final powerpoint decision science
DESCRIPTION
transportation modelsTRANSCRIPT
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TRANSPORTATION(SPECIALLY STRUCTURED LINEAR PROGRAMS)
SUBMITTED BY:
Transportation 1
Escober, Markde la Cruz, Edghan Bryan
Transportation 2
Kali, Norhasim Lagasca, MichelleRonidel, Roderick
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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.
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TRANSPORTATION METHOD
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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
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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:
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 P 4 P 8 P 8
Plant X 16 24 16
Plant Y 8 16 24
Amount Available/Week
Plant Location Truckloads
W San Rafael 56
X San Ildefonso 82
Y San Miguel 77
Total 215
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Gravel plants, road construction projects and transportation costs for Bulacan Gravel
Company
Project A
(Marilao)
72 loads required
Project B
(Bocaue)
102 loads
required
Project C
(Balagtas)
41 loads
required
Plant X ( San Ildefonso)
82 loads available
Plant W (San Rafael)
56 loads available
Plant Y (San Miguel)
77 loads available
P4
P24
P16
P16P8
P24
P16
P8
P8
Gravel plants, road construction projects and transportation costs for Bulacan Gravel
Company
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The objective is to minimize the total transportation cost:
There are three “origin constraints” which say that Bulacan cannot ship out more gravel than they have:
There are three “destination constraints”, which say that each project must receive the gravel it requires:
4WA + 8WB + 8WC + 16XA + 24XB + 16XC + 8YA + 16YB + 24YC
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
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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
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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
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Project A
Plant Y
Plant X
Project B
Project C
Plant W
Plant Capacit
y
ProjectRequireme
nts
56
72 102 41
77
82
FromTo
215215
A B
C
D
E
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
215215
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
X1X2
X3
X4 X5 X6
X7 X8 X9
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+-
4
X1
WA12
3
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
215215
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
56
16 66
36 41
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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
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Total Cost of the Initial Solution
Is this the best 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
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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
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
215215
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
56
16 66
36 41
_
+ _
+
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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
- P 4
Improvement index for square WB = WB – WA + XA – XB
= P8 - P4 + P16 - P24
WB = -P4
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Improvement index for WC = WC – WA + XA – XB + YB – YC
= P8 – 4P + P16 -P24 + P16 - P 24
WC = - P 12
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Improvement index for XC= XC – XB + YB – YC
= P16 - P24 + P16 - P24
XC = -P16
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Improvement index for YA = YA - XA + XB – YB
= P8 - P16 + P24 - P16
YA = P 0
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
215215
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
56
16 66
36 41
-4 -12
-16
0
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`
24 16
2416
66
36
41
_
+ _
+
XB XC
YB YC
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`
41
XB XC
YB YC
25
77 0
0 + 4166 - 41
36 + 41 41 - 41
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102
n
41
56
82
77
215215
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
56
16 25
77
41
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Shipping Quantity Unit Total
Assignments Shipped x Cost = Cost
WA 56 P 4 P224
XA 16 16 256
XB 25 24 600
XC 41 16 656
YB 77 16 1,232
P 2,968
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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
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`
25
WA WB
XA XB
31
41 0
0 + 2556 - 25
16 + 25 = 25 - 25
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102
n
41
56
82
77
215215
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
31
41
25
77
41
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Shipping Quantity Unit Total
Assignments Shipped x Cost = Cost
WA 31 P 4 P124
WB 25 8 200
XA 41 16 656
XC 41 16 656
YB 77 16 1,232
P 2,868
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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 +YC - YB + WB - WA + XA - XC + 24 – 16 + 8 – 4 + 16 – 16 = + 12
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102
n
41
56
82
77
215215
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
31
41
56
46
41
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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
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Shipping Quantity Unit Total
Assignments Shipped x Cost = Cost
WB 56 P 8 P448
XA 41 16 656
XC 41 16 656
YA 31 8 248
YB 46 16 736
P 2,744Total Transportation Cost
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R i = value assigned to row i
K j = value assigned to row j
C i j = cost in square ij ( the square at the intersection
of row i and column j
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Project A
Project B
Project C
Plant Capacit
y
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
R2
R3
Ri
Kj K1 K2 K3
R1
215
215
56
16
66
36
41
4 8 8
16
24
16
16
24
8
WA WB WC
XA XB XC
YA YB YC
From
To
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R1R2
R2
R3
R3
+
+
+
+
+
K1
K1
K2
K2
K3
= 4
= 16
= 24
= 16
= 24
To solve the five equations, then we proceed as follows. If R1 = 0, then
R1+ K1= 4
0 + K1 = 4
K1 = 4
R2 + K1 = 16
R2 + 4 = 16
R2 = 12
R2 + K2 = 24
12 + K2 = 24
K2 = 12
R3 + K2 = 16
R3 + 12 = 16
R3 = 4
R3 + K3 = 24
4 + K3 = 24
K3 = 20
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Project A
Project B
Project C
Plant Capacit
y
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
R2 =12
R3 = 4
Ri
Kj K1 = 4 K2 = 12 K3 = 20
R1 = 0
215
215
56
16
66
36
41
4 8 8
16
24
16
16
24
8
WA WB WC
XA XB XC
YA YB YC
From
To
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Unused Square
C ij – Ri – Kj Improvement Index
12 C12 - R1 – K2
8 – 0 – 12- 4
13 C13 - R1 - K3
8 – 0 - 20-12
23 C23 – R2 - K3
16 – 12 - 20-16
31 C31 – R3 – K1
8 – 4 - 40
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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.
