Transcript No Slide Title

Chapter 3
Applications of Linear and
Integer Programming
Models - 2
1
3.5 Applications of Integer
Linear Programming Models
Many real life problems call for at least one integer
decision variable.
 There are three types of Integer models:

Pure integer (AILP)
 Mixed integer (MILP)
 Binary (BILP)

2
The use of binary variables in
constraints
• AAny decision situation that can be modeled by
“yes”/“no”, “good”/“bad” etc., falls into the binary
category.

To illustrate
1 If a new health care plan is adopted
X  0 If it is not

1 If a new police station is built downtown
X  0 If it is not

1 If a particular constraint must hold
X  0 If it is not

3
The use of binary variables in
constraints

Example
A decision is to be made whether each of three plants should be
built (Yi = 1) or not built (Yi = 0)
Requirement
Binary Representation

At least 2 plants must be built
If plant 1 is built, plant 2 must not be built
If plant 1 is built, plant 2 must be built
One, but not both plants must be built
Both or neither plants must be built
Plant construction cannot exceed $17 million
given the costs to build plants are $5, $8, $10 million
Y1 + Y2 + Y3 2
Y1 + Y2 1
Y1 – Y2 
Y1+ Y2 = 1
Y1 – Y2 =0
5Y1+8Y2+10Y3 17
4
The use of binary variables in
constraints

Example - continued

Two products can be produced at a plant.
• Product 1 requires 6 pounds of steel and product 2 requires 9 pounds.
• If a plant is built, it should have 2000 pounds of steel available.

The production of each product should satisfy the steel
availability if the plant is opened, or equal to zero if the plant is
not opened.
6X1 + 9X2 2000Y1
If the plant is built Y1 = 1.
The constraint becomes
6x1 + 9X2 2000
If the plant is not built Y1 = 0.
The constraint becomes
6x1 + 9X2 0, and thus,
X1 = 0 and X2 = 0
5
3.5.1 Personnel Scheduling Models
Assignments of personnel to jobs under
minimum required coverage is a typical integer
problems.
 When resources are available over more than
one period, linking constraint link the resources
available in period t to the resources available in
a period t+1.

6
7
8
Sunset Beach Lifeguard Assignments


The City of Sunset Beach staffs lifeguards 7 days a
week.
Regulations require that city employees work five days.

Insurance requirements mandate 1 lifeguard per 8000
average daily attendance on any given day.

The city wants to employ as few lifeguards as possible.
9
Sunset Beach Lifeguard Assignments

Problem Summary
Schedule lifeguard over 5 consecutive days.
 Minimize the total number of lifeguards.
 Meet the minimum daily lifeguard requirements
Sun. Mon. Tue. Wed. Thr. Fri. Sat.
8
6
5
4
6
7
9

10
Sunset Beach Lifeguard Assignments

Decision Variables
Xi = the number of lifeguards scheduled to begin on day “ i ”
for i=1, 2, …,7 (i=1 is Sunday)

Objective Function
Minimize the total number of lifeguard scheduled

Constraints
Ensure that enough lifeguards are scheduled each day.
11
Sunset Beach Lifeguard Assignments
To ensure that enough lifeguards are scheduled for
each day, identify which workers are on duty.
For example: …
12
Sunset Beach Lifeguard Assignments
Who works on Friday ?
Who works on Saturday ?
X2
X3
X4
X5
X6
X3
X4
X5
X6
X1
Mon Tue. Wed. Thu. Fri. Sat Sun.
Repeat this procedure for each day of the week, and
build the constraints accordingly.
13
Sunset Beach Lifeguard Assignments
Min X1 +
S.T. X1 +
X1 +
X1 +
X1 +
X1 +
X2 + X3 + X4 + X5 + X6 + X7
X4 + X5 + X6 + X7  8
X2 +
X5 + X6 + X7  6
X2 + X3 +
X6 + X7  5
X2 + X3 + X4 +
X7  4
X2 + X3 + X4 + X5
6
X2 + X3 + X4 + X5 + X6
7
X3 + X4 + X5 + X6 + X7  9
All the variables are non negative integers
14
Sunset Beach Lifeguard Assignments
15
Sunset Beach Lifeguard Assignments
16
Sunset Beach Lifeguard Assignments
OPTIMAL ASSIGNMENTS
LIFEGUARDS
DAY
SUNDAY
MONDAY
TUESDAY
WEDNESDAY
THURSDAY
FRIDAY
SATURDAY
Note: An alternate
optimal solution exists.
PRESENT
REQUIRED
BEGIN SHIFT
9
8
6
5
6
7
9
8
6
5
4
6
7
9
1
0
1
1
3
2
2
TOTAL LIFEGUARDS
10
17
3.5.2 Project selection Models
These models involve a “go/no-go” situations, that
can be modeled using binary variables.
 Typical elements in such models are:

