v3 = 9, v4 = 23.
Saturday, February 23, 2019Saturday, February 23, 2019

9. Dr.G.SureshKumar@KLUniversity
Solution
Computing dij = ui + vj – cij for non-basic cells. d11 = u1 + v1 – c11 = – 21 = -14 d12 = u1 + v2 – c12 = – 16 = -8 d13 = u1 + v3 – c13 = – 25 = -26 d23 = u2 + v3 – c23 = – 14 = -5 d31 = u3 + v1 – c31 = – 32 = -6 d34 = u3 + v4 – c34 = – 41 = -9
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10. Dr.G.SureshKumar@KLUniversity
Solution
All dij = ui + vj – cij ≤ 0. Therefore, the current basic solution is optimal. Hence the optimal solution is x14 = 11, x21 = 6, x22 = 3, x24 = 4, x32 = 7 and x33 = 12. The minimum transportation cost = 13x x6 + 18x3 + 23x4 + 27x7 + 18x12 = Rs 796.
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11. Dr.G.SureshKumar@KLUniversity
Assignment Problem
It is a special case of transportation problem. Here all availabilities and requirements are unity. Let m be the number of jobs as well as the operators, and tij be the processing time of the ith job if it is assigned to the jth operator. Here the objective is to assign the jobs to the operators such that the total processing time is minimized.
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12. Dr.G.SureshKumar@KLUniversity
Assignment Table
Operators
Job 1 2 … j m
t11
t12
t1j
t1m . i
ti1
tij
tim
tm1
tm2
tmj
tmm
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13. Dr.G.SureshKumar@KLUniversity
Mathematical Model
Xij = 1, for ith job assigned to jth operator
Xij = 0, otherwise.
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14. Dr.G.SureshKumar@KLUniversity
Hungarian Method
Step I : (A) Row reduction: Subtract the minimum entry of each row from all the entries of the respective row in the cost matrix. (B) Column reduction: After completion of row reduction, subtract the minimum entry of each column from all the entries of the respective column.
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15. Dr.G.SureshKumar@KLUniversity
Step II : Zero assignment
(A) Starting with first row of the matrix received in first step, examine the rows one by one until a row containing exactly one zero is found. Then an experimental
assignment indicated by ‘ ’ is marked to that zero.
Now cross all the zeros in the column in which the assignment is made.
This procedure should be adopted for each row assignment.
(B) When the set of rows has been completely examined, an identical procedure is applied successively to columns.
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16. Dr.G.SureshKumar@KLUniversity
Starting with column 1, examine all columns until a column containing exactly one zero is found. Then make an experimental assignment in that position and cross other zeros in the row in which the assignment was made.
Continue these successive operations on rows and columns until all zero’s have either been assigned or crossed-out.
Now there are two possibilities:
(a) Either all the zeros are assigned or crossed out, i.e., we get the maximal assignment. or
(b) At least two zeros are remained by assignment or by crossing out in each row or column. In this situation we try to exclude some of the zeros by trial and error method.
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17. Dr.G.SureshKumar@KLUniversity
This completes the second step. After this step we can get two situations. (i) Total assigned zero’s = n The assignment is optimal. (ii) Total assigned zero’s < n Use step III and onwards. Step III: Draw of minimum lines to cover zero’s In order to cover all the zero’s at least once you may adopt the following procedure. (i) Marks (√) to all rows in which the assignment has not been done. (ii) See the position of zero in marked (√) row and then mark (√) to the corresponding column.
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18. Dr.G.SureshKumar@KLUniversity
(iii) See the marked (√) column and find the position of assigned zero’s and then mark (√) to the corresponding rows which are not marked till now. (iv) Repeat the procedure (ii) and (iii) till the completion of marking. (v) Draw the lines through unmarked rows and marked columns. Note: If the above method does not work then make an arbitrary assignment and then follow step IV.
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19. Dr.G.SureshKumar@KLUniversity
Step IV: Select the smallest element from the uncovered elements. (i) Subtract this smallest element from all those elements which are not covered. (ii) Add this smallest element to all those elements which are at the intersection of two lines. Step V: Thus we have increased the number of zero’s. Now, modify the matrix with the help of step II and find the required assignment. Repeat this procedure till to obtain an optimal assignment.
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20. Dr.G.SureshKumar@KLUniversity
Problem#1
Four jobs are to be done on four different machines. The cost (in rupees) of producing ith job on the jth machine is given below and assign the jobs to different machines so as to minimize the total cost. M1 M2 M3 M4 J1 15 11 13 J2 17 12 J3 14 10 J4 16
Machines
Jobs
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21. Dr.G.SureshKumar@KLUniversity
Solution
Step#1: subtract the smallest element 11 of row 1 from all its element. Similarly, subtract 12, 10, 11 from rows 2, 3, and 4 respectively. The reduced table is below. M1 M2 M3 M4 J1 4 2 J2 5 1 J3 J4 6
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22. Dr.G.SureshKumar@KLUniversity
Solution
Now subtract the smallest element 4 of column 1 from all its element and subtract 1 from column 4. The reduced table is below. M1 M2 M3 M4 J1 2 3 J2 1 J3 5 J4
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23. Dr.G.SureshKumar@KLUniversity
Solution
Assign for single zero in a row and cross the zeros in the column. After assigned zeros in rows continue same in columns. The resultant table as follows. M1 M2 M3 M4 J1 2 3 J2 1 J3 5 J4
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24. Dr.G.SureshKumar@KLUniversity
Solution
The optimal assignment is J1 – M2, J2 – M4, J3 – M1 and J4 – M3. Total cost = = 49
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25. Dr.G.SureshKumar@KLUniversity
Problem#2
A firm plans to begin production of three new products on its three plants. The unit cost of producing i at plant j is as given below. Determine the assignment schedule that minimizes the total cost.
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26. Dr.G.SureshKumar@KLUniversity
Problem#3
A company has six jobs to be processed by six mechanics. The following table gives the return in rupees when the ith job is assigned to the jth mechanics. How should the job be assigned to the mechanics so as to maximize the overall return?
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27. Dr.G.SureshKumar@KLUniversity
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