3.1 Solved Problem Set.pdf

8/10/2019 3.1 Solved Problem Set.pdf

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Linear Programming - ExamplesInstructors - . Krca and H. Sral

Em505, Fall 2012 - Chapter 3 Page 1

Revised - 2

PRODUCT MIX PROBLEM

Five types of electronic equipment (A, B, C, D, and E) areproduced by using different components with different

quantities.

Component usages:

Product

Component A B C D E Availability(unit)

Resistor 3 4 0 3 5 2000Capacitator 2 3 2 2 6 1800

Transformer 1 1 1 1 2 750

Speaker 1 1 0 0 2 500

Transistor 2 3 2 4 8 2250

Demand for the products(units):

Product A B C D EMax Sales 200 150 50 400 500

Profit ($)/unit 2 4 8 6 2

OBJECTIVE:Max total profit

D.V.s: Quantity of each product (A, B, C, D, E) to be produced.

CONSTRAINTS:

1.Component availability should not be exceeded (Capacity).

2.Demand for each product should not be exceeded.


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xj: quantity of each product (A, B, C, D, and E) to be produced

(j=1, 2, 3, 4, and 5).

Max Z= 2x1+4x2+8x3+6x4+2x5

S.t.:

3x1+4x2+0x3+3x4+5x5 2000 (Resistor)

2x1+3x2+2x3+2x4+6x5 1800 (Capacitator)

1x1+1x2+1x3+1x4+2x5 750 (Transformer)

1x1+1x2+0x3+0x4+2x5 500 (Speaker)

2x1+3x2+2x3+4x4+8x5 2250 (Transistor)

x1200

x2150

x350

x4400

x5500

xj0


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LINDO:

MAX 2 X1 + 4 X2 + 8 X3 + 6 X4 + 2 X5

SUBJECT TO

RES) 3 X1 + 4 X2 + 3 X4 + 5 X5


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LP OPTIMUM FOUND AT STEP 4

OBJECTIVE FUNCTION VALUE

1) 3500.000

VARIABLE VALUE REDUCED COST

X1 50.000000 0.000000

X2 150.000000 0.000000

X3 50.000000 0.000000

X4 400.000000 0.000000

X5 0.000000 6.000000

ROW SLACK OR SURPLUS DUAL PRICES

RES) 50.000000 0.000000

CAP) 350.000000 0.000000

TRF) 100.000000 0.000000

SPK) 300.000000 0.000000

TRS) 0.000000 1.000000

DEM1) 150.000000 0.000000

DEM2) 0.000000 1.000000

DEM3) 0.000000 6.000000

DEM4) 0.000000 2.000000

DEM5) 500.000000 0.000000

NO. ITERATIONS= 4


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RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE

COEF INCREASE DECREASE

X1 2.000000 0.666667 1.500000

X2 4.000000 INFINITY 1.000000

X3 8.000000 INFINITY 6.000000

X4 6.000000 INFINITY 2.000000

X5 2.000000 6.000000 INFINITY

RIGHTHAND SIDE RANGES

ROW CURRENT ALLOWABLE ALLOWABLE

RHS INCREASE DECREASE

RES 2000.000000 INFINITY 50.000000

CAP 1800.000000 INFINITY 350.000000

TRF 750.000000 INFINITY 100.000000

SPK 500.000000 INFINITY 300.000000

TRS 2250.000000 33.333332 100.000000

DEM1 200.000000 INFINITY 150.000000

DEM2 150.000000 33.333332 100.000000

DEM3 50.000000 50.000000 16.666666

DEM4 400.000000 25.000000 16.666666

DEM5 500.000000 INFINITY 500.000000


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FEED MIX PROBLEM

20,000 broilers (chicken) fed for 8 weeks

Each consume 1 kg feed (in 8 weeks)

Chicken needs Calcium, Protein, and Fiber.Calcium Protein Fiber TL/kg

Limestone .380 - - .04

Corn .001 .09 .02 .15Soybean .002 .50 .08 .40

- Calcium should be between 0.8 to 1.2% of the mix

- Protein should be at least 22% of the mix

- Fiber should not exceed 5% of the mix

Limestone Corn Soybeans

Feed


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x1= kg of limestone in the mix

x2= kg of corn in the mix

x3= kg of soybeans in the mix

Min Z= 0.04x1+0.15x2+0.40x3S.t.

