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LP Formulation Set 2. Problem (From Hillier and Hillier) . Strawberry shake production Several ingredients can be used in this product . Ingredient calories from fat Total calories Vitamin Thickener Cost

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slide2

Problem (From Hillier and Hillier)

Strawberry shake production

Several ingredients can be used in this product.

Ingredient calories from fatTotal caloriesVitaminThickener Cost

( per tbsp) (per tbsp) (mg/tbsp) (mg/tbsp)( c/tbsp)

Strawberry flavoring 1 50 20 3 10

Cream 75 100 0 8 8

Vitamin supplement 0 0 50 1 25

Artificial sweetener 0 120 0 2 15

Thickening agent 30 80 2 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.

slide3

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

slide4

Constraints

Objective Function

Min Z = 10X1 + 8X2 + 25X3 +15X4 + 6X5

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 +2X4 + 2.5X5= 15

Non-negativity

X1 , X2 , X3 , X4 ,X5  0

slide5

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)

1 400 600

2 600 800

3 300 375

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

slide6

Agricultural planning : narrative

Crop Max desired Water consumption Net return

(Acres) (Acre feet / Acre) ($/Acre)

1 600 3 1000

2 500 2 750

3 325 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 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

slide7

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

slide8

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

slide9

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

slide10

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

slide11

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

slide12

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 sourcesof pollution: Open furnaceand Blast furnace

Threetypes 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)

Threepollution 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

slide13

Controlling air pollution : narrative

Pollutant Taller FilterBetter fuel

smokestacks

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

Particulate 12 9 25 20 17 13

Sulfur 35 42 18 31 56 49

Hydrocarb. 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 FilterBetter fuel

smokestacks

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

Cost 12 9 25 20 17 13

slide14

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

slide15

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.

slide16

Controlling air pollution : Formulation

Pollutant Taller FilterBetter fuel

smokestacks

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

Particulate 12 9 25 20 17 13

Sulfur 35 42 18 31 56 49

Hydrocarb. 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 ????

slide17

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

slide18

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.

slide19

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: C1  0.3 (C1+C2+C3+C4)

slide20

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

slide21

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

slide22

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

slide23

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 1

0 40 2

1 60 3

2 90 4.5

3 10 .5

NPV 45 2.25

slide24

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

slide25

Capital budgeting : Formulation

x1 = proportion of project 1 invested by us.

x2 = proportionof project 2 invested by us.

x3 = proportionof 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

slide26

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.

slide27

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 Requirement

12am-4am 11

4am-8am 15

8am-12pm 31

12pm-4pm 17

4pm-8pm 25

8pm-12am 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.

slide28

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, 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.

slide29

Personnel scheduling problem : Pictorial representation

Period Shift

1 2 3 4 5 6

11

15

31

17

25

19

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

slide30

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

slide31

Personnel problem : constraints and objective function

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.

slide33

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.

slide34

Refresh

We need to lease warehouse space. The estimated required space ( in 1000 sqft) 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.

slide35

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

slide36

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

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