Lecture 7
Business Analytics
Lecture 7

Shelby Shelving Model

Graphical representation of linear programs

Sensitivity Analysis and Shadow Prices
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Lecture 7
Business Analytics
Bland Brewery Problem
“Raw” Components
Corn
Products
Beer
Hops
Malt
Ale
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Lecture 7
Business Analytics
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Bland Brewery Problem (Continued)
• Profitability
1 Barrel of Beer
1 Barrel of Ale
$23
$13
• Mixing Quantities
• Availability
1 Barrel of Beer
1 Barrel of Ale
Corn
480 lbs
Corn
15 lbs
5 lbs
Hops
160 ozs
Hops
4 ozs
4 ozs
Malt
1,190 lbs
Malt
20 lbs
35 lbs
How much of beer and ale to produce, if we want to
maximize profits?
Business Analytics
Bland Brewery Model: Standard Notation



Decision Variables
Let A = # of barrels of ale to produce, and
B = # of barrels of beer to produce.
Note: Use suggestive (mnemonic) variable names for
readability
Objective Function
Profit in $ = 13A + 23B
Constraints
Corn Availability: 5A + 15B ≤ 480
Hops Availability: 4A + 4B ≤ 160
Malt Availability: 35A + 20B ≤ 1190
Non-negativity: A, B ≥ 0
Lecture 7
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Lecture 7
Business Analytics
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Bland Brewery Linear Program
Objective Function
max 13 A + 23 B (Profit)
Coefficients
subject to
(corn) 5A + 15B ≤ 480
(hops) 4A + 4B ≤ 160
Right hand sides
(malt) 35A + 20B ≤ 1190
(nonnegativity)
A, B ≥ 0
Variables
Lecture 7
Business Analytics
Terminology

Feasible and Infeasible Solutions
 A production plan (A,B) that satisfies all of the constraints is
called a feasible solution
 For example, in the Bland Brewery LP, the solution (A=10,
B=10) is feasible.

Constraints
Corn Availability: 5 x 10 + 15 x 10 = 200 ≤ 480
Hops Availability: 4 x 10 + 4 x 10 = 80 ≤ 160
Malt Availability: 35 x 10 + 20 x 10 = 550 ≤ 1190
Non-negativity: 10, 10 ≥ 0
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Lecture 7
Business Analytics
Terminology (continued)

Feasible and Infeasible Solutions
 The production plan (A=40, B=10) is not feasible, i.e. it is
infeasible because the hops and malt constraints are violated

Constraints
Corn Availability: 5 x 40 + 15 x 10 = 350 ≤ 480
Hops Availability: 4 x 40 + 4 x 10 = 200 > 160
Malt Availability: 35 x 40 + 20 x 10 = 1600 > 1190
Non-negativity: 40, 10 ≥ 0
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Business Analytics

Lecture 7
Optimal Solution
 For a maximization (respectively, minimization) problem, an
optimal solution is a feasible solution that has the largest
(respectively, smallest) objective function value among all
feasible solutions
 The optimal solution for the Bland Brewery production model
is (A=12, B=28). The optimal objective function value is $800.
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Lecture 7
Business Analytics
Assumptions in a Linear Program

Continuity: the decision variables are continuous, i.e., fractional
values are allowed

Proportionality: for example, it takes twice as much hops to
make twice as much beer or ale; there are no economies of
scale
 Additivity: profit is the sum of the profit contributions from ale
and beer
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Lecture 7
Business Analytics
Assumptions in a Linear Program



In short, the objective function and constraints must be linear
with respect to decision variables
Linear functions of A and B:
 13A + 23B
 0.5A + (2/3)B
Non-Linear functions of A and B:
 13A2 + 23AB
 log(A) + cos(B)
 max(A,0)
 IF(A< 5,0,10)
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Lecture 7
Business Analytics
Assumptions in a Linear Program
.
Allowable variations:

Objective function can be maximized or minimized

Constraints can be ≥, ≤, or =

Noninteger or integer coefficients and right-hand sides are
allowed

Negative or positive coefficients and right-hand sides are allowed
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Lecture 7
Business Analytics
Shelby Shelving Case

A small company that produces two types of shelves for grocery stores:
Model S and Model LX

Three steps in manufacturing process:
 Stamping, forming and assembly
 Capacity of Stamping and Forming machines: 800 hours a month
 Capacity of model S assembly department: 1900 units/month
 Capacity of model LX assembly department: 1400 units/month

Machine Requirements (Hours/per unit):
 Stamping: 0.3h/u for model S and 0.3h/u for model LX
 Forming: 0.25h/u for model S and 0.5h/u for model LX

Selling Price: Model S $1800/unit; Model LX $2100/unit

Current monthly production: 400 units of S and 1400 unit of LX
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Lecture 7
Business Analytics
Shelby Shelving: Data and Objective

Fixed Cost = 385,000

Variable Costs

Variable Costs
Model S
Model LX
Selling Price
1,800
2,100
Direct Materials
1,000
1,200
Direct Labor
175
210
Variable
Overhead
500
445
Objective to be maximized = Net Profit
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Lecture 7
Business Analytics
Shelby Shelving: Decision Model

Decision Variables:
Let S = # of Model S shelves to produce, and
LX = # of Model LX shelves to produce.
Model S
Model LX
Selling Price
1,800
2,100
Direct Materials
1,000
1,200
Direct Labor
175
210
Variable
Overhead
500
445
Profit
Contribution
125
245
= 1,800-(1,000+175+500)

