Chapter 6 - Scheduling Algorithm

8/10/2019 Chapter 6 - Scheduling Algorithm

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Chapter VI

Scheduling ModelsPrepared by:

Engr. Romano A. GabrilloMEngg-MEM


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Scheduling Problems

Scheduling problem is a special type of integer

programming problem where assignees are

being assigned to perform tasks.

It involves the following problems:

Production Problems

Transshipment Problems

2


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3RAG 2007

Production Problems

Production problems involve a single productwhich is to be manufactured over a number ofsuccessive time periods to meet prespecified

demands.

The objective is to determine a production

schedule which will meet all future demands atminimum total cost (which is total productioncost plus total storage cost, as total shipping costis presumed fixed).


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4RAG 2007

Solution to Production Problems

Production problems may be converted intotransportation problems by considering the timeperiods during which production can take place assources, and the time periods in which units will be

shipped as destinations.

The production capacities are taken to be the supplies.Therefore, xijdenotes the number of units to be

produced. cijis the unit production cost during timeperiod i plus the cost of storing a unit of product fromtime period i until time period j. Since units cannot beshipped prior to being produced, cijis madeprohibitively largefor i > j to force the corresponding

xijto be zero.


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5RAG 2007

Example No. 1

An industrial firm must plan for each of the four

seasons over the next year. The companys production

capacities and the expected demands (all in units) are as

follows:

Spring Summer Fall Winter

Demand 250 100 400 500

Regular

Capacity

200 300 350

Overtime

Capacity

100 50 100 150


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6RAG 2007

Regular production costs for the firm are P7.00 perunit; the unit cost of overtime varies seasonally, being

P8.00 in spring and fall, P9.00 in summer, and P10.00in winter.

The company has 200 units of inventory on January 1,

but as it plans to discontinue the product at the end ofthe year, it wants no inventory after the winter season.Units produced on regular shifts are not available forshipment during the season of production; generally,they are sold during the following season. Those that

are not added to inventory are carried forward at a costof P0.70 per unit per season. In contrast, unitsproduced on overtime shifts must be shipped in the

same season as produced. Determine a productionschedule that meets all demands at minimum total cost.


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7RAG 2007

Input the Supply and Demand

Supply ui

Demand

vj

150

Overtime

(Spring) 100

Overtime

(Summer)

50

Overtime

(Fall) 100

Initial

Inventory 200

250 100 400 500 200

Overtime

(Winter)

Sources

Regular

(Spring) 200

Regular

(Summer) 300

Regular

(Fall) 350

DestinationSpring Summer Fall Winter Dummy


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9RAG 2007

Input the Cost for Overtime Capacity

Supply ui

10000 7.00 7.70 8.40

10000 10000 7.00 7.70

10000 10000 10000 7.00

8.00 10000 10000 10000

10000 9.00 10000 10000

10000 10000 8.00 10000

10000 10000 10000 10.00

Demand

vj

150

Overtime

(Spring) 100

Overtime

(Summer) 50

Overtime

(Fall) 100

Initial

Inventory 200

250 100 400 500 200

Overtime

(Winter)

Sources

Regular

(Spring) 200

Regular

(Summer) 300

Regular

(Fall) 350

Destination

Spring Summer Fall Winter Dummy


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10RAG 2007

Input the Cost for Initial Inventory

Supply ui

10000 7.00 7.70 8.40

10000 10000 7.00 7.70

10000 10000 10000 7.00

0.00 0.70 1.40 2.10

8.00 10000 10000 10000

10000 9.00 10000 10000

10000 10000 8.00 10000

10000 10000 10000 10.00

Demand

vj

150

Overtime

(Spring) 100

Overtime

(Summer) 50

Overtime

(Fall) 100

Initial

Inventory 200

250 100 400 500 200

Overtime

(Winter)

Sources

Regular

(Spring) 200

Regular

(Summer) 300

Regular

(Fall) 350

Destination



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11RAG 2007

Input the Cost for the Dummy

Supply

Since all units in initial inventory already have been produced, a positiveallocation from initial inventory to the dummy must be avoided. This is doneby assigning a prohibitively large number 10000 as the associated unit cost.

