The document contains biographical and contact information for Dr. Atif Shahzad, along with slides from one of his lectures on optimization models and the simplex method. The slides cover converting linear programs to standard form, basic and non-basic variables, basic feasible solutions, optimality and feasibility conditions, and working through an example using the simplex method in 4 steps - choosing an entering variable, leaving variable, updating the pivot row and other rows, and iterating until optimality is reached.

1. _____________________Dr. Atif Shahzad
BE, MECHANICAL ENGINEERING
UNIVERSITY OF ENGINEERING & TECHNOLOGY, TAXILA, PAKISTAN, 2000
MCS, SOFTWARE ENGINEERING
SZABIST,, ISLAMABAD, PAKISTAN, 2003
MS, AUTOMATION & PRODUCTION SYSTEMS
ECOLE CENTRALE DE NANTES, NANTES, FRANCE, 2007
PhD, AUTOMATION & APPLIED INFORMATICS
UNIVERSITE DE NANTES, NANTES, FRANCE, 2011
EMAIL: atifshahzad@Gmail.com
TEL: +92-333-5219846, +92-51-5179755
LINKEDIN: pk.linkedin.com/in/dratifshahzad
1 Dr. Atif Shahzad 15-Oct-15

2. OPTIMIZATION
MODELS
COURSE OBJECTIVES
2Dr. Atif Shahzad
15-Oct-15

3. OPTIMIZATION
MODELS
TODAY’S LECTURE
Simplex method
3Dr. Atif Shahzad
15-Oct-15

4. 4 Dr. Atif Shahzad 15-Oct-15

5. LP Problem
1–5
15-Oct-15Dr. Atif Shahzad

6. Converting LP to standard form
1
All Inequalities
to be
EQUATIONS
2
.
All variables
To be
NON-NEGATIVE
1–6
15-Oct-15Dr. Atif Shahzad

7. LP Standard Form
Inequalities to Equations
 Slack Variable: Variable added to a  constraint to convert it to an
equation (=).
 A slack variable represents unused resources
 A slack variable contributes nothing to the objective function value.
 Surplus Variable: Variable subtracted from a  constraint to convert
it to an equation (=).
 A surplus variable represents an excess above a constraint
requirement level.
 Surplus variables contribute nothing to the calculated value of the
objective function.
1–7
15-Oct-15Dr. Atif Shahzad

8. LP Standard Form
Dealing with unrestricted variables
 Unrestricted Variable:A variable that is unbounded and can take
any value.
 BUT, standards form requires that the variables be nonnegative
 So, What to do??
 Write an unrestricted variable 𝑥 as:
 𝑥 = 𝑥+- 𝑥−, where 𝑥+, 𝑥−≥ 0
 If 𝑥+> 0 and 𝑥−= 0 then 𝑥+ represents a slack
 If 𝑥−
> 0 and 𝑥+
= 0 then 𝑥−
represents a surplus
1–8
15-Oct-15Dr. Atif Shahzad

9. Basic & Non-Basic Variables
 Consider a System A𝑥 = 𝑏 of 𝑚 linear equations in 𝑛
variables
 Assume 𝑛 ≥ 𝑚
 Set 𝑛 − 𝑚 variables equal to ZERO
(Non-Basic Variables)
 Solve for remaining 𝑚 variables
(Basic Variables)
1–9
15-Oct-15Dr. Atif Shahzad

10. Basic & Non-Basic Variables
Example
𝑥1 + 𝑥2 = 3
−𝑥2 + 𝑥3 = −1
 𝑛 = 3, 𝑚 = 2, 𝑛 − 𝑚 =1
 If NBV={𝑥3} then BV={𝑥1, 𝑥2}
 Setting 𝑥3 =0, we get {𝑥1, 𝑥2} ={2,1}
 {𝑥1, 𝑥2, 𝑥3}={2,1,0} is a basic solution.
 If NBV={𝑥2} then BV={𝑥1, 𝑥3} THEN ??
 If NBV={𝑥1} then BV={𝑥2, 𝑥3} THEN ??
1–10
15-Oct-15Dr. Atif Shahzad

11. Basic Feasible Solution
 Basic Solution(BS) : This solution is obtained by setting any
n variables (among m+n variables) equal to zero and
solving for remaining m variables, provided the determinant
of the coefficients of these variables is non-zero.
 Such m variables are called basic variables and remaining
n zero valued variables are called non basic variables.
 Basic Feasible Solution(BFS) : It is a basic solution which
also satisfies the non negativity restrictions, i.e., all variables
are nonnegative.
1–11
15-Oct-15Dr. Atif Shahzad

