For LastMinute Tutor
SOLUTION ASSIGNMNT 2. CH4
Problem #1, Textbook
First we develop t he Transportation Tableau
Demand➔ 80 170 230 140
1- Formulation:
- Limits on supply (Material send out from a source equal its supply
X1,1 +X1,2+X1,3+X1,4 = 230
X2,1 +X2,2+X2,3+X2,4 = 130
X3,1 +X3,2+X3,3+X3,4 = 240
-Requirement to meet the demands
X1,1 +X2,1+X3,1 = 80
……………………..
………………………….
X1,4 +X2,4+X3,4 =140
-Non-negativity Condition/restriction
Xi,j ≥ 0
-Objective Function
Minimize Z= 25X1,1+ 20X1,2 +16X1,3+ …… 17X3.4
SOLUTION USING POM SOFTWARE
Optimal solution value = $7800
1- Solve the problem
Destination 1 Destination 2 Destination 3 Destination 4
Source 1 80 160 -- 10
Source 2 - - -- 130
Source 3 - 10 230 240
2)- Optimum Shipping Schedule: X1,1 =80, X1,= 160, X2,2= 130, X3,3 =10, X 3,3 =240
Busiest route is X3,3
3)- Optimum cost = $7800
Problem #2A, Textbook
After adding a Dummy Demand point (Supply is>Demand),The Transportation Tableau is:
Boston New York Chicago Indianapolis Supply
Detroit 15 20 16 21 230
Pittsburgh 25 13 5 11 130
Buffalo 15 15 7 17 240
Data
COSTS Destination 1 Dummy
Destination 2
Destination 3
Destination 4 Supply
Source 1 15 0 20 16 21 250
Source 2 25 0 13 5 11 130
Source 3 15 0 15 7 17 240
Demand 50 30 170 230 140 620 \ 620
Solution for the problem is:
problem#1, Chapter 4 solution
Optimal solution value = $7350 Destination 1 Dummy Destination 2 Destination 3 Destination 4
Source 1 50 30 160 10
Source 2 130
Source 3 10 230
Created by POM-QM for Windows
Problem #2B, Textbook In this case total supply < Total demand ➔ we create a Dummy source with (Demand -Supply)
50 units and with Ci,j a large number (Here we used $50/unit). The Transportation Tableau is:
problem#1, Chapter 4
Destination 1 Dummy Destination 2 Destination 3 Destination 4 SUPPLY
Source 1 15 0 20 16 21 250
Source 2 25 0 13 5 11 130
Source 3 15 0 15 7 17 240
Source 4 50 50 50 50 50 50
Demand 80 0 170 280 140
And the solution is:
problem#1, Chapter 4 solution
Optimal solution value = $9970 Destination 1 Dummy Destination 2 Destination 3 Destination 4
Source 1 80 0 170 0
Source 2 40 90
Source 3 240
Source 4 50
Created by POM-QM for Windows
Solution: HW problems from handout
Problem 1
1)- The Transportation matrix is: Destinations Sources
Albuquerque Boston Cleveland Total supply
Des Moines
X1,1 $5 X1,2 $4
X3,1 $3 100
Evansville
X2,1 $8 X2,2 $4
X2,3 $3
300
Fort Lauderdale
X3,1 $9 X3,2 $7
X3,3 $5 300
Demands➔
300 200 200 700
2) Formulation : Minimize ∑( Ci,j)*(Xi,j) where i= 1,2, & 3, J=1, 2, & 3 Objective function
X1.1 +X1,2 +X1,3 = 100
X2.1 +X2,2 +X2,3 = 300 Source Constraint
X3.1 +X3,2 +X3,3 = 300
X1.1 +X2,1 +X3,1 = 300
X1.2 +X2,2 +X3,2 = 200 Demand constraint (we have to meet demand)
X1.3 +X3,2 +X3,3 = 200
Xi,j ≥ 0 for all values of I and j Non-negativity constraint
3- Solution to the problem
Optimal cost = $3,900
Destinations Destination 1 Destination 2 Destination 3 (Albaq…) (Boston) (Cleveland) Sources
Source 1 100 **** ****
Source 2 **** 200 100
Source 3 200 **** 100
Final Solution:
X1,1 =200, X2,2 =200, X2,3 =100, X3,1 = 200, X3,3 = 100 (these are called Basic variables)
All other Xij = 0 (These are called non-basic variables)
Case 1A: Since total demand is less than total supply (by 50 units), to develop a Balanced Transportation Model, we need to add a DUMMY sink with demand =50 unit and shipment cost to the
dummy sink of $0, to the network. Albuquerque
100 300 Des Moines
150
300 Boston Evansville Cleveland 300 200
Fort Lauderdale
50
The transportation matrix:
Sinks Sources
D1 D2 D3 Dummy Total supply
S1 $5 $4 $3 0 100
S2 $8 $4 $3 0 300
S3 $9 $7 $5 0 300
Demands = 300 150 200 50 700
Formulation: 2) Formulation : Minimize ∑( Ci,j)*(Xi,j) where i= 1,2, & 3, J=1, 2, 3, & 4 Objective function X1.1 +X1,2 +X1,3 +X1,4 = 100 X2.1 +X2,2 +X2,3 +X2,4 = 300 Source Constraint
X3.1 +X3,2 +X3,3+X3,4 = 300