Supply ui

10000 7.00 7.70 8.40 0

10000 10000 7.00 7.70 0

10000 10000 10000 7.00 0

0.00 0.70 1.40 2.10 10000

8.00 10000 10000 10000 0

10000 9.00 10000 10000 0

10000 10000 8.00 10000 0

10000 10000 10000 10.00 0

Demand

vj

150

Overtime

(Spring) 100

Overtime

(Summer) 50

Overtime

(Fall) 100

InitialInventory

200

250 100 400 500 200

Overtime

(Winter)

Sources

Regular

(Spring) 200

Regular

(Summer) 300

Regular

(Fall) 350

Destination



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Excel Solver Installation

The Excel Solver is part of the Solver Add-in of

Microsoft Excel which you can find in the Data Menu.

It is first being installed in the computer by clicking theOffice Toolbar (for Office 2007) and click Excel

Options.

On the left side of the Options Dialog-Box, select Add-

Ins. On the Add-Ins name, search for Solver Add-in

and click Go. The Excel Solver will now be installed

on your computer. 12


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13RAG 2007

Solve Using Excel Solver

Input the Transportation Tableau on the

Spreadsheet.

Spring Summer Fall Winter Dummy Supply

R-Spring 10000 7 7.7 8.4 0 200

R-Summer 10000 10000 7 7.7 0 300

R-Fall 10000 10000 10000 7 0 350

Initial Inventory 0 0.7 1.4 2.1 10000 200

O-Spring 8 10000 10000 10000 0 100

O-Summer 10000 9 10000 10000 0 50

O-Fall 10000 10000 8 10000 0 100

O-Winter 10000 10000 10000 10 0 150

Demand 250 100 400 500 200


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14RAG 2007

Input Another Cell Range Just Below It

The Row and Column Total must have the

formula for the sum of the respective row and

column.

Spring Summer Fall Winter Dummy TotalR-Spring 0 0 0 0 0 0

R-Summer 0 0 0 0 0 0

R-Fall 0 0 0 0 0 0

Initial Inventory 0 0 0 0 0 0

O-Spring 0 0 0 0 0 0

O-Summer 0 0 0 0 0 0

O-Fall 0 0 0 0 0 0

O-Winter 0 0 0 0 0 0

Total 0 0 0 0 0


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15RAG 2007

Create the Cell for the Costs (z)

Create a cell just below the new range of cells

with the formula as the sumproduct of range of

cells above with the range of cells below

excluding the row and column for the row total,column total, demand and supply.

Example:

=SUMPRODUCT(B2:F9,B14:F21)


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16RAG 2007

The Excel Solver

On the Data Menu, Choose Solver and the

Solver Parameters Dialog Box will appear:


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17RAG 2007

Input Parameters on the Solver

Dialog Box

Set the Target Cell as the Cell corresponding for thecost (z).

Set the Equal to Option to Minimum, as with theminimization program of the transportation tableau.

Set the Range of cells that you have created below thetransportation tableau as the range of cells in the ByChanging Cells Option.

Add the Constraints on the Subject by the Constraints

Option by setting the Row Total equal to total Supplyand the Column Total equal to the Demand Supply.Click Ok if done.


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18RAG 2007

Getting the Optimal Solution

Click the Options Button and the SolverOptions Dialog Box will appear.

On the Solver Options Dialog Box, select

Assume Linear and Assume NonnegativeCheckboxes. Click Ok and you will be movedback the Solver Dialog Box.

Click Solve to get the optimal solution.The values in the Range of Cells below are the

optimal values of the variables and costs z is theoptimal solution to the problem.


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19RAG 2007

Final Answer

Spring Summer Fall Winter Dummy Total

R-Spring 0 100 0 100 0 200

R-Summer 0 0 300 0 0 300

R-Fall 0 0 0 350 0 350

Initial Inventory 200 0 0 0 0 200O-Spring 50 0 0 0 50 100

O-Summer 0 0 0 0 50 50

O-Fall 0 0 100 0 0 100

O-Winter 0 0 0 50 100 150Total 250 100 400 500 200

costs 7790


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20RAG 2007

Transshipment Problems

A transshipment problem, like a transportationproblem, involves sources, having supplies, anddestinations, having demands. In addition, it alsoinvolves junctions, through which goods can beshipped.

Unit shipping costs are given between all directly

accessible locations, and the objective is to develop atransportation schedule that will meet all demands atminimum total cost.


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21RAG 2007

Solution to Transshipment Problems Transshipmet problems may be converted into

transportation problems by making every junctionbotha source and a destination. Each junction is assigned asupply equal to its original supply (or zero, if thejunction did not originally coincide with a source) plus

the total number of units in the system, same thing withthe demand. These assignments allow for thepossibility that all units may pass through a givenjunction.