12. Basic Feasible Solution
 BFS are of two types:
Degenerate BFS: If one or more basic variables
are zero.
Non-Degenerate BFS: All basic variables are non-
zero.
 Optimal BFS: BFS which optimizes the objective function.
1–12
15-Oct-15Dr. Atif Shahzad

13. Optimality & Feasibility Conditions
Conditions
FEASIBILITY
For both the maximization and
the minimization problems, the
leaving variable is the basic
variable associated with the
smallest nonnegative ratio (with
strictly positive denominator).
OPTIMALITY
The entering variable in a
maximization (minimization)
problem is the nonbasic
variable having the most
negative (positive) coefficient
in the z-row.
Tiesarebrokenarbitrarily.
Tiesarebrokenarbitrarily.
The optimum is reached at the
iteration where all the z-row
coefficients of the non basic
variables are nonnegative
(non-positive ).
1–14
15-Oct-15Dr. Atif Shahzad

14. Simplex Method
Algorithm
 Step 1. Determine a starting basic feasible solution.
 Step 2. Select an entering variable using the optimality
condition. Stop if there is no entering variable; the last
solution is optimal. Else, go to step 3.
 Step 3. Select a leaving variable using the feasibility
condition.
 Step 4. Determine the new basic solution by using the
appropriate Gauss-Jordan computations. Go to step 2.
1–15
15-Oct-15Dr. Atif Shahzad

15. Example
max z = 13𝑥1 + 11𝑥2
4𝑥1 + 5𝑥2 ≤ 1500
5𝑥1 + 3𝑥2 ≤ 1575
𝑥1 + 2𝑥2 ≤ 420
𝑥1, 𝑥2 ≥ 0
16
Dr. Atif Shahzad 15-Oct-15

16. Step 0:
Standard Form
Convert all the inequality constraints into equalities by the use of slack
variables.
Let S1, S2 , S3 be three slack variables.
Model can rewritten as:
Z - 13x1 -11x2 = 0
Subject to constraints:
4x1+5x2 + S1 = 1500
5x1+3x2 + S2 = 1575
x1+2x2 +S3 = 420
x1, x2, S1, S2, S3 > 0
17
Dr. Atif Shahzad 15-Oct-15

17. Step 1
Initial BFS
 Find the Initial BFS.
 Choose Basic variables
 S1, S2, S3
 One Feasible solution that satisfies all the
constraints is:
x1= 0, x2= 0, S1= 1500,
S2= 1575, S3= 420 and Z=0.
So, NBV={𝑥1, 𝑥2} & BV={𝑆1, 𝑆2, 𝑆3}
And BFS is {𝑥1, 𝑥2, 𝑆1, 𝑆2, 𝑆3}={0,0,1500,1575,420}
1–18
15-Oct-15Dr. Atif Shahzad

18. Step 1:
Tableaux
Row No. Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500
2 S2 0 5 3 0 1 0 1575
3 S3 0 1 2 0 0 1 420
Set up an initial table.
19
Dr. Atif Shahzad 15-Oct-15

19. Step 2:
Entering Variable
Row
NO.
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500
2 S2 0 5 3 0 1 0 1575
3 S3 0 1 2 0 0 1 420
Choose the most negative number from row 0 (i.e Z row).
Therefore, x1 is an entering variable.
Iteration 1
20
Dr. Atif Shahzad 15-Oct-15

20. Step 3:
Leaving variable
Row
NO.
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500 375
2 S2 0 5 3 0 1 0 1575 315
3 S3 0 1 2 0 0 1 420 420
Calculate Ratio = Sol col. / 𝑥1 col. (𝑥1 > 0)
Choose minimum Ratio. That variable(i.e., S2) is
a leaving variable.
Iteration 1
21
Dr. Atif Shahzad 15-Oct-15

21. Step 4:
Find New BFS (Pivoting)
Row
NO.
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500 375
2 S2 0 5 3 0 1 0 1575 315
3 S3 0 1 2 0 0 1 420 420
• x1 becomes basic variable
• S2 becomes non basic variable.
Iteration 1
22
Dr. Atif Shahzad 15-Oct-15

22. Step 4:
Find New BFS (New Pivot Row)
Row
NO.
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500 375
2 S2 0 5 3 0 1 0 1575 315
3 S3 0 1 2 0 0 1 420 420
0 5 3 0 1 0 15750 1 3/5 0 1/5 0 375
5
New Pivot row =
Current Pivot row
Pivot Element
=
Iteration 1
23
Dr. Atif Shahzad 15-Oct-15

