# 4. LP & Transportation Model

```Linear Programming
C–1
Linear Programming
 A mathematical technique to
help plan and make decisions
necessary to allocate resources
 Will find the minimum or
maximum value of the objective
 Guarantees the optimal solution
to the model formulated
C–2
Requirements of an
LP Problem
1. LP problems seek to maximize or
minimize some quantity (usually
profit or cost) expressed as an
objective function
2. The presence of restrictions, or
constraints, limits the degree to
which we can pursue our
objective
C–3
Requirements of an
LP Problem
3. There must be alternative courses
of action to choose from
4. The objective and constraints in
linear programming problems
must be expressed in terms of
linear equations or inequalities
C–4
Formulating LP Problems
The product-mix problem at Shader Electronics
 Two products
1. Shader Walkman, a portable CD/DVD
player
Internet-connected color TV
 Determine the mix of products that will
produce the maximum profit
C–5
Formulating LP Problems
Hours Required
to Produce 1 Unit
Department
Electronic
Assembly
Profit per unit
Walkman Watch-TVs
(X1)
(X2)
4
2
\$7
3
1
\$5
Available Hours
This Week
240
100
Decision Variables:
X1 = number of Walkmans to be produced
X2 = number of Watch-TVs to be produced
C–6
Formulating LP Problems
Objective Function:
Maximize Profit = \$7X1 + \$5X2
There are three types of constraints
 Upper limits where the amount used is ≤
the amount of a resource
 Lower limits where the amount used is ≥
the amount of the resource
 Equalities where the amount used is =
the amount of the resource
C–7
Formulating LP Problems
First Constraint:
Electronic
time used
is ≤
Electronic
time available
4X1 + 3X2 ≤ 240 (hours of electronic time)
Second Constraint:
Assembly
time used
is ≤
Assembly
time available
2X1 + 1X2 ≤ 100 (hours of assembly time)
C–8
Graphical Solution
 Can be used when there are two
decision variables
1. Plot the constraint equations at their
limits by converting each equation to
an equality
2. Identify the feasible solution space
3. Create an iso-profit line based on the
objective function
4. Move this line outwards until the
optimal point is identified
C–9
Graphical Solution
X2
100 –
Number of Watch-TVs
–
80 –
Assembly (constraint B)
–
60 –
–
40 –
Electronics (constraint A)
–
20 – Feasible
–
|–
0
region
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of Walkmans
C – 10
Graphical Solution
X2
100 –
Number of Watch-TVs
–
80 –
–
60 –
–
40 –
\$210 = \$7X1 + \$5X2
(0, 42)
–
20 –
(30, 0)
–
|–
0
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of Walkmans
C – 11
Graphical Solution
X2
100 –
Number of Watch-TVs
–
\$350 = \$7X1 + \$5X2
80 –
\$280 = \$7X1 + \$5X2
–
60 –
\$210 = \$7X1 + \$5X2
–
40 –
–
\$420 = \$7X1 + \$5X2
20 –
–
|–
0
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of Walkmans
C – 12
Graphical Solution
X2
100 –
Number of Watch-TVs
–
Maximum profit line
80 –
–
60 –
Optimal solution point
(X1 = 30, X2 = 40)
–
40 –
–
\$410 = \$7X1 + \$5X2
20 –
–
|–
0
|
|
20
|
|
40
|
|
60
|
|
80
|
|
100
X1
Number of Walkmans
C – 13
Corner-Point Method
X2
100 –
Number of Watch-TVs
2
–
80 –
–
60 –
–
3
40 –
–
20 –
–
1
|–
0
|
|
20
|
|
40
|
4
|
60
|
|
80
|
|
100
X1
Number of Walkmans
C – 14
Solving Minimization
Problems
 Formulated and solved in much the
same way as maximization
problems
 In the graphical approach an isocost line is used
 The objective is to move the isocost line inwards until it reaches the
lowest cost corner point
C – 15
Minimization Example
X1 = number of tons of black-and-white chemical
produced
X2 = number of tons of color picture chemical
produced
Minimize total cost = 2,500X1 + 3,000X2
Subject to:
X1
X2
X1 + X2
X1, X2
≥ 30 tons of black-and-white chemical
≥ 20 tons of color chemical
≥ 60 tons total
≥ \$0 nonnegativity requirements
C – 16
Minimization Example
Table B.9
X2
60 –
X1 + X2 = 60
50 –
Feasible
region
40 –
30 –
b
20 –
a
10 –
|–
0
X1 = 30
|
10
|
20
X2 = 20
|
30
|
40
|
50
|
60
X1
C – 17
Minimization Example
Total cost at a = 2,500X1
+ 3,000X2
= 2,500 (40) + 3,000(20)
= \$160,000
Total cost at b = 2,500X1
+ 3,000X2
= 2,500 (30) + 3,000(30)
