Computer Output Solver

1
Computer Output Solver MPL
2

• Trial and Error
• Solve by hand. Can be used for small problems
  Linear Models, but is impractical for large
  complex problems b/c numerous calculations
  required
• Graphical Method
• Solve by hand. Can be used for Linear models of
  Two variables
• Computer
• Used for all sizes and types of programs Special
  solution procedures have been developed for
  specific mathematical models.
• Simplex Method
• Solve by hand. Used for All Linear models

How do you SOLVE a Model?
3
Computer Programs designed to solve LP problems
are now widely available Free Student/Trial
Versions of MPL (Mathematical Programming
Language) for Windows with CPLEX Solver http//ww
w.maximal-usa.com/ Linear programming solvers
are now part of many spreadsheet packages, such
as Solver in Microsoft Excel.
4
MPL and Excel Solver use the simplex method to
find a solution!
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5
MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max Demand of
A 2x1 3x2 lt 19 Available P x1 x2 lt 8
Available Q x1,x2 gt0 Non-negativity
6
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2
2x1 3x2 lt 19
1
x1
1
2
3
4
5
6
7
8
9
10
7

• The Solver add-in is not installed on the lab
  machines.
• To install follow thisRight-click the empty
  blue bar that contains "Home""Insert".."View"Sel
  ect "Customize quick-access toolbar."Select the
  "Add-ins" Tab on the leftClick the "Go" button
  on the bottom of the windowCheck the "Solver
  Add-in" boxClick "OK"-OR-
• Click on the round symbol in upper left corner of
  the screen.
• Click on Excel Options
• Click on Add-Ins
• Click on Solver (it is at the top of the list)
• Click OK
• The solver takes 2 minutes to install. It should
  appear on the right end of the "Data" tab in
  Excel when complete.

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You dont need non-negativity conditions because
Solver already assumes non-negativity To solve
go to Tools, Solver. Make sure in Options
that Non-negativity is checked and Press Solve
9
MPL is installed on 4 computers in ISOM room G23.
Facing the room, they are the back row of
computers on the left.
10
You dont need non-negativity conditions because
MPL already assumes non-negativity
11
Push this button to solve the model
12
Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
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1E30 stands for Infinity
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1E020 stands for Infinity
17
Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
18
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2
2x1 3x2 lt 19
1
x1
1
2
3
4
5
6
7
8
9
10
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Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
22
The RANGE OF OPTIMALITY for each PARAMETER
provides the range of values over which the
current solution will remain optimal.
22
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MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max Demand of
A 2x1 3x2 lt 19 Available P x1 x2 lt 8
Available Q x1,x2 gt0 Non-negativity
24
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2x13x2lt19
2
C
1
x1
24
B
A
1
2
3
4
5
6
7
8
9
10
25
x2
If the slope of the objective function is
changed, the optimal point will change when the
slope of the objective function line is parallel
to a binding constraint
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x13x2lt19
2
1
x1
25
1
2
3
4
5
6
7
8
9
10
26
x2
If the slope of the objective function is
changed, the optimal point will change when the
slope of the objective function line is parallel
to a binding constraint
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x13x2lt19
2
1
x1
26
1
2
3
4
5
6
7
8
9
10
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This means that we are determining the range for
EACH parameter holding the other parameters
CONSTANT!

• For EACH parameter
• Determine the Binding Constraints (the
  constraints that pass through the optimal point)
• Find the slopes of the binding constraints
• In the objective function at the optimal value,
  Replace the Parameter in question with c Max
  c1x1 c2x2 and Find the slope.
• Set up the following inequality
• Solve for c Remember when you multiple
  inequalities you
• -Have to multiple both sides
• -Multiplication/Division by a (-) reverses the
  