Net Profit ($) = 125S + 245LX - 385,000
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Lecture 7
Business Analytics
Shelby Shelving: Constraints
(S assembly)
S
≤ 1900
(LX assembly)
LX ≤ 1400
(Stamping) 0.3 S + 0.3 LX ≤ 800
(Forming) 0.25 S + 0.5 LX ≤ 800
(Nonnegativity)
S, LX ≥ 0
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Business Analytics
Shelby Shelving: Complete LP formulation

Linear Program
max 125 S + 245 LX - 385,000
(Net Profit)
subject to:
(S assembly)
S
≤ 1900
(LX assembly)
LX ≤ 1400
(Stamping) 0.3 S + 0.3 LX ≤ 800
(Forming) 0.25 S + 0.5 LX ≤ 800
(Nonnegativity)
S, LX ≥ 0
Lecture 7
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Lecture 7
Business Analytics
Spreadsheet Solution

Optimal Solution: S = 1900 and LX = 650

Very different from the solution to part A: S = 400 and LX = 140
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Lecture 7
Business Analytics
Shelby Shelving Case

A small company that produces two types of shelves for grocery stores:
Model S and Model LX

Three steps in manufacturing process:
 Stamping, forming and assembly
 Capacity of Stamping and Forming machines: 800 hours a month
 Capacity of model S assembly department: 1900 units/month
 Capacity of model LX assembly department: 1400 units/month

Machine Requirements (Hours/per unit):
 Stamping: 0.3h/u for model S and 0.3h/u for model LX
 Forming: 0.25h/u for model S and 0.5h/u for model LX

Selling Price: Model S $1800/unit; Model LX $2100/unit

Current monthly production: 400 units of S and 1400 units of LX
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Lecture 7
Business Analytics
Shelby Shelving: Decision Model


Decision Variables:
Let S = # of Model S shelves to produce, and
LX = # of Model LX shelves to produce.
Constraints:
(S assembly)
S
≤ 1900
(LX assembly)
LX ≤ 1400
(Stamping) 0.3 S + 0.3 LX ≤ 800
(Forming) 0.25 S + 0.5 LX ≤ 800
(Nonnegativity)
S, LX ≥ 0
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Lecture 7
Business Analytics
Shelby Shelving: Complete LP formulation

Linear Program
max 125 S + 245 LX - 385,000
(Net Profit)
subject to:
(S assembly)
S
≤ 1900
(LX assembly)
LX ≤ 1400
(Stamping) 0.3 S + 0.3 LX ≤ 800
(Forming) 0.25 S + 0.5 LX ≤ 800
(Nonnegativity)
S, LX ≥ 0
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Business Analytics
Lecture 7
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Graphing the Shelby Shelving Problem
2500
LX model
2000
1500
1000
(S=1000, LX=500) – a solution
500
0
500
1000
1500
2000
2500
3000 S model
Business Analytics
Lecture 7
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Constraint Representation and Feasibility: Assembly Constraints
2500
LX model
S=1900 (S assembly constraint)
2000
1500
LX=1400 (LX assembly constraint)
1000
Feasible
500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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Constraint Representation and Feasibility: Forming Constraint
2500
LX model
S=1900 (S assembly)
2000
1500
Infeasible
LX=1400 (LX assembly)
1000
Feasible
500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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Constraint Representation and Feasibility: Stamping Constraint
2500
LX model
S=1900 (S assembly)
2000
1500
LX=1400 (LX assembly)
1000
500
0
Feasible Region for
the Shelby Shelving
500
1000
1500
2000
2500
NOTE: The stamping constraint is redundant!
3000S model
Business Analytics
Lecture 7
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Isoprofit Lines
2500
LX model
Profit for (S=1000, LX = 500) = 125x1000+245x500 =
247,500
2000
1500
1000
(S=1000, LX=500)
500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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Isoprofit Lines
2500
LX model
Profit for (S=1000, LX = 500) = 125x1000+245x500 =
247,500
2000
1500
1000
(S=1000, LX=500)
500
Profit = 247,500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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Optimal Solution
2500
LX model
2000
Direction of increasing profit
1500
1000
(S=1900, LX=650) Optimal solution:
This is as far as we
can go in the direction
of increasing profit!
500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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What if profit coefficient for S is increased to 150?
2500
LX model
Profit for (S=1000, LX = 500) = 150x1000+245x500 =
272,500
2000
1500
1000
(S=1000, LX=500)
500
Profit = 272,500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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Optimal Solution for New Profit Values
2500
LX model
2000
New direction of increasing profit
1500
1000
Same Optimal Solution:
(S=1900, LX=650)
500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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What if profit coefficient for LX is increased to 500?
2500
LX model
Profit for (S=1000, LX = 500) = 125x1000+500x500 =
375,000
2000
1500
1000
(S=1000, LX=500)
500
Profit = 375,000
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
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Optimal Solution for New Profit Values
2500
LX model
2000
1500
New direction of increasing profit
New Optimal Solution:
(S=400, LX=1400)
1000
Old Optimal Solution:
(S=1900, LX=650)
500
0
500
1000
1500
2000
2500
3000S model
Business Analytics
Lecture 7
Graphical Solution Summary

Optimal solutions are at “corner points” of the constraint set, defined by
a set of “binding constraints”
 Binding constraints – these are constraints that limit the
improvement in the objective function, e.g. use all resources
available
 Non-binding constraints – these are constraints that do not limit
improvement, e.g. have “left over” resources
 Changes in objective function coefficients do not necessarily change the
optimal solution (i.e. the same constraints may be binding)
 When the optimal solution does change as a result of changing
objective function coefficients, it “jumps” from one corner point to
another corner point, so small changes in parameters may lead to large
shifts in the decision variables
 Small changes in the binding constraints will result in changes in the
optimal solution; small changes in non-binding constraints will not affect
the optimal solution.
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