Budget
 Space
 Priority conditions

18
Salem City Council – Project Selection
The Salem City Council needs to decide how to
allocate funds to nine projects such that public
support is maximized.
 Data reflect costs, resource availabilities, concerns
and priorities the city council has.

19
Salem City Council – Project Selection
Survey results
Project
X1 Hire seven new police officers
X2 Modernize police headquarters
X3 Buy two new police cars
X4 Give bonuses to foot patrol officers
X5 Buy new fire truck/support equipment
X6 Hire assistant fire chief
X7 Restore cuts to sport programs
X8 Restore cuts to school music
X9 Buy new computers for high school
Cost (1000)
$
400.00
$
350.00
$
50.00
$
100.00
$
500.00
$
90.00
$
220.00
$
150.00
$
140.00
Jobs
7
0
1
0
2
1
8
3
2
Points
4176
1774
2513
1928
3607
962
2829
1708
3003
20
Salem City Council – Project Selection

Decision Variables:
Xj- a set of binary variables indicating if a project j is
selected (Xj=1) or not (Xj=0) for j=1,2,..,9.

Objective function:
Maximize the overall point score of the funded projects

Constraints:
See the mathematical model.
21
Salem City Council – Project Selection

The Mathematical Model
(Xi = 0,1 for i=1, 2…, 9)
Max 4176X1+1774X2+ 2513X3+1928X4+3607X5+962X6+2829X7+1708X8+3003X9
S.T.
The maximum amounts of funds to be allocated is $900,000
400X1+ 350X2+
50X3+ 100X4+ 500X5+ 90X6+ 220X7+ 150X8+ 140X9  900
The number of new jobs created must be at least 10
7X1+
X3 +
2X5+
X6 +
8X7+
3X8+
2x9
The number of police-related activities selected is at most 3 (out of 4)
X1 +
X2 +
X3 +
 3
X4
Either police car or fire truck be purchased
X3 +
X5
= 1
Sports funds and music funds must be restored / not restored together
X7 -
Sports funds and music funds must be
restored before computer equipment
is purchased
 10
X8
= 0
X9  0
X7 x8 -
x9

22
Salem City Council – Project selection
=SUMPRODUCT(B4:B12,E4:E1
=SUMPRODUCT(B4:B12,C4:C12)
2)
=SUMPRODUCT(B4:B12,D4:D12)
=SUM(B4:B7)
=B6+B8
=B10-B11
=B10-B12
=B11-B12
23
3.5.3 Supply Chain Management



Supply chain management models integrate the
manufacturing process and the distribution of goods to
customers.
The overall objective of these models is to minimize total
system costs
The requirements concern (among others)


Appropriate production levels
Maintaining a transportation system to satisfy demand in timely
manner.
24
Globe Electronics, Inc.
Globe Electronics, Inc. manufactures two styles of
remote control cable boxes, G50 and H90.
 Globe runs four production facilities and three
distribution centers.
 Each plant operates under unique conditions, thus
has a different fixed operating cost, production
costs, production rate, and production time
available.

25
Globe Electronics, Inc.
Demand has decreased, and management is
contemplating closing one or more of its facilities.
 Management wishes to:

– Develop an optimal distribution policy.
– Determine which plant to close (if any).
26
Globe Electronics, Inc.