• inequality also Inversion reverses inequality

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Given the objective function at the optimal
value, 5x17x246 Lets 1st examine change in
the 1st parameter, c1 c1x17x246 Find
slope c1x17x246 7x2-c1x146 x2-c1x16.57
7 Slope-c1/7
What are the binding constraints? 2x13x2lt19 x1x2
lt8 What are their slopes? 2x13x219 x2-(2/3)x1
6.33 Slope -2/3 x1x28 x2-x18 Slope -1
-1 lt -c1/7 lt -2/3 -1 lt -c1/7 lt -2/3 -7 -7
-7 7 gt c1 gt 4.67 -or- 4.67 lt c1 lt 7
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29
Holding c2 constant, c1 can change between 4.67
and 7 without affecting the optimal solution
(5,3). What happens to the value of the
objective function? It WILL change!
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30
Holding c1 constant, c2 can change between 5 and
7.5 without affecting the optimal solution (5,3).
What happens to the value of the objective
function? It WILL change!
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c1 upper bound 52 7, lower bound
5-0.334.67 The range of optimality for c1 is
4.67, 7 c2 upper bound 70.57.5, lower bound
7-25 The range of optimality for c2 is 5,
7.5 Managers should focus on those objective
coefficients that have a narrow range of
optimality and coefficients near the endpoints of
the range.
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Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
34
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2
2x1 3x2 lt 19
1
x1
1
2
3
4
5
6
7
8
9
10
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Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
38
If a decision variable has a value of 0 in
optima, the reduced cost is the amount the
variable's objective function coefficient would
have to improve (increase for maximization
problems, decrease for minimization problems)
before this variable could assume a positive
value. The reduced cost for a decision variable
with a positive value in optima is 0.
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If a decision variable has a value of 0 in
optima, the reduced cost is the amount the
variable's objective function coefficient would
have to improve (increase for maximization
problems, decrease for minimization problems)
before this variable could assume a positive
value. The reduced cost for a decision variable
with a positive value in optima is 0.
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Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
42
Slack and surplus variables represent the
difference between the left and right sides of
the constraints. To find the slack/surplus value
at each corner point plug the corner point into
each constraint and solve for the slack/surplus
variable
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In an optimal solution, the slack (or surplus)
variables associated with the binding constraints
are zero. (The binding constraints are
tight) In an optimal solution, the slack (or
surplus) variables associated with the
non-binding constraints are positive. (The
non-binding constraints are not tight)
Whats the relationship between slack/surplus
and shadow prices??if slack/surplusgt0, then the
corresponding constraint is non-binding,
therefore, the shadow price is zero.
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MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max Demand of
A 2x1 3x2 lt 19 Available P x1 x2 lt 8
Available Q x1,x2 gt0 Non-negativity
45
MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Let
s1 denotes the slack for constraint 1 Let s2
denotes the slack for constraint 2 Let s3 denotes
the slack for constraint 3 Max Profit 5x1 7x2
0s1 0s2 0s3 s.t. x1 s1 6 Max Demand of
A 2x1 3x2 s2 19 Available P x1 x2 s3
8 Available Q x1,x2,s1,s2,s3 gt0 Non-negativity
46
MODEL Point E (5,3) x1 s1 6 5 s1 6 s1
6-51 2x1 3x2 s2 19 2(5) 3(3) s2 19
s2 19-190 x1 x2 s3 8 5 3 s3 8 s3
8-80
47
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
(0,6.33,6,0,1.67)
6
A
(5,3,1,0,0)
5
4
Feasible Region
3
E
(6,2,0,1,0)
2
2x1 3x2 lt 19
D
1
(6,0,0,7,2)
(0,0,6,19,8)
x1
C
B
1
2
3
4
5
6
7
8
9
10
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Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
51
MODEL Let x1 denotes the quantity of A produced
Let x2 denotes the quantity of B produced. Max
Profit 5x1 7x2 s.t. x1 lt 6 Max
Demand of A 2x1 3x2 lt 19 Available P x1 x2
lt 8 Available Q x1,x2 gt0 Non-negativity
We will be examining changes to the RIGHT HAND
SIDE values of the CONSTRAINTS
52
The improvement in the value of the optimal
solution per unit increase in the right-hand side
is called the SHADOW PRICE. What does shadow
price mean? The shadow price of constraint i
(resource i) represents the price we should be
willing to pay to obtain one more unit of
resource i.
How do changes in the RHS values of the
constraints affect the optimal solution?
53
Changing a RHS value results in a parallel shift
of the changed constraint. This may affect both
the optimal solution and the optimal value of the
objective function.
How do changes in the RHS values of the
constraints affect the optimal solution?
54
Remember that a positive shadow price means
an IMPROVEMENT which in a maximization problem
increases the obj. function value and decreases
the obj. function value in a minimization
problem!
55

