Production Mix Problem

1
Production Mix Problem

• Graphical Solution

lrg
40
27-inch sets
30
20-inch sets
(10,20)
(Optimal Product Mix!)
Profit 8010 12020 3200/mo
20
Electronics
Feasible Region
10
Cabinetry
Profit
0
med
0 10 20 30
40 50 60
2
Product Mix Problem

• Primal Problem Formulation

Production Capacity Constraints 6med 15lrg
lt 360 4med 5lrg lt 140 med gt 0
lrg gt 0
(hours/set)sets hours
Market Constraints med lt 15 lrg lt 40
(sets)
Objective Function Maximize Profit 80med
120lrg
(/set)sets
3
Primal 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
4
Work-Scheduling Problem

• Formulate as Linear Programming Problem

Define variables
Let xi number of employees beginning work on
day i, i 1,,7
Write objective function
Min z x1 x2 x3 x4 x5 x6 x7
Impose constraints
x1 x4 x5 x6 x7 gt 17
(Monday) x1 x2 x5 x6 x7 gt
13 (Tuesday) x1 x2 x3 x6
x7 gt 15 (Wednesday) x1 x2 x3 x4
x7 gt 19 (Thursday) x1 x2 x3 x4 x5
gt 14 (Friday) x2 x3 x4
x5 x6 x7 gt 16 (Saturday) x3
x4 x5 x6 x7 gt 11 (Sunday) xi gt 0 (i 1,
, 7) (Non-negativity)
5
Work-Scheduling Problem

• (LINGO Model with Integer Constraint)

MIN _at_SUM( EMPLOYEES NHIRED) _at_FOR( NEEDS(
I) _at_SUM( EMPLOYEES( J) CONSTRAINTS( I, J)
NHIRED( J)) gt NREQUIRED( I)) ! We want
NHIRED to be integer _at_FOR(EMPLOYEES(I)
_at_GIN(NHIRED(I))) END
6
Work-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
7
Allocation of Scarce Resources, II

• Transportation Problems

Powerco, Electrical Power Company
Power Transmission Costs (/million kwh)
To Supply From City 1 City 2 City
3 City 4 (million kwh) Plant 1 8 6
10 9
35 Plant 2 9 12 13 7
50 Plant 3 14 9 16
5 40 Demand 45 20
30 30
8
Transportation Problem

• Powerco Power Plant Formulation

Define Variables Let Xij number of (million
kwh) produced at plant i and sent to city j.
Objective Function Min z 8X11 6X12
10X13 9X14 9X21 12X22
13X23 7X24 14X31 9X32
16X33 5X34
Supply Constraints X11 X12 X13 X14 lt
35 (Plant 1) X21 X22 X23 X24 lt 50
(Plant 2) X31 X32 X33 X34 lt 40 (Plant
3)
9
Transportation Problem

• Powerco Power Plant Formulation (Contd.)

Demand Constraints X11 X21 X31 gt 45 (City
1) X12 X22 X32 gt 20 (City 2) X13 X23
X33 gt 30 (City 3) X14 X24 X34 gt 30
(City 4)

Nonnegativity Constraints Xij gt 0 (i1,..,3
j1,..,4)
Balanced? Total Demand 45 20 30 30
125 Total Supply 35 50 40 125 Yes,
Balanced Transportation Problem! (If problem is
unbalanced, add dummy supply or demand as
required.)
10
Transportation Problem

• Powerco 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
_at_SUM( ROUTES COST QUANTITY)

11
Transportation Problem

• Powerco Power Plant LINGO Formulation (Contd.)

! The demand constraints _at_FOR( CUSTOMER( J)
DEM _at_SUM( PLANT( I) QUANTITY( I, J))gt DEMAND(
J)) ! The supply constraints _at_FOR( PLANT( I)
SUP _at_SUM( CUSTOMER(J) QUANTITY(I,J))ltCAPACITY(
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 ENDDATA end

12
Transportation Problem

• Powerco 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 Quantities are Integers!
13
Transportation Problem

• Powerco Power Plant LINGO Solution (Contd.)

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)
14
Transportation Problem

• Powerco Power Plant Sensitivity Analysis

Row 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 ? P3 e.g. ?C2
1, ?P1 1 ? ?z 91 (-3)1 6
15
Transportation Problem

• Powerco Power Plant Sensitivity Analysis

Row 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 ? P3 e.g. ?C2
1, ?P1 1 ? ?z 91 (-3)1 6
16

• Powerco Power Plant Sensitivity Analysis
  (Contd.)

Variable Value Reduced Cost
QUANTITY( 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
Nonbasic Basic Basic Nonbasic Basic Nonbasic Basic
Nonbasic Nonbasic Basic Nonbasic Basic



17
Inventory Problems as Transportation Problems

• Sailco Corporation Manufacturer of Sailboats

Demand 1st Quarter 40 (Sailboats) 2nd Quarter
60 3rd Quarter 75 4th Quarter 25
Supply (Initial inventory 10) Production
(Storage _at_ 20/sailboat/quarter) Quarter
Regular Overtime 1st 40_at_
400 150_at_440 2nd 40_at_400 150_at_440 3rd 40
_at_400 150_at_440 4th 40_at_400 150_at_440
18
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
19
Inventory Problems as Transportation Problems

• Sailco Corporation Formulation (Contd.)

Demand Nodes Node Description Demand,
Sailboats 1 1st quarter 40 2
2nd quarter 60 3 3rd quarter
75 4 4th quarter 25 5
Dummy 570 Total
770
20
Inventory Problems as Transportation Problems

• Sailco Corporation Formulation (Costs)

Cost Coefficients (/Sailboat) Supply
Demand 1 2 3 4 Dummy I
1 0 20 40 60
0 R 2 400 420
440 460 0 OT 3
450 470 490
510 0 R 4 M
400 420 440
0 OT 5 M 450
470 490 0 R 6
M M 400
420 0 OT 7 M
M 450 470
0 R 8 M M
M 400 0 OT 9
M M M
450 0
M is a very large, arbitrary, positive number.
21
Assignment Problems
Personnel Assignment Time (hours) Job
1 Job 2 Job 3 Job 4 Person 1 14 5
8 7 Person 2 2 12 6
5 Person 3 7 8 3
9 Person 4 2 4 6 10
22
Assignment 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 14X11 5X12
10X44
Personnel Constraints X11 X12 X13 X14
1 X21 X22 X23 X24 1 X31 X32 X33 X34
1 X41 X42 X43 X44 1
Demand Constraints X11 X21 X31 X41 1 X12
X22 X32 X42 1 X13 X23 X33 X43
1 X14 X24 X34 X44 1
Binary Constraints Xij 0 or Xij 1
23
Assignment Problem Algorithm
Row Minimum 5 2 3
2
14 5 8 7 2 12 6
5 7 8 3 9 2
4 6 10

Subtract Row Minimum from Each Row
9 0 3 2 0 10 4
3 4 5 0 6 0 2
4 8 0 0 0 2
Column Minimum
24
Assignment Problem Algorithm
Subtract Column Minimum from Each Column
9 0 3 0 0 10 4
1 4 5 0 4 0 2
4 6

Subtract Minimum uncrossed value from uncrossed
values and add to twice-crossed values
Solution
10 0 3 0 0 9 3
0 5 5 0 4 0 1
3 5
0
0 0 0
Draw lines to cross out zeros and read solution
from zeros
25
Hungarian Method
Step 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.