= \$165,000
Lowest total cost is at point a
C – 18
LP Applications
Production-Mix Example
Department
Product
XJ201
XM897
TR29
BR788
Wiring Drilling
.5
1.5
1.5
1.0
3
1
2
3
Assembly
2
4
1
2
Inspection
.5
1.0
.5
.5
Unit Profit
\$ 9
\$12
\$15
\$11
Department
Capacity
(in hours)
Product
Minimum
Production Level
Wiring
Drilling
Assembly
Inspection
1,500
2,350
2,600
1,200
XJ201
XM897
TR29
BR788
150
100
300
400
C – 19
LP Applications
X1 = number of units of XJ201 produced
X2 = number of units of XM897 produced
X3 = number of units of TR29 produced
X4 = number of units of BR788 produced
Maximize profit = 9X1 + 12X2 + 15X3 + 11X4
subject to
.5X1 + 1.5X2 + 1.5X3 + 1X4
3X1 + 1X2 + 2X3 + 3X4
2X1 + 4X2 + 1X3 + 2X4
.5X1 + 1X2 + .5X3 + .5X4
X1
X2
X3
X4
≤ 1,500 hours of wiring
≤ 2,350 hours of drilling
≤ 2,600 hours of assembly
≤ 1,200 hours of inspection
≥ 150 units of XJ201
≥ 100 units of XM897
≥ 300 units of TR29
≥ 400 units of BR788
C – 20
The Simplex Method
 Real world problems are too
complex to be solved using the
graphical method
 The simplex method is an algorithm
for solving more complex problems
 Developed by George Dantzig in the
late 1940s
 Most computer-based LP packages
use the simplex method
C – 21
NLP in Facility Location
• Consider an existing network with m
facilities
• It is desired to add n new facilities to the
network
• Let’s
– (ai, bi) denote the coordinates of existing
facility ith
– (Xi, Yi) denote the coordinates of the to-befound new facility ith that minimize the total
distribution cost
C – 22
NLP in Facility Location (cont.)
• Let’s
– gij denote the load or flow of activity from a
new facility ith to an existing facility jth
– fij denote the load or flow of activity between
new facilities ith and jth
– cij denote the cost per unit travel between new
facilities
– dij denote the cost per unit travel between new
facilities ith to an existing facility jth
C – 23
NLP in Facility Location (cont.)
• NLP Model
 c f | X
n
Minimize
n
i 1 j 1
n
m
ij ij

i

 X j |  | Yi  Y j |

  d ij g ij | X i  a j |  | Yi  b j |
i 1 j 1
C – 24
Example of NLP Application
• บริ ษทั หนึ่งมีศูนย์ บริ การ 4 แห่ ง และ Warehouse 1 แห่ ง ตั้งอย่ ทู ี่
จดุ Coordinate (X, Y) คือ
(8,20),(8,10),(10,20),(16,30) และ (35,20) ตามลาดับ
บริษทั ต้ องการสร้ าง Warehouse อีก 2 แห่ ง ซึ่งต้ องตั้งห่ างกันตาม
แนวแกน Y และ X ไม่ น้อยกว่ า 5 หน่ วย ปริมาณงานระหว่ างนับเป็ น
Trip ระหว่ าง Facilities มีดงั ตาราง และต้ นทุนการขนส่ งระหว่ าง
Warehouse ทีส่ ร้ างใหม่ เท่ ากับ 5 ต่ อระยะทาง 1 หน่ วย และ
ระหว่ าง Warehouse ทีส่ ร้ างใหม่ กบั Facilities เดิมเท่ ากับ10
ต่ อระยะทาง 1 หน่ วย จงกาหนดตาแหน่ งทีต่ ั้งของ Warehouse ใหม่
ทัง้ 2 แห่ ง
C – 25
Example of NLP Application
WH1
WH2
E1
7
3
E2
7
2
WH1
WH2
WH1
WH2
E3
5
4
E4
4
5
E5
2
2
2
1
C – 26
Transportation Models
C – 27
Transportation Modeling
 An interactive procedure that
finds the least costly means of
moving products from a series
of sources to a series of
destinations
 Can be used to help resolve
distribution and location
decisions
C – 28
Transportation Modeling
 A special class of linear
programming
 Need to know
1. The origin points and the capacity
or supply per period at each
2. The destination points and the
demand per period at each
3. The cost of shipping one unit from
each origin to each destination
C – 29
Transportation Problem
To
From
Albuquerque
Boston
Cleveland
Des Moines
\$5
\$4
\$3
Evansville
\$8
\$4
\$3
Fort Lauderdale
\$9
\$7
\$5
C – 30
Transportation Problem
Des Moines
(100 units
capacity)
Albuquerque
(300 units
required)
Cleveland
(200 units
required)
Boston
(200 units
required)
Evansville
(300 units
capacity)
Fort Lauderdale
(300 units
capacity)
C – 31
Transportation Matrix
Figure C.2
To
From
Albuquerque
\$5
Des Moines
Evansville
Fort Lauderdale
Warehouse
requirement
Boston
300
Cleveland
\$4
\$3
\$8
\$4
\$3
\$9
\$7
\$5
200
Cost of shipping 1 unit from Fort
Lauderdale factory to Boston warehouse
200
Factory
capacity
100
300
300
Des Moines
capacity
constraint
Cell
representing
a possible
source-todestination
shipping
assignment
(Evansville
to Cleveland)
700
Cleveland
warehouse demand
Total demand