Data
Production costs, Times, Availability
Plant
Philadelphia
St. Louis
New Orleans
Denver
Fixed Cost
per Month
40
35
20
30
Production Cost / unit
G50
H90
10
14
12
12
8
10
13
15
Production Time (hr/unit)
G50
H90
0.06
0.06
0.07
0.08
0.09
0.07
0.05
0.09
Available hr
per Month
640
960
480
640
Monthly Demand Projection
G50
G90
Demand
Cincinnati Kansas CitySan Francisco
2000
5000
3000
6000
5000
7000
27
Globe Electronics, Inc.

Transportation Costs per 100 units
Cincinnati
Philadelphia
$200
Kansas
City
300
St.Louis
New Orleans
Denver
100
200
300
100
200
100
San
Francisco
500
400
300
100
At least 70% of the demand in each distribution center
must be satisfied.
 Unit selling price

• G50 = $22;
H90 = $28.
28
Globe Electronics, Inc.
The Globe problem
Ordering raw material
Scheduling personnel
Production
1. Production level for
each product
in each plant.
2. Distribution plan.
Distribution centers
1. Storage
2. Sale and Dissemination
to retail establishments
29
Globe Electronics, Inc. - Variables
Transportation variables
Philadelphia
St. Louis
New Orleans
Denver
1
1
Cincinnati
2
Kansas City
3
San Francisco
2
3
4
30
Globe Electronics, Inc. - Variables
Production variables in each plant
Philadelphia
1
GG1211
Cincinnati
1
G13 GG1112
2
St. Louis
G13 G12
G11 G13G12 G 2
Kansas City
13
G12
G
11
G13
New Orleans 3
G11
G12 G13
G311
12
G11 SanGFrancisco
G13
Denver
4
G
G
11
12
G11
G13
G
G
11
12
G13
G
12
Total production of G50 in Philadelphia = GP = G
G11
+
G
+
G
11
12
13
31
Globe Electronics, Inc. - Variables
Shipment variables to each distribution center
Philadelphia
St. Louis
New Orleans
Denver
1
1
Cincinnati
2
Kansas City
3
San Francisco
2
3
4
Total shipment of H90 to Cincinnati = HC = H11 + H21 + H31 +H41
32
Globe Electronics
Model No. 1:
All The Plants Remain Operational
33
Globe Electronics – all plants opened

Objective function
Max Gross Profit = 22(Total G50)+28(Total H90) – Total
Production Cost – Total transportation Cost =
Max 22G + 28H
G = total number of
G50 produced
H = total number of
H90 produced
34
Globe Electronics – all plants opened

Objective function
Max Gross Profit = 22(Total G50)+28(Total H90) – Total
Production Cost – Total transportation Cost =
Max 22G + 28H
Production costs
– 10GP – 12GSL – 8GNO – 13GD
– 14HP – 12HSL – 10HNO – 15HD
– 2G11 – 3G12 – 5G13
– 1G21 – 1G22 – 4G23
– 2G31 – 2G32 – 3G33
– 3G41 – 1G42 – 1G43
– 2H11 – 3H12 – 5H13
– 1H21 – 1H22 – 4H23
– 2H31 – 2H32 – 3H33
– 3H41 – 1H42 – 1H43
Transportation costs
35
Globe Electronics – all plants opened

Constraints:
Ensure that the amount shipped from a plant equals the amount
produced in a plant (summation constraints).
For G50
G11 + G12 + G13 = GP
G21 + G22 + G23 = GSL
G31 + G32 + G33 = GNO
G41 + G42 + G43 = GD
For H90
H11 + H12 + H13 = HP
H21 + H22 + H23 = HSL
H31 + H32 + H33 = HNO
H41 + H42 + H43 = HD
The amount received by a distribution center is equal to all the
shipments made to this center (summation constraints).
For G50
For H90
G11 + G21 + G31 + G41 = GC
H11 + H21 + H31 + H41 = HC
G12 + G22 + G32 + G42 = GKC
H12 + H22 + H32 + H42 = HKC
G13 + G23 + G33 + G43 = GSF
H13 + H23 + H33 + H43 = HSF
36
Globe Electronics – all plants opened