X1.1 +X2,1 +X3,1 = 300
X1.2 +X2,2 +X3,2 = 200 Demand constraint (we have to meet demand)
X1.3 +X3,2 +X3,3 = 150
X1,4 +X2,4+ X3,4 = 50
Xi,j ≥ 0 for all values of I and j Non-negativity constraint
Case 1B. In this case (see network below) S1&S3 are pure source, S2 is a transshipment node and the rest are pure demand nodes. S=total supply =700 . Then the Transportation Matrix is: Abaqerquue Boston Cleveland Evansville
Sinks Sources
D1 D2 D3 S2 Dummy Total supply
S1 (des Moines) $5 $4 $999 $2.25 $0 100
S2 (Evansville) $8 $4 $3 $0 $0 300+700
S3 (Fort Laud. )
$9 $7 $5 $999 $0 300
Demands = 300 200 200 0+700 0 700
S2
S3
S1
D2
D3
D1
Dummy
100 300
150
300
200
300
Formulation: Similar to Case 4A
Solution: The output of the software (for this case ) is: Optimal cost = $3,650
(Albuquerque) ( Boston) ( Cleveland) (Evansville) Destination 1 Destination 2 Destination 3 Destination 4 Des Moines 100 ---- ---- ---- Evansville ---- 200 100 700 Fort laud… 200 ---- 100 ----
Final Solution: X1,1 =100. X2,2 =200. X2.3 =100. X3,1= 200. X3,3 =100, X2,4 =700 ( This is not a real
shipment) Total Cost =$3900
PROBLEM #2:
Alternative A: The transportation table (with New Orleans as the third source) is as follows
Destination 1 Destination 2 SUPPLY Source 1 14 11 600 Source 2 9 12 900 Source 3 9 10 500 DEMAND 800 1200 . Formulation:
Minimize Z= ∑)Ci,j)*(Xi,j) for x=1, 2, 3. And J=1, 2
Subject to: X1,1 +X2,1+X3,1 =800 X1,2 +X2,2+X3,2 = 1200
X1,1+X1,2=600 X2,1+X2,2 =900 X3,1 +X3,2 =500
Xi,j ≥ 0 , all i and j
Solution (QM output)
From To Shipment Cost per unit Shipment cost
Source 1 Destination 2 600 11 6600 Source 2 Destination 1 800 9 7200
S2
S3
Dummy
S3
S1
D2
D1
S2
Source 2 Destination 2 100 12 1200 Source 3 Destination 2 500 10 5000 Total Cost = $20,000
Alternative B: The transportation table (with Houston as the third source) is as follows
Destination 1 Destination 2 SUPPLY Source 1 14 11 600 Source 2 9 12 900 Source 3 7 9 500 DEMAND 800 1200
Formulation:
He same the case before except, we have to use shipping cost (Ci,j) to Houston instead of Ci,j for New Orlean.
Solution(QM output):
From To Shipment Cost per unit Shipment cost
Source 1 Destination 2 600 11 6600 Source 2 Destination 1 800 9 7200 Source 2 Destination 2 100 12 1200 Source 3 Destination 2 500 9 4500 Total Cost = $19,500
CONCLUSION: Select Houston as the third source
CASE 2A
in problem 2 the Company picked “Houston” as the new source. The new network will be as follows:
Let,
S= Total supply = 600 + 900 + 500 = 2000. This value (∑supply) will be added to each
transshipment points’ Supply and demand. In the above network: In the above network;
a)- Houston is a pure source
Atlanta LA
Tulsa
New Source
Say Houston
NY
b)- LA and NY are pure sinks c)- Atlanta and Tulsa are transshipment points ➔ The transportation Matrix will look like :
LA NY ATLANTA TULSA Total Supply
ATLANTA
8
5 0 999 600+S = 2600
TULSA
4
7 999 0 900 + S =2900
HOUSTON
4
6 4.5 4.5
500
800 1200 0 + S =2000
0 + S =2000
Solution: Now use the software to determine minimum cost shipping schedule. The input
matrix (transportation matrix) is :
Destination 1 Destination 2 Destination 3 Destination 4 SUPPLY
Source 1 8 5 0 999 2600
Source 2 4 7 999 0 2900
Source 3 4 6 4.5 4.5 500
DEMAND 800 1200 2000 2000
Solution (QM output) From To Shipment Cost per unit Shipment cost
Source 1 Destination 2 600 5 3000 Source 1 Destination 3 2000 0 0 Source 2 Destination 1 800 4 3200 Source 2 Destination 2 100 7 700 Source 2 Destination 4 2000 0 0 Source 3 Destination 2 500 6 3000 Total transportation cost = $9,900
Total cost (Transportation and production) =
$9900+ (600 *$6/unit) + ((800+100)*5) +(500 * 3) = $19500
Conclusion: The new policy will not reduce the total cost since in previous analysis we
Determined that using Houston as the 3rd warehouse will result in a total
cost = $19500.