Production Mix ProblemProduction Mix ProblemGraphical Solution

0 10 20 30 40 50 60 med

lrg

40

30

20

10

0

ElectronicsCabinetryProfit

(10,20) (Optimal Product Mix!)

Profit = 80*10 + 120*20 = $3200/mo

27-inch sets

20-in

ch s

ets

Feasible Region

Product Mix ProblemProduct Mix ProblemPrimal Problem Formulation

Production Capacity Constraints:6*med + 15*lrg < = 3604*med + 5*lrg < = 140med > = 0; lrg > = 0

Market Constraints:med < = 15lrg < = 40

Objective Function:Maximize: Profit = 80*med + 120*lrg

(hours/set)*sets = hours

(sets)

($/set)*sets = $

Primal ProblemPrimal Problem (Interpretation of Dual Prices)

Row Dual Price (Marginal value of an 1 1.000000 additional unit of 2 2.666667 the resource) 3 16.00000 4 0.0000000 5 0.0000000

Work-Scheduling ProblemWork-Scheduling ProblemFormulate as Linear Programming Problem

Let xi = number of employees beginning work on day i, i = 1,…,7

Define variables:

Write objective function:Min z = x1 + x2 + x3 + x4 + x5 + x6 + x7

Impose constraints:x1 + x4 + x5 + x6 + x7 > 17 (Monday)x1+ x2 + x5 + x6 + x7 > 13 (Tuesday)x1+ x2 + x3 + x6 + x7 > 15 (Wednesday)x1+ x2 + x3 + x4 + x7 > 19 (Thursday)x1+ x2 + x3 + x4 + x5 > 14 (Friday) x2 + x3 + x4 + x5 + x6 + x7 > 16 (Saturday) x3 + x4 + x5 + x6 + x7 > 11 (Sunday)

xi > 0 (i= 1, …, 7) (Non-negativity)

Work-Scheduling ProblemWork-Scheduling Problem(LINGO Model: with Integer Constraint)

MIN = @SUM( EMPLOYEES: NHIRED);

@FOR( NEEDS( I): @SUM( EMPLOYEES( J): CONSTRAINTS( I, J) * NHIRED( J)) >= NREQUIRED( I));

! We want NHIRED to be integer;@FOR(EMPLOYEES(I): @GIN(NHIRED(I)));END

Work-Scheduling ProblemWork-Scheduling Problem(LINGO Solution: with Integer Constraint)

Optimal solution found at step: 8 Objective value: 23.00000 Branch count: 1 Variable Value

NHIRED( MONDAY) 7.000000 1.000000 NHIRED( TUESDAY) 3.000000 1.000000 NHIRED( WEDNESDAY) 2.000000 1.000000 NHIRED( THURSDAY) 7.000000 1.000000 NHIRED( FRIDAY) 1.000000 1.000000 NHIRED( SATURDAY) 3.000000 1.000000 NHIRED( SUNDAY) 0.0000000 1.000000

Allocation of Scarce Resources, IIAllocation of Scarce Resources, II

Transportation Problems

Powerco, Electrical Power Company

Power Transmission Costs: ($/million kwh)

To SupplyFrom City 1 City 2 City 3 City 4 (million kwh)

Plant 1 $8 $6 $10 $9 35Plant 2 $9 $12 $13 $7 50Plant 3 $14 $9 $16 $5 40

Demand 45 20 30 30

Transportation ProblemTransportation Problem

Powerco Power Plant: FormulationDefine Variables:

Let Xij = number of (million kwh) produced at plant i and sent to city j.

Objective Function:Min z = 8*X11 + 6*X12 + 10*X13 + 9*X14 + 9*X21 + 12*X22 + 13*X23 + 7*X24 + 14*X31 + 9*X32 + 16*X33 + 5*X34

Supply Constraints:X11 + X12 + X13 + X14 < = 35 (Plant 1) X21 + X22 + X23 + X24 < = 50 (Plant 2) X31 + X32 + X33 + X34 < = 40 (Plant 3)

Transportation Transportation ProblemProblem

Powerco Power Plant: Formulation (Cont’d.)Demand Constraints:

X11 + X21 +X31 > = 45 (City 1) X12 + X22 +X32 > = 20 (City 2) X13 + X23 +X33 > = 30 (City 3) X14 + X24 +X34 > = 30 (City 4)

Nonnegativity Constraints:Xij > = 0 (i=1,..,3; j=1,..,4)

Balanced?Total Demand = 45 + 20 + 30 + 30 = 125Total Supply = 35 + 50 + 40 = 125

Yes, Balanced Transportation Problem! (If problem is unbalanced, add dummy supply or demand as required.)