1)Calcium:(0.38x1+0.001x2+0.002x3 )/20000 0.008

0.38x1+0.001x2+0.002x3 1600.38x1+0.001x2+0.002x3 240

2)Protein:0.09x2+0.50x3 4400

3)Fiber:0.02x2+0.08x3 1000

4) Demand:

x1+x2+x3 =20000

xj0 j=1,2,3


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MIN 0.04 X1 + 0.15 X2 + 0.4 X3

SUBJECT TO

CAMIN) 0.38 X1 + 0.001 X2 + 0.002 X3 >= 160

CAMAX) 0.38 X1 + 0.001 X2 + 0.002 X3 = 4400

FIBER) 0.02 X2 + 0.08 X3


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Alternative formulation: Prepare 1 kg. (portion)

x1= fraction(kg) of limestone in the 1 kg mix

x2= fraction(kg) of corn in the 1 kg mix

x2= fraction(kg) of soybeans in the 1 kg mix

Min Z= 0.04x1+0.15x2+0.40x3S.t.

1) Calcium:0.38x1+0.001x2+0.002x3 0.080.38x1+0.001x2+0.002x3 0.012

2) Protein:0.09x2+0.50x3 0.22

3) Fiber:0.02x2+0.08x3 0.05

4) Demand:

x1+x2+x3 =1

xj0 j=1,2,3


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TRIMLOSS- CUTTING STOCK PROBLEM

STOCK size L PARTS (N)

lj: size of part j

dj: units demanded for partj

Paper rolls L=20 ft.

Order (part) Size (lj) ft. Demand(dj) units

1 5 150

2 7 200

3 9 300

Patterns

Size (ft) 1 2 3 4 5 6

5 0 2 2 4 1 0

7 1 1 0 0 2 0

9 1 0 1 0 0 2

Loss 4 3 1 0 1 2

xk: number of rolls cut by pattern k

Min Z= 4x1+3x2+1x3+0x4+1x5+2x6

S.t.

0x1+2x2+2x3+4x4+1x5+0x6 150 demand for 5



xk 0


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Microsoft Excel 9.0 Answer ReportWorksheet: [Book1]Sheet1Report Created: 31.10.2003 10:18:45

Target Cell (Min)

Cell NameOriginalValue Final Value

$H$4objLHS 0 400

Adjustable Cells


$B$3 X x1 0 0$C$3 X x2 0 0

$D$3 X x3 0 0$E$3 X x4 0 12.5$F$3 X x5 0 100$G$3 X x6 0 150

Constraints

Cell Name Cell Value Formula Status Slack

$H$5 5' LHS 150 $H$5>=$I$5 Binding 0$H$6 7' LHS 200 $H$6>=$I$6 Binding 0$H$7 9' LHS 300 $H$7>=$I$7 Binding 0

x1 x2 x3 x4 x5 x6 LHS RHS

X 0 0 0 0 0 0

obj 4 3 1 0 1 2 0

5' 0 2 2 4 1 0 0 150

7' 1 1 0 0 2 0 0 200

9' 1 0 1 0 0 2 0 300


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Microsoft Excel 9.0 Sensitivity Report

Worksheet: [Book1]Sheet1

Report Created: 31.10.2003 10:18:45

Adjustable Cells

Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$B$3 X x1 0 2.5 4 1E+30 2.5

$C$3 X x2 0 2.5 3 1E+30 2.5

$D$3 X x3 0 0 1 1E+30 0

$E$3 X x4 12.5 0 0 0 0

$F$3 X x5 100 0 1 5 1

$G$3 X x6 150 0 2 0 2

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$H$5 5' LHS 150 0 150 1E+30 50

$H$6 7' LHS 200 0.5 200 100 200

$H$7 9' LHS 300 1 300 1E+30 300


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WORK (worker) SCHEDULING

Consider a post office. Each worker at the office works 5

consecutive days and takes next 2 days off.

