# Lect3n ```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.
1
• 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.
2
• 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
3
Management is seeking a
production schedule that will
increase the company’s
profit.
4
SOLUTION
• Decisions variables:
– 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
5
The Linear Programming Model
Max 8X1 + 5X2
(Weekly profit)
subject to
2X1 + 1X2 &lt; = 1200 (Plastic)
3X1 + 4X2 &lt; = 2400 (Production Time)
X1 + X2 &lt; = 800
(Total production)
X1 - X2 &lt; = 450
(Mix)
Xj&gt; = 0, j = 1,2
(Nonnegativity)
6
X2
1200
The plastic constraint:
2X1+X2&lt;=1200
The
Plastic constraint
Total production constraint:
X1+X2&lt;=800
Infeasible
600
Production
Time
3X1+4X2&lt;=2400
Production mix
constraint:
X1-X2&lt;=450
Feasible
600
800
Interior points.
Boundary
points.
• There are three
types
of feasible
Extreme points.
points
X1
7
We now demonstrate the search for an optimal solu
Start at some arbitrary profit, say profit = \$2,000...
1200
X2
Then increase the profit, if possible...
...and continue until it becomes infeasible
Profit
=\$5040
4,
Profit
= \$ 3,
000
2,
800
600
X1
400
600
800
8
1200
X2
Let’s take a closer look at
the optimal point
Infeasible
800
600
Feasible
Feasible
region
region
X1
400
600
800
9
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).
10
Chapter 3:
Linear Programming
Sensitivity Analysis
Sensitivity Analysis
What if there is uncertainly about one or more
values in the LP model?
1. Raw material changes,
2. Product demand changes,
3. Stock price
Sensitivity analysis allows a manager to ask
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?
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:
1. An Objective Function Coefficient (OFC)
2. A Right Hand Side (RHS) value of a constraint
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.
• 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
Graphical Sensitivity Analysis
We can use the graph of an LP to see what
happens when:
1. An OFC changes, or
2. A RHS changes
Recall the Flair Furniture problem
Sensitivity to Coefficients
x1 + x2 &lt; 8
Max 5x1 + 7x2
Example 1:
Max
5x1 + 7x2
s.t.
x1
6
2x1 + 3x2 &lt; 19
x1 + x 2 &lt; 8
x1 &gt; 0 and x2 &gt; 0
x1 &lt; 6
Optimal solution:
x1 = 5, x2 = 3
2x1 + 3x2 &lt; 19
Feasible
region
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:
Example 1:
-1 &lt; -c1/7 &lt; -2/3
Max
5x1 + 7x2
Multiplying by -1, 1 &gt; c1/7 &gt; 2/3 s.t.
x1
&lt; 6
Multiplying by 7,
7 &gt;
c1
&gt; 14/3
2x1 + 3x2 &lt; 19
x 1 + x2 &lt; 8
x1 &gt; 0 and x2 &gt; 0
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:
Example 1:
-1 &lt; -5/c2 &lt; -2/3
Example 1:
Max
5x1 + 7x2
Max
5x + 7x
Multiplying by -1, 1 &gt; 5/c2 &gt; 2/3
s.t.
x
&lt; 6
s.t.
x1
&lt; 6
Inverting,
1 &lt; c2/5 &lt; 3/2
2x
+ 3x &lt; 19
2x1 + 3x 2 &lt; 19
xx1 ++ xx2&lt;&lt; 8 8
,
xx1 &gt;&gt; 00 and
and xx2&gt; &gt;0 0
Multiplying by 5
5 &lt; c2 &lt; 15/2

1
2
1
1
1
2
1
2
2
Sensitivity to Coefficients
Graphical solution of Example 1
1200
Cutting &amp; Dyeing
0,1062
FinishingLine
Max 7S + 9D
7/10S+1D&lt;=630 (Cutting
&amp; dyeing)
1/2S+5/6D &lt;=600 Sewing
1S+2/3D&lt;=708 Finishing
1/10S+1/4D&lt;=135 Inspection
&amp; Packaging
1000
800
0, 630
600
400
200
708,0
0
0
200
400
600
800
900,0
1000
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 &lt;=slope of the objective function &lt;= slope of line A
i.e. the slope of the objective function should be in between these two
lines
•
•
•
•
•
•
•
•
•
•
•
•
•
7/10S+D=630 (C&amp;D)
D=-7/10S+630
S+2/3D=708 (Finishing)
D=-3/2S+1062
-3/2&lt;=slope of objective function &lt;=-7/10
-3/2&lt;=-Cs/Cd&lt;=-7/10
if we put profit contribution of delux bag same i.e 9
-3/2&lt;=-Cs/9&lt;=-7/10
Cs&gt;=3*9/2
Cs&gt;=27/2
Cs&gt;=13.5
Cs&gt;=63/10
Cs&gt;=6.3
• 6.3 &lt;=Cs&lt;=13.5 (limits for Cs with same optimal solution)
• Similarly the keeping profit contribution of S bag constant.
Cs=10
• 6.67&lt;=Cd&lt;=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&lt;=-Cs/Cs&lt;=-7/10
• now -Cs/Cd=-1.625 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.
• 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
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.
Example
• 6.3 &lt;=Cs&lt;=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= 13.49-10 =3.49
• the increase in percentage is
1.5*100/3.49=42.86% of allowable increase
Example
•
•
•
•
•
•
•
6.67&lt;=Cd&lt;=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= 32.14 of
allowable decrease.
• sum of allowable increase and decrease is 42.86% +
32.14% &lt;100% hence optimal solution is still valid
S=540 and D-252.
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?
Effect of change of the righ hand side
of the constraint.
and dyeing constraint
• 7/10S+1d&lt;=640 ???
• Feasible region extended, find extreme point
using intersection of two lines
• S=527.5 ,D=270.75 Max 10S+9D
• Profit= 7711.75 which is 7711.75-7668.00 =43.75
• Increase in profit/hr =
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.
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.
Sensitivity Analysis for
a Minimization Problem
Burn-Off makes a “miracle” diet drink
Decision: How much of each of 4
ingredients to use?
Objective: Minimize cost of ingredients
Data
Units of Chemical per Ounce of Ingredient
Ingredient
X
A
3
B
4
C
8
D
10
&gt; 280 units
Y
5
3
6
6
&gt; 200 units
Z
10
25
20
40
&lt; 1050 units
Chemical
\$ per ounce of ingredient
\$0.40
\$0.20 \$0.60
\$0.30
Requirement
Min 0.40A + 0.20B + 0.60C + 0.30D
(\$ of cost)
Subject to the constraints
A+B+C+D
&gt; 36 (min daily ounces)
3A + 4B + 8C + 10D
&gt; 280 (chem x min)
5A + 3B + 6C + 6D
&gt; 200 (chem y min)
10A + 25B + 20C + 40D &lt; 1050 (chem z max)
A, B, C, &gt; 0
```