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Project A
Project B
Project C
Plant Capacit
y
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
R2
R3
Ri
Kj K1 K2 K3
R1
215
215
56
16
25
77
41
4 8 8
16
24
16
16
24
8
WA WB WC
XA XB XC
YA YB YC
From
To
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Stone square 11:
R1+ K1= 4
0 + K1 = 4
K1 = 4
Stone square 21:
R2+ K1= 16
R2+ 4 = 16
R2 = 12
Stone square 22:
R2 + K2 = 24
12 + K2 = 24
K2 = 12
Stone square 23:
R2 + K3 = 16
12 + K3 = 16
K3 = 4
Stone Square 32:
R3 + K2 = 16
R3 + 12 = 16
R3 = 4
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Unused Square
C ij – Ri – Kj Improvement Index
12 C12 - R1 – K2
8 – 0 - 12- 4
13 C13 - R1 – K3
8 – 0 - 4+ 4
31 C31 – R3 – K1
8 – 4 - 40
33 C33 – R3 – K3
24 – 4 - 4+16
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Project A
Project B
Project C
Plant Capacit
y
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
R2 =12
R3 = 8
Ri
Kj K1 = 4 K2 = 8 K3 = 4
R1 = 0
215
215
31
41
25
77
41
4 8 8
16
24
16
16
24
8
WA WB WC
XA XB XC
YA YB YC
From
To
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Unused Square
C ij – Ri – Kj Improvement Index
13 C13 - R1 – K3
8 – 0 - 4+ 4
22 C22 – R2 – K2
24 – 12 - 8+ 4
31 C31 – R3 – K1
8 – 8 - 4- 4
33 C33 – R3 – K3
24 – 8 - 4+12
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Project A
Project B
Project C
Plant Capacit
y
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41
56
82
77
R2
R3
Ri
Kj K1 K2 K3
R1
215
215
31
41
56
46
41
4 8 8
16
24
16
16
24
8
WA WB WC
XA XB XC
YA YB YC
From
To
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Stone square 12:
R1+ K2= 8
0 + K2 = 8
K2 = 8
Stone square 21:
R2+ K1= 16
R2 + 0 = 16
R2 = 16
Stone square 23:
R2 + K3 = 16
16 + K3 = 16
K3 = 0
Stone square 31:
R3 + K1 = 8
R3 + 0 = 8
R3 = 8
Stone Square 32:
R3 + K2 = 16
8 + K2 = 16
K2 = 8
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Unused Square
C ij – Ri – Kj Improvement Index
11 C11 - R1 – K1
4 – 0 - 0+ 4
13 C13 – R1 – K3
8 – 0 - 0+ 8
22 C22 – R2 – K2
24 – 16 - 80
33 C33 – R3 – K3
24 – 8 - 0+16
Is this the Optimal Solution?
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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.
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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 .
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Project A
Project B
Project C
Dummy D
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41 20
Plant Capacit
y
76
82
77
72
82
16
41
4 8 8
16
24
16
16
24
8
WA WB WC
XA XB XC
YA YB YC
From
To
0
4
20
0
0
235
235
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
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Project A
Project B
Project C
Dummy D
Project Requirem
ents
Plant Y
Plant X
Plant W
72 102 41 20
Plant Capacit
y
76
82
77
76
21
72
5
4 8 8
16
24
16
16
24
8
WA WB WC
XA XB XC
YA YB YC
From
To
0
20
0
0
235
235
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
41
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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
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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
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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.
