taylmod bff

318

PROBLEM SUMMARY

1. Balanced transportation


3. Short answer, discussion

4. Unbalanced transportation





9. Unbalanced transportation, multiple optimal

10. Sensitivity analysis (B–9)


12. Unbalanced transportation, degenerate







19. Shortage costs (B–17)












31. Unbalanced transportation, production scheduling(B–30)



34. Shortage costs

35. Multiperiod scheduling

36. Unbalanced assignment, LP formulation

37. Assignment

38. Assignment

39. Assignment

40. Unbalanced assignment, multiple optimal

41. Assignment, multiple optimal

42. Assignment


44. Balanced assignment

45. Balanced assignment (B–9)

46. Unbalanced assignment



49. Unbalanced assignment, prohibited assignment


51. Assignment

52. Unbalanced assignment (maximization)

PROBLEM SOLUTIONS

1. Using the VAM initial solutions:

Module B: Transportation and Assignment Solution Methods

TaylMod-Bff.qxd 4/21/06 8:50 PM Page 318

319


320

TaylMod-Bff.qxd 4/21/06 8:50 PM 20

321


322

$1290

b.) Solutions achieved using minimum cost cell method is optimal

40

80

30


323


324


325

�

��


326


327


328


329


330


331


332


333


From ToA B C D Supply

From

+7 22 250 17 +10 30 170 181 250 7 170 18 420

520 15 +18 35 90 20 +7 252 520 15 90 20 610

+17 28 +8 21 130 16 210 143 130 16 210 14 340

+20 40 +33 65 180 25 +27 50Dummy 180 25 180

Demand 520 250 400 380 1550

334

19.

20.

Optimal, TC = $26,430

Transportation cost = $21,930Shortage cost = $ 4,500

$26,430

From ToFrom NC-E NC-SW NC-P NC-W VA-SW VA-C VA-T Dummy Supply

15.2 14.3 13.9 13.5 14.7 16.5 18.7 0GA-1 2 10 12

12.8 11.3 10.6 12.0 12.7 13.2 15.6 0GA-2 6 4 10

12.4 10.8 9.4 11.3 13.1 12.8 14.5 0SC-1 1 6 7 14

18.2 19.4 18.2 17.9 20.5 20.7 22.7 0FL-1 9 6 15

19.3 20.2 19.5 20.2 21.2 21.3 23.5 0FL-2 5 7 12

Demand 9 7 6 8 10 9 7 7 63

Total Cost = $823,000


335

21.


FromTo

A B C D

North

South

East

West

12 23 35 17

26 15 21 27

18 20 22 31

29 24 35 10

15 10 23 16

0 0 0 0

Central

Dummy

Demand

Supply

250 250

340340

200 200

110 310

210 210

200

40 60 190 290

400 400 400 400 1,600

–3

336

21. VAM initial solution:

22.

23.


337

22. VAM initial solution:

Allocate 40 students to cell (South, C):

Optimal (with multiple optimal solutions)Total travel time = 20,700 minutes

FromTo

A B C D

North

South

East

West

12 23 35 17

26 15 21 27

18 20 22 31

29 24 35 10

15 10 23 16

0 0 0 0

Central

Dummy

Demand

Supply

250 250

340300 40

200 200

150 310

210 210

160

100 190 290

400 400 400 400 1,600

FromTo

A B C D

North

South

East

West

12 23 35 17

26 15 21 27

18 20 22 31

29 24 35 10

15 10 23 16Central

Demand

Supply

250 250

340300

310

210 210

310

100 4010 140 290

350 350 350 350 1,400

–7

24.


Demand 350 350 350 350 1,400

FromTo

A B C D

North

South

East

West

12 23 35 17

26 15 21 27

18 20 22 31

29 24 35 10

15 10 23 16Central

Supply

250 250

340300 40

310

210 210

310

100 50 140 290

–3

338

Allocate 40 students to cell (South, C):

Allocate 100 students to cell (East, A):

Optimal. Total travel time = 21,200 minutes.

The overall travel time increased by 500 minutes,which divided by all 1,400 students is only an increaseof .357 minutes per student. This does not seem to bea significantly large increase.

