# Applications of Linear and Integer Programming Models - 2

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Applications of Linear and Integer Programming Models - 2
Chapter 3 Applications of Linear and Integer Programming Models - 2

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)

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

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 Y1 + Y2 + Y3 ³ 2 If plant 1 is built, plant 2 must not be built Y1 + Y2 £ 1 If plant 1 is built, plant 2 must be built Y1 – Y2 £ 0 One, but not both plants must be built Y1+ Y2 = 1 Both or neither plants must be built Y1 – Y2 =0 Plant construction cannot exceed \$17 million given the costs to build plants are \$5, \$8, \$10 million 5Y1+8Y2+10Y3 £ 17

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

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.

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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.

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.

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.

Sunset Beach Lifeguard Assignments
To ensure that enough lifeguards are scheduled for each day, identify which workers are on duty. For example: …

Sunset Beach Lifeguard Assignments
Who works on Friday ? Who works on Saturday ? X2 X3 X3 X4 X5 X6 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.

Sunset Beach Lifeguard Assignments
Min X1 + X2 + X3 + X4 + X5 + X6 + X7 S.T. X1 + X4 + X5 + X6 + X7 ³ 8 X1 + X X5 + X6 + X7 ³ 6 X1 + X2 + X X6 + X7 ³ 5 X1 + X2 + X3 + X4 + X7 ³ 4 X1 + X2 + X3 + X4 + X5 ³ 6 X2 + X3 + X4 + X5 + X6 ³ 7 X3 + X4 + X5 + X6 + X7 ³ 9 All the variables are non negative integers

Sunset Beach Lifeguard Assignments

Sunset Beach Lifeguard Assignments

Sunset Beach Lifeguard Assignments
OPTIMAL ASSIGNMENTS LIFEGUARDS DAY PRESENT REQUIRED BEGIN SHIFT SUNDAY 9 8 1 MONDAY 6 TUESDAY 5 WEDNESDAY 4 THURSDAY 3 FRIDAY 7 2 SATURDAY 10 Note: An alternate optimal solution exists. TOTAL LIFEGUARDS

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

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.

Salem City Council – Project Selection
Survey results X1 X2 X3 X4 X5 X6 X7 X8 X9

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.

Salem City Council – Project Selection
The Mathematical Model Max 4176X1+1774X X3+1928X4+3607X5+962X6+2829X7+1708X8+3003X9 S.T. 400X1+ 350X X3+ 100X4+ 500X X6+ 220X7+ 150X8+ 140X9 £ 900 7X1+ X X5+ X X7+ 3X x9 ³ 10 X1+ X2+ X3+ X £ 3 X3+ X = 1 X7 - X8 = 0 X X9 ³ 0 x8 - x9 ³ 0 (Xi = 0,1 for i=1, 2…, 9) The maximum amounts of funds to be allocated is \$900,000 The number of new jobs created must be at least 10 The number of police-related activities selected is at most 3 (out of 4) Either police car or fire truck be purchased Sports funds and music funds must be restored / not restored together Sports funds and music funds must be restored before computer equipment is purchased

Salem City Council – Project selection
=SUMPRODUCT(B4:B12,E4:E12) =SUMPRODUCT(B4:B12,C4:C12) =SUMPRODUCT(B4:B12,D4:D12) =SUM(B4:B7) =B6+B8 =B10-B11 =B10-B12 =B11-B12

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.

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.

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).

Globe Electronics, Inc. Data Production costs, Times, Availability
Monthly Demand Projection

Globe Electronics, Inc. Transportation Costs per 100 units
At least 70% of the demand in each distribution center must be satisfied. Unit selling price G50 = \$22; H90 = \$28. City Francisco Cincinnati Kansas San Philadelphia \$200 300 500 St.Louis 100 400 New Orleans 200 Denver

Globe Electronics, Inc. The Globe problem Ordering raw material
Scheduling personnel Production Production level for each product in each plant. Distribution plan. Distribution centers 1. Storage 2. Sale and Dissemination to retail establishments

Globe Electronics, Inc. - Variables
Transportation variables 1 Philadelphia G11, H11 G12, H12 1 Cincinnati G13, H13 2 St. Louis G22, H22 2 G31, H31 Kansas City 3 New Orleans G41, H41 3 San Francisco 4 Denver

