LP Formulation Set 2 Agricultural planning : narrative Three farming communities are developing a joint agricultural production plan for the coming year. Production capacity.
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Transcript LP Formulation Set 2 Agricultural planning : narrative Three farming communities are developing a joint agricultural production plan for the coming year. Production capacity.
LP Formulation
Set 2
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
1
2
3
Land (Acres)
400
600
300
Water (Acres Feet)
600
800
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
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June-2013
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Agricultural planning : narrative
Crop
1
2
3
Max desired
(Acres)
600
500
325
Water consumption
(Acre feet / Acre)
3
2
1
Net return
($/Acre)
1000
750
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 300
out 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
LP-Formulation
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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
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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
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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
LP-Formulation
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
LP-Formulation
300
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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
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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
cost / pound
Sales price
/ 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
M4 : exactly 10%
2.5
7
C
M1 : 70% of total
2
5.5
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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.
LP-Formulation
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SAVE-IT Company : Mixture Specification
A1: weight of solid waste 1 in grade A
A1, A2, A1, A4, B1, B2, B3, B4, C1, C2, C3, C4
Mixture Specifications:
Grade A:
A1 0.3 (A1+A2+A3+A4)
A2 0.4 (A1+A2+A3+A4)
A3 0.5 (A1+A2+A3+A4)
A3 = 0.2 (A1+A2+A3+A4)
Grade B:
B1 0.5(B1+B2+B3+B4)
B2 0.1(B1+B2+B3+B4)
B4 = 0.1(B1+B2+B3+B4)
Grade C:
LP-Formulation
C1 0.3 (C1+C2+C3+C4)
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SAVE-IT Company : Material Availability and ussage
Availability of material
A1+B1+C1 3000
A2+B2+C2 2000
A3+B3+C3 4000
A4+B4+C4 1000
At least half of the material treated
A1+B1+C1 1500
A2+B2+C2 1000
A3+B3+C3 2000
A4+B4+C4 500
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SAVE-IT Company : Treatment and Processing
Costs, and Profit
Spend all the treatment budget
3(A1+B1+C1)+6(A2+B2+C2)+4(A3+B3+C3)+5(A4+B4+C4) = 30000
Maximize profit Z
(8.5-3)(A1+A2+A3+A4)+(7-2.5) (B1+B2+B3+B4)+(5.5-2)
(C1+C2+C3+C4)
– 3(A1+B1+C1)-6(A2+B2+C2)-4(A3+B3+C3)-5(A4+B4+C4))
A1, A2, A1, A4, B1, B2, B3, B4, C1, C2, C3, C4 0
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SAVE-IT Company : Treatment and Processing
Costs, and Profit
Pounds of Material j in Product i
A
B
C
1
750
2250
0
2 1000
0
0
3
250
2250
0
4
500
500
0
SUM
3000
1000
2500
1000
Processing $/Unit Available
3
3000
6
2000
4
4000
5
1000
Sum
2500
5000
0
Cost
5.5
4.5
3.5
A1
A3
B1
B2
C1
A4
0.3
0.5
0.5
0.1
0.7
0.2
0
-1000
-250
-500
0
0
<=
<=
<=
<=
<=
=
0
0
0
0
0
0
B4
A2
0.1
0.4
0
0
=
>=
0
0
LP-Formulation
6250
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At Least Half
1500
1000
2000
500
Budget
30000
June-2013
30000
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Problem (From Hillier and Hillier)
Strawberry shake production
Several ingredients can be used in this product.
Ingredient
calories from fat Total calories Vitamin
( per tbsp)
(per tbsp) (mg/tbsp)
Strawberry flavoring
1
50
20
Cream
75
100
0
Vitamin supplement
0
0
50
Artificial sweetener
0
120
0
Thickening agent
30
80
2
Thickener Cost
(mg/tbsp) ( c/tbsp)
3
10
8
8
1
25
2
15
25
6
This beverage has the following requirements
Total 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.
LP-Formulation
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Decision variables
Decision Variables
X1 : tbsp of strawberry
X2 : tbsp of cream
X3 : tbsp of vitamin
X4 : tbsp of Artificial sweetener
X5 : tbsp of thickening
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Constraints
Objective Function
Min Z = 10X1 + 8X2 + 25 X3 + 15 X4 + 6 X5
Calories
50X1 + 100 X2 + 120 X4 + 80 X5 380
50X1 + 100 X2 + 120 X4 + 80 X5 420
Calories from fat
X1 + 75 X2 + 30 X5 0.2(50X1 + 100 X2 + 120 X4 + 80 X5)
Vitamin
20X1 + 50 X3 + 2 X5 50
Strawberry and sweetener
X1 2 X4
Thickeners
3X1 + 8X2 + X3 + 2 X4 + 2.5 X5 = 15
Non-negativity
X1 , X2 , X3 , X4 , X5 0
LP-Formulation
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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
LP-Formulation
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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
0
1
2
3
NPV
LP-Formulation
Project 1
40
60
90
10
45
5% of Project 1
2
3
4.5
.5
2.25
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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
LP-Formulation
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Capital budgeting : Formulation
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
LP-Formulation
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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 45
Year 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 80
Year 3 : 200x1 + 310x2 + 230 x3 80
x1 , x2, x3 0.