Demand for workers:

1 2 3 4 5 6 7

Monday T W Th F Sa Sunday

17 13 15 19 14 16 11

Minimize the total number of workers hired.

xj: number of workers that start on day j

Min Z = x1+x2+x3+x4+x5+x6+x7

S.t.

M: x1 +x4 +x5 +x6 +x7 17T: x1 +x2 +x5 +x6 +x7 13

W: x1 +x2 +x3 +x6 +x7 15

Th: x1 +x2 +x3 +x4 +x7 19

F: x1 +x2 +x3 +x4 +x5 14

Sa: +x2 +x3 +x4 +x5 +x6 16

Su: +x3 +x4 +x5 +x6 +x7 11

xj0 j=1,2,..,7


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Microsoft Excel 9.0 Answer ReportWorksheet: [Book1]Sheet1

Report Created: 9/25/00 5:17:02 PM

Target Cell (Min)


$J$3Obj(min) 0 22.33333333

Adjustable Cells


$B$2 X1 0 6.333333333$C$2 X2 0 5$D$2 X3 0 0.333333333$E$2 X4 0 7.333333333$F$2 X5 0 0$G$2 X6 0 3.333333333$H$2 X7 0 0

Constraints

Cell Name Cell Value Formula Status Slack

$J$4 M 17 $J$4>=$L$4 Binding 0

$J$5 T 14.66666667 $J$5>=$L$5 NotBinding 1.666666667$J$6 W 15 $J$6>=$L$6 Binding 0$J$7 Th 19 $J$7>=$L$7 Binding 0

$J$8 F 19 $J$8>=$L$8NotBinding 5

$J$9 Sa 16 $J$9>=$L$9 Binding 0$J$10 Su 11 $J$10>=$L$10 Binding 0

X1 X2 X3 X4 X5 X6 X76.33 5 0.33 7.33 0 3.33 0

Obj

(min) 1 1 1 1 1 1 1 22.33M 1 1 1 1 1 17.00 >= 17T 1 1 1 1 1 14.67 >= 13W 1 1 1 1 1 15.00 >= 15Th 1 1 1 1 1 19.00 >= 19F 1 1 1 1 1 19.00 >= 14Sa 1 1 1 1 1 16.00 >= 16Su 1 1 1 1 1 11.00 >= 11


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Integer:

Microsoft Excel 9.0 Answer Report

Worksheet: [worksch1int.xls]Sheet1

Report Created: 14.10.2003 18:32:59

Target Cell (Min)

Cell Name Original Value Final Value

$J$3Obj(min) 23 23

Adjustable Cells

Cell Name Original Value Final Value

$B$2 X1 7 7

$C$2 X2 4 5

$D$2 X3 0 1

$E$2 X4 8 6

$F$2 X5 0 0

$G$2 X6 4 4

$H$2 X7 0 0

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$B$2 X1 6.333333 0 1 0 1$C$2 X2 5 0 1 0 0

$D$2 X3 0.333333 0 1 0 0$E$2 X4 7.333333 0 1 0.5 1$F$2 X5 0 0.33333333 1 1E+30 0.33333333$G$2 X6 3.333333 0 1 0.5 1$H$2 X7 0 0 1 1E+30 0

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$J$4 M 17 0.33333333 17 0.5 2.5$J$5 T 14.66667 0 13 1.66666667 1E+30$J$6 W 15 0.33333333 15 11 1$J$7 Th 19 0.33333333 19 5 1$J$8 F 19 0 14 5 1E+30$J$9 Sa 16 0.33333333 16 0.5 2.5$J$10 Su 11 0 11 1.66666667 0.33333333


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FINANCIAL PLANNING

-2 Million dollars available to invest now.-3 investment opportunities A, B, C with cash inflows and

outflows over a 3-year horizon. You can invest once for each

alternative at the beginning and the partial investment for an

alternative is possible.