Globe Electronics, Inc. - Variables
Production variables in each plant 1 G11, H11 G11 Philadelphia G12 G12, H12 G11 G12 1 Cincinnati G13 G13, H13 G12 G13 G11 2 St. Louis G22, H22 G13 G12 G13 G12 G31, H31 G11 2 G13 Kansas City G12 G11 3 G13 New Orleans G11 G12 G13 G11 3 San Francisco G12 G11 4 G13 Denver G11 G12 G11 G13 G11 G12 G11 G13 Total production of G50 in Philadelphia = GP = G12 G11 G12 G13

Globe Electronics, Inc. - Variables
Shipment variables to each distribution center 1 Philadelphia G11, H11 G12, H12 1 Cincinnati G13, H13 2 St. Louis G22, H22 G31, H31 2 Kansas City 3 New Orleans 3 San Francisco 4 Denver Total shipment of H90 to Cincinnati = HC = H11 + H21 + H31 +H41

Globe Electronics Model No. 1: All The Plants Remain Operational

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

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 – 10GP – 12GSL – 8GNO – 13GD – 14HP – 12HSL – 10HNO – 15HD Production costs – 2H11 – 3H12 – 5H13 – 1H21 – 1H22 – 4H23 – 2H31 – 2H32 – 3H33 – 3H41 – 1H42 – 1H43 – 2G11 – 3G12 – 5G13 – 1G21 – 1G22 – 4G23 – 2G31 – 2G32 – 3G33 – 3G41 – 1G42 – 1G43 Transportation costs

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 G11 + G21 + G31 + G41 = GC G12 + G22 + G32 + G42 = GKC G13 + G23 + G33 + G43 = GSF For H90 H11 + H21 + H31 + H41 = HC H12 + H22 + H32 + H42 = HKC H13 + H23 + H33 + H43 = HSF

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 ³ GC £ HC ³ HC £ 5000 Kansas City GKC ³ GKC £ HKC ³ HKC £ 6000 San Francisco GSF ³ GSF £ HSF ³ HSF £ 7000

Globe Electronics – all plants opened
Constraints: Production time used at each plant cannot exceed the time available: .06GP HP £ GSL+ .08HSL £ GNO + .07HNO £ GD HD £ 640 All the variables are non negative

Globe Electronics – all plants opened spreadsheet
=F10*F9+F19*F18-SUMPRODUCT(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

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.

Globe Electronics Model No
Globe Electronics Model No. 2: The number of plants that remain operational in each city is a decision variable.

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.

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 HP £ 640YP GSL+ .08HSL £ 960YSL .09GNO + .07HNO £ 480YNO .05GD HD £ 640YD

Globe Electronics – which plant
remains opened? =F10*F9+F19*F18-SUMPRODUCT(G23:G26,F5:F8) - SUMPRODUCT(H23:H26,F14:F17)-SUMPRODUCT(C5:E8,C23:E26)-SUMPRODUCT(C14:E17,C23:E26)-SUMPRODUCT(F23:F26,A5:A8)

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.

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.

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

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 \$ should be spent on advertising in any one of the trade publications.

Vertex Software, Inc. Circulation and advertising costs
Publication Frequency Circulat. Cost/Ad Building Today 5 day/week 400,000 Full pg.: \$800 Half pg.: \$500 Only B&W Lumber Weekly Weekly 250,000 B&W pg.: \$1500 Color pg.: \$4000 Timber World Monthly ,000 B&W pg.: \$2000 Color pg.: \$6000 Key reader attitudes Percentage of Readership Attribute Rating Bldng. Lumbr Timber Computer data-base user % Large Firm (>2M sales) Location (city / suburb) Age of firm (>5 years)

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.

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

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.

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

Vertex Software, Inc. The Mathematical Model
Max X X X X X X6 S.T. 800X X X X X X6 £ 90000 X1 + X £ X1 + X ³ X3 + X £ 13 X3 + X ³ 4 X5 + X6 = 3 X1 + X3 + X4 + X5 + X6 ³ 50 X X6 ³ 8 800X X £ 1500X X4 ³ 2000X X6 £ All variables non-negative Budget # of Building Today ads # of Lumber Weekly ads Timber World ads Full Page Colored Maximum spent In each magazine

Vertex Software, Inc.