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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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 2
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 3
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 4
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 5
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 6
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 7
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 8
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 9
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 10
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
1x1+1x2+0x3+0x4+2x5+0x6 200 demand for 7
1x1+0x2+1x3+0x4+0x5+2x6 300 demand for 9
xk 0
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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 11
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
Cell NameOriginalValue Final Value
$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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 12
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 13
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 14
Microsoft Excel 9.0 Answer ReportWorksheet: [Book1]Sheet1
Report Created: 9/25/00 5:17:02 PM
Target Cell (Min)
Cell NameOriginalValue Final Value
$J$3Obj(min) 0 22.33333333
Adjustable Cells
Cell NameOriginalValue Final Value
$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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 15
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 16
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 17
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 18
LP OPTIMUM FOUND AT STEP 9
OBJECTIVE FUNCTION VALUE
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
ROW SLACK OR SURPLUS DUAL PRICES
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 19
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
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
RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 20
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 21
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
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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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 23
LP OPTIMUM FOUND AT STEP 6
OBJECTIVE FUNCTION VALUE
1) 67156.03
ROW SLACK OR SURPLUS DUAL PRICES
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
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
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
RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
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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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
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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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
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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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
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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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
Em505, Fall 2012 - Chapter 3 Page 29
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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Linear Programming - ExamplesInstructors - . Krca and H. Sral
[ ]
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 ?
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