-A period of the horizon is 6-month long.

-Cash inflows from previous investments.

-Cash borrowing (credit) at 7% for six months.-Cash lending at 6% for six months.

-Total credit borrowed cannot exceed $2M in a period.

Cash flows ($M):

t 0 1 2 3 4 5 6Cash inflow 2.00 0.50 0.40 0.38 0.36 0.34 0.30A -3.00 -1.00 -1.80 0.40 1.80 1.80 5.50B -2.00 -0.50 1.50 1.50 1.50 0.20 -1.00C -2.00 -2.00 -1.80 1.00 1.00 1.00 6.00

Let:

XA, XB, XC: fraction of investment A,B,C bought respectively

Bt: cash borrowed at the beginning of time t

Lt: cash lent at the beginning of time t

Event happen at the beginning of each period


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OUT FLOW = IN FLOW

[T0] 3*XA+2*XB+2*XC+L0 = B0+2

[T1] 1.07*B0+L1+XA+0.5*XB+2*XC = 1.06*L0+B1+0.5

[T2] 1.07*B1+L2+1.8*XA+1.8*XC = 1.06*L1+B2+0.40+1.5*XB

[T3] 1.07*B2+L3 = 1.06*L2+B3+0.38+0.4*XA+1.5*XB+XC

[T4] 1.07*B3+L4 = 1.06*L3+B4+0.36+1.8*XA+1.5*XB+XC

[T5] 1.07*B4+L5 = 1.06*L4+B5+1.8*XA+0.2*XB+XC+0.34

[T6] 1.07*B5+L6+XB = 1.06*L5+0.3+5.5*XA+6*XC

[BR0] B02

[BR1] B12

[BR2] B22

[BR3] B32

[BR4] B42

[BR5] B52

[FRA] XA1

[FRB] XB1[FRC] XC1

MAX Z = L6

All variables are nonnegative

Solution:

XA=0.69 XB=0.655 XC=0

t 0 1 2 3 4 5 6Bt 1.38 2.00 2.00 0.50 0 0 0Lt 0 0 0 0 2.05 3.89 7.57


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1) 7.567981

VARIABLE VALUE REDUCED COST

L6 7.567981 0.000000

XA 0.690919 0.000000

XB 0.655769 0.000000

XC 0.000000 0.400744

L0 0.000000 0.017826

B0 1.384296 0.000000

L1 0.000000 0.365497

B1 2.000000 0.000000

L2 0.000000 0.062540

B2 2.000000 0.000000L3 0.000000 0.011236

B3 0.499978 0.000000

L4 2.052331 0.000000

B4 0.000000 0.010600

L5 3.890279 0.000000

B5 0.000000 0.010000


T0) 0.000000 1.907424

T1) 0.000000 1.782640

T2) 0.000000 1.336927

T3) 0.000000 1.202252T4) 0.000000 1.123600

T5) 0.000000 1.060000

T6) 0.000000 1.000000

BR0) 0.615704 0.000000

BR1) 0.000000 0.352128

BR2) 0.000000 0.050517

BR3) 1.500022 0.000000

BR4) 2.000000 0.000000

BR5) 2.000000 0.000000

FRA) 0.309081 0.000000

FRB) 0.344231 0.000000

FRC) 1.000000 0.000000

NO. ITERATIONS= 9


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L6 1.000000 INFINITY 1.000000XA 0.000000 2.941575 0.211771

