1. Linear Programming (6S) and Transportation Problem (8S)
Lecture 3
Linear Programming (6S)
and
Transportation Problem (8S)

2. Linear Programming George Dantzig – 1914 -2005
Concerned with optimal allocation of limited resources such as
Materials
Budgets
Labor
Machine time
among competitive activities
under a set of constraints
George Dantzig –

3. Linear Programming Example
Variables
Objective function
Maximize 60X1 + 50X2
Subject to
4X1 + 10X2 <= 100
2X1 + 1X2 <= 22
3X1 + 3X2 <= 39
X1, X2 >= 0
Constraints
Non-negativity Constraints
What is a Linear Program?
A LP is an optimization model that has
continuous variables
a single linear objective function, and
(almost always) several constraints (linear equalities or inequalities)

4. Linear Programming Model
Decision variables
unknowns, which is what model seeks to determine
for example, amounts of either inputs or outputs
Objective Function
goal, determines value of best (optimum) solution among all feasible (satisfy constraints) values of the variables
either maximization or minimization
Constraints
restrictions, which limit variables of the model
limitations that restrict the available alternatives
Parameters: numerical values (for example, RHS of constraints)
Feasible solution: is one particular set of values of the decision variables that satisfies the constraints
Feasible solution space: the set of all feasible solutions
Optimal solution: is a feasible solution that maximizes or minimizes the objective function
There could be multiple optimal solutions

5. Another Example of LP: Diet Problem
Energy requirement : 2000 kcal
Protein requirement : 55 g
Calcium requirement : 800 mg
Food
Energy(kcal)
Protein(g)
Calcium(mg)
Price per serving($)
Oatmeal
110 4 2 3
Chicken
205 32 12 24
Eggs
160 13 54
Milk 8
285 9
Pie
420 22
Pork
260 14 80

6. Example of LP : Diet Problem
oatmeal: at most 4 servings/day
chicken: at most 3 servings/day
eggs: at most 2 servings/day
milk: at most 8 servings/day
pie: at most 2 servings/day
pork: at most 2 servings/day
Design an optimal diet plan which minimizes the cost per day

7. Step 1: define decision variables
x1 = # of oatmeal servings
x2 = # of chicken servings
x3 = # of eggs servings
x4 = # of milk servings
x5 = # of pie servings
x6 = # of pork servings
Step 2: formulate objective function
In this case, minimize total cost
minimize z = 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6

8. Meet energy requirement Meet protein requirement
Step 3: Constraints
Meet energy requirement
110x x x x x x6 2000
Meet protein requirement
4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55
Meet calcium requirement
2x1 + 12x2 + 54x x4 + 22x5 + 80x6  800
Restriction on number of servings
0x14, 0x23, 0x32, 0x48, 0x52, 0x62

9. So, how does a LP look like?
minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6
subject to
110x x x x x x6 2000
4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55
2x1 + 12x2 + 54x x4 + 22x5 + 80x6  800
0x14
0x23
0x32
0x48
0x52
0x62

10. Guidelines for Model Formulation
Understand the problem thoroughly.
Describe the objective.
Describe each constraint.
Define the decision variables.
Write the objective in terms of the decision variables.
Write the constraints in terms of the decision variables
Do not forget non-negativity constraints

11. Transportation Problem
Objective:
determination of a transportation plan of a single commodity
from a number of sources
to a number of destinations,
such that total cost of transportation is minimized
Sources may be plants, destinations may be warehouses
Question:
how many units to transport
from source i
to destination j
such that supply and demand constraints are met, and
total transportation cost is minimized

12. A Transportation Table
Table 8S.1
Warehouse 1 2 3 4
Factory 4 7 7 1
Factory 1
can supply 100
units per period
100 1 12 3 8 8 2
200 8 10 16 5
150 3
Total supply
capacity per
period
450
Demand 80 90
120
160
450
Warehouse B’s demand is 90
units per period
Total demand
per period

