1. Linear Programming We are to learn two topics today: LP formulation
determining LP solutions
Graphical approach
Simplex method – will be taken up in next lecture
Tutorial Questions: Ch 2, 3,6,9,12,16,20
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2. LP formulation Note that Chapter 4 has presented 8 LP formulations
Here, we pick up the following three examples for discussions
Example 1 (Investment)
Example 2 (Marketing)
Example 3 (Transportation)
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3. Example 1 (Investment) Please refer to the handouts(p112)
Spend 5 mins to read through them
Question asked was to formulate its LP problem!
Solution:
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4. How to achieve it?
From lecture 1, we know LP formulation involve the following steps:
Step 1: define decision variables
Step 2: define the objective function
Step 3: state all the resource constraints
Step 4: define non-negativity constraints
Overall LP formulation
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5. Step 1: define decision variables
Question: how much to invest in four alternative choices:
Let,
x1 = amount invested in municipal bonds ($)
x2 = amount invested in certificates of deposit ($)
x3 = amount invested in treasury bills ($)
x4 = amount invested in growth stock fund($)
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6. Step 2: define the objective function
Expected profits are:
8.5% for x1 = 0.085x1,
5% for x2 = 0.05x2
6.5% for x3 = 0.065x3
13% for x4 = 0.13x4
Thus, objective function is
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7. Objective function maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4
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8. Step 3: state all the resource constraints
Four constraints
Not more than 20%for x1
X2 should not exceed total of other investments
At least 30% spent for x2 and x3
X2 and x3 should be greater than x1 and x4, by ratio of at least 1.2 to 1
Note: total investment is $70,000
All constraints are:
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9. Not more than 20%for x1 Given that total investment is $70,000
20% of it = $14,000
Thus this constraint could be expressed as
x1  14,000
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10. X2 should not exceed total of other investments
Total 4 investments, x1, x2, x3 and x4
Thus, other investments would be
x1, x3 and x4
This constraint is then expressed as:
x2  x1 + x3 + x4 Or
x2 - x1 - x3- x4  0
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11. At least 30% spent for x2 and x3
That is,
30% of $70,000 = $21,000 Or
x2 + x3  21,000
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12. That is x2 + x3  1.2(x1 + x4) or -1.2x1 + x2 + x3 - 1.2 x4  0
X2 and x3 should be greater than x1 and x4, by ratio of at least 1.2 to 1
That is
x2 + x3  1.2(x1 + x4) or
-1.2x1 + x2 + x x4  0
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13. total investment is $70,000 That is, x1 + x2 + x3 + x4 = 70,000
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14. Over constraints x1  14,000 x2 - x1 - x3- x4  0 x2 + x3  21,000
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15. Step 4: define non-negativity constraints
That is,
x1, x2, x3, x4  0
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16. Overall LP formulation
maximize Z = $0.085x x x x4
subject to
x1  14,000
x2 - x1 - x3- x4  0
x2 + x3  21,000
-1.2x1 + x2 + x x4  0
x1 + x2 + x3 + x4 = 70,000
x1, x2, x3, x4  0
(at least 20% on x1)
(x2 spent not more than total investment)
(at least 20% on x2 and x3)
(x2+x3 has at least 1.2 ratio to x1+x4
(total investment is $70,000)
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Note: it is up to you if you like to arrange in this format or let it be the way in
according to how constraints were stated!