• The shadow price for a non-binding constraint is
  0
• Graphically, the shadow price of the RHS of a
  binding constraint is determined by
• Add 1 to the right hand side value of the
  constraint
• 2. Resolve for the optimal solution in terms of
  the same two binding constraints.
• 3. Shadow price new value of the objective
  function original value of the objective
  function
• Max problem
• shadow price znew zold
• Min problem
• shadow price zold znew

How do we compute the SHADOW PRICE?
55
56
x2
Constraint 1 Since x1 lt 6 is not a binding
constraint, its shadow price is 0 (its slack is
1) . So, Springfield Inc. does not want to pay
more money to increase one more unit of the
demand of A.
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x13x2lt19
2
1
x1
1
2
3
4
5
6
7
8
9
10
57
Constraint 2 (a binding constraint) Increase the
RHS value of the 2nd constraint from 19 to 20 (by
1 unit) and resolve for the optimal point
determined by the two binding constraints 2x1
3x2 20 and x1 x2 8. The solution is x1
4, x2 4, Z 48. Hence, the shadow price
znew-zold48-462 This means that 1 more unit of
material P will increase the total profit by 2.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x13x2lt19
2
1
x1
1
2
3
4
5
6
7
8
9
10
58
Constraint 3 Constraint 3 (a binding
constraint) Change the RHS value of the third
constraint from 8 to 9 and resolve for the
optimal point determined by the two
binding constraints 2x1 3x2 19 and x1 x2
9. The solution is x1 8, x2 1, z
47. The shadow price is znew - zold
47-461. This means that 1 more unit of material
Q will increase the total profit by 1.
x2
10
Max 5x17x2
9
x1 lt 6
8
x1 x2 lt 8
Optimal point (5,3) Optimal obj. function value
46
7
6
5
4
Feasible Region
3
2x13x2lt19
2
1
x1
1
2
3
4
5
6
7
8
9
10
59
Unfortunately, MS Excel Solver gives Dual Price
though it is called Shadow Price. Remember!!!!
For maximization Prob. DualShadow price. For
minimization prob. Dual-Shadow price.
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Since this is a maximization problem, a positive
shadow price means the objective function
increases
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Provides information about The objective
function The optimal value Ranges of
optimality The decision variables The optimal
solution Reduced cost The constraints Slack
or surplus Shadow price RHS ranges of
feasibility
63
The range of values for the RHS over which the
shadow price is applicable, which means the
optimal solution is formed by the same set of
constraints As the RHS changes, other
constraints will become binding and limit the
change in the value of the objective function.
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1E020 stands for Infinity
65
Constraint 1 upper bound 6(1E30) infinity
lower bound 6-15 The range of feasibility for
constraint 1 is (5, infinity) Constraint
2 upper bound 19524 lower bound
19-118 The range of feasibility for constraint 2
is (18, 24) Constraint 3 upper bound 81/38
1/3 lower bound 8-1 2/36 1/3 The range of
feasibility for constraint 3 is (6 1/3, 8 1/3)
66
x2
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2
2x1 3x2 lt 19
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
67
x2
10
If we increase the RHS of constraint 1 by 1 unit,
it does not affect the optimal solution or
optimal value, hence shadow price 0
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2
2x13x2lt19
1
C
x1
B
A
1
2
3
4
5
6
7
8
9
10
68
x2
10
If we increase the RHS of constraint 1 by 2
units, it does not affect the optimal solution or
optimal value, hence shadow price 0
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2
2x13x2lt19
1
x1
A
B
1
2
3
4
5
6
7
8
9
10
69
x2
10
If we increase the RHS of constraint 1 by
infinity, it does not affect the optimal solution
or optimal value, hence shadow price 0
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2
2x13x2lt19
1
x1
A
C/B
1
2
3
4
5
6
7
8
9
10
70
x2
If we decrease the RHS of constraint 1 by 1 unit,
the constraint x1lt 6 will start to become binding
and therefore, the optimal solution will be
formed by a new set of constraints. The shadow
price of the constraint x1lt 6 will start to
become non-zero.
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