XB 0.000000 0.132797 0.618110

XC 0.000000 0.400744 INFINITY

L0 0.000000 0.017827 INFINITY

B0 0.000000 0.017958 0.422384

L1 0.000000 0.365497 INFINITY

B1 0.000000 INFINITY 0.352128

L2 0.000000 0.062540 INFINITY

B2 0.000000 INFINITY 0.050517

L3 0.000000 0.011236 INFINITY

B3 0.000000 0.011236 0.032693

L4 0.000000 0.010600 0.234414

B4 0.000000 0.010600 INFINITYL5 0.000000 0.010000 0.236207

B5 0.000000 0.010000 INFINITY



RHS INCREASE DECREASE

T0 2.000000 1.567052 2.376805

T1 0.500000 0.841235 1.891358

T2 0.400000 1.295682 0.904893

T3 0.380000 0.499978 1.500022

T4 0.360000 INFINITY 2.052331T5 0.340000 INFINITY 3.890279

T6 0.300000 INFINITY 7.567981

BR0 2.000000 INFINITY 0.615704

BR1 2.000000 0.604248 1.151110

BR2 2.000000 0.970764 0.323569

BR3 2.000000 INFINITY 1.500022

BR4 2.000000 INFINITY 2.000000

BR5 2.000000 INFINITY 2.000000

FRA 1.000000 INFINITY 0.309081

FRB 1.000000 INFINITY 0.344231

FRC 1.000000 INFINITY 1.000000


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MULTI-PERIOD PLANNING PROBLEM

-A bicycle company will be manufacturing mens and womensmodels for its 10 speed bicycles during the next 2 months.

-Management wants to develop a production schedule indicating howmany bicycles of each model should be produced in each month.

-Demand forecasts call for 150 mens and 125 womens models to beshipped during the first month and 200 mens and 150 womensduring the second month.

Model Productioncost

Labor Requirements CurrentinventoryProduction Assembly

Mens $120 2.0 hrs 1.5 hrs 20Womens $90 1.6 hrs 1.0 hrs 30

-Last month the company used a total of 1000 hours of labor.Companys policy will not allow the total hours of labor to increase

or decrease by more than 100 hours from month to month.

-The company charges 2% of the unit production cost for a unitinventory per period. The company would like to have at least 25units of each model at the end of the 2 months.

What is the minimum cost production schedule?


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Let: for i=1,2 and t=1,2

Xit= amount of product i produced in month t

Iit= amount of inventory of product i at the end of month tLt= total hours of labor used in month t

INFLOW = OUTFLOW (material balance)

MIN Z= 120X11+ 120X12+ 90X21+ 90X22 + 2.4I11+ 2.4I12+ 1.8I21+ 1.8I22

S.T.

P1MEN) X11- I11= 130

P2MEN) X12 + I11- I12= 200

P1WMN) X21- I21= 95

P2WMN) X22+ I21- I22= 150

MINVMEN) I12>= 25

MINVWMN) I22>= 25

P1LABOR) 3.5 X11+ 2.6 X21= L1

P2LABOR) 3.5 X12+ 2.6 X22= L2

MINL1) L1>= 900

MAXL1) L1= - 100

MAXL2) L2- L1


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Adjustable Cells

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$B$2 x11 192.93 0.000 120 0.023 2.40$C$2 x12 162.07 0.000 120 2.400 0.02$D$2 x21 95.00 0.000 90 ###### 0.02$E$2 x22 175.00 0.000 90 0.017 92.69$F$2 I11 62.93 0.000 2.4 0.023 2.40$G$2 I12 25.00 0.000 2.4 ###### 123.60$H$2 I21 0.00 0.017 1.8 ###### 0.02$I$2 I22 25.00 0 1.8 ###### 92.69$J$2 L1 922.25 0 0 ###### 0.69$K$2 L2 1022.25 0 0 0.686 70.63

Constraints

Final Shadow Constraint Allowable AllowableCell Name Value Price R.H. Side Increase Decrease

$L$6 P1Mens 130.00 118.800 130 ###### 12.71

$L$7 P2Mens 200.00 121.200 200 ###### 12.71$L$8 P2InvMens 25.00 123.600 25 ###### 12.71$L$9 P1Wmns 95.00 89.109 95 ###### 17.12$L$10 P2Wmns 150.00 90.891 150 ###### 17.12$L$11 P2InvWmns 25.00 92.691 25 ###### 17.12$L$12 P1Labor 0.00 -0.343 0 ###### 44.50$L$13 P2Labor 0.00 0.343 0 ###### 44.50$L$14 P1LbrLB 922.25 0.000 900 22.250 #########$L$15 P2LbrLB 100.00 0.000 -100 ###### #########$L$16 P1LbrUB 922.25 0.000 1100 ###### 177.75$L$17 P2LbrUB 100.00 -0.343 100 44.500 200.00

x11 x12 x21 x22 I11 I12 I21 I22 L1 L2192.93 162.07 95.00 175.00 62.93 25.00 0.00 25.00 922.25 1022.25