The cost of transporting 1 unit from a junction(considered as a source) to itself (considered as adestination) is zero. Those units that do not passthrough a junction under the optimal schedule will

appear as allocations from the junction to itself.


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22RAG 2007

Example No. 2

A corporation must transport 70 units of a product

from location 1 to locations 2 and 3, in the amounts of

45 and 25 units, respectively. Air freight charges cij(in

dollars per unit) between locations served by the aircarrier are given below where dotted lines signify that

service is not available. Determine a shipping schedule

that allocates the required number of goods to each

destination at a minimum total freight cost. Noshipment need be flown directly; shipments through

intermediate points are allowed.


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23RAG 2007

i j 1 2 3 4

1 38 56 34

2 38

27

3 56 27 19

4 34 19

Table 2

1 3

4

2

$27

$27$38

$38

$56

$56

$19

$19

$34$34Figure 2

-45

+70 -25


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24RAG 2007

Create the Transportation Tableau

Location 4 is a pure junction.

Locations 2 and 3 serve as both destinations andjunctions.

Location 1 serves both as a source and a junction. Therefore, we have 1, 2, 3, and 4 as the sources and 1,

2, 3, and 4 as destinations but we delete 1 as adestination since it could never be optimal to ship

goods from location 1 and have them return as somelater time, only to be shipped out again, the problemcan be simplified by not allowing shipments to location1, thereby restricting it to being solely a source.


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25RAG 2007

Solution Using Excel Solver

2 3 4 Supply1 38 56 34 70

2 0 27 10000 70

3 27 0 19 70

4 10000 19 0 70

Demand 115 95 70

2 3 4 Total

1 45 0 25 70

2 70 0 0 703 0 70 0 70

4 0 25 45 70

Total 115 95 70

Costs 3035


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26RAG 2007

Assignment Problems

Assignment problems involve schedulingworkers to jobs on one-to-one basis (moregenerally, they involve permutations of a set of

objects).

The objective is to schedule every worker to a

job so that all jobs are completed in theminimum total time (or, to find the permutationthat has the greatest total value).


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27RAG 2007

Solution to Assignment Problems

Assignment problems can be converted into

transportation problems by considering the

workers as sources and the jobs as destinations,

where all supplies and demands are equal to 1.

A solution procedure more efficient than the

general transportation algorithm is the

Hungarian Method (which will be discussed

later).


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28RAG 2007

Example No. 3

For the Figure below, determine a shipping

schedule that meets all demands at a minimum

total cost.

1

2 4 6

53

+70

+95

+15

-30

-30

-45

$3

$4

$2

$3

$7

$2

$4

$8


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29RAG 2007

Solution

Locations 1 and 2 are sources.

Locations 5 and 6 are destinations

Location is both a source and a junction. Location 4 serves both as a destination and a

junction.

Because the total supply is 180 units but totaldemand is only 105, location 7 is created as a

dummy destination.


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30RAG 2007

Set up the Transportation Tableau

3 4 5 6 dummy 7 supply

1 3 10000 8 10000 0 95

2 2 7 10000 10000 0 70

3 0 3 4 4 0 195

4 10000 0 10000 2 0 180

demand 180 210 30 45 75

3 4 5 6 dummy 7 total

1 0

2 0

3 04 0

total 0 0 0 0 0

costs 0


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31RAG 2007

Solve Using Excel Solver

3 4 5 6 dummy 7 supply

1 3 10000 8 10000 0 95

2 2 7 10000 10000 0 70

3 0 3 4 4 0 195

4 10000 0 10000 2 0 180

demand 180 210 30 45 75

3 4 5 6 dummy 7 total

1 20 0 0 0 75 95

2 70 0 0 0 0 70

3 90 30 30 45 0 1954 0 180 0 0 0 180

total 180 210 30 45 75

costs 590


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32RAG 2007

Hungarian Method

Step 1

In each row of the Hungarian Tableau, locate the smallestelement and subtract it from every element in that row. Repeatthis of each column (the column minimum is determined after

the row subtractions). The revised cost matrix will have at leastone zero in every row and column.

Step 2

Determine whether there exists feasible assignment involvingonly zero costs in the revised cost matrix. In other words, find ifthe revised matrix has n zero entries no two of which are in thesame row or column. If such an assignment exists, it is optimal.If no such exists, go to Step 3.


