Linear Programming

MIN: 50x1 + 83x2 + 130x3

+ 61y1 + 97y2 + 145y3

• Demand Constraints

x1 + y1 = 3,000 } model 1

x2 + y2 = 2,000 } model 2

x3 + y3 = 900 } model 3

• Resource Constraints

2x1 + 1.5x2 + 3x3 <= 10,000 } wiring

1x1 + 2.0x2 + 1x3 <= 5,000 } harnessing

• Nonnegativity Conditions

x1, x2, x3, y1, y2, y3 >= 0

Make / buy decision : Constraints

y1 = 3,000- x1

y2 = 2,000-x2

y3 = 900-x3

MIN: 50x1 + 83x2 + 130x3 + 61(3,000- x1)

+ 97(2,000-x2) + 145(900-x3)

y1 = 3,000- x1>=0 y2 = 2,000-x2>=0 y3 = 900-x3>=0

x1 <= 3,000x2 <= 2,000x3 <= 900

Problem ( HHL 3.21, HL 3.4.17)

Men, women, and children gloves.

Material and labor requirements for each type and the corresponding profit are given below. Glove Material (sq-feet) Labor (hrs) ProfitMen 2 .5 8Women 1.5 .75 10Children 1 .67 6

Total available material is 5000 sq-feet.

We can have full time and part time workers.Full time workers work 40 hrs/w and are paid $13/hrPart time workers work 20 hrs/w and are paid $10/hrWe should have at least 20 full time workers.The number of full time workers must be at least twice of that of part times.

Decision variables

X1 : Volume of production of Men’s glovesX2 : Volume of production of Women’s glovesX3 : Volume of production of Children’s gloves

Y1 : Number of full time employeesY2 : Number of part time employees

Constraints

Row material constraint2X1 + 1.5X2 + X3 5000

Full time employeesY1 20

Relationship between the number of Full and Part time employeesY1 2 Y2

Labor Required.5X1 + .75X2 + .67X3 40 Y1 + 20Y2

Objective FunctionMax Z = 8X1 + 10X2 + 6X3 - 520 Y1 - 200 Y2

Non-negativityX1 , X2 , X3 , Y1 , Y2 0

Constraints

2X1 + 1.5X2 + X3 5000Y1 20Y1 - 2 Y2 0.5X1 + .75X2 + .67X3 - 40 Y1 - 20Y2 0Max Z = 8X1 + 10X2 + 6X3 - 520 Y1 - 200 Y2

2 1.5 1 0 <= 50001 0 >= 201 -2 0 >= 0

0.5 0.75 0.67 -40 -20 0 <= 08 10 6 -520 -200 0

X1 X2 X3 Y1 Y2

2 1.5 1 5000 <= 50001 25 >= 201 -2 0 >= 0

0.5 0.75 0.67 -40 -20 0 <= 08 10 6 -520 -200 4500

2500 0 0 25 12.5X1 X2 X3 Y1 Y2

Problem ( HHL Problem 3.23, HL 3.4.19)

Strawberry shake productionSeveral ingredients can be used in this product.

Ingredient calories from fat Total calories Vitamin Thickener Cost( per tbsp) (per tbsp) (mg/tbsp) (mg/tbsp) ( c/tbsp)

Strawberry flavoring 1 50 20 3 10Cream 75 100 0 8 8Vitamin supplement 0 0 50 1 25Artificial sweetener 0 120 0 2 15Thickening agent 30 80 2 25 6

This beverage has the following requirementsTotal calories between 380 and 420.No more than 20% of total calories from fat.At least 50 mg vitamin.At least 2 tbsp of strawberry flavoring for each 1 tbsp of artificial sweetener.Exactly 15 mg thickeners.Formulate the problem to minimize costs.

Decision variables

Decision VariablesX1 : tbsp of strawberryX2 : tbsp of creamX3 : tbsp of vitaminX4 : tbsp of Artificial sweetenerX5 : tbsp of thickening

Constraints

Calories 50X1 + 100 X2 + 120 X4 + 80 X5 38050X1 + 100 X2 + 120 X4 + 80 X5 420Calories from fat X1 + 75 X2 + 30 X5 .2(50X1 + 100 X2 + 120 X4 + 80 X5)

Vitamin20X1 + 50 X3 + 2 X5 50Strawberry and sweetenerX1 2 X4 Thickeners3X1 + 8X2 + X3 + 2 X4 + 2.5 X5 = 15

Objective FunctionMin Z = 10X1 + 8X2 + 25 X3 + 15 X4 + 6 X5 Non-negativityX1 , X2 , X3 , X4 , X5 0

Agricultural planning : narrative

Three farming communities are developing a joint agricultural production plan for the coming year.Production capacity of each community is limited by their land and water.

Community Land (Acres) Water (Acres Feet)1 400 6002 600 8003 300 375

The crops suited for this region include sugar beets, cotton, and sorghum. These are the three being considered for the next year.

