transportation models

1. TRANSPORTATION
(SPECIALLY STRUCTURED LINEAR PROGRAMS)
SUBMITTED BY:
Transportation 1
Escober, Mark
de la Cruz, Edghan Bryan
Transportation 2
Kali, Norhasim
Lagasca, Michelle
Ronidel, Roderick

2. TRANSPORTATION AND ASSIGNMENT PROBLEM
 Transportation problem is concerned with selecting routes in a
product – distribution networks among manufacturing plants to
distribution warehouse or among regional warehouses to local
distribution outlets.
 Assignment problem involves assigning employees to tasks,
salesperson to towns, contract to bidders or jobs to plants.

3. TRANSPORTATION METHOD

4. The Transportation Problem:
(demand equals supply)
 Let us consider the case of the Bulacan Gravel Company , which has received
a contract to supply gravel for three road projects located in the towns of
Marilao, Bocaue and Balagtas. Construction engineers have estimated the
amount of gravel which be needed at three construction projects:
Weekly Requirement
Project Location Truckloads
A Marilao 72
B Bocaue 102
C Balagtas 41
Total 215

5. The Transportation Problem:
(demand equals supply)
 The Bulacan Gravel Company has three gravel plants located in the towns of San
Rafael, San Ildefonso and San Miguel. The gravel required for the construction
projects can be supplied by these three plants. Bulacan’s chief dispatcher has
calculated the amounts of gravel which can be supplied by each plant:
Amount Available/Week
Plant Location Truckloads
W San Rafael 56
X San Ildefonso 82
Y San Miguel 77
Total 215
 The company has computed the delivery costs from each plant to each project site.
Cost per Truckload
From To project A To project B To project C
Plant W P4 P8 P8
Plant X 16 24 16
Plant Y 8 16 24

6. Gravel plants, road construction projects and transportation costs for Bulacan Gravel
Company
P4 Plant W (San Rafael)
Project A 56 loads available
(Marilao)
72 loads required
P8
P16
Project B
(Bocaue) P24 Plant X ( San Ildefonso)
102 loads 82 loads available
required
P8
P8
P16
P16
Project C
Plant Y (San Miguel) P24 (Balagtas)
77 loads available 41 loads
required

7.  The objective is to minimize the total transportation cost:
4WA + 8WB + 8WC + 16XA + 24XB + 16XC + 8YA + 16YB + 24YC
 There are three “origin constraints” which say that Bulacan cannot ship out
more gravel than they have:
WA + WB + WC < 56: Plant W
XA + XB + XC < 82 : Plant X
YA + YB + YC < 77: Plant Y
 There are three “destination constraints”, which say that each project must
receive the gravel it requires:
WA + XA + YA > 72: Project A
WB + XB + YB > 102: Project B
WC +XC + YC > 41:Project C

8.  Using it all together, we get the final linear programming model:
minimize
4WA + 8WB + 8WC + 16XA + 24XB + 16XC + 8YA + 16YB + 24YC
subject to
WA + WB + WC < 56: Plant W
XA + XB + XC < 82: Plant X
YA + YB + YC < 77: Plant Y
WA + XA + YA > 72: Project A
>
WB + XB + YB 102: Project B
WC + XC + YC > 41: Project C
All Variables > 0

9. STEP 1: Set up the transportation table
STEP 2: Develop an initial solution
STEP 3: Test the solution for improvement
STEP 4: Develop the improved solution

10. C
Plant
To Project Project Project
From Capacit
A B C
y
Plant
56
W
A Plant X E 82 B
Plant Y 77
Project 215
Requireme 72 102 41 215
nts
D

11. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 56
X1 X2 X3
1 2 1
XA XB XC
Plant X 6 4 6 82
X4 X5 X6
1 2
YA 8 YB YC
Plant Y 6 4 77
X7 X8 X9
Project
72 102 41 215
Requirem 215
ents

12. 1 WA 4 2
+
-
X1 3

13. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 56
56
1 2 1
XA XB XC
Plant X 6 4 6 82
16 66
1 2
YA 8 YB YC
Plant Y 6 4 77
36 41
Project
72 102 41 215
Requirem 215
ents

14. From To Quantity,
Plant Project truckloads/week
W A 56
X A 16
X B 66
Y B 36
Y C 41
215
Used Squares = Total Rim Requirements – 1
5=6-1

15.  Total Cost of the Initial Solution
Source - destination Quantity Unit Total
Combination Shipped x Cost = Cost
WA 56 P 4 224.00
XA 16 16 256.00
XB 66 24 1,584.00
YB 36 16 576.00
YC 41 24 984.00
Total Transporation
Cost P 3,624.00
 Is this the best solution?