23. Step 4:
Find New BFS (New Other Rows)
Row
NO.
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 -13 -11 0 0 0 0
1 S1 0 4 5 1 0 0 1500
2 S2 0 1 3/5 0 1/5 0 375
3 S3 0 1 2 0 0 1 420
1 0 -16/5 0 13/5 0 4095
-13 0 1 3/5 0 1/5 0 375
Iteration 1
New Other row Current row its Pivot col Coeff.x New pivot row
= 1 -13 -11 0 0 0 0
−
−
24
Dr. Atif Shahzad 15-Oct-15

24. Step 4:
New Tableaux
Row
No.
Basic
Varia
ble
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240
2 x1 0 1 3/5 0 1/5 0 315
3 S3 0 0 7/5 0 -1/5 1 105
Iteration 1
• x1 enters and becomes basic variable.
• S2 leaves and becomes non basic variable.
• We have a new table.
25
Dr. Atif Shahzad 15-Oct-15

25. Step 2:
Entering Variable
Row
NO.
Basic
Varia
ble
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240
2 x1 0 1 3/5 0 1/5 0 315
3 S3 0 0 7/5 0 -1/5 1 105
Iteration 2
26
Dr. Atif Shahzad 15-Oct-15

26. Step 3:
Leaving variable
Row
NO.
Basic
Variab
le
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2
Calculate Ratio = Sol col. / 𝑥2 col. (𝑜𝑛𝑙𝑦 𝑤ℎ𝑒𝑟𝑒 𝑥2 > 0)
Choose minimum Ratio. That variable(i.e., S3) is
a leaving variable.
27
Dr. Atif Shahzad 15-Oct-15

27. Step 3:
Leaving variable
Row
No.
Basic
Variab
le
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2
• x2 becomes basic variable
• S3 becomes non basic variable.
28
Dr. Atif Shahzad 15-Oct-15

28. Row
No.
Basic
Variab
le
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2Step 4:
Find New BFS (New Pivot Row)
0 0 7/5 0 -1/5 1 750 0 1 0 -1/7 5/7 375/7
7/5
New Pivot row =
Current Pivot row
Pivot Element
=
29
Dr. Atif Shahzad 15-Oct-15

29. Row
No.
Basic
Variab
le
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 0 -16/5 0 13/5 0 4095
1 S1 0 0 13/5 1 -4/5 0 240 277/3
2 x1 0 1 3/5 0 1/5 0 315 525
3 S3 0 0 7/5 0 -1/5 1 105 75
Iteration 2Step 4:
Find New BFS (New Other Row)
-16/5 0 0 1 0 -1/7 5/7 375/7
New Other row Current row its Pivot col Coeff. New pivot row
= 1 0 -16/5 0 13/5 0 4095
−
−
x
30
Dr. Atif Shahzad 15-Oct-15

30. Row
No.
Basic
Variab
le
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
0 Z 1 0 0 0 15/7 16/7 4335
1 S1 0 0 0 1 -3/7 -13/7 45
2 x1 0 1 0 0 2/7 -3/7 270
3 S3 0 0 1 0 -1/7 5/7 75
Iteration 2Step 4:
New Tableaux
• x2 enters and becomes basic variable.
• S3 leaves and becomes non basic variable.
• We have a new table.
31
Dr. Atif Shahzad 15-Oct-15

31. Step 4:
RESULT
Next Table is :
Row
NO.
Basic
Variab
le
Coefficients of: Sol.
Z x1 x2 S1 S2 S3
0 Z 1 0 0 0 15/7 16/7 4335
1 S1 0 0 0 1 -3/7 -13/7 45
2 x1 0 1 0 0 2/7 -3/7 270
3 x2 0 0 1 0 -1/7 5/7 75
Optimal Solution is : x1= 270, x2= 75, Z= 4335
32
Dr. Atif Shahzad 15-Oct-15

32. Iterations Done
x1 x2 S1 S2 S3
Sol.
Z -13 -11 0 0 0
S1 4 5 1 0 0 1500
S2 5 3 0 1 0 1575
S3 1 2 0 0 1 420
Z 0 -3 1/5 0 2 3/5 0 4095
S1 0 2 3/5 1 - 4/5 0 240
x1 1 3/5 0 1/5 0 315
S3 0 1 2/5 0 - 1/5 1 105
Z 0 0 0 2 1/7 2 2/7 4335
S1 0 0 1 - 3/7 -1 6/7 45
x1 1 0 0 2/7 - 3/7 270
x2 0 1 0 - 1/7 5/7 75
Optimal Solution is :
x1= 270, x2= 75, Z= 4335
33
Dr. Atif Shahzad 15-Oct-15