Demand 350 350 350 350 1,400

FromTo

A B C D

North

South

East

West

12 23 35 17

26 15 21 27

18 20 22 31

29 24 35 10

15 10 23 16Central

Supply

250 250

340200 140

100 310

210 210

210

150 140 290


Demand

FromTo

1 2 3 4

A

B

C

D

9 8 11 12 7 8 0

10 10 8 6

8 6 6 5

4 6 9 5

12 10 8 9

9

7

8

6

7

4

10

7

0

0

0

0E

5 6

8 18 26

13 27 40

20 20

40

25 15 4 1 45

25 15 30 18 27 35 21 171

SupplyDummy

Demand

FromTo

1 2 3 4

A

B

C

D

9 8 11 12 7 8 0

10 10 8 6

8 6 6 5

4 6 9 5

12 10 8 9

9

7

8

6

7

4

10

7

0

0

0

0E

5 6

8 18 26

13 27 40

20 20

5 35 40

25 2 17 1 45

25 15 30 18 27 35 21 171

SupplyDummy

0

0

0

0

5 35

339

23. The initial solution is determined using VAM, as follows.

Optimal, Total Profit = $1,528 with multiple optimalsolutions at (B,2), (E,4) and (B, Dummy)

Alternate, allocate 13 cases to (B,2)

25.


340

Alternate, allocate 17 cases to (E,4)

Demand

FromTo

1 2 3 4

A

B

C

D

9 8 11 12 7 8 0

10 10 8 6

8 6 6 5

4 6 9 5

12 10 8 9

9

7

8

6

7

4

10

7

0

0

0

0E

5 6

25 1 26

13 27 40

20 20

5 35 40

25 2 17 1 45

25 15 30 18 27 35 21 171

SupplyDummy

24. If Easy Time purchased all the baby food demanded at each store from the distributortotal profit would be $1,246, which is less thanbuying it from the other locations asdetermined in problem 25. This profit iscomputed by multiplying the profit at eachstore by the demand. In order to determine ifsome of the demand should be met by thedistributor a new source (F) must be added tothe transportation tableau from problem 25.This source represents the distributor and hasan available supply of 150 cases, the totaldemand from all the stores. The tableau andoptimal solution is shown as follows.

Demand

FromTo

1 2 3 4

A

B

C

D

9 8 11 12 7 8 0

10 10 8 6

8 6 6 5

4 6 9 5

9 8 9 9

9

7

8

7

7

4

10

8

7

0

0

0

0

12 10 8 9 6 0E

F

5 6

8 18 26

27 13 40

20 20

35 5 40

25 15 5 45

22 128 150

25 15 30 18 27 35 171 321

SupplyDummy

Optimal, Total profit = $1,545 with multipleoptimal solutions

26.


This new solution results in a greater profit than thesolution in problem 25 ($1,545 > $1,528). Thus, someof the demand (specifically at store 3) should be metfrom the distributors.

25. Solve the model as a linear programming model to obtain the shadow prices. Among the5 purchase locations, the store at Albany hasthe highest shadow price of $3. The sensitivityrange for supply at Albany is 25 ≤ q1 ≤ 43.Thus, as much as 17 additional cases can bepurchased from Albany which would increaseprofit by $51 for a total of $1,579.

341

Demand

FromTo

A B C D

1

2

3

4

36 40 32 43 29 0

28 27 29 40

34 35 41 29

41 42 35 27

25 28 40 34

38

31

36

38

0

0

0–6

0

31 30 43 38 40 0

5

6

E

2 5 7

10 10

3 5 8

0 8 8

9 0 9

3 3 6

9 6 12 8 10 3 48

SupplyDummy

27. VAM initial solution (degenerate):

27.

28.

29.


342

Allocate “0” to cell (4,dummy):

Allocate 3 units to cell (3,dummy):

2 5 7

10 10

3 5 8

8 0 8

9 0 9

3 3 6

9 6 12 8 10 3 48Demand

FromTo

A B C D

1

2

3

4

36 40 32 43 29 0

28 27 29 40

34 35 41 29

41 42 35 27

25 28 40 34

38

31

36

38

0

0

0

–5

0

31 30 43 38 40 0

5

6

E SupplyDummy

Demand

FromTo

A B C D

1

2

3

4

36 40 32 43 29 0

28 27 29 40

34 35 41 29

41 42 35 27

25 28 40 34

38

31

36

38

0

0

0

–3

0

31 30 43 38 40 0

5

6

E

2 5 7

10 10

0 5 3 8

8 0 8

9 0 9

6 6

9 6 12 8 10 3 48

SupplyDummy


343

Allocate “0” to cell (2,B):