Transportation ProblemTransportation ProblemPowerco Power Plant: LINGO Formulation

model:! A 3 Plant, 4 Customer Transportation Problem;SETS: PLANT / P1, P2, P3/ : CAPACITY; CUSTOMER / C1, C2, C3, C4/ : DEMAND; ROUTES( PLANT, CUSTOMER) : COST, QUANTITY;ENDSETS

! The objective; [OBJ] MIN = @SUM( ROUTES: COST * QUANTITY);

Transportation ProblemTransportation ProblemPowerco Power Plant: LINGO Formulation (Cont’d.)

! The demand constraints;@FOR( CUSTOMER( J): [DEM]@SUM( PLANT( I): QUANTITY( I, J))>= DEMAND( J));! The supply constraints;@FOR( PLANT( I): [SUP]@SUM( CUSTOMER(J): QUANTITY(I,J))<=CAPACITY( I));! Here are the parameters;DATA: CAPACITY = 35, 50, 40 ; DEMAND = 45, 20, 30, 30; COST = 8, 6, 10, 9,

9, 12, 13, 7, 14, 9, 16, 5;

ENDDATAend

Transportation ProblemTransportation ProblemPowerco Power Plant: LINGO Solution

Optimal solution found at step: 7 Objective value: 1020.000

QUANTITY( P1, C1) 0.0000000 QUANTITY( P1, C2) 10.00000 QUANTITY( P1, C3) 25.00000 QUANTITY( P1, C4) 0.0000000 QUANTITY( P2, C1) 45.00000 QUANTITY( P2, C2) 0.0000000 QUANTITY( P2, C3) 5.000000 QUANTITY( P2, C4) 0.0000000 QUANTITY( P3, C1) 0.0000000 QUANTITY( P3, C2) 10.00000 QUANTITY( P3, C3) 0.0000000 QUANTITY( P3, C4) 30.00000

All Quantitiesare Integers!

Transportation ProblemTransportation ProblemPowerco Power Plant: LINGO Solution (Cont’d.)

Row Slack or Surplus Dual Price OBJ 1020.000 1.000000 DEM( C1) 0.0000000 -9.000000 DEM( C2) 0.0000000 -9.000000 DEM( C3) 0.0000000 -13.00000 DEM( C4) 0.0000000 -5.000000 SUP( P1) 0.0000000 3.000000 SUP( P2) 0.0000000 0.0000000 SUP( P3) 0.0000000 0.0000000

All constraints are binding!(Typical of Balanced Transportation Problems; results in simple algorithms)

Transportation ProblemTransportation ProblemPowerco Power Plant: Sensitivity AnalysisRow Dual Price

OBJ 1.00000 DEM( C1) -9.00000 (-v1) DEM( C2) -9.00000 (-v2) DEM( C3) -13.00000 (-v3) DEM( C4) -5.00000 (-v4) SUP( P1) 3.00000 (-u1) SUP( P2) 0.00000 (-u2) SUP( P3) 0.00000 (-u3)

Changes in total cost due to changes in demand and supply: z = v1* C1 + v2* C2 + v3* C3 + v4* C4

+ u1* P1 + u2* P2 + u3* P3e.g. C2 = 1, P1 = 1 z = 9*1 + (-3)*1 = $6

Transportation ProblemTransportation ProblemPowerco Power Plant: Sensitivity AnalysisRow Dual Price

OBJ 1.00000 DEM( C1) -9.00000 (-v1) DEM( C2) -9.00000 (-v2) DEM( C3) -13.00000 (-v3) DEM( C4) -5.00000 (-v4) SUP( P1) 3.00000 (-u1) SUP( P2) 0.00000 (-u2) SUP( P3) 0.00000 (-u3)

Changes in total cost due to changes in demand and supply: z = v1* C1 + v2* C2 + v3* C3 + v4* C4

+ u1* P1 + u2* P2 + u3* P3e.g. C2 = 1, P1 = 1 z = 9*1 + (-3)*1 = $6

Powerco Power Plant: Sensitivity Analysis (Cont’d.)Powerco Power Plant: Sensitivity Analysis (Cont’d.)Variable Value Reduced CostQUANTITY( P1, C1) 0.00 2.00 (c11)QUANTITY( P1, C2) 10.00 0.00 (c12) QUANTITY( P1, C3) 25.00 0.00 (c13)QUANTITY( P1, C4) 0.00 7.00 (c14)QUANTITY( P2, C1) 45.00 0.00 (c21)QUANTITY( P2, C2) 0.00 3.00 (c22)QUANTITY( P2, C3) 5.00 0.00 (c23)QUANTITY( P2, C4) 0.00 2.00 (c24)QUANTITY( P3, C1) 0.00 5.00 (c31)QUANTITY( P3, C2) 10.00 0.00 (c32)QUANTITY( P3, C3) 0.00 3.00 (c33)QUANTITY( P3, C4) 30.00 0.00 (c34)z = c11* Q11 + c12* Q12 + c13* Q13 + c14* Q14 + c21* Q21 + c22* Q22 + c23* Q23 + c24* Q24 + c31* Q31 + c32* Q32 + c33* Q33 + c34* Q34