33. Example 2
Max. Z = 3x1+5x2+4x3
Subject to constraints:
2x1+3x2 < 8
2x2+5x3 < 10
3x1+2x2+4x3 < 15
x1, x2, x3 > 0
34
Dr. Atif Shahzad 15-Oct-15

34. Cont…
Let S1, S2, S3 be the three slack variables.
Modified form is:
Z - 3x1-5x2-4x3 =0
2x1+3x2 +S1= 8
2x2+5x3 +S2= 10
3x1+2x2+4x3+S3= 15
x1, x2, x3, S1, S2, S3 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8,
S2= 10, S3= 15 and Z=0.
35
Dr. Atif Shahzad 15-Oct-15

35. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 x3 S1 S2 S3
Z 1 -3 -5 -4 0 0 0 0
S1 0 2 3 0 1 0 0 8 8/3
S2 0 0 2 5 0 1 0 10 5
S3 0 3 2 4 0 0 1 15 15/2
Therefore, x2 is the entering variable and S1 is the
departing variable.
36
Dr. Atif Shahzad 15-Oct-15

36. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 x3 S1 S2 S3
Z 1 1/3 0 -4 5/3 0 0 40/3
x2 0 2/3 1 0 1/3 0 0 8/3 -
S2 0 -4/3 0 5 -2/3 1 0 14/3 14/15
S3 0 5/3 0 4 -2/3 0 1 29/3 29/12
Therefore, x3 is the entering variable and S2 is the
departing variable.
37
Dr. Atif Shahzad 15-Oct-15

37. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 x3 S1 S2 S3
Z 1 -11/15 0 0 17/15 4/5 0 256/15
x2 0 2/3 1 0 1/3 0 0 8/3 4
x3 0 -4/15 0 1 -2/15 1/5 0 14/15 -
S3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41
Therefore, x1 is the entering variable and S3 is the
departing variable.
38
Dr. Atif Shahzad 15-Oct-15

38. Cont…
Basic
Variable
Coefficients of: Sol.
Z x1 x2 x3 S1 S2 S3
Z 1 0 0 0 45/41 24/41 11/41 765/41
x2 0 0 1 0 15/41 8/41 -10/41 50/41
x3 0 0 0 1 -6/41 5/41 4/41 62/41
x1 0 1 0 0 -2/41 -12/41 15/41 89/41
Optimal Solution is : x1= 89/41, x2= 50/41, x3=62/41, Z= 765/41
39
Dr. Atif Shahzad 15-Oct-15

39. Example 3:
Minimization
Min.. Z = x1 - 3x2 + 2x3
Subject to constraints:
3x1 - x2 + 3x3 < 7
-2x1 + 4x2 < 12
-4x1 + 3x2 + 8x3 < 10
x1, x2, x3 > 0
40
Dr. Atif Shahzad 15-Oct-15

40. Cont…
Convert the problem into maximization problem
Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -
Z
Subject to constraints:
3x1 - x2 + 3x3 < 7
-2x1 + 4x2 < 12
-4x1 + 3x2 + 8x3 < 10
x1, x2, x3 > 0
41
Dr. Atif Shahzad 15-Oct-15

41. Cont…
Let S1, S2 and S3 be three slack variables.
Modified form is:
Z’ + x1 - 3x2 + 2x3 = 0
3x1 - x2 + 3x3 +S1 = 7
-2x1 + 4x2 + S2 = 12
-4x1 + 3x2 + 8x3 +S3 = 10
x1, x2, x3 > 0
Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 =
10
42
Dr. Atif Shahzad 15-Oct-15

42. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 1 -3 2 0 0 0 0
S1 0 3 -1 3 1 0 0 7 -
S2 0 -2 4 0 0 1 0 12 3
S3 0 -4 3 8 0 0 1 10 10/3
Therefore, x2 is the entering variable and S2 is the
departing variable.
43
Dr. Atif Shahzad 15-Oct-15

43. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 -1/2 0 2 0 3/4 0 9
S1 0 5/2 0 3 1 1/4 0 10 4
x2 0 -1/2 1 0 0 1/4 0 3 -
S3 0 -5/2 0 8 0 -3/4 1 1 -
Therefore, x1 is the entering variable and S1 is the
departing variable.
44
Dr. Atif Shahzad 15-Oct-15

