1. THE GALAXY INDUSTRY PRODUCTION PROBLEM -
Galaxy manufactures two toy models:
Space Ray.
Zapper.
Resources are limited to
1200 pounds of special plastic.
40 hours of production time per week.

2. Marketing requirement
Total production cannot exceed 800 dozens.
Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 450.
Technological input
Space Rays requires 2 pounds of plastic and
3 minutes of labor per dozen.
Zappers requires 1 pound of plastic and
4 minutes of labor per dozen.

3. Current production plan calls for:
Producing as much as possible of the more profitable product, Space Ray ($8 profit per dozen).
Use resources left over to produce Zappers ($5 profit
per dozen).
The current production plan consists of:
Space Rays = 550 dozens
Zapper = 100 dozens
Profit = 4900 dollars per week

4. Management is seeking a production schedule that will increase the company’s profit.

5. SOLUTION Decisions variables: Objective Function:
X1 = Production level of Space Rays (in dozens per week).
X2 = Production level of Zappers (in dozens per week).
Objective Function:
Weekly profit, to be maximized

6. The Linear Programming Model
Max 8X1 + 5X2 (Weekly profit)
subject to
2X1 + 1X2 < = (Plastic)
3X1 + 4X2 < = (Production Time)
X1 + X2 < = (Total production)
X X2 < = (Mix)
Xj> = 0, j = 1,2 (Nonnegativity)

7. There are three types of feasible pointsX2
The plastic constraint:
2X1+X2<=1200
1200
The Plastic constraint
Total production constraint:
X1+X2<=800
Infeasible
600
Production mix constraint:
X1-X2<=450
Production Time
3X1+4X2<=2400
Feasible X1
600
800
Interior points.
There are three types of feasible points
Boundary points.
Extreme points.

8. We now demonstrate the search for an optimal solution
Start at some arbitrary profit, say profit = $2,000...
Then increase the profit, if possible... X2
1200
...and continue until it becomes infeasible
Profit =$5040
Profit = $ 2, 3, 4,
800
Recall the feasible Region
600 X1
400
600
800

9. X2 1200 800 600 X1 400 600 800 Let’s take a closer look at
the optimal point
800
Infeasible
600
Feasible
region
Feasible
region X1
400
600
800

10. Summary of the optimal solution
Space Rays = 480 dozens
Zappers = 240 dozens
Profit = $5040
This solution utilizes all the plastic and all the production hours.
Total production is only 720 (not 800).
Space Rays production exceeds Zapper by only 240 dozens (not 450).

11. Chapter 3: Linear Programming Sensitivity Analysis

12. What if there is uncertainly about one or more values in the LP model?
Sensitivity Analysis
What if there is uncertainly about one or more values in the LP model?
Raw material changes,
Product demand changes,
Stock price
Sensitivity analysis allows a manager to ask certain hypothetical questions about the problem, such as:
How much more profit could be earned if 10 more hours of labour were available?
Which of the coefficient in model is more critical?

13. Sensitivity Analysis
Sensitivity analysis allows us to determine how “sensitive” the optimal solution is to changes in data values.
Is the optimal solution sensitive to changes in input parameters?
This includes analyzing changes in:
An Objective Function Coefficient (OFC)
A Right Hand Side (RHS) value of a constraint

14. Limit of Range of optimality
Max Ax + BY
Keeping x, Y same how the object function behaves if A, B are changed.
The optimal solution will remain unchanged as long as An objective function coefficient lies within its range of optimality There are no changes in any other input parameters
If the OFC changes beyond that range a new corner point becomes optimal.

15. Generally, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines.
Binding constraint
Are the constraints that restrict the feasible region

16. Graphical Sensitivity Analysis
We can use the graph of an LP to see what happens when:
An OFC changes, or
A RHS changes
Recall the Flair Furniture problem

17. Feasible region Sensitivity to Coefficients Example 1: Max 5x1 + 7x2
s.t x
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
x1 + x2 < 8
Max 5x1 + 7x2
x1 < 6
Optimal solution:
x1 = 5, x2 = 3
2x1 + 3x2 < 19
Feasible
region

18. Compute the range of optimality for c1 in Example 1.
Sensitivity to Coefficients Range of optimality for c1
Compute the range of optimality for c1 in Example 1.
The slope of an objective function line, Max c1x1 + c2x2, is -c1/c2.
The slope of the binding third constraint, x1 + x2 = 8, is -1.
The slope of the binding second constraint, 2x1 + 3x2 = 19, is -2/3.
Find the range of values for c1 (with c2 staying 7) such that the objective-function line slope lies between that of the two binding constraints:
-1 < -c1/7 < -2/3
Multiplying by -1, > c1/7 > 2/3
Multiplying by 7, > c1 > 14/3
Example 1:
Max x1 + 7x2
s.t x < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0

19. Likewise, compute the range of optimality for c2 in Example 1.
Sensitivity to Coefficients Range of optimality for c2
Likewise, compute the range of optimality for c2 in Example 1.
The slope of the binding third constraint is -1.
The slope of the binding second constraint is -2/3.
Find the range of values for c2 (with c1 staying 5) such that the objective-function line slope lies between that of the two binding constraints:
-1 < -5/c2 < -2/3
Multiplying by -1, > 5/c2 > 2/3
Inverting, < c2/5 < 3/2 ,
Multiplying by < c2 < 15/2
Example 1:
Max x1 + 7x2
s.t x < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Example 1:
Max x1 + 7x2
s.t x < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0

