Assignment Method

AssignmentProblem(A Special Case of Transportation Problem)

Linear Programming

Assignment Problem• involves determining the most efficient

assignment of people to projects, salespeople to territories, contracts to bidders, jobs to machines, and so on.

• Objective: to minimize total costs or time of performing the tasks at hand.

• One important characteristic of assignment problems is that only one job (or worker) is assigned to one machine (or project).

Assignment Problem / Hungarian Method

• 2 characteristics• Number of rows = number of columns• In the optimal solution, there will be one and only one

assignment in a given row or column of the given assignment table

Assignment ProblemMachine

Job X Y ZA $25 $31 $35

B 15 20 24

C 22 19 17• The Kellum Machine Shop does custom metalworking for a number of local

plants. Kellum currently has 3 jobs to be done. Kellum also has 3 machines on which to do the work. Any one of the jobs can be processed completely on any one of the machines. Furthermore, the cost of processing any job on any machine is known. The assignment of jobs to machines must be a one-to-one basis; that is, each job must be assigned exclusively to one and only one machine. The objective is to assign the jobs to the machines so as to minimize total cost.

Assignment Problem (Naïve Solution)Machine

Job X Y ZA $25 $31 $35

B 15 20 24

C 22 19 17

62 Machine

Job X Y Z

A $25 $31 $35

B 15 20 24

C 22 19 17

68 Machine

Job X Y Z

A $25 $31 $35

B 15 20 24

C 22 19 17

69 Machine

Job X Y Z

A $25 $31 $35

B 15 20 24

C 22 19 17

77 Machine

Job X Y Z

A $25 $31 $35

B 15 20 24

C 22 19 17

77 Machine

Job X Y Z

A $25 $31 $35

B 15 20 24

C 22 19 17

63 Machine

Job X Y Z

A $25 $31 $35

B 15 20 24

C 22 19 17

Possible Solutions

3 Steps Using the Assignment Method

Step 1. Determine the opportunity-cost table.Machine

Job X Y ZA $25 $31 $35

B 15 20 24

C 22 19 17

ComputationsX Y Z

25 – 15 = 10 31 – 19 = 12 35 – 17 = 1815 – 15 = 0 20 – 19 = 1 24 – 17 = 722 – 15 = 7 19 – 19 = 0 17 – 17 = 0

Step 1. Determine the opportunity-cost table.Computations Minimums

X Y Z10 12 18 100 1 7 07 0 0 0

MachineJob X Y ZA 0 2 8

B 0 1 7

C 7 0 0

Computations

X Y Z

10 – 10 = 0 12 – 10 = 2 18 – 10 = 8

0 – 0 = 0 1 – 0 = 1 7 – 0 = 7

7 – 0 = 7 0 – 0 = 0 0 – 0 = 0

Step 2: Determine whether an optimal assignment can be made

• Assign jobs to the machines so as to minimize total costs.

• Straight line method • Optimal assignment: number of lines = number of rows / columns


B 0 1 7

C 7 0 0

Note: Highlight all zeros.Cover all the zeros using as minimum lines as possible.

Step 3: Revise the total opportunity-cost table.

a.) All numbers not covered by a straight line LESS the smallest number not covered by a straight line

b.) ADD this same lowest number to the numbers lying at the intersection of any two lines

MachineJob X Y ZA 0 2 8B 0 1 7C 7 0 0


• As the minimum number of lines necessary to cover all zeros is three, and as the number is equal to the number of rows, an optimal assignment can be made.



B 0 0 6

C 8 0 0



MachineJob X Y Z

A $25 $31 $35

B 15 20 24

C 22 19 17

Assignment CostA to X $ 25

B to Y 20

C to Z 17

Total $ 62

Maximization Problem

AutomobileBuyer Ford Dodge Buick Volkswagen Toyota

Amy $ 3, 000 $ 2, 500 $ 3, 300 $ 2, 600 $ 3, 100

Bert 3, 500 3, 000 2, 800 2, 800 3, 300

Carl 2, 800 2, 900 3, 900 2, 300 3, 600

Dolly 3, 300 3, 100 3, 400 2, 900 3, 500

Edgar 2, 800 3, 500 3, 600 2, 900 3, 000

Heidi wishes to determine which bid to accept from each of the five bidders so that each of them can purchase one vehicle while the total of the five accepted bids is a maximum.

Identical to the minimization problem•We need to convert each of the bids into a regret value.

Convert each of the bids into a regret value.


Amy $ 3, 000 $ 2, 500 $ 3, 300 $ 2, 600 $ 3, 100

Bert 3, 500 3, 000 2, 800 2, 800 3, 300

Carl 2, 800 2, 900 3, 900 2, 300 3, 600

Dolly 3, 300 3, 100 3, 400 2, 900 3, 500

Edgar 2, 800 3, 500 3, 600 2, 900 3, 000


Amy 500 1000 600 300 500

Bert 0 500 1100 100 300

Carl 700 600 0 600 0

Dolly 200 400 500 0 100

Edgar 700 0 300 0 600

Note: Highlight all zeros.Cover all the zeros using as minimum lines as possible.

Convert each of the bids into a regret value. (cont’d.)


Amy 500 1000 600 300 500

Bert 0 500 1100 100 300

Carl 700 600 0 600 0

Dolly 200 400 500 0 100

Edgar 700 0 300 0 600


Amy 500 1000 500 300 400

Bert 0 500 1000 100 200

Carl 800 700 0 700 0

Dolly 200 400 400 0 0

Edgar 700 0 200 0 500

a.) All numbers not covered by a straight line LESS the smallest number not covered by a straight line

ADD this same lowest number to the numbers lying at the intersection of any two lines

Revise the Regret Value-TableAutomobile

Buyer Ford Dodge Buick Volkswagen Toyota

Amy 500 1000 500 300 400

Bert 0 500 1000 100 200

Carl 800 700 0 700 0

Dolly 200 400 400 0 0

Edgar 700 0 200 0 500


Amy 200 700 200 0 100

Bert 0 500 1000 100 200

Carl 800 700 0 700 0

Dolly 200 400 400 0 0

Edgar 700 0 200 0 500

Revise the Regret Value-Table

Automobile

Buyer Ford Dodge Buick Volkswagen

Toyota

Amy $ 3, 000 $ 2, 500 $ 3, 300 $ 2, 600 $ 3, 100

Bert 3, 500 3, 000 2, 800 2, 800 3, 300

Carl 2, 800 2, 900 3, 900 2, 300 3, 600

Dolly 3, 300 3, 100 3, 400 2, 900 3, 500

Edgar 2, 800 3, 500 3, 600 2, 900 3, 000

Original Data Buyer Bid accepted

Bid Price

Amy Volkswagen $ 2, 600

Bert Ford 3, 500

Carl Buick 3, 900

Dolly Toyota 3, 500

Edgar Dodge 3, 500

Total $ 17, 000


Amy 200 700 200 0 100

Bert 0 500 1000 100 200

Carl 800 700 0 700 0

Dolly 200 400 400 0 0

Edgar 700 0 200 0 500

Thank you!

Assignment Method

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