13. LP Formulation of Transportation Problem
Minimize total cost of transportation
minimize 4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32
+16x33+5x34
Subject to
x11+x12+x13+x14=100
x21+x22+x23+x24=200
x31+x32+x33+x34=150
x11+x21+x31=80
x12+x22+x32=90
x13+x23+x33=120
x14+x24+x34=160
xij>=0, i=1,2,3; j=1,2,3,4
Supply constraint for factories
Demand constraint of warehouses

14. Solution in Management Scientist
Total transportation cost = 4(80) + 7(0) + 7(10)+ 1(10) + 12(0) + 3(90) + 8(110) + 8(0) + 8(0) +10(0) + 16(0) +5 (150) = $2300

15. Assignment Problem Special case of transportation problem
When # of rows = # of columns in the transportation tableau
All supply and demands =1
Objective: Assign n jobs/workers to n machines such that the total cost of assignment is minimized
Plenty of practical applications
Job shops
Hospitals
Airlines, etc.

16. Cost Table for Assignment Problem
Aircraft (j) 1 2 3 4 6 9 7 10 5 11 8
Pilot (i)

17. LP Formulation of Assignment Problem
minimize x11+4x12+6x13+3x14 + 9x21+7x22+10x23+9x24 + 4x31+5x32+11x33+7x34 + 8x41+7x42+8x43+5x44
subject to
x11+x12+x13+x14=1
x21+x22+x23+x24=1
x31+x32+x33+x34=1
x41+x42+x43+x44=1
x11+x21+x31+x41=1
x12+x22+x32+x42=1
x13+x23+x33+x43=1
x14+x24+x34+x44=1
xij = 1, if pilot i is assigned to aircraft j, i=1,2,3,4; j=1,2,3,4
0 otherwise

18. Product Mix Problem Formulate this problem as a linear program
Floataway Tours has $420,000 that can be used to purchase new rental boats for hire during the summer.
The boats can be purchased from two
different manufacturers.
Floataway Tours would like to purchase at least 50 boats.
They would also like to purchase the same number from Sleekboat as from Racer to maintain goodwill.
At the same time, Floataway Tours wishes to have a total seating capacity of at least 200.
Formulate this problem as a linear program

19. Product Mix Problem Maximum Expected Daily
Boat Builder Cost Seating Profit
Speedhawk Sleekboat $ $ 70
Silverbird Sleekboat $ $ 80
Catman Racer $ $ 50
Classy Racer $ $110

20. Product Mix Problem Define the decision variables
x1 = number of Speedhawks ordered
x2 = number of Silverbirds ordered
x3 = number of Catmans ordered
x4 = number of Classys ordered
Define the objective function
Maximize total expected daily profit:
Max: (Expected daily profit per unit) x (Number of units)
Max: 70x1 + 80x2 + 50x x4

21. Product Mix Problem Define the constraints
(1) Spend no more than $420,000:
6000x x x x4 < 420,000
(2) Purchase at least 50 boats:
x1 + x2 + x3 + x4 > 50
(3) Number of boats from Sleekboat equals number of boats from Racer:
x1 + x2 = x3 + x4 or x1 + x2 - x3 - x4 = 0
(4) Capacity at least 200:
3x1 + 5x2 + 2x3 + 6x4 > 200
Nonnegativity of variables:
xj > 0, for j = 1,2,3,4

22. Product Mix Problem - Complete Formulation
Max 70x1 + 80x2 + 50x x4
s.t.
6000x x x x4 < 420,000
x1 + x2 + x3 + x4 > 50
x1 + x2 - x3 - x4 = 0
3x1 + 5x2 + 2x3 + 6x4 > 200
x1, x2, x3, x4 > 0