Information regarding the maximum desired production of each product, water consumption , and net profit are given below

Agricultural planning : narrative

Crop Max desired Water consumption Net return(Acres) (Acre feet / Acre) ($/Acre)

1 600 3 10002 500 2 7503 325 1 250

Because of the limited available water, it has been agreed that every community will plant the same proportion of its available irritable land. For example, if community 1 plants 200 of its available 400 acres, then communities 2 and 3 should plant 300out of 600, and 150 out of 300 acres respectively. However, any combination of crops may be grown at any community.

Goal : find the optimal combination of crops in each community, in order to maximize total return of all communities

Agricultural planning : decision variables

x11 = Acres allocated to Crop 1 in Community 1






……………..

xij = Acres allocated to Crop i in Community j

i for crop j for community, we could have switched them

Note that x is volume not portion, we could have had it as portion

Agricultural planning : Formulation

Land

x11+x21+x31 400

x12+x22+x32 600

x13+x23+x33 300

Water

3x11+2x21+1x31 600

3x12+2x22+1x32 800

3x13+2x23+1x33 375

Agricultural planning : Formulation

Crops

x11+ x12 + x13 600

x21 +x22 +x23 500

x31 +x32 +x33 320

Proportionality of land use

x11+x21+x31 x12+x22+x32

400 600

x11+x21+x31 x13+x23+x33

400 300

Agricultural planning : all variables on LHS

Proportionality of land use

600(x11+x21+x31 ) - 400(x12+x22+x32 ) = 0

300(x11+x21+x31 ) - 400(x13+x23+x33 ) = 0

600x11+ 600 x21+ 600 x31 - 400x12- 400 x22- 400 x32 = 0

300x11+ 300 x21+ 300 x31 - 400x13- 400 x23- 400 x33 = 0

x11, x21,x31, x12, x22, x32, x13, x23, x33 0

Controlling air pollution : narrative

This is a good example to show that the statement of a problem could be complicated. But as soon as we define the correct decision variables, things become very clear

Two sources of pollution: Open furnace and Blast furnace

Three types of pollutants: Particulate matter, Sulfur oxides, and hydrocarbons. ( Pollutant1, Pollutant2, Pollutant3). Required reduction in these 3 pollutants are 60, 150, 125 million pounds per year. ( These are RHS)

Three pollution reduction techniques: taller smokestacks, Filters, Better fuels. ( these are indeed our activities). We may implement a portion of full capacity of each technique.

If we implement full capacity of each technique on each source, their impact on reduction of each type of pollutant is as follows

Controlling air pollution : narrative

Pollutant Taller Filter Better fuel smokestacks

B.F. O.F B.F. O.F. B.F. O.F.

Particulate 12 9 25 20 17 13Sulfur 35 42 18 31 56 49Hydrocarb. 37 53 28 24 29 20

The cost of implementing full capacity of each pollutant reduction technique on each source of pollution is as follows



Cost 12 9 25 20 17 13

Controlling air pollution : Decision Variables

How many techniques??

How many sources of pollution??

How many constraints do we have in this problem???

How many variables do we have

Technique i source j

Controlling air pollution : Decision Variables

x11 = Proportion of technique 1 implemented of source 1


x21 = Proportion of technique 2 implemented of source 1.



x32 = Proportion of technique 3 implemented of source 2.

Controlling air pollution : Formulation



Particulate 12 9 25 20 17 13Sulfur 35 42 18 31 56 49Hydrocarb. 37 53 28 24 29 20

Min Z= 12x11+9x12+ 25x21+20x22+ 17x31+13x32

Particulate; 12x11+9x12+ 25x21+20x22+ 17x31+13x32 60

Sulfur; 35x11+42x12+ 18x21+31x22+ 56x31+49x32 150

Hydrocarbon; 37x11+53x12+ 28x21+24x22+ 29x31+20x32 125

x11, x12, x21, x22, x31, x32 ????

x11, x12, x21, x22, x31, x32 ????

SAVE-IT Company : Narrative

A reclamation center collects 4 types of solid waste material,

treat them, then amalgamate them to produce 3 grades of

product. Techno-economical specifications are given below

Grade Specifications Processing Sales price cost / pound / pound

M1 : 30% of total

A M2 : 40% of total 3 8.5

M3 : 50% of total M4 : exactly 20%

M1 : 50% of total

B M2 : 10% of total 2.5 7 M4 : exactly 10%

C M1 : 70% of total 2 5.5


Availability and cost of the solid waste materials M1, M2, M3,

and M4 per week are given below

Material Pounds available / week Treatment cost / pound

M1 3000 3

M2 2000 6

M3 4000 4

M4 1000 5

Due to environmental considerations, a budget of

$30000 / week should be used to treat these material.

Furthermore, for each material, at least half of the pounds

per week available should be collected and treated.


1. Mixture Specifications

2. Availability of material

3. At least half of the material treated

4. Spend all the treatment budget

5. Maximize profit Z

Chapter 3

We are done with chapter 3.If you need more example, read the following slides.Also solve at least two problems ( among previous or later slides) using excel. If you need more reading; read sections 3.1 to 3.6 inclusive. That is pages 24-72.If you still need more problem, pickup one problem in each of the following pages.91, 92,93,94,95,96, 97,98, 99, 100, 101, 102. Do not solve the excel part or software part of the problems you pick.