3
Feasible Region
C/D
2
2x13x2lt19
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
71
x2
10
If we decrease the RHS of constraint 1 by 2
units, it does affect the optimal solution and
optimal value, hence shadow price ? 0
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (4,3.67) Optimal obj. function
value 45.67
7
6
E
5
4
D
3
Feasible Region
2
2x13x2lt19
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
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Constraint 1 upper bound 6(1E30) infinity
lower bound 6-15 The range of feasibility for
constraint 1 is (5, infinity) Constraint
2 upper bound 19524 lower bound
19-118 The range of feasibility for constraint 2
is (18, 24) Constraint 3 upper bound 81/38
1/3 lower bound 8-1 2/36 1/3 The range of
feasibility for constraint 3 is (6 1/3, 8 1/3)
73
x2
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
Optimal point (5,3) Optimal obj. function value
46
7
6
E
5
4
Feasible Region
3
D
2
2x13x2lt19
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
74
x2
10
If we increase the RHS of constraint 2 by 1 unit,
it improves the optimal value by 2 (from 46 to
48) hence shadow price 2
Max 5x17x2
9
x1lt6
8
x1x2lt8
Optimal point (4,4) Optimal obj. function value
48
7
E
6
5
4
D
3
Feasible Region
2x13x2lt20
2
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
75
x2
10
If we increase the RHS of constraint 2 by 5
units, it improves the optimal value by 10 (from
46 to 56) hence shadow price 2
Max 5x17x2
9
x1lt6
8
D/E
x1x2lt8
Optimal point (0,8) Optimal obj. function value
56
7
6
5
4
Feasible Region
2x13x2lt24
3
2
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
76
x2
If we increase the RHS of constraint 2 by more
than 5 units it does not improve the objective
any more and the optimal solution is formed by
two new constraints, hence shadow price is no
longer applicable
10
Max 5x17x2
9
x1lt6
8
E
x1x2lt8
Optimal point (0,8) Optimal obj. function value
56
7
6
5
4
Feasible Region
3
2x13x2lt30
2
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
77
x2
If we decrease the RHS of constraint 2 by 1 unit
it makes the objective function value worse by 2
units Hence shadow price 2
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
7
Optimal point (6,2) Optimal obj. function value
44
6
E
5
4
3
Feasible Region
2
2x13x2lt18
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
78
x2
If we decrease the RHS of constraint 2 by 3 unit
it makes the objective function value worse by
(46-39.33) 6.67 According to the shadow price,
a 3 unit decrease should have resulted in 236
unit change in the obj. function value Hence
shadow price is not applicable since the new
optimal solution is formed by two different
constraints. Therefore, we are beyond the
allowable decrease for the RHS
10
Max 5x17x2
9
x1 lt 6
8
x1x2lt8
7
Optimal point (6,1.33) Optimal obj. function
value 39.33
6
E
5
4
3
Feasible Region
2
2x13x2lt16
D
C
1
x1
B
A
1
2
3
4
5
6
7
8
9
10
79
Olympic Bike is introducing two new lightweight
bicycle frames, the Deluxe and the Professional,
to be made from special aluminum and steel
alloys. The anticipated unit profits are 10 for
the Deluxe and 15 for the Professional. A
supplier delivers 100 pounds of the aluminum
alloy and 80 pounds of the steel alloy weekly.
How many Deluxe and Professional frames
should Olympic produce each week?
80
Verbal Statement of the Objective
Function Maximize total weekly profit. Verbal
Statement of the Constraints Total weekly usage
of aluminum alloy lt 100 pounds. Total weekly
usage of steel alloy lt 80 pounds. Definition of
the Decision Variables x1 number of Deluxe
frames produced weekly. x2 number of
Professional frames produced weekly.
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85

• Question
• Suppose the profit on deluxe frames is decreased
  to 8. Should the company still produce deluxe
  frames? If so, what is new total profit?

86

• Answer
• The output states that the solution remains
  optimal as long as the objective function
  coefficient of x1 is between 7.5 and 22.5. Since
  8 is within this range, the optimal solution will
  not change. The optimal profit will change 8x1
  15x2 8(15) 15(17.5) 382.5.