23. Marketing Application: Media Selection
Advertising Media
# of potential customers reached
Cost ($) per advertisement
Max times available per month
Exposure Quality Units
Day TV
1000
1500 15 65
Evening TV
2000
3000 10 90
Daily newspaper
400 25 40
Sunday newspaper
2500 4 60
Radio
300
100 30 20
Advertising budget for first month = $30000
At least 10 TV commercials must be used
At least customers must be reached
Spend no more than $18000 on TV adverts
Determine optimal media selection plan

24. Media Selection Formulation
Step 1: Define decision variables
DTV = # of day time TV adverts
ETV = # of evening TV adverts
DN = # of daily newspaper adverts
SN = # of Sunday newspaper adverts
R = # of radio adverts
Step 2: Write the objective in terms of the decision variables
Maximize 65DTV+90ETV+40DN+60SN+20R
Step 3: Write the constraints in terms of the decision variables
DTV
<= 15
ETV 10 DN 25 SN 4 R 30
1500DTV +
3000ETV
400DN
1000SN
100R
30000
>=
18000
1000DTV
2000ETV
1500DN
2500SN
300R
50000
Availability of Media
Budget
TV Constraints
Customers reached
DTV, ETV, DN, SN, R >= 0

25. Applications of LP Product mix planning Distribution networks
Truck routing
Staff scheduling
Financial portfolios
Capacity planning
Media selection: marketing

26. Graphical Solution of LPs
Consider a Maximization Problem
Max x1 + 7x2
s.t x < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1, x2 > 0

27. Graphical Solution Example
Constraint #1 Graphed x2 8 7 6 5 4 3 2 1
x1 < 6
(6, 0) x1

28. Graphical Solution Example
Constraint #2 Graphed x2 8 7 6 5 4 3 2 1
(0, 6 1/3)
2x1 + 3x2 < 19
(9 1/2, 0) x1

29. Graphical Solution Example
Constraint #3 Graphed x2
(0, 8) 8 7 6 5 4 3 2 1
x1 + x2 < 8
(8, 0) x1

30. Graphical Solution Example
Combined-Constraint Graph x2
x1 + x2 < 8 8 7 6 5 4 3 2 1
x1 < 6
2x1 + 3x2 < 19 x1

31. Graphical Solution Example
Feasible Solution Region x2 8 7 6 5 4 3 2 1
Feasible
Region x1

32. Graphical Solution Example
Objective Function Line x2 8 7 6 5 4 3 2 1
(0, 5)
Objective Function
5x1 + 7x2 = 35
(7, 0) x1

33. Graphical Solution Example
Optimal Solution x2
Objective Function
5x1 + 7x2 = 46 8 7 6 5 4 3 2 1
Optimal Solution
(x1 = 5, x2 = 3) x1

34. Graphical Linear Programming
Set up objective function and constraints in mathematical format
Plot the constraints
Identify the feasible solution space
Plot the objective function
Determine the optimum solution

35. Possible Outcomes of a LP
A LP is either
Infeasible – there exists no solution which satisfies all constraints and optimizes the objective function
or, Unbounded – increase/decrease objective function as much as you like without violating any constraint
or, Has an Optimal Solution
Optimal values of decision variables
Optimal objective function value

36. Infeasible LP – An Example
minimize 4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16x33+5x34
Subject to
x11+x12+x13+x14=100
x21+x22+x23+x24=200
x31+x32+x33+x34=150
x11+x21+x31=80
x12+x22+x32=90
x13+x23+x33=120
x14+x24+x34=170
xij>=0, i=1,2,3; j=1,2,3,4
Total demand exceeds total supply

37. Unbounded LP – An Example
maximize 2x1 + x2
subject to
-x1 + x2  1
x1 - 2x2  2
x1 , x2  0
x2 can be increased indefinitely without violating any constraint
=> Objective function value can be increased indefinitely

38. Multiple Optima – An Example
maximize x x2
subject to
2x1 + x2  4
x1 + 2x2  3
x1 , x2  0
x1= 2, x2=0, objective function = 2
x1= 5/3, x2=2/3, objective function = 2