17. Example 2 (Marketing) Please read the handout (118)
Read it for 5 mins and then try to formulate it!
How to start?
Again we following the four steps
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18. 4 procedural steps Step 1: define decision variables
Step 2: define the objective function
Step 3: state all the resource constraints
Step 4: define non-negativity constraints
Overall LP formulation
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19. Step 1: define decision variables
Question: how to achieve a max exposure to three of possible media
Let
x1 = number of television commercials
x2 = number of radio commercials
x3 = number of newspaper aids
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20. Step 2: define the objective function
Exposure are: (refer to the attached table)
20,000 for x1
12,000 for x2
9,000 for x3
Thus the objective function is
maximize Z = ,000x1 + 12,000x2 + 9,000x3
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21. Audience exposure rate
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22. Step 3: state all the resource constraints
We have the following resources:
Budget limit $100,000
Television time for four commercials
Radio time for 10 commercials
Newspaper space for 7 ads
Resources for no more than 15 commercials and/or ads.
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23. Budget limit $100,000
All associated cost to each variables has a budget of $100,000
Thus,
$15,000x1 + 6,000x 2+ 4,000x3  100,000
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24. Television time for four commercials
That is
x1  4 (max of 4 TV ads)
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25. Radio time for 10 commercials
That is,
x2  10 (max of 10 radio ads)
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26. Newspaper space for 7 ads
That is,
x3  7 (max of 7 newspaper ads)
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27. Resources for no more than 15 commercials and/or ads
That is
x1 + x2 + x3  15 (max of 15 ads can
be subscribed)
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28. Step 4: define non-negativity constraints
That is,
x1, x2, x3  0
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29. Overall LP formulation
maximize Z = 20,000x1 + 12,000x2 + 9,000x3
subject to
$15,000x1 + 6,000x 2+ 4,000x3  100,000
x1  4
x2  10
x3  7
x1 + x2 + x3  15
x1, x2, x3  0
(note it is advisable
to include explanation
here!)
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30. Example 3 (Transportation)
Please read the handout (121)
Read it for 5 mins and then try to formulate it!
How to start?
Again we following the four steps
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31. 4 procedural steps Step 1: define decision variables
Step 2: define the objective function
Step 3: state all the resource constraints
Step 4: define non-negativity constraints
Overall LP formulation
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32. Step 1: define decision variables
We are to determine all combination of shipping from warehouse i to store j, i=1,2,3;j=A,B,C
That is, xij, where i is warehouse, j is store or
X1A= no of TV shipped from warehouse I to A store
X1B = no of TV shipped from warehouse 1 to B store
X1C = no of TV shipped from warehouse 1 to C store
X2A = no of TV shipped from warehouse 2 to A store
X2B = no of TV shipped from warehouse 2 to B store
X2C = no of TV shipped from warehouse 2 to C store
X3A = no of TV shipped from warehouse 3 to A store
X3B = no of TV shipped from warehouse 3to B store
X3C = no of TV shipped from warehouse 3 to C store OR
Xij = no of TV shipped from warehouse i to j store, i=1,2,3;j=A,B,C
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33. Step 2: define the objective function
Here, we attempt to min the shipping cost, i.e.
Cost
Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A +
15x3B + 17x3C
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34. Step 3: state all the resource constraints
Our resources are:
Warehouse supply of televisions sets: Retail store demand for television sets:
1- Cincinnati A - New York 150
2- Atlanta B - Dallas 250
3- Pittsburgh C - Detroit 200
total total 600
So, the resource constraints are:
x1A + x1B+ x1  300
x2A+ x2B + x2C  200
x3A+ x3B + x3C  200
x1A + x2A + x3A = 150
x1B + x2B + x3B = 250
x1C + x2C + x3C = 200
How to get these?
Note, how this
Problem really looks
Like?
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35. Transportation Problem
Total supply
300
200
200
700
150
250
200
Total demand
600
Note: total supply is greater than total demand.
Thus, we set demand constraint to = and supply to <=
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36. Step 4: define non-negativity constraints
That is,
xij  0, i=1,2,3; j=A,B,C
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37. Overall LP formulation
Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C
Subject to
x1A + x1B+ x1C  300
x2A+ x2B + x2C  200
x3A+ x3B + x3C  200
x1A + x2A + x3A = 150
x1B + x2B + x3B = 250
x1C + x2C + x3C = 200
xij  0, i=1,2,3; j=A,B,C
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38. LP - Graphical approach
Here, we now try to solve an LP formulation using a graphical approach, that is
Plot all equations onto a graph and determine Z values of decision variables that satisfying all resource constraints
How it works?
Types of LP solutions
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39. LP - Graphical approach
Consider a simple LP problem:
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40 ………(e1)
4x2 + 3x2  120 ……..(e2)
x1  ……...(e3)
x2  ………(e4)
Objective function
Resource
Constraints
Solution steps of graphical approach
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40. Solution steps of graphical approach
Step 1: take one resource constraint and treat it as an equation and draw them onto a graph
Step 2: determine the region that satisfying the inequality equation of step 1
Step 3: repeat steps 1-2 for all resource constraints
Step 4: super imposing objective function to determine the solution set for decision variables
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41. Step 1: Consider 1x1 + 2x2  40 ……… (1) And teating it as
How to draw them ?
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42. 1x1 + 2x2 = 40 How to plot? Consider any points on x and y axis
in turn, then we have :
x1=0, x2=20 and
x2=0, x1=40
Step2:
Where is 1x1 + 2x2  40 ?
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43. Determine 1x1 + 2x2  40 This region We take a point Is known
as feasible
region!
We take a point
at (0,0) to check
which side satisfying
this constraint!
(0,0)
At (0,0), 1*0 + 0*0  40 ? Yes, then all points on the side of (0,0) are feasible
region
Repeat these steps for equation e2
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44. Constraint e2
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When combine this to the one we obtained before, then we have ….