Obj(min) 120 120 90 90 2.4 2.4 1.8 1.8 67156.03

ConstraintsP1Mens 1 -1 130.00 = 130P2Mens 1 1 -1 200.00 = 200P2InvMens 1 25.00 >= 25P1Wmns 1 -1 95.00 = 95P2Wmns 1 1 -1 150.00 = 150P2InvWmns 1 25.00 >= 25P1Labor -3.5 -2.6 1 0.00 = 0P2Labor -3.5 -2.6 1 0.00 = 0P1LbrLB 1 922.25 >= 900P2LbrLB -1 1 100.00 >= -100

P1LbrUB 1 922.25


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1) 67156.03


P1MEN) 0.000000 -118.800003

P2MEN) 0.000000 -121.199997

P1WMN) 0.000000 -89.108574

P2WMN) 0.000000 -90.891426

MINVMEN) 0.000000 -123.599998

MINVWMN) 0.000000 -92.691429

P1LABOR) 0.000000 -0.342857

P2LABOR) 0.000000 0.342857

MINL1) 22.250000 0.000000

MAXL1) 177.750000 0.000000

MINL2) 200.000000 0.000000MAXL2) 0.000000 0.342857

NO. ITERATIONS= 6





X11 120.000000 0.023088 2.400000

X12 120.000000 2.400000 0.023088

X21 90.000000 INFINITY 0.017151

X22 90.000000 0.017151 92.691429

I11 2.400000 0.023088 2.400000

I12 2.400000 INFINITY 123.599998

I21 1.800000 INFINITY 0.017148

I22 1.800000 INFINITY 92.691429

L1 0.000000 INFINITY 0.685714

L2 0.000000 0.685714 70.628571



RHS INCREASE DECREASEP1MEN 130.000000 101.571426 12.714286

P2MEN 200.000000 101.571426 12.714286

P1WMN 95.000000 136.730774 17.115385

P2WMN 150.000000 136.730774 17.115385

MINVMEN 25.000000 101.571426 12.714286

MINVWMN 25.000000 136.730774 17.115385

P1LABOR 0.000000 44.500000 355.500000

P2LABOR 0.000000 44.500000 355.500000

MINL1 900.000000 22.250000 INFINITY

MAXL1 1100.000000 INFINITY 177.750000

MINL2 -100.000000 200.000000 INFINITY

MAXL2 100.000000 44.500000 200.000000


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REGIONAL PLANNING

One of the interesting social experiments is the system of kibbutzim, orcommunal farming communities in Israel.

It is common for groups of kibbutzim to join together to share common

technical services and to coordinate their production. The example

concerns one such group of three kibbutzim, called Southern

Confederation of Kibbutzim (SCK).

Overall planning for SCK is done in its technical office. The office

currently is planning agricultural production for the coming year.

The agricultural output of each kibbutz is limited both by:

- amount of available irrigable land

- quantity of water allocated for irrigation (by a national government

official)

Kibbutz Usable land

(acres)

Water allocation

(acre feet)

1 400 600

2 600 800

3 300 375


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The crops suited for this region include sugar beets, cotton, and sorghum,

and these are considered for the upcoming season. These crops differ

primarily in their expected net return/acre and their consumption of

water.

In addition, the Ministry of Agriculture has set a maximum quota for the

total acreage that can be devoted to each of these crops by the SCK.

Crop Maximum

quota (acres)

Water

consumption

(acre feet/acre)

Net Return

($/acre)

Sugar beet 600 3 400

Cotton 500 2 300

Sorghum 325 1 100

-The three kibbutzim belonging to the SCK have agreed that every kibbutz

will plant the same proportion of its available irrigable land.