Constraints


The amount shipped to each distribution center is at least
70% of its projected demand.
The amount shipped to each distribution center does not
exceed its demand.
• Cincinnati:
GC  1400
HC  3500
GC  2000
HC  5000
• Kansas City
GKC  2100
HKC  4200
GKC  3000
HKC  6000
• San Francisco
GSF  3500
HSF  4900
GSF  5000
HSF  7000
37
Globe Electronics – all plants opened

Constraints:

Production time used at each plant cannot exceed the time
available:
.06GP + .0 6HP  640
.07GSL+ .08HSL  960
.09GNO + .07HNO 480
.05GD + .09HD 640
All the variables are non negative
38
Globe Electronics – all plants opened spreadsheet
=F10*F9+F19*F18SUMPRODUCT(G23:G26,F5:F8)
SUMPRODUCT(H23:H26,F14:F17
)SUMPRODUCT(C5:E8,C23:E26)SUMPRODUCT(C14:E17,C23:E26
)-SUM(F23:F26)
=$I23*$F5+$J23*$F14
Drag to L24:L26
39
Globe Electronics 1 - Summary





The optimal value of the objective function is $356,571.43
Note that the fixed cost of operating the plants was not
included in the objective function because all the plants
remain operational.
Subtracting the fixed cost of $125,000 results in a net
monthly profit of $231,571.43
Rounding down several non-integer solution values results
in an integral solution with total profit of $231,550.
This solution may not be optimal, but it is very close to it.
40
Globe Electronics Model No. 2:
The number of plants that remain
operational in each city is a
decision variable.
41
Globe Electronics – which plant
remains opened?
High set up costs raise the question:
Is it optimal to leave all the plants operational?
 Using binary variables the optimal solution
provides suggestions for:

Production levels for each product in each plant,
 Transportation pattern from each plant to distribution
center,
 Which plant remains operational.

42
Globe Electronics – which plant
remains opened?
• Binary Decision Variables
Yi = a binary variable that describes the number of
operational plants in city i.
• Objective function
Subtract the following conditional set up costs from the previous
objective function:
40,000YP + 35,000YSL + 20,000YND + 30,000YD
• Constraints
Change the production constraints
.06GP + .0 6HP  640YP
.09GNO + .07HNO 480YNO
.07GSL+ .08HSL  960YSL
.05GD + .09HD 640YD
43
Globe Electronics – which plant
remains opened?
=F10*F9+F19*F18SUMPRODUCT(G23:G26,F5:F8) SUMPRODUCT(H23:H26,F14:F17)
-SUMPRODUCT(C5:E8,C23:E26)SUMPRODUCT(C14:E17,C23:E26)
-SUMPRODUCT(F23:F26,A5:A8)
44
Globe Electronics 2 - Summary



The Philadelphia plant should be closed, while the other
plants work at capacity.
Schedule monthly production according to the quantities
shown in the Excel output.
The net monthly profit will be $266,083 (after rounding
down the non-integer variable values), which is $34,544
per month greater than the optimal monthly profit
obtained when all four plants are operational.
45
Appendix 3.4 (CD): Advertising Models
46
Appendix 3.4 (CD): Advertising Models


Many marketing situations can be modeled by linear
programming models.
Typically, such models consist of:





Budget constraints,
Deadlines constraints,
Choice of media,
Exposure to target population.
The objective is to achieve the most effective advertising
plan.
47
Vertex Software, Inc.

Vertex Software has developed a new software product,
LUMBER 2000.

A marketing plan for this product is to be developed for the
next quarter.


The product will be promoted using black and white and colored full
page ads.
Three publications are considered:
• Building Today
• Lumber Weekly
• Timber World
48
Vertex Software, Inc.

Requirements

A maximum of one ad should be placed in any one issue of any of
the publication during the quarter.

At least 50 full-page ads should appear during the quarter.

at least 8 color ads should appear during the quarter.

One ad should appear in each issue of Timber World.

At least 4 weeks of advertising should be placed in each of the
Building Today and Lumber Weekly publications.

