LP Formulation Set 2
Jan 19, 2016
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 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.
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Decision variables
Decision VariablesX1 : tbsp of strawberryX2 : tbsp of creamX3 : tbsp of vitaminX4 : tbsp of Artificial sweetenerX5 : tbsp of thickening
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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
5Ardavan Asef-Vaziri June-2013LP-Formulation
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
6Ardavan Asef-Vaziri June-2013LP-Formulation
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
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Agricultural planning : decision variables
x11 = Acres allocated to Crop 1 in Community 1
x21 = Acres allocated to Crop 2 in Community 1
x31 = Acres allocated to Crop 3 in Community 1
x12 = Acres allocated to Crop 1 in Community 2
x22 = Acres allocated to Crop 2 in Community 2
x32 = Acres allocated to Crop 3 in Community 2
……………..
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
8Ardavan Asef-Vaziri June-2013LP-Formulation
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
9Ardavan Asef-Vaziri June-2013LP-Formulation
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
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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
11Ardavan Asef-Vaziri June-2013LP-Formulation
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
12Ardavan Asef-Vaziri June-2013LP-Formulation
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
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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
Pollutant Taller Filter Better fuel
smokestacksB.F. O.F B.F. O.F. B.F. O.F.
Cost 12 9 25 20 17 13
14Ardavan Asef-Vaziri June-2013LP-Formulation
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
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Controlling air pollution : Decision Variables
x11 = Proportion of technique 1 implemented of source 1
x12 = 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 1
x32 = Proportion of technique 3 implemented of source 2.
16Ardavan Asef-Vaziri June-2013LP-Formulation
Controlling air pollution : Formulation
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 ????
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
17Ardavan Asef-Vaziri June-2013LP-Formulation
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
18Ardavan Asef-Vaziri June-2013LP-Formulation
SAVE-IT Company : Narrative
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.
19Ardavan Asef-Vaziri June-2013LP-Formulation
SAVE-IT Company : Narrative
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
20Ardavan Asef-Vaziri June-2013LP-Formulation
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
21Ardavan Asef-Vaziri June-2013LP-Formulation
Capital budgeting : Narrative representation
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
22Ardavan Asef-Vaziri June-2013LP-Formulation
Capital budgeting : Narrative representation
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
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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 25
Year 1 : Investment in year 0 + Investment in year 1 45
Capital budgeting : Formulation
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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
25Ardavan Asef-Vaziri June-2013LP-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
26Ardavan Asef-Vaziri June-2013LP-Formulation
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
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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
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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
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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
30Ardavan Asef-Vaziri June-2013LP-Formulation
Personnel scheduling problem : excel solution
31Ardavan Asef-Vaziri June-2013LP-Formulation
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.
32Ardavan Asef-Vaziri June-2013LP-Formulation
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.
33Ardavan Asef-Vaziri June-2013LP-Formulation
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
34Ardavan Asef-Vaziri June-2013LP-Formulation
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
35Ardavan Asef-Vaziri June-2013LP-Formulation
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
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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
37Ardavan Asef-Vaziri June-2013LP-Formulation
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