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DEGENERACY IN ESTABLISHING AN INITIAL SOLUTION
From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
35 45 35
55
25
35
115115
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
35
16 25
35
20
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From
ToProject A Project B Project C
Plant Capacity
Project Requirem
ents
Plant Y
Plant X
Plant W
35 45 35
55
25
35
115115
4 8
16
824
16
8
24
16
WA WB WC
XA
YA
XB XC
YB YC
35
16 25
35
20
0
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From
ToA B C
Plant Capacity
Project Requirem
ents
Y
X
W
30 25 45
25
35
40
100100
9 8
6
7 9
4
5
8
6
TRANSPORTATION TABLE FOR SUPER FERRY SHIPPING
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Shipping Assignments
Quantity Shipped
x Unit Cost
= Total Cost
WC 25 5 P125
WX 15 6 90
WC 20 4 80
YAYB
1525
76
105150
P550Total Transportation Cost
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0
Shipping Assignments
Quantity Shipped
x Unit Cost
= Total Cost
WA 25 9 P225
XA 5 6 30
XBXC
255
84
20020
YC 40 9 360P835Total Transportation Cost
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0
Shipping Assignments
Quantity Shipped
X Unit Cost
= Total Cost
WA 15 9 P135
WC 10 5 50
XC 35 4 140
YAYB
1525
76
105150
P580Total Transportation Cost
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0
Opportunity Costs for the first allocation
First Allocation using the VAM (Vogel Approximation Method)
Row W
Row X
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0
Opportunity costs for the
Second allocation
Second Allocation
Using the VAM
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0
Opportunity costs for the
Third allocation
Second Allocation
Using the VAM
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•THE ASSIGNMENT PROBLEM
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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.
Job X Y ZTotal
A P25 P31 P35 91
B 15 20 24 59
C 22 19 17 58
Total 62 70 76 208
Machine
•THE ASSIGNMENT PROBLEM
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STEP 1: Determine the opportunity- cost table
Job X Y Z
A 10 12 18
B 0 1 7
C 7 0 0
Column X Column Y Column Z
25-15=10 31-19 = 12 35 – 17 = 18
15-15 =0 20 – 19 = 1 24 – 17 = 7
22-15 = 7 19 – 19 = 0 17 – 17 = 0
Computations
Step 1; part a
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0
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
Computations:Step 1; part b
Line 2
Line 1
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0
STEP 3: Revise the total opportunity-cost table
0 2 – 1 = 1 8 – 1 = 7
0 1 – 1 = 0 7 – 1 = 6
8 0 0
a) Subtract
lowest number
from all uncovered
numberLine 1
Line 2
7 + 1 =
b) Add same smallest number to
numbers lying at the intersection
of two lines
Computations:
Job X Y Z
A 0 1 7
B 0 0 6
C 8 0 0
revised opportunity
cost table
Line 2
Line 1
Line 3
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0
Total Cost:
Assignment Cost
A to X P25
B to Y 20
C to Z 17
P62
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•MAXIMIZATION PROBLEMS
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0
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 requiredto service each of the three areas, respectively. The company recently hired 65 newsalespersons to cover these areas, and based on their experience and prior salesperformance, 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, Malolosestimates the following annual revenues per salesperson:
Dave wishes to determine how many salespersons of each type to assign to the threeregions so that total annual revenues will be maximized.
Type Number of Salespersons
availableAngat Norzagaray Baliuag
A 24 P100,000 120,000 130,000
B 27 90,000 106,000 126,000
C 14 84,000 98,000 120,000
Annual Revenues
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0
evaluation of unused squares
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0
evaluation of unused squares
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0
Salesperson Type
Territory
Number
Assigned
Annual Revenue
A Angat 9 P900, 000
A Norzagaray
15 1,800,000
B Angat 11 990,000
B Baliuag 16 2,016,000
C Baliuag 14 1,680,000P7,386,000
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0
STEP 1: Select the highest and second-highest revenuealternatives from among those not already allocated. Thedifference between these two revenues will be theopportunity cost for the row or column.
STEP 2: Scan these opportunity-cost figures and identify therow or columns with the largest opportunity cost.
STEP 3: Allocate as many units as possible to this row orcolumn in the square with the greatest revenue.
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0
ASSIGNMENT FOR METHOD OF
MAXIMIZATION PROBLEMS
Josefa Santos manages the Aves Car Rental Agency. This year, sheplans to purchase five new automobiles to replace five oldervehicles. The older vehicles are to be sold at auction. Josefa hassolicited bid from five individuals, each of whom wishes to purchaseonly one vehicle but has agreed to make a sealed bid on each ofthe five. The bids are as follows:
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
Automobile
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0
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
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
Automobile
Automobile
•REGRET VALUES FOR AUTOMOBILE
•OPTIMAL ASSIGNMENT OF BID AWARDS
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0
Buyer Bid Accepted Bid Price
Amalia Mitsubishi P 2,600
Berto Ford 3,500
Carlos Kia 3,900
Dolores Toyota 3,500
Edgardo Honda 3,500P 17,000
Reference: Quantitative Approaches to Management, 8th edition; Richard I. Levin, et. al
•OPTIMAL SOLUTION TO Aves CAR RENTAL AGENCY
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Thank You