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33RAG 2007

Hungarian Method

Step 3

Cover all zeroes in the revised cost matrix with as fewhorizontal and vertical lines as possible. Eachhorizontal line must pass through an entire row, each

vertical line must pass through an entire column; thetotal number of lines in this minimal covering will besmaller than n. Locate the smallest number in the costmatrix not covered by a line. Subtract this numberfrom every element not covered by a line and add it toevery element covered by two lines.

Step 4

Return to Step 2


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34RAG 2007

Hungarian Tableau

The Hungarian method, uses only the cost matrix,

as input:

Jobs

Workers

cnncncncnn

ncccc

ncccc

ncccc

n

...321

..................

3...3332313

2...2322212

1...1312111

...321


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35RAG 2007

Example No. 4

A 400-meter medley relay involves four differentswims, who successively swim 100 meters of the

backstroke, breaststroke, butterfly, and freestyle. A

coach has six very fast swimmers whose expected time

(in seconds) in the individual events are given below:

Event 1

(Backstroke0

Event 2

(breaststroke)

Event 3

(butterfly)

Event 4

(freestyle)

Swimmer 1 65 73 63 57

Swimmer 2 67 70 65 58

Swimmer 3 68 72 69 55

Swimmer 4 67 75 70 59

Swimmer 5 71 69 75 57

Swimmer 6 69 71 66 59


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36RAG 2007

Solution Using the Hungarian

Method

How should the coach assign swimmers to the

relay so as to minimize the sum of their times?

Solution: Set the Hungarian Tableau

1 2 3 4 5 6

Swimmer 1 65 73 63 57 0 0Swimmer 2 67 70 65 58 0 0

Swimmer 3 68 72 69 55 0 0

Swimmer 4 67 75 70 59 0 0

Swimmer 5 71 69 75 57 0 0

Swimmer 6 69 71 66 59 0 0

Events

Tableau 4A


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37RAG 2007

Step 1 of the Hungarian Method

Subtracting 0 from every row of Tableau 4a and

then subtracting 65, 69, 55, 0, and 0 from the

columns 1 through 6, respectively generates

tableau 4b below:

1 2 3 4 5 6

Swimmer 1 0 4 0 2 0 0

Swimmer 2 2 1 2 3 0 0Swimmer 3 3 3 6 0 0 0

Swimmer 4 2 6 7 4 0 0

Swimmer 5 6 0 12 2 0 0

Swimmer 6 4 2 3 4 0 0

Events

Tableau 4B


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38RAG 2007

Step 2 of the Hungarian Method

The smallest uncovered element is 1, subtract 1

from every uncovered element and add 1 to

every element covered by 2 lines.

1 2 3 4 5 6

Swimmer 1 0 4 0 2 1 1

Swimmer 2 1 0 1 2 0 0

Swimmer 3 3 3 6 0 1 1

Swimmer 4 1 5 6 3 0 0

Swimmer 5 6 0 12 2 1 1

Swimmer 6 3 1 2 3 0 0

Events

Tableau 4C


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39RAG 2007

Step 3 and 4 of the Hungarian

Method

Tableau 4c also does not contain a feasible zero-

cost assignment. Repeating step 3 of the

Hungarian method, we determine 1 is again the

smallest uncovered element as in Tableau 4d:

1 2 3 4 5 6

Swimmer 1 0 5 0 2 2 2

Swimmer 2 0 0 0 1 0 0

Swimmer 3 3 4 6 0 2 2

Swimmer 4 0 5 5 2 0 0

Swimmer 5 5 0 11 1 1 1

Swimmer 6 2 1 1 2 0 0

Events

Tableau 4D


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Optimal Value

Tableau 4d which does contain a feasible zero-costassignment, as indicated thus, the optimal allocation isswimmer 1 to event 1 (backstroke), swimmer 2 to event3 (butterfly), swimmer 3 to event 4 (freestyle), and

swimmer 5 to event 2 (breaststroke); swimmers 4 and 6are not entered in the medley.

The minimum total time (in seconds) is calculated from

Tableau 4a.

z = c11 + c23 + c34 + c52 = 65+65+55+69= 254s


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Schedule of Final Exam

December 6, 2013

Coverage: Chapters 5-6

Friday, 3-5pm

Bring your own laptop with Excel Solver!

41


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End of Module

Thank you very much for listening!

Chapter 6 - Scheduling Algorithm

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