16.  Choose the unused square to be evaluated
 Beginning with the selected unused square, traced a closed path
(moving horizontally and vertically only)
 Assign plus (+) and minus (-) signs alternately at each corner square of
the closed path
 Determine the net change in costs as a result of the changes made in
tracing the path
 Repeat the above steps until an improvement index has been
determined for each unused squares

17. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 56
56 _ +
1 2 1
XA XB XC
Plant X 6 4 6 82
16 + 66 _
1 2
YA 8 YB YC
Plant Y 6 4 77
36 41
Project
72 102 41 215
Requirem 215
ents

18. Addition to cost: From plant W to project B P 8
From plant X to project A 16 P24
Reduction to cost: From plant W to project A P 4
From plant X to project B 24 28
-P4
Improvement index for square WB = WB – WA + XA – XB
= P8 - P4 + P16 - P24
WB = -P4

19. Improvement index for WC = WC – WA + XA – XB + YB – YC
= P8 – 4P + P16 -P24 + P16 - P 24
WC = - P 12

20. Improvement index for XC= XC – XB + YB – YC
= P16 - P24 + P16 - P24
XC = -P16

21. Improvement index for YA = YA - XA + XB – YB
= P8 - P16 + P24 - P16
YA = P 0

22. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 56
56 -4 -12
1 2 1
XA XB XC
Plant X 6 4 6 82
16 66 -16
1 2
YA 8 YB YC
Plant Y 6 4 77
0 36 41
Project
72 102 41 215
Requirem 215
ents

23. 24 16
XB XC
`
6
_
6 +
16 24
YB YC
3 4
+ _
6 1

24. XB XC
66 - 41 0 + 41
`
2 4
5 1
YB YC
36 + 41 41 - 41
7
7 0

25. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 56
56
1 2 1
XA XB XC
Plant X 6 4 6 82
16 25 41
1 2
YA 8 YB YC
Plant Y 6 n 4 77
77
Project
72 102 41 215
Requirem 215
ents

26. Shipping Quantity Unit Total
Assignments Shipped x Cost = Cost
WA 56 P4 P224
XA 16 16 256
XB 25 24 600
XC 41 16 656
YB 77 16 1,232
P 2,968

27. Unused Closed Computation of
Squares path Improvement Index
WB + WB - WA + XA - XB + 8 - 4 + 16 - 24 = - 4
WC + WC - WA + XA - XC + 8 - 4 + 16 - 16 = + 4
YA + YA - XA + XB - YB + 8 - 16 + 24 - 16 = 0
YC + YC - XC + XB - YB + 24 - 16 + 24 -16 = + 16

28. WA WB
56 - 25 0 + 25
`
3 2
1 5
XA XB
16 + 25 = 25 - 25
4
1 0

29. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 56
31 25
1 2 1
XA XB XC
Plant X 6 4 6 82
41 41
1 2
YA 8 YB YC
Plant Y 6 n 4 77
77
Project
72 102 41 215
Requirem 215
ents

30. Shipping Quantity Unit Total
Assignments Shipped x Cost = Cost
WA 31 P4 P124
WB 25 8 200
XA 41 16 656
XC 41 16 656
YB 77 16 1,232
P 2,868

31. Unused Closed Computation of
Squares path Improvement Index
WC +WC - WA + XA - XC +8 – 4 + 16 – 16 = + 4
XB +XB - WB + WA - XA +24 – 8 + 4 – 16 = + 4
YA +YA - WA + WB – YB +8 – 4 + 8 – 16 = - 4
YC - YB + WB - WA +
YC + XA - XC + 24 – 16 + 8 – 4 + 16 – 16 = + 12

32. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 56
56
1 2 1
XA XB XC
Plant X 6 4 6 82
41 41
1 2
YA 8 YB YC
Plant Y 6 n 4 77
31 46
Project
72 102 41 215
Requirem 215
ents

33. Unused Closed Computation of
Squares path Improvement Index
WA +WA – YA + YB - WB +4 – 8 + 16 – 8 = + 4
WC +WC – XC + XA – YA + YB - WB +8 – 16 + 16 – 8 + 16 – 8 = + 8
XB +XB – YB + YA - XA +24 – 16 + 8 – 16 = 0
YC +YC – XC + XA - YA +24 – 16 + 16 – 8 = +16

34. Shipping Quantity Unit Total
Assignments Shipped x Cost = Cost
WB 56 P8 P448
XA 41 16 656
XC 41 16 656
YA 31 8 248
YB 46 16 736
Total Transportation Cost P 2,744

37. R i = value assigned to row i
K = value assigned to row j
j
C i j = cost in square ij ( the square at the intersection
of row i and column j

38. Kj K1
Ri K2 K3
Plant
To Project Project Project
Capacit
From A B C
y
WA 4 WB 8 WC 8
R1 Plant W 5
56
6
1 2 1
XA XB 4 XC 6
R2 Plant X 1
6
6 82
6 6
1 2
YA 8 YB 6 YC
R3 Plant Y 3 4
4 77
6 1
Project 215
Requirem 72 102 41
215
ents

39. R1 + K1 To solve the five equations, then we proceed as follows. If R1 = 0, then
=4
R2 + K1 R1+ K1= 4 R2 + K2 = 24
= 16
R2 +K = 24 0 + K1 = 4 12 + K2 = 24
2
R3 +K = 16 K1 = 4 K2 = 12
+ 2 R3 + K3 = 24
R3 K3 = 24 4 + K3 = 24
K3 = 20
R2 + K1 = 16 R3 + K2 = 16
R2 + 4 = 16 R3 + 12 = 16
R2 = 12 R3 = 4

40. Kj K1 = 4 K2 = 12 K3 = 20
Ri
To Plant
Project Project Project
Capacit
From A B C
y
WA 4 WB 8 WC 8
R1 = 0 56
Plant W 5
6
1 2 1
R2 =12 XA 6 XB 4 XC 6
Plant X 1 6 82
6 6
1 2
R3 = 4 YA 8 YB 6 YC 4
Plant Y 3 4 77
6 1
Project 215
Requirem 72 102 41
215
ents

41. Unused C ij – Ri – Kj Improvement
Square Index
12 C12 - R1 – K2 -4
8 – 0 – 12
13 C13 - R1 - K3 -12
8 – 0 - 20
23 C23 – R2 - K3 -16
16 – 12 - 20
31 C31 – R3 – K1 0
8–4-4

42.  Traced a close path for the cell having the largest negative
improvement index
 Placed plus and minus signs at alternative corners of the path
beginning with a plus sign at the unused square.
 The smallest stone in a negative position on the close path
indicates the quantity that can be assigned to the unused
square.
 Finally, the improvement indices for the new solution are
calculated.

43. Kj K1 K2 K3
Ri
To Plant
Project Project Project
Capacit
From A B C
y
WA 4 WB 8 WC 8
R1 56
Plant W 5
6
1 2 1
R2 XA 6 XB 4 XC 6
Plant X 1 2 4 82
6 5 1
1 2
R3 YA 8 YB 6 YC 4
Plant Y 7 77
7
Project 215
Requirem 72 102 41
215
ents

44. Stone square 11: Stone square 22:
R1+ K1= 4 R2 + K2 = 24
0 + K1 = 4 12 + K2 = 24 Stone Square 32:
K1 = 4 K2 = 12 R3 + K2 = 16
R3 + 12 = 16
Stone square 21: Stone square 23: R3 = 4
R2+ K1= 16 R2 + K3 = 16
R2+ 4 = 16 12 + K3 = 16
R2 = 12 K3 = 4