2 5 7

0 10 10

5 3 8

8 0 8

9 0 9

6 6

9 6 12 8 10 3 48Demand

FromTo

A B C D

1

2

3

4

36 40 32 43 29 0

28 27 29 40

34 35 41 29

41 42 35 27

25 28 40 34

38

31

36

38

0

0

0

0

31 30 43 38 40 0

5

6

E SupplyDummy

28. Initial Transportation tableau (cost = $100s)with VAM initial solution

Optimal solution; total cost = $180,645

FromTo

Jan Feb March April

Jan

Feb

March

April

12 12.15 12.30 12.45 12.6 12.75 0

M 11.4 11.55 11.7

M M 10.83 10.98

M M M 10.29

M M M M

11.85

11.13

10.44

9.78

12.0

11.28

10.59

9.93

0

0

0

0May

May June

180 30 90 300

260 40 300

300 300

210 90 300

280 20 300

300 300

180 260 340 210 400 320 90 1,800

SupplyDummy

Demand

M M M M M 9.29 0June

Optimal; total transport time = 1,275 hours

30.


32. Cost = $1000s

344

FromTo

Jan Feb Mar April May June Dummy Supply

Rj

Oj

Rj

Oj

RF

OF

RM

OM

RA

OA

RM

OM

June

May

April

March

February

January

12 12.15

14.4

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

14.55 14.70

12.3 12.45

14.85 15.0

12.6 12.75

15.15

0

0

0

0

0

0

0

0

0

0

0

0

90 200

300300

300

300

200

200

300

130

200

300

200

300

200

300

200

300

200

300

3,000

11.4

13.68 13.83

11.55

10.83

13.0

12.35

10.29

13.15

10.98

13.98

11.7 11.85

14.13 14.28

11.13

13.3

10.44

12.5

9.78

11.74

12.0

11.28

13.45

10.59

12.65

9.93

11.89

9.29

11.15

Demand 410 320 500 620 430 380 340

60

70

12080

300

110

20 120

120180

Total cost = $301, 004

29. Optimal transportation tableau (costs = $100s).Multiple optimal solutions exist.

31.

From ToFrom Denver St. Paul Louisville Akron Topeka Dummy Supply

3.7 4.6 4.9 5.5 4.3 0Sac. 13 5 18

3.4 5.1 4.4 5.9 5.2 0Bak. 8 2 5 15

3.3 4.1 3.7 2.9 2.6 0San. 10 10

1.9 4.2 2.7 5.4 3.9 0Mont. 12 12

6.1 5.1 3.8 2.5 4.1 0Jack. 15 5 20

6.6 4.8 3.5 3.6 4.5 0Ocala 15 15

Demand 20 15 15 15 20 5 90

Total cost, Z = $278,000

It is cheaper for National Foods to continue to operate its own trucking firm.

33. Increasing the supply at Sacramento, Jacksonville and Ocala to 25 tons would have little effect, reducing the over-all monthly shipping cost to $276,000, which is still higher than the $245,000 the company is currently spendingwith its own trucks.

Alternatively, increasing the supply at San Antonio and Montgomery to 25 tons per month reduces the monthlyshipping cost to $242,500 which is less than the company’s cost with their own trucks.


345

From To

From Hong Kong Singapore Taipei Supply

300 210 340

L.A. 150 300 450

490 520 610

Savannah 400 200 600

360 320 500

Galveston 350 350

Order 800 920 1100

Shortages 200 200

Orders 600 500 500 1600

Z = $723,500

Penalty costs = 200 x $800 = $160,000

Period of Production

Period of Use 1 2 3 4 Capacity

0 2 4 6Beginning Inventory 300 300

20 22 24 26Regular 8,700 300 9,000

1 25 27 29 31Overtime 1000 1,000

27 29 31 33Subcontract 3,000

M 20 22 24Regular 10,000 10,000

2 M 25 27 29Overtime 700 800 1,500

M 27 29 31Subcontract 200 3,000

M M 20 22Regular 12,000 12,000

3 M M 25 27Overtime 2,000 2,000

M M 27 29Subcontract 1,000 2,000 3,000

M M M 20Regular 12,000 12,000

4 M M M 25Overtime 2,000 2,000

M M M 27Subcontract 3,000 3,000

Demand 9,000 12,000 16,000 19,000

Total cost; Z = $1,198,500; multiple optimal solutions exist

35.

34.


346

36. 37.


347

38.

39.

40.


348

41.