No more than $ 40000 should be spent on advertising in any one of
the trade publications.
49
Vertex Software, Inc.
Circulation and advertising costs
Publication
Building Today
Frequency
5 day/week
Circulat.
400,000
Lumber Weekly
Weekly
250,000
Timber World
Monthly
200,000
Cost/Ad
Full pg.:
Half pg.:
$800
$500
Only B&W
B&W pg.: $1500
Color pg.: $4000
B&W pg.: $2000
Color pg.: $6000
Key reader attitudes
Attribute
Computer data-base user
Large Firm (>2M sales)
Location (city / suburb)
Rating
.50
.25
.15
Bldng.
60%
40
60
Percentage of Readership
Lumbr
Timber
80
90
80
80
60
80
Age of firm (>5 years)
.10
20
40
50
50
Vertex Software, Inc.

Solution
The requirements are:
• Stay within a $90,000 budget for print advertising.
• Place no more than 65 ads(=5 x 13 weeks) and no less than 20 ads
(=5 X 4 weeks) in Building Today.
• Place no more than 13 and no less than 4 ads in Lumber Weekly.
• Place exactly 3 ads in Timber World.
• Place at least 50 full-page ads.
• Place at least 8 color ads.
• Spend no more than $40,000 on advertisement in any one of the trade
publications.
51
Vertex Software, Inc.

Variables






X1 = number of full page B&W ads placed in
Building Today
X2 = number of half page B&W ads placed in
Building Today
X3 = number of full page B&W ads placed in
Lumber Weekly
X4 = number of full page color ads placed in
Lumber Weekly
X5 = number of full page B&W ads placed in
Timber World
X6 = number of full page color ads placed in
Timber World
52
Vertex Software, Inc.

The Objective Function

The objective function measures the effectiveness of the
promotion operation (to be maximized).

It depends on the number of ads in each publication, as
well as on the relative effectiveness per ad.

A special technique (external to this problem) is applied to
evaluate this relative effectiveness.
53
Vertex Software, Inc.
=SUMPRODUCT($B$6:$B$9,C6:C9)
Drag to cells D11 and E11
=C$11*C$13*$B17
Drag across to D17:E17
then down to C19:E19.
Then delete formulas in
cells C17,D19, and E19
54
Vertex Software, Inc.
• The Mathematical Model
Max 102000X1+40800X2+91250X3+182500X4+82000X5+164000X6
S.T.
800X1 + 500X2+ 1500X3+ 4000X4+ 2000X5 + 6000X6
X1 +
X2
X1 +
X2
X3 +
X4
X3 +
X4
X5 +
X6
X1 +
X3 +
X4 +
X5 +
X6
X4 +
X6 ³ 8
800X1 + 500X2
1500X3 + 4000X4
2000X5 + 6000X6
All variables non-negative





=




Budget
90000
65
# of Building
20
Today ads
13
# of Lumber
4
Weekly ads
3
Timber World ads
50
Full Page
40000 Colored
40000
40000
Maximum spent
In each magazine
55
Vertex Software, Inc.
VERTEX SOFTWARE, INC.
Totals
Cost
Expsoure Units
Publication
Page Size
Style
Cost Per Ad Exposure Units
Ads
Building Today
Full
Half
B&W
B&W
$800
102000
$500
40800
Totals for Building Today
50
0
50
$40,000
$0
$40,000
5100000
0
5100000
Lumber Weekly
Full
Full
B&W
Color
$1,500
91250
$4,000
182500
Totals for Lumber Weekly
5
8
13
$7,500
$32,000
$39,500
456250
1460000
1916250
Timber World
Full
Full
B&W
Color
$2,000
82000
$6,000
164000
Totals for Timber World
2
1
3
$4,000
$6,000
$10,000
164000
164000
328000
TOTALS
66
$89,500
7344250
Size Totals
Full Page
Half Page
66
0
$89,500
$0
7344250
0
Style Totals
B&W
Color
57
9
$51,500
$38,000
5720250
1624000
LIMITS
Budget
Max Build Today
Min Build Today
Max Lum Week
Min Lum Week
# Timber World
Min Full Page
Min Color
Max Any Pub
$90,000
65
20
13
4
3
50
8
$40,000
56
Copyright
2002 John Wiley & Sons, Inc. All rights
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contained herein.
57