-Any combination of the crops may be grown at any of the kibbutzim.

The technical office is to plan how many acres to devote to each crop at the

respective kibbutzim while satisfying the given restrictions.

The objective is to maximize the total net return to SCK.

Max total profit

St.

1) Land availability

2) Water availability

3) Quota

4) Same proportion


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Let: kibbutz (i=1,2,3), crop (j=1,2,3)

xij= number of acres allocated in kibbutz i to crop j

K= proportion of land planted by each kibbutz

Max Z = 400(x11+x21+x31) + 300(x12+x22+x32) + 100(x13+x23+x33)

St:

1) Land availability:

[L1] x11+x12+x13=400K

[L2] x21+x22+x23=600K

[L3] x31+x32+x33=300K

2) Water availability:

[W1] 3x11+2x12+1x13600

[W2] 3x21+2x22+1x23800

[W3] 3x31+2x32+1x33375

3) Quota:

[Q1] x11+x21+x31600

[Q2] x12+x22+x32500

[Q3] x13+x23+x33325

4) Same proportion:

[EQ] K1

xij0 i=1,2,3; j=1,2,3


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Solution: K=0.5833 Z=$253,333.3

CropKibbutz (acres)

1 2 3 Total

Sugar beet 133.33 100 25 258.33Cotton 100 250 150 500

Sorghum 0 0 0 0

Total 233.33 350 175 758.33

Variable Value Reduced Cost

X11 133.3333 0.000000

X21 100.0000 0.000000

X31 25.00000 0.000000

X12 100.0000 0.000000X22 250.0000 0.000000

X32 150.0000 0.000000

X13 0.000000 33.33333

X23 0.000000 33.33333

X33 0.000000 33.33333

K 0.5833333 0.000000

Row Slack or Surplus Dual Price

L1 0.000000 0.000000

L2 0.000000 0.000000

L3 0.000000 0.000000

W1 0.000000 133.3333

W2 0.000000 133.3333W3 0.000000 133.3333

Q1 341.6667 0.000000

Q2 0.000000 33.33333

Q3 325.0000 0.000000

EQ 0.4166667 0.000000

Righthand Side Ranges:

Row Current Allowable Allowable

RHS Increase Decrease

L1 0.0 96.29630 48.14815

L2 0.0 92.85714 54.16667

L3 0.0 16.25000 65.00000W1 600.0000 144.4444 167.7419

W2 800.0000 162.5000 144.4444

W3 375.0000 195.0000 29.54545

Q1 600.0000 INFINITY 341.6667

Q2 500.0000 162.5000 325.0000

Q3 325.0000 INFINITY 325.0000

EQ 1.000000 INFINITY 0.4166667


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FACILITY LOCATION

We want to locate a new facility which will interact with the

existing ones. The cost of interaction is proportional to thedistance.

Let there be N existing facilities and (ai, bi) be their coordinates.

(x,y): coordinate of the new facility

di(x,y)= distance from facility i to the new facility located at (x,y)

Euclidean: [ ](straight line)

S. Euclidean: [ ]

3

2

4

?

(2,1)

(11,2)

(14,7)

(4,11)

(x,y)


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Rectilinear: | | | |(Manhattan)

| | | |

|aix|= Max (ai-x , x-ai)

viai-x

and

vix-ai i=1,,N

similarly:

vibi-y

and

uiy-bi i=1,,N

3

2

4

?

(2,1)

(11,2)

(14,7)

(4,11)

(x,y)


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[ ]

S.t.

viai-x i=1,,N

vix-ai i=1,,N

uibi-y i=1,,N

uiy-bi i=1,,N

vi0 and ui0 i=1,,N

Alternative formulation:

free means urs

free = free+- free

-and

free

+0, free

-0

[ ] +[ ]St.

wi: weight of facility i ?

3.1 Solved Problem Set.pdf

Documents

Transcript of 3.1 Solved Problem Set.pdf