and total supply
C – 32
Northwest-Corner Rule
 Start in the upper left-hand cell (or
northwest corner) of the table and allocate
units to shipping routes as follows:
1. Exhaust the supply (factory capacity) of each
row before moving down to the next row
2. Exhaust the (warehouse) requirements of
each column before moving to the next
column
3. Check to ensure that all supplies and
demands are met
C – 33
Northwest-Corner Rule
1. Assign 100 tubs from Des Moines to Albuquerque
(exhausting Des Moines’s supply)
2. Assign 200 tubs from Evansville to Albuquerque
(exhausting Albuquerque’s demand)
3. Assign 100 tubs from Evansville to Boston
(exhausting Evansville’s supply)
4. Assign 100 tubs from Fort Lauderdale to Boston
(exhausting Boston’s demand)
5. Assign 200 tubs from Fort Lauderdale to
Cleveland (exhausting Cleveland’s demand and
Fort Lauderdale’s supply)
C – 34
Northwest-Corner Rule
To
From
(D) Des Moines
(E) Evansville
(A)
Albuquerque
100
200
Warehouse
requirement
300
(C)
Cleveland
\$5
\$4
\$3
\$8
\$4
\$3
\$9
(F) Fort Lauderdale
(B)
Boston
100
100
200
\$7
200
200
\$5
Factory
capacity
100
300
300
700
Means that the firm is shipping 100
bathtubs from Fort Lauderdale to Boston
C – 35
Northwest-Corner Rule
Computed Shipping Cost
Route
From
To
D
E
E
F
F
A
A
B
B
C
Tubs Shipped
Cost per Unit
100
200
100
100
200
\$5
8
4
7
5
Total Cost
\$
500
1,600
400
700
\$1,000
Total: \$4,200
This is a feasible solution
but not necessarily the
lowest cost alternative
C – 36
Intuitive Lowest-Cost Method
1. Identify the cell with the lowest cost
2. Allocate as many units as possible to
that cell without exceeding supply or
demand; then cross out the row or
column (or both) that is exhausted by
this assignment
3. Find the cell with the lowest cost from
the remaining cells
4. Repeat steps 2 and 3 until all units
have been allocated
C – 37
Intuitive Lowest-Cost Method
To
From
(A)
Albuquerque
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement
300
(B)
Boston
(C)
Cleveland
\$5
\$4
\$8
\$4
\$3
\$9
\$7
\$5
200
100
200
\$3
Factory
capacity
100
300
300
700
First, \$3 is the lowest cost cell so ship 100 units from
Des Moines to Cleveland and cross off the first row as
Des Moines is satisfied
C – 38
Intuitive Lowest-Cost Method
To
From
(A)
Albuquerque
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement
300
(B)
Boston
\$5
\$4
\$8
\$4
\$9
\$7
200
(C)
Cleveland
100
100
\$3
\$3
\$5
200
Factory
capacity
100
300
300
700
Second, \$3 is again the lowest cost cell so ship 100 units
from Evansville to Cleveland and cross off column C as
Cleveland is satisfied
C – 39
Intuitive Lowest-Cost Method
To
From
(A)
Albuquerque
(D) Des Moines
(E) Evansville
\$5
\$4
\$8
\$4
200
\$9
(F) Fort Lauderdale
Warehouse
requirement
(B)
Boston
300
(C)
Cleveland
100
100
\$7
200
\$3
\$3
\$5
200
Factory
capacity
100
300
300
700
Third, \$4 is the lowest cost cell so ship 200 units from
Evansville to Boston and cross off column B and row E
as Evansville and Boston are satisfied
C – 40
Intuitive Lowest-Cost Method
To
From
(A)
Albuquerque
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement
300
300
(B)
Boston
\$5
\$4
\$8
\$4
200
\$9
(C)
Cleveland
100
100
\$7
200
\$3
\$3
\$5
200
Factory
capacity
100
300
300
700
Finally, ship 300 units from Albuquerque to Fort
Lauderdale as this is the only remaining cell to complete
the allocations
C – 41
Intuitive Lowest-Cost Method
To
From
(A)
Albuquerque
(D) Des Moines
(E) Evansville
(F) Fort Lauderdale
Warehouse
requirement
300
300
(B)
Boston
\$5
\$4
\$8
\$4
200
\$9
(C)
Cleveland
100
100
\$7
200
\$3
\$3
\$5
200
Factory
capacity
100
300
300
700
Total Cost = \$3(100) + \$3(100) + \$4(200) + \$9(300)
= \$4,100
C – 42
Intuitive Lowest-Cost Method
To
From
(A)
Albuquerque
(B)
Boston
\$5
This
is Moines
a feasible solution,
(D) Des
and an improvement over
\$8
the previous solution, but
(E) Evansville
not
necessarily the lowest 200
cost alternative \$9
(F) Fort Lauderdale
Warehouse
requirement
300
300
200
\$4
\$4
(C)
Cleveland
100
100
\$7
\$3
\$3
\$5
200
Factory
capacity
100
300
300
700
Total Cost = \$3(100) + \$3(100) + \$4(200) + \$9(300)
= \$4,100
C – 43
```