lp formulation set 2

LP FormulationSet 2

2Ardavan Asef-Vaziri June-2013LP-Formulation

Problem (From Hillier and Hillier)

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 6This 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

Objective FunctionMin Z = 10X1 + 8X2 + 25 X3 + 15 X4 + 6 X5 Calories 50X1 + 100 X2 + 120 X4 + 80 X5 38050X1 + 100 X2 + 120 X4 + 80 X5 420Calories from fat X1 + 75 X2 + 30 X5 0.2(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 = 15Non-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 250Because 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 1x21 = Acres allocated to Crop 2 in Community 1x31 = Acres allocated to Crop 3 in Community 1x12 = Acres allocated to Crop 1 in Community 2x22 = Acres allocated to Crop 2 in Community 2x32 = Acres allocated to Crop 3 in Community 2……………..xij = Acres allocated to Crop i in Community ji for crop j for community, we could have switched themNote that x is volume not portion, we could have had it as portion


Agricultural planning : Formulation

Land x11+x21+x31 400x12+x22+x32 600x13+x23+x33 300

Water3x11+2x21+1x31 6003x12+2x22+1x32 8003x13+2x23+1x33 375



Cropsx11+ x12 + x13 600x21 +x22 +x23 500x31 +x32 +x33 320

Proportionality of land usex11+x21+x31 x12+x22+x32

400 600x11+x21+x31 x13+x23+x33

400 300


Agricultural planning : all variables on LHS

Proportionality of land use600(x11+x21+x31 ) - 400(x12+x22+x32 ) = 0300(x11+x21+x31 ) - 400(x13+x23+x33 ) = 0

600x11+ 600 x21+ 600 x31 - 400x12- 400 x22- 400 x32 = 0300x11+ 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

smokestacksB.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 haveTechnique i source j


Controlling air pollution : Decision Variables

x11 = Proportion of technique 1 implemented of source 1x12 = Proportion of technique 1 implemented of source 2

x21 = Proportion of technique 2 implemented of source 1.x22 = Proportion of technique 2 implemented of source 2

x31 = Proportion of technique 3 implemented of source 1x32 = Proportion of technique 3 implemented of source 2.


Controlling air pollution : Formulation

Min Z= 12x11+9x12+ 25x21+20x22+ 17x31+13x32 Particulate; 12x11+9x12+ 25x21+20x22+ 17x31+13x32 60Sulfur; 35x11+42x12+ 18x21+31x22+ 56x31+49x32 150Hydrocarbon; 37x11+53x12+ 28x21+24x22+ 29x31+20x32 125 x11, x12, x21, x22, x31, x32 ???? x11, x12, x21, x22, x31, x32 ????



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


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 belowGrade 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 belowMaterial Pounds available / week Treatment cost / poundM1 3000 3M2 2000 6M3 4000 4M4 1000 5Due 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 Specifications2. Availability of material3. At least half of the material treated4. Spend all the treatment budget5. Maximize profit Z


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 30 40 80 901 60 80 602 90 80 203 10 70 60NPV 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 45MOur total available money to invest in years 0, 1, 2 is 65MOur total available money to invest in years 0, 1, 2, 3 is 80MTo 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 = proportion of project 1 invested by us.x2 = proportion of project 2 invested by us.x3 = proportion of project 3 invested by us.Maximize NPV Z = 45x1 + 70 x2 + 50 x3

subject to Year 0 : 40 x1 + 80 x2 + 90 x3 25Year 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 114 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 / weekProduction 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 followsWeek 1 2 3 4 5 6Orders 105 170 230 180 150 250It 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.


Refresh

We need to lease warehouse space. The estimated required space ( in 1000 sq ft) is given below. Month 1 2 3 4 5Space required 30 20 40 10 50

If the leasing cost was fixed the best strategy was to lease as needed. But this is not the caseLeasing period (months) 1 2 3 4 5Cost per sq-feet leased 65 100 135 160 190Now it may be more economical to lease for more than one month and take advantage of the lower rates for longer periods.

Find the optimal leasing strategy to minimize leasing costs.


Decision Variables

Xij spaced leased in month i and kept until month j months. i = 1, 2, 3, 4, 5. j= i, i+1, …, 5

Min z = 65X11 +100 X12 +135 X13 +160 X14+190 X15

+ 65X22+100 X23 +135 X24 +160 X25 + 65X33+100 X34 +135 X35

+ 65X44+100 X45

+ 65X55


Constraints

X11 + X12 + X13 + X14+ X15 30,000

X12 + X13 + X14+ X15 + X22+ X23 + X24 + X25 20,000

X13 + X14+ X15 + X23 + X24 + X25 + X33+ X34 + X35 40,000

X14+ X15 + X24 + X25 + X34 + X35 + X44+ X45 10,000

X15 + X25 + X35 + X45 + X55 50,000

X11 , X12 , X13 , X14 , X15 , X22 , X23 , X24 , X25 , X33 , X34 , X35

X44 , X45 , X55 0


excel; Format 1

x11 x12 x13 x14 x15 x22 x23 x24 x25 x33 x34 x35 x44 x45 x551 1 1 1 1 0 0 0 0 0 0 0 0 0 0 30 >= 300 1 1 1 1 1 1 1 1 0 0 0 0 0 0 30 >= 200 0 1 1 1 0 1 1 1 1 1 1 0 0 0 40 >= 400 0 0 1 1 0 0 1 1 0 1 1 1 1 0 30 >= 100 0 0 0 1 0 0 0 1 0 0 1 0 1 1 50 >= 50

65 100 135 160 190 65 100 135 160 65 100 135 65 100 65


excel; Format 2

x11 x12 x13 x14 x15x22 x23 x24 x25

x33 x34 x35x44 x45

x55

0 0 0 0 30 30 >= 300 0 0 0 0 30 >= 200 0 10 0 0 40 >= 400 0 0 0 0 30 >= 100 0 0 0 20 50 >= 50

65 100 135 160 190100000 65 100 135 160100000 100000 65 100 135100000 100000 100000 65 100100000 100000 100000 100000 65

7650


excel; Best Format (Ctrl)

x11 x12 x13 x14 x15x22 x23 x24 x25

x33 x34 x35x44 x45

x55

0 0 0 0 30 30 >= 300 0 0 0 30 >= 20

10 0 0 40 >= 400 0 30 >= 10

20 50 >= 50

65 100 135 160 19065 100 135 160

65 100 13565 100

657650

lp formulation set 2

Documents

ardavan asefvaziri

community ji

community 1x21

community 1x31

community 1x12

community 2x22

community 2x32

tbsp of thickening