44. Cont…
Basic
Variable
Coefficients of: Sol.
Z’ x1 x2 x3 S1 S2 S3
Z’ 1 0 0 13/5 1/5 8/10 0 11
x1 0 1 0 6/5 2/5 1/10 0 4
x2 0 0 1 3/5 1/5 3/10 0 5
S3 0 0 0 11 1 -1/2 1 11
Optimal Solution is : x1= 4, x2= 5, x3= 0,
Z’ = 11 Z = -11
45
Dr. Atif Shahzad 15-Oct-15

45. Example 4
Max.. Z = 3x1 + 4x2
Subject to constraints:
x1 - x2 < 1
-x1 + x2 < 2
x1, x2 > 0
46
Dr. Atif Shahzad 15-Oct-15

46. Cont…
Let S1 and S2 be two slack variables
.
Modified form is:
Z -3x1 - 4x2 = 0
x1 - x2 +S1 = 1
-x1 + x2 +S2 = 2
x1, x2, S1, S2 > 0
Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2
and Z=0.
47
Dr. Atif Shahzad 15-Oct-15

47. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -3 -4 0 0 0
S1 0 1 -1 1 0 1 -
S2 0 -1 1 0 1 2 2
Therefore, x2 is the entering variable and S2 is the
departing variable.
48
Dr. Atif Shahzad 15-Oct-15

48. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -7 0 0 4 8
S1 0 0 0 1 1 3 -
x2 0 -1 1 0 1 2 -
x1 is the entering variable, but as in x1 column every no. is less than equal to
zero, ratio cannot be calculated. Therefore given problem is having a
unbounded solution.
49
Dr. Atif Shahzad 15-Oct-15

49. Example :
Solving in a compact form
max z = 40𝑥1 + 88𝑥2
2𝑥1 + 8𝑥2 ≤ 60
5𝑥1 + 2𝑥2 ≤ 60
𝑥1, 𝑥2 ≥ 0
50
Dr. Atif Shahzad 15-Oct-15

50. Step 0:
Standard Form
Convert all the inequality constraints into equalities by the use of slack
variables.
Let S1, S2 be three slack variables.
Model can rewritten as:
z - 40x1 -88x2 = 0
Subject to constraints:
2x1+8x2 + S1 = 60
5x1+2x2 + S2 = 60
x1, x2, S1, S2> 0
51
Dr. Atif Shahzad 15-Oct-15

51. 1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
52
Dr. Atif Shahzad 15-Oct-15

52. 1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
53
Dr. Atif Shahzad 15-Oct-15

53. 1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
60
2
60
5
54
Dr. Atif Shahzad 15-Oct-15

54. 1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
60
2
60
5
55
Dr. Atif Shahzad 15-Oct-15

55. 1
0
0
−40
2
5
−88
8
2
0
1
0
0
0
1
0
60
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
𝑅1 + 8𝑅3
𝑅2 − 0.4𝑅3
56
Dr. Atif Shahzad 15-Oct-15

56. 1
0
0
0
0
5
−72
7.2
2
0
1
0
8
−0.4
1
480
36
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
𝑅1 + 8𝑅3
𝑅2 − 0.4𝑅3
57
Dr. Atif Shahzad 15-Oct-15

57. 1
0
0
0
0
5
−72
7.2
2
0
1
0
8
−0.4
1
480
36
60
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
𝑅1 + 10𝑅2
𝑅3 − 2
7.2
𝑅2
58
Dr. Atif Shahzad 15-Oct-15

58. 1
0
0
0
0
5
0
7.2
0
10
1
− 1
3.6
8
−0.4
1
0.9
840
36
50
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
𝑅1 + 10𝑅2
𝑅3 − 2
7.2
𝑅2
59
Dr. Atif Shahzad 15-Oct-15

59. 1
0
0
0
0
5
0
7.2
0
10
1
− 1
3.6
8
−0.4
1
0.9
840
36
50
𝑧 𝑥1 𝑥2 𝑆1 𝑆2 𝑏
𝑅1 + 10𝑅2
𝑅3 − 2
7.2
𝑅2
𝒛 𝒙 𝟏 𝒙 𝟐
𝟖𝟒𝟎 𝟓𝟎
𝟓
𝟑𝟔
𝟕.𝟐
60
Dr. Atif Shahzad 15-Oct-15

60. QUESTIONS

61. Dr. Atif Shahzad
THANK YOU FOR YOUR INTEREST
15-Oct-15