NonbasicBasicBasicNonbasicBasicNonbasicBasicNonbasicNonbasicBasicNonbasicBasic

Inventory Problems as Inventory Problems as Transportation Problems Transportation Problems

Sailco Corporation: Manufacturer of Sailboats Demand:

1st Quarter 40 (Sailboats)2nd Quarter 603rd Quarter 754th Quarter 25

Supply: (Initial inventory: 10)Production: (Storage @ $20/sailboat/quarter)

Quarter Regular Overtime1st 40@$400 150@$4402nd 40@$400 150@$4403rd 40@$400 150@$4404th 40@$400 150@$440

Inventory Problems as Transportation Problems Inventory Problems as Transportation Problems Sailco Corporation: Formulation

Supply Nodes:Node Description Capacity, Sailboats 1 Initial Inventory 10 2 1st quarter, regular 40 3 1st quarter, overtime 150 4 2nd quarter, regular 40 5 2nd quarter, overtime 150 6 3rd quarter, regular 40 7 3rd quarter, overtime 150 8 4th quarter, regular 40 9 4th quarter, overtime 150

Total 770

Inventory Problems as Transportation ProblemsInventory Problems as Transportation Problems

Sailco Corporation: Formulation (Cont’d.)

Demand Nodes:Node Description Demand, Sailboats 1 1st quarter 402 2nd quarter 603 3rd quarter 754 4th quarter 25 5 Dummy 570

Total 770

Inventory Problems as Transportation Problems Inventory Problems as Transportation Problems Sailco Corporation: Formulation (Costs)

Cost Coefficients: ($/Sailboat)Supply Demand 1 2 3 4 DummyI 1 0 20 40 60 0R 2 400 420 440 460 0OT 3 450 470 490 510 0R 4 M 400 420 440 0OT 5 M 450 470 490 0R 6 M M 400 420 0OT 7 M M 450 470 0R 8 M M M 400 0 OT 9 M M M 450 0

M is a very large, arbitrary, positive number.

Assignment ProblemsAssignment ProblemsPersonnel Assignment:

Time (hours)Job 1 Job 2 Job 3 Job 4

Person 1 14 5 8 7Person 2 2 12 6 5Person 3 7 8 3 9Person 4 2 4 6 10

Assignment Problem FormulationAssignment Problem Formulation

Define Variables: Let Xij = 1 if ith person is assigned to jth job

Xij = 0 if ith person is not assigned to jth job

Objective Function:Min z = 14*X11 + 5*X12 + … + 10*X44

Personnel Constraints:X11 + X12 + X13 + X14 = 1X21 + X22 + X23 + X24 = 1X31 + X32 + X33 + X34 = 1X41 + X42 + X43 + X44 = 1

Demand Constraints:X11 + X21 + X31 + X41 = 1X12 + X22 + X32 + X42 = 1X13 + X23 + X33 + X43 = 1X14 + X24 + X34 + X44 = 1

Binary Constraints:Xij = 0 or Xij = 1

Assignment Problem AlgorithmAssignment Problem Algorithm

14 5 8 72 12 6 57 8 3 92 4 6 10

Row Minimum 5 2 3 2

9 0 3 20 10 4 34 5 0 60 2 4 8

0 0 0 2Column Minimum

Subtract Row Minimum from Each Row:

Assignment Problem AlgorithmAssignment Problem Algorithm

9 0 3 00 10 4 14 5 0 40 2 4 6

Subtract Column Minimum from Each Column:

10 0 3 00 9 3 05 5 0 40 1 3 5

Subtract Minimum uncrossed value from uncrossed values and add to twice-crossed values:

Draw lines to cross out zeros and read solution from zeros

0 0

0 0

Solution:

Hungarian MethodHungarian MethodStep 1: Find the minimum element in each row of the m x m cost matrix. Construct a new matrix by subtracting from each cost the minimum cost in its row. For this new matrix, find the minimum cost in each column. Construct a new matrix (called the reduced cost matrix) by subtracting from each cost the minimum cost in its column.

Step 2: Draw the minimum number of lines (horizontal and/or vertical) that are needed to cover all the zeros in the reduced cost matrix. If m lines are required, an optimal solution is available among the covered zeros in the matrix. If fewer than m lines are needed, proceed to Step 3.

Step 3: Find the smallest nonzero element (call its value k) in the reduced cost matrix that is uncovered by the lines drawn in Step 2. Now subtract k from each uncovered element of the reduced cost matrix and add k to each element that is covered by two lines. Return to Step 2.