45. Unused C ij – Ri – Kj Improvement
Square Index
12 C12 - R1 – K2 -4
8 – 0 - 12
13 C13 - R1 – K3 +4
8–0-4
31 C31 – R3 – K1 0
8–4-4
33 C33 – R3 – K3 +16
24 – 4 - 4

46. Kj K1 = 4 K2 = 8 K3 = 4
Ri
To Plant
Project Project Project
Capacit
From A B C
y
WA 4 WB 8 WC 8
R1 = 0 56
Plant W 3 2
1 5
1 2 1
R2 =12 XA 6 XB 4 XC 6
Plant X 4 4 82
1 1
1 2
R3 = 8 YA 8 YB 6 YC 4
Plant Y 7 77
7
Project 215
Requirem 72 102 41
215
ents

47. Unused C ij – Ri – Kj Improvement
Square Index
13 C13 - R1 – K3 +4
8–0-4
22 C22 – R2 – K2 +4
24 – 12 - 8
31 C31 – R3 – K1 -4
8–8-4
33 C33 – R3 – K3 +12
24 – 8 - 4

48. Kj K1 K2 K3
Ri
To Plant
Project Project Project
Capacit
From A B C
y
WA 4 WB 8 WC 8
R1 56
Plant W 5
6
1 2 1
R2 XA 6 XB 4 XC 6
Plant X 4 4 82
1 1
1 2
R3 YA 8 YB 6 YC 4
Plant Y 3 4 77
1 6
Project 215
Requirem 72 102 41
215
ents

49. Stone square 12: Stone square 23:
R1+ K2= 8 Stone Square 32:
R2 + K3 = 16
0 + K2 = 8 R3 + K2 = 16
16 + K3 = 16
K2 = 8 8 + K2 = 16
K3 = 0
K2 = 8
Stone square 21: Stone square 31:
R2+ K1= 16 R3 + K1 = 8
R2 + 0 = 16 R3 + 0 = 8
R2 = 16 R3 = 8

50. Unused C Ri – Kj Improvement
Is this the Optimal Solution? Square
ij –
Index
11 C11 - R1 – K1 +4
4–0-0
13 C13 – R1 – K3 +8
8–0-0
22 C22 – R2 – K2 0
24 – 16 - 8
33 C33 – R3 – K3 +16
24 – 8 - 0

51. 1. For each solution, compute the R and K values for the table
2. Calculate the improvement indices for all unused squares
3. Select the unused square with the most negative index
4. Trace the closed path for the unused square having the most negative
index.
5. Develop an improved solution
6. Repeat steps 1 to 5 until an optimal solution has been found.

52.  Demand Less than Supply
Considering the original Bulacan Gravel Company problem, suppose that
plant W has a capacity of 76 truckloads for week rather than 56. The
company would be able to supply 235 truckloads per week. However,
the project requirements remain the same .

53. Plant
To Project Project Project Dummy
Capacit
From A B C D
y
WA 4 WB 8 8 0
Plant W WC 76
7 4
2
1 2 1
XA 6 XB 4 XC 6 0
Plant X 8 82
2
1 2
YA 8 YB 6 YC 4
0
Plant Y 1 4 2 77
6 1 0
Project 235
Requirem 72 102 41 20
235
ents
Total Cost:
72 x P4 = P 288
4 x P8 = 32
82 x P24= 1,968
16 x P16 = 256
41 x P24 = 984
20x P0 = 0
P 3,528

54. Plant
To Project Project Project Dummy
Capacit
From A B C D
y
WA 4 WB 8 8 0
Plant W WC 76
7
6
1 2 1
XA 6 XB 4 XC 6 0
Plant X 2 4 2 82
1 1 0
1 2
YA 8 YB 6 YC 4
0
Plant Y 7 77
5
2
Project 235
Requirem 72 102 41 20
235
ents
Total Cost:
76 x P8 = P 608
21 x P24= 504
41 x P16= 656
20 x P0 = 0
72 x P8 = 576
5x P16 = 80
P 2,424