87

• Question
• If the unit profit on deluxe frames were 6
  instead of 10, would the company change the
  production plan?

88

• Answer
• The output states that the solution remains
  optimal as long as the objective function
  coefficient of x1 is between 7.5 and 22.5. Since
  6 is outside this range, the optimal solution
  would change.

89

• Question
• If Olympic Bike can get an extra supply of
  aluminum alloy for 2 per pound or an extra
  supply of steel alloy for 1.2 per pound. What
  would you recommend Olympic Bike to purchase?

90

• Answer
• The shadow price for aluminum alloy is 3.125,
  which means additional pound of aluminum alloy
  will contribute 3.125 increase in total profit.
  The shadow price for steel alloy is 1.25, which
  means addition pound of steel alloy will increase
  the total profit by 1.25. Given the purchase
  cost for the additional material, the net profit
  for each additional pound of aluminum alloy is
  3.125-21.125 while 1.25-1.20.05 for steel
  alloy. So the company should buy aluminum alloy.

91
End of Lecture 5
92

• Read Chapter 3
• Complete Homework Problems

93
LEO Dairy. Inc. produces three products liquid
milk, yogurt and milk powder. In February, LEO
Dairy has 100 tons of raw milk available. It
takes one unit of raw milk to produce one unit of
liquid milk, two units of raw milk to produce one
unit of yogurt, and eight units of raw milk to
produce one unit of milk powder. The maximum
demand for yogurt is 20 tons per month. The
maximum demand for milk powder is 10 tons per
month. The demand of liquid milk is unlimited. In
order to keep the market share, LEO Dairy at
least produces 5 tons of liquid milk each month.
The profit of the one ton liquid milk is 1,000,
one ton yogurt is 2,200 and one ton milk powder
is 10,000 LEO Dairy wants to maximize the total
profit in February. Solve in Excel. Decision
Variables Let x1 denote the number of tons of
liquid milk produced Let x2 denote the number of
tons of yogurt produced Let x3 denote the number
of tons of milk powder produced
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Q1 If the liquid milk profit now is 1080 per
ton, should we produce more liquid milk? Q2 If
the profit of milk powder now is 8000 per ton,
should we change our production plan? Q3 (I am
tricky) If we have 1500 Investment which can be
used in one of the following options 1. Invest
in TV commercial and increase the demand of
milk powder by 1.5 tons. 2. Purchase 2 tons of
raw milk. 3. Increase the demand of yogurt by 4
units Q4. Would it be beneficial if we decrease
the level of liquid milk market share?
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• You already developed the model, Solved using the
  Graphical Method, determined the feasible region,
  corner points, optimal solution and optimal
  objective function value. Also you graphed the
  objective function. Now, apply sensitivity
  analysis and determine the range of optimality
  and the dual prices Compute the Slack/Surplus
  graphically, and also Solve in Excel.
• HighTec Inc. assembles two different models of
  personal computers, Deskpro and Portable. The
  Deskpro generates a profit contribution of
  50/unit and the Portable generates a profit
  contribution of 40/unit. HighTec Inc. is
  currently interested in developing a weekly
  production schedule for both products. For next
  weeks production, a maximum of 150 hours of
  assembly time can be made available. Each unit
  of the Deskpro requires 3 hours of assembly time
  and each unit of the Portable requires 5 hours of
  assembly time. HighTec Inc. only has 20 Portable
  display components in inventory thus, no more
  than 20 units of the Portable may be assembled.
  Finally, only 300 square feet of warehouse space
  can be made available for new production.
  Assembly of each Deskpro requires 8 square feet
  of warehouse space similarly, each Portable
  requires 5 square feet.

95
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MODEL Let d units Deskpro Let punits
Portable Max Profit 50d40p s.t. 3d5p lt 150
(Assembly Time) plt20 (Portable Display)
8d5p lt 300 (Warehouse
Capacity) d,pgt0 (Non-negativity) S
OLUTION Produce 30 Deskpro and 12 Portable for a
optimal value of 1980.
8d5p lt 300
p lt 20
A
E
3d5p lt 150
Feasible Region
D
B
C
d
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