20. Sensitivity to Coefficients Graphical solution of Example 1
Max 7S + 9D
7/10S+1D<=630 (Cutting
& dyeing)
1/2S+5/6D <=600 Sewing
1S+2/3D<=708 Finishing
1/10S+1/4D<=135 Inspection
& Packaging
Optimal solution:
S=540,D=252
1. objective function
2. Cutting Line
3.Finishing Line
This point will be an optimal solution as long as:
slope of line A <=slope of the objective function <= slope of line A
i.e. the slope of the objective function should be in between these two lines

21. 6.3 <=Cs<=13.5 (limits for Cs with same optimal solution)
7/10S+D=630 (C&D)
D=-7/10S+630
S+2/3D=708 (Finishing)
D=-3/2S+1062
-3/2<=slope of objective function <=-7/10
-3/2<=-Cs/Cd<=-7/10
if we put profit contribution of delux bag same i.e 9
-3/2<=-Cs/9<=-7/10
Cs>=3*9/2
Cs>=27/2
Cs>=13.5
Cs>=63/10
Cs>=6.3
6.3 <=Cs<=13.5 (limits for Cs with same optimal solution)

22. 6.67<=Cd<=14.29 (range of optimality)
Similarly the keeping profit contribution of S bag constant. Cs=10
6.67<=Cd<=14.29 (range of optimality)
If both the Cs, Cd are changed simultaneously (i.e S bags to 13, D bags to 8)
Calculate the slope again:
-cs/cd =-13/8=-1.625
-3/2<=-Cs/Cs<=-7/10
now -Cs/Cd= which is less than -3/2 which is not acceptable according to above equation hence this means if we change the both cofficient than 540 and 252 would not be the optimal solution.

23. Multiple changes
The range of optimality is valid only when a single objective function coefficients changes.
When more than one variable changes we turn to the
100% rule

24. 100% RuLE
For each increase (decrease) in an objective function coefficient, calculate (and express as a percentage) the ratio of the change in the coefficient to the maximum possible increase (decrease) as determined by the limits of the range of optimality.
Sum all these percent changes. If the total is less than 100 percent, the optimal solution will not change. If this total is greater than or equal to 100%, the optimal solution may change.

25. Example 6.3 <=Cs<=13.5 (limits for Cs with same optimal solution
Suppose:
S is changed from 10$ to 11.50$;
Range of optimality from the sensitivity analysis :
allowable upper limit for S= 13.49;
Value of S =$10
Allowable increase in S= =3.49
the increase in percentage is 1.5*100/3.49=42.86% of allowable increase

26. Example 6.67<=Cd<=14.29 (range of optimality)
For D allowable lower limit is 6.66
allowable Value of D=9
allowable Decrease =9-6.66=2.33
D is reduced from $9 to $8.25
for the present case D is reduced from $9 to $8.25
for present case =0.75/2.33*100= of allowable decrease.
sum of allowable increase and decrease is 42.86% % <100% hence optimal solution is still valid S=540 and D-252.

27. Effect of change of the right hand side of the constraint.
Any change in a right hand side of a binding constraint will change the optimal solution.
Any change in a right-hand side of a nonbinding constraint that is less than its slack or surplus, will cause no change in the optimal solution.
Keeping all other factors the same, how much would the optimal value of the objective function (for example, the profit) change if the right-hand side of a constraint changed by one unit?

28. Effect of change of the righ hand side of the constraint.
suppose if additional 10 hrs is added to cutting and dyeing constraint
7/10S+1d<=640 ???
Feasible region extended, find extreme point using intersection of two lines
S=527.5 ,D= Max 10S+9D
Profit= which is =43.75
Increase in profit/hr =

29. Dual Price
It is the improvement in the optimal solution per unit increase in the RHS of constraint
If dual price is negative this means value of objective function will not improved rather it would get worse ,if value at rhs of the constrain is increased by 1 unit . for minimization problem it means cost will increase by 10.

30. Characteristics of RHS Changes
The constraint line shifts, which could change the feasible region
Slope of constraint line does not change
Corner point locations can change
The optimal solution can change

31. Reduced Cost is:
The minimum amount by which the OFC of a variable should change to cause that variable to become non-zero.
The amount by which the objective function value would change if the variable were forced to change from 0 to 1.

32. Sensitivity Analysis for a Minimization Problem
Burn-Off makes a “miracle” diet drink
Decision: How much of each of ingredients to use?
Objective: Minimize cost of ingredients

33. Data Units of Chemical per Ounce of Ingredient A B C D X 3 4 8 10
Requirement A B C D X 3 4 8 10
> 280 units Y 5 6
> 200 units Z 25 20 40
< 1050 units
$ per ounce of ingredient
$0.40
$0.20
$0.60
$0.30

34. A + B + C + D > 36 (min daily ounces)
Min 0.40A B C D ($ of cost)
Subject to the constraints
A + B + C + D > 36 (min daily ounces)
3A + 4B + 8C + 10D > 280 (chem x min)
5A + 3B + 6C + 6D > 200 (chem y min)
10A + 25B + 20C + 40D < 1050 (chem z max)
A, B, C, > 0