42.


349

43.

44. a) It actually can be solved by either method—transportation or assignment. However, if theassignment method is used there will be 9 desti-nations, with each city repeated 3 times.

b) The solution with the transportation method.

From ToFrom Athens Columbia Nashville Supply

165 90 1301 1 1

75 210 3202 1 1

180 170 1403 1 1

220 80 604 1 1

410 140 805 1 1

150 170 1906 1 1

170 110 1507 1 1

105 125 1608 1 1

240 200 1559 1 1

Demand 3 3 3 9

Total mileage, Z = 985; multiple optimal solutions exist.


350

45. This changes the solution and increases the totalmileage. The new assignments are:

Officials 3, 6 and 7 – AthensOfficials 1, 2 and 8 – ColumbiaOfficials 4, 5 and 9 – Nashville

Total mileage, Z = 1,220

46. Solution Summary:

Al’s – Parent’s BrunchBon Apetít – Post-game PartyBon Apetít – Lettermen’s DinnerDivine – Booster Club LuncheonEpicurean – Contributor’s DinnerUniversity – Alumni Brunch

Total Cost (Z) = $103,600


351

ABCDEF

0.07.15.04.12.19

.20

.13

.080

.10

.23

0.22.06.18.26.03

.01

.10

.04

.16 0.13

000000

000000

1 2 3 4 D1 D2

ABCDEF

0.04.12.04.12.06

.20

.10

.05 0.10.20

0.19.03.18.26

0

.01

.07

.01

.16 0.10

.03 0 0.03.03

0

.030

0.03

.030

1 2 3 4 D1 D2

Annie – 1. BackstrokeBeth – dummyCarla – dummyDebbie – 2. BreaststrokeErin – 4. FreestyleFay – 3. Butterfly

Total time = 10.61 minutes

40.47. 48.


352

123456

000000

000000

HomeFurnishings China Appliance Jewelry D1 D2

41. Subract all sales values from highest sales, $630.

29070

360270180350

470260

M410440310

20110280

060

140

340180210480320270

123456

000000

000000


2200

290200110280

2100

M15018050

20110280

060

140

1600

5030014090

123456

0200

2000

0200

2000


2000

27020090

260

1900

M15016030

0110260

040

120

1400

3030012070

Opportunity cost table:

123456

30500

5000

30500

5000


2000

24020060

230

1900

M150130

0

0110230

01090

14000

3009040

123456

0500

2000

0500

2000


1700

24017060

230

1600

M12013030

0140260

040

120

11000

2709040

Employee 1 – dummy 0Employee 2 – Home Furnishing 560Employee 3 – Jewelry 420Employee 4 – Appliances 630Employee 5 – dummy 0Employee 6 – China 320

$1,930 total sales

49.

TaylMod-Bff.qxd 4/21/06 8:51 PM 2

353

50.


354

51.

Multiple optimal solutions:1-B 42-A 43-F 84-D 55-C 66-E 9

36 nights

1-B 42-C 53-F 84-D 55-A 56-E 9

36 nights

GA FL FL VA PA NY NJ

A 38 44 44 35 29 45 37

B 35 30 30 37 19 25 28

C 54 47 47 38 45 36 50

D 42 34 34 30 33 29 51

E 23 27 27 31 20 20 26

F 32 27 27 28 20 22 43

G 28 40 40 26 28 38 39


A 5 8 8 4 0 16 0

B 12 4 4 0 9 0 6

C 14 4 4 0 0 0 6

D 11 0 0 1 6 2 16

E 1 2 2 11 2 2 0

F 10 0 0 6 0 2 15

G 0 9 9 0 4 14 7


A 7 10 10 6 0 16 2

B 14 6 6 18 0 6 3

C 16 6 6 2 9 0 8

D 11 0 0 1 4 0 16

E 1 2 2 11 0 0 0

F 10 2 2 8 0 2 17

G 0 9 9 0 2 12 7

52. First subtract all the %’s from 100, and make 2 Florida columns:

Opportunity cost table:


355


A 4 7 7 3 0 15 0

B 11 3 3 15 0 5 0

C 14 4 4 0 10 0 7

D 11 0 0 1 7 2 17

E 0 1 1 10 2 2 0

F 10 0 0 6 1 2 16

G 0 9 9 0 5 14 8

A–NJB–PAC–NYD–FLE–GAF–FLG–VAZ=498

Average success rate = �49

7

8� = 71.1%


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