45. Constraints e1 to e2 Interception point (x1,x2)=(24,8) (0,20) (30,0)
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When add e3 and e4, we have ……

46. Common region of e1 to e4
Now, we proceed to Step 4, that is impose Objective function Z onto
this graph ………. How to plot it?
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47. maximize Z=$40x1 + 50x2
To plot this equation Z, we can simply assign an aribitrary number for Z,
Say, Z = 800
Then =$40x1 + 50x2
When x2 = 0, then x1= 20 ….. A plotting point (20,0)
When x1 = 0, then x2 = 16 ….a plotting point (0,16)
When plotting it on the graph, we would have ….
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Where is the
optimal solution?

48. Optimal solution
The optimal solution is at the corner point of the feasible region!
To obtain the optimal point, we use a ruler move toward the right hand side and the last corner point
Solution
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For max problem
if min then it is on the left hand side

49. Optimal point This is the solution (x1,x2)=(24,8)
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Let check to see if it is the optimal point/solution ….

50. Checking!
Optimal
solution
X1=0
X2-0
Z=0
More exercise! ……..
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51. Another example Consider Minimize Z = $6x1 + 3x2 subject to
2x1 + 4x2  16 …………. (e1)
4x1 + 3x2  24 …………...(e2)
x1  …………..(e3)
x2  …………..(e4)
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Feasible region for constraints e1 to e4 ………

52. Resource constraints Feasible region Imposing function Z And also
x1  0, x2  0
2x1 + 4x2  16
4x1 + 3x2  24
Imposing inequalities
Note, Min Z is at the most left
Hand side of the Z function
Imposing function Z
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53. Types of LP solutions Two types of LP solution:
The solution – the unique solution
Multiple solutions
Question:
Do all LP problems have a solution?
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54. The solution – unique solution
It is referred to ….there is only one optimal solution Z has satisfying all LP formulation …. i.e. only one (corner) point from the feasible region
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55. Multiple solutions
It refers to that we have more than one set of solutions satisfying to the LP formulation. That is, the optimal function Z lies onto one of resource constraints that forms the feasible region …… and that all points (x1,x2) on that region are optimal solutions to the LP problem
For Max Z,
all points between
here are optimal
solutions
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56. LP solutions LP solution does not exit if our LP problems are:
Infeasible problems
Unbound problems
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57. An Infeasible Problem
Every possible solution violates at least one constraint:
maximize Z = 5x1 + 3x2
subject to
4x1 + 2x2  8
x1  4
x2  6
x1, x2  0
Here, we cannot
find a region that
has satisfying all
constraints, i.e.
no feasible region!
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58. An Unbounded Problem
Value of objective function increases indefinetly:
maximize Z = 4x1 + 2x2
subject to
x1  4
x2  2
x1, x2  0
Note: there is no
bounded feasible region,
thus it cannot find Max Z.
But it has solution if Z is a Min problem! (Why?)
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59. (to p3)
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60. (to p30)