55.  Demand Greater than Supply
 Assume that project A will require 10 additional truckloads per week
and that project C estimates additional requirements of 20 truckloads.
The total project requirements now would be equal to 245 truckloads,
as opposed to the 215 available from the plants.
Total Cost:
56 x P4 = P224
26 x P16= 416
56 x P24= 1,344
46 x P16= 736
31 x P24= 744
30 x P 0= 0
P 3,464

56. Total Cost:
56 x P 8 = P 448
21 x P16 = 336
61 x P16 = 976
61 x P 8 = 488
16 x P 16 = 256
30 x P 0 = 0
P 2,504

57.  There maybe an excessive number of stone squares in a solution; the
number of stone squares is greater than the number of rim
requirements minus 1.
 There may be an insufficient number of stone squares in a solution.

58.  DEGENERACY IN ESTABLISHING AN INITIAL SOLUTION
To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 55
35 20
1 2 1
XA XB XC
Plant X 6 4 6 25
16 25
1 2
YA 8 YB YC
Plant Y 6 4 35
35
Project
35 45 35 115
Requirem 115
ents

59. To Plant
Project A Project B Project C
From Capacity
WA 4 WB 8 WC 8
Plant W 55
35 20
1 2 1
XA XB XC
Plant X 6 4 6 25
16 25 0
1 2
YA 8 YB YC
Plant Y 6 4 35
35
Project
35 45 35 115
Requirem 115
ents

60. To Plant
A B C
From Capacity
9 8 5
W 25
6 8 4
X 35
7 6 9
Y 40
Project
30 25 45 100
Requirem 100
ents
TRANSPORTATION TABLE FOR SUPER FERRY SHIPPING

61. Shipping Quantity x Unit = Total
Assignments Shipped Cost Cost
WC 25 5 P125
WX 15 6 90
WC 20 4 80
YA 15 7 105
YB 25 6 150
Total Transportation Cost P550

62. Shipping Quantity x Unit = Total
Assignments Shipped Cost Cost
WA 25 9 P225
XA 5 6 30
XB 25 8 200
XC 5 4 20
YC 40 9 360
Total Transportation Cost P835
0

63. Shipping Quantity X Unit = Total
Assignments Shipped Cost Cost
WA 15 9 P135
WC 10 5 50
XC 35 4 140
YA 15 7 105
YB 25 6 150
Total Transportation Cost P580
0

64. Opportunity Costs for the first allocation
Row W
Row X
First Allocation using the VAM (Vogel Approximation Method)
0

65.  Opportunity costs for the
Second allocation
 Second Allocation
Using the VAM
0

66.  Opportunity costs for the
Third allocation
 Second Allocation
Using the VAM
0

67. •THE ASSIGNMENT PROBLEM

68. •THE ASSIGNMENT PROBLEM
 The Meycauayan Machine Shop does custom metalworking for a
number of local plants. Meycauayan currently has three jobs to be done
(let us symbolize them A, B and C). Meycauayan also has three
machines on which to do the work (X, Y, and Z).
 Anyone of the jobs can be processed completely on any of the
machines. Furthermore, the cost of processing any job on any machine
is known. The assignment of jobs to machine must be on one-to-one
basis; that is, each job must be assigned exclusively to one and only
one machine. T jobs to the objective is to assign the jobs to the
machines so as to minimize the cost.
Machine Total
Job X Y Z
A P25 P31 P35 91
B 15 20 24 59
C 22 19 17 58
Total 62 70 76 208

69. STEP 1: Determine the opportunity- cost table
Column X Column Y Column Z
25-15=10 31-19 = 12 35 – 17 = 18 Step 1; part a
15-15 =0 20 – 19 = 1 24 – 17 = 7
22-15 = 7 19 – 19 = 0 17 – 17 = 0
Job X Y Z
 Computations A 10 12 18
B 0 1 7
C 7 0 0

70. Step 1; part b Computations:
X Y Z
Row A 10 – 10 = 0 12 – 10 = 2 18 – 10 = 8
Row B 0–0=0 1–0=1 7–0=7
Row C 7–0=7 0–0=0 0–0=0
STEP 2: Determine whether an optimal assignment can be made
Line 1
Line 2
0