LP-Formulation
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June-2013
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Personnel scheduling problem : Narrative
representation
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:
Period
12am-4am
4am-8am
8am-12pm
12pm-4pm
4pm-8pm
8pm-12am
Requirement
11
15
31
17
25
19
The union 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.
LP-Formulation
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June-2013
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Personnel scheduling problem : Narrative
representation
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, 8pm4am.
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.
LP-Formulation
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June-2013
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Personnel scheduling problem : Pictorial
representation
Period
1
2
Shift
3
4
5
6
11
12 am to 4 am
4 am to 8 am
15
8 am to 12 pm
31
12 pm to 4 pm
17
4 pm to 8 pm
25
8 pm to 12 am
19
LP-Formulation
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June-2013
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Personnel scheduling problem : Decision variables
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
LP-Formulation
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June-2013
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Personnel problem : constraints and objective
function
Min Z = x1+ x2+ x3+ x4+ x5+ x6
12 am - 4 am :
4 am - 8 am :
8 am - 12 pm :
+x6 11
x1
15
x1 +x2
31
+x2 + x3
12 pm - 4 pm :
+x3 + x4 17
4 pm - 8 pm :
+x4 + x5
25
+x5 + x6 19
8 pm - 12 am :
x1 , x2, x3, x4, x5, x6 0.
LP-Formulation
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June-2013
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Personnel scheduling problem : excel solution
LP-Formulation
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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.
LP-Formulation
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Refresh
We need to lease warehouse space. The estimated required
space ( in 1000 sq ft) is given below.
Month
1
2
3
4
5
Space 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 case
Leasing period (months) 1
2
3
4
5
Cost per sq-feet leased
65
100
135 160
190
Now 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.
LP-Formulation
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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
LP-Formulation
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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
LP-Formulation
Ardavan Asef-Vaziri
June-2013
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excel; Format 1
x11 x12 x13 x14 x15 x22 x23 x24 x25 x33 x34 x35 x44 x45 x55
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
1
1
1
0
1
1
1
1
1
1
0
0
0
0
0
0
1
1
0
0
1
1
0
1
1
1
1
0
0
0
0
0
1
0
0
0
1
0
0
1
0
1
1
65 100 135 160 190
LP-Formulation
65 100 135 160
65 100 135
65 100
Ardavan Asef-Vaziri
30
30
40
30
50
>=
>=
>=
>=
>=
30
20
40
10
50
65
June-2013
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excel; Format 2
x11
x12
x22
x13
x23
x33
x14
x24
x34
x44
x15
x25
x35
x45
x55
0
0
0
0
0
0
0
0
0
0
0
0
10
0
0
0
0
0
0
0
30
0
0
0
20
65
100000
100000
100000
100000
100
65
100000
100000
100000
135
100
65
100000
100000
160
135
100
65
100000
190
160
135
100
65
30
30
40
30
50
>=
>=
>=
>=
>=
30
20
40
10
50
7650
LP-Formulation
Ardavan Asef-Vaziri
June-2013
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excel; Best Format (Ctrl)
x11
x12
x22
x13
x23
x33
x14
x24
x34
x44
x15
x25
x35
x45
x55
0
0
0
0
0
10
0
0
0
0
30
0
0
0
20
65
100
65
135
100
65
160
135
100
65
190
160
135
100
65
30
30
40
30
50
>=
>=
>=
>=
>=
30
20
40
10
50
7650
LP-Formulation
Ardavan Asef-Vaziri
June-2013
35
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
LP-Formulation
Ardavan Asef-Vaziri
June-2013
36
Controlling air pollution : narrative
Pollutant
Particulate
Sulfur
Hydrocarb.
Taller
smokestacks
B.F. O.F
12
35
37
9
42
53
Filter
Better fuel
B.F. O.F.
B.F. O.F.
25
18
28
17
56
29
20
31
24
13
49
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
smokestacks
B.F. O.F
B.F. O.F.
B.F. O.F.
Cost
LP-Formulation
12
9
25
20
Ardavan Asef-Vaziri
17
June-2013
13
37
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
LP-Formulation
Ardavan Asef-Vaziri
June-2013
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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.
LP-Formulation
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Controlling air pollution : Formulation
Pollutant
Particulate
Sulfur
Hydrocarb.
Taller
smokestacks
B.F. O.F
12
35
37
9
42
53
Filter
Better fuel
B.F. O.F.
B.F. O.F.
25
18
28
17
56
29
20
31
24
13
49
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 ????
LP-Formulation
Ardavan Asef-Vaziri
June-2013
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