Capital budgeting : Narrative representation

We are an investor, and there are 3 investment projects offered to the public.

We may invest in any portion of one or more projects.

Investment requirements of each project in each year ( in millions of dollars) is given below. The Net Present Value (NPV) of total cash flow is also given.

Year Project 1 Project 2 Project 3

0 40 80 90

1 60 80 60

2 90 80 20

3 10 70 60

NPV 45 70 50


If we invest in 5% of project 1, then we need to invest 2, 3, 4.5, and 0.5 million dollars in years 0, 1, 2, 3 respectively. The NPV of our investment would be also equal to 5% of the NPV of this project, i.e. 2.25 million dollars.

Year Project 1 5% of Project 10 40 21 60 32 90 4.53 10 .5NPV 45 2.25


Based on our budget forecasts,

Our total available money to invest in year 0 is 25M.

Our total available money to invest in years 0 and 1 is 45M

Our total available money to invest in years 0, 1, 2 is 65M

Our total available money to invest in years 0, 1, 2, 3 is 80M

To clarify, in year 0 we can not invest more than 25M.

In year 1 we can invest 45M minus what we have invested in year 0.

The same is true for years 2 and 3.

The objective is to maximize the NPV of our investments

x1 = proportionproportion of project 1 invested by us.



Maximize NPV Z = 45x1 + 70 x2 + 50 x3

subject to

Year 0 : 40 x1 + 80 x2 + 90 x3 25

Year 1 : Investment in year 0 + Investment in year 1 45

Capital budgeting : Formulation

Investment in year 0 = 40 x1 + 80 x2 + 90 x3 Investment in year 1 = 60 x1 + 80 x2 + 60 x3

Year 1 : 60 x1 + 80 x2 + 60 x3 + 40 x1 + 80 x2 + 90 x3 45Year 1 : 100x1 + 160 x2 + 150 x3 45

Year 2 : 90x1 + 80x2 + 20 x3 + 100x1 + 160 x2 + 150 x3 65

Year 2 : 190x1 + 240x2 + 170 x3 65

Year 3 : 10x1 + 70x2 + 60 x3 + 190x1 + 240x2 + 170 x3 80Year 3 : 200x1 + 310x2 + 230 x3 80

x1 , x2, x3 0.

Capital budgeting : Formulation

An airline reservations office is open to take reservations by telephone 24 hours per day, Monday through Friday.The number of reservation officers needed for each time period is shown below.

The union contract requires all employees to work 8 consecutive hours. Therefore, we have shifts of 12am-8am, 4am-12pm, 8am-4pm, 12pm-8pm, 4pm-12am, 8pm-4am. Hire the minimum number of reservation agents needed to cover all requirements.

Personnel scheduling problem : Narrative representation

Period Requirement12am-4am 114am-8am 158am-12pm 3112pm-4pm 174pm-8pm 258pm-12am 19

The union contract requires all employees to work 8 consecutive hours.

We have shifts of

12am-8am, 4am-12pm, 8am-4pm, 12pm-8pm, 4pm-12am, 8pm-4am.

Hire the minimum number of reservation agents needed to cover all requirements.

If there were not restrictions of 8 hrs sifts, then we could hire as required, for example 11 workers for 4 hors and 15 workers for 4 hours.

Personnel scheduling problem : Narrative representation

Personnel scheduling problem : Pictorial representation

12 am to 4 am

4 am to 8 am

8 am to 12 pm

12 pm to 4 pm

4 pm to 8 pm

8 pm to 12 am

Period Shift1 2 3 4 5 6

11

15

31

17

25

19

x1 = Number of officers in 12 am to 8 am shift

x2 = Number of officers in 4 am to 12 pm shift

x3 = Number of officers in 8 am to 4 pm shift

x4 = Number of officers in 12 pm to 8 pm shift

x5 = Number of officers in 4 pm to 12 am shift

x6 = Number of officers in 8 pm to 4 am shift

Personnel scheduling problem : Decision variables

Min Z = x1+ x2+ x3+ x4+ x5+ x6

12 am - 4 am : x1 +x6 11

4 am - 8 am : x1 +x2 15

8 am - 12 pm : +x2 + x3 31

12 pm - 4 pm : +x3 + x4 17

4 pm - 8 pm : +x4 + x5 25

8 pm - 12 am : +x5 + x6 19

x1 , x2, x3, x4, x5, x6 0.

Personnel problem : constraints and objective function

Personnel scheduling problem : excel solution

Aggregate Production Planning : Narrative

PM Computer Services assembles its own brand of computers.

Production capacity in regular time is 160 computer / week

Production capacity in over time is 50 computer / week

Assembly and inspection cost / computer is $190 in regular

time and $260 in over time.

Customer orders are as follows

Week 1 2 3 4 5 6

Orders 105 170 230 180 150 250

It costs $10 / computer / week to produce a computer in one

week and hold it in inventory for another week.

The Goal is to satisfy customer orders at minimum cost.

Linear Programming

Documents

Transcript of Linear Programming