71. STEP 3: Revise the total opportunity-cost table
Computations:
0 2–1=1 8–1=7 a) Subtract
lowest number
0 1–1=0 7 – 1 = 6 from all uncovered
number
7+1= 8 0 0 Line 1
b) Add same smallest number to
Line 2
numbers lying at the intersection
of two lines
Job X Y Z revised opportunity
cost table
A 0 1 7
B 0 0 6 Line 2
C 8 0 0 Line 3
Line 1
0

72. Total Cost:
Assignment Cost
A to X P25
B to Y 20
C to Z 17
P62
0

73. •MAXIMIZATION PROBLEMS

74. TRANSPORTATION FOR MAXIMIZATION PROBLEMS
 Juan dela Cruz is Marketing Vice President with the Malolos Company. Malolos is
planning to expand its sales of computer software into new towns – Angat, Norzagaray,
and Baliuag. It has been determined that 20, 15 , and 30 salespersons will be required
to service each of the three areas, respectively. The company recently hired 65 new
salespersons to cover these areas, and based on their experience and prior sales
performance, the new hires have been classified as Type A, B, or C salesperson.
Depending upon the regions to which each of the three types are assigned, Malolos
estimates the following annual revenues per salesperson:
Annual Revenues
Type Number of
Salespersons Angat Norzagaray Baliuag
available
A 24 P100,000 120,000 130,000
B 27 90,000 106,000 126,000
C 14 84,000 98,000 120,000
 Dave wishes to determine how many salespersons of each type to assign to the three
regions so that total annual revenues will be maximized.
0

75.  evaluation of unused squares
0

76.  evaluation of unused squares
0

77. Salesper Territor Numbe Annual
son Type y r Revenue
Assign
ed
A Angat 9 P900, 000
A Norzaga 15 1,800,000
ray
B Angat 11 990,000
B Baliuag 16 2,016,000
C Baliuag 14 1,680,000
P7,386,000
0

78. STEP 1: Select the highest and second-highest revenue
alternatives from among those not already allocated. The
difference between these two revenues will be the
opportunity cost for the row or column.
STEP 2: Scan these opportunity-cost figures and identify the
row or columns with the largest opportunity cost.
STEP 3: Allocate as many units as possible to this row or
column in the square with the greatest revenue.
0

79. ASSIGNMENT FOR METHOD OF
MAXIMIZATION PROBLEMS
Josefa Santos manages the Aves Car Rental Agency. This year, she
plans to purchase five new automobiles to replace five older
vehicles. The older vehicles are to be sold at auction. Josefa has
solicited bid from five individuals, each of whom wishes to purchase
only one vehicle but has agreed to make a sealed bid on each of
the five. The bids are as follows:
Automobile
Buyer Ford Honda Kia Mitsubishi Toyota
Amalia P 3,000 P 2,500 P 3,300 P 2,600 P 3,100
Berto 3,500 3,000 2,800 2,800 3,300
Carlos 2,800 2,900 3,900 2,300 3,600
Dolores 3,300 3,100 3,400 2,900 3,500
Edgardo 2,800 3,500 3,600 2,900 3,000
0

80. •REGRET VALUES FOR AUTOMOBILE
Automobile
Buyer Ford Honda Kia Mitsubishi Toyota
Amalia 500 1000 600 300 500
Berto 0 500 1100 100 300
Carlos 700 600 0 600 0
Dolores 200 400 500 0 100
Edgardo 700 0 300 0 600
•OPTIMAL ASSIGNMENT OF BID AWARDS
Automobile
Buyer Ford Honda Kia Mitsubishi Toyota
Amalia 200 700 200 0 100
Berto 0 500 1000 100 200
Carlos 800 700 0 700 0
Dolores 200 400 400 0 0
Edgardo 700 0 200 0 500
0

81. •OPTIMAL SOLUTION TO Aves CAR RENTAL AGENCY
Buyer Bid Accepted Bid
Price
Amalia Mitsubishi P 2,600
Berto Ford 3,500
Carlos Kia 3,900
Dolores Toyota 3,500
Edgardo Honda 3,500
P 17,000
Reference: Quantitative Approaches to Management, 8th edition; Richard I. Levin, et. al
0

82. Thank You