chapter 8 transportation

TRANSPORTATION MODEL




Introduction



• The transportation problem is a special form of the general linear programming technique called the network flow problems.

• Networks are activities or jobs to be performed which can be represented by joining one start event to one end event.



Example of a network

End

Activity



Start



Transportation Problem Defined

• Transportation problem involves the distribution of goods from sources of supply (e.g. factories, warehouses) to a number of destinations of demand (e.g. retail – shops)

• Usually, the capacity of goods at each source and the requirements or demands for the goods at each destination are given or known.



Developing A Transportation Table

Example

ABC Office Supplier has three depots located in various towns. It receives order for a total of 15 special filing cabinets from four customers.

Customer Order

A 3

B 3

C 4

D 5

Total 15



In total the available filing cabinets in three depots are

Depot Available

X 2

Y 6

Z 7

Total 15



The management wishes to minimize delivery costs by dispatching the filing cabinets from appropriate depot for each customer. The transportation costs from each depot to each customer per filing cabinet are given in Table 1.





Table 1 - Transportation Costs per Cabinet for ABC Office Supplier

To Requirements (customer)

From A B C D

Depot X RM13 RM11 RM15 RM20

Depot Y 17 14 12 13

Depot Z 18 18 15 12





Our first step is setting-up a transportation table i.e to summarize all relevant data and to keep track of algorithm computations. Using the above information, the transportation table is shown in Table 2.





Table 2 – Transportation Table for ABC Office Supplier



To



From

Customer

A



Customer

B



Customer

C



Customer

D



Depot

Available



Depot X



13

11



15



20



2



Depot Y



17



14



12



13



6



Depot Z



18



18



15



12



7

Customer

Requirements

3

3

4

5

15









Step-by-step explanation

1. we starts in the upper-left hand corner, distribute 2 (supply capacity) units cabinet to customer A. Since depot X has only 2 units cabinet available and demand (requirement) from customer A is 3 units, move down to the second row in the same column.

2. Now, deliver 1 unit cabinet (availability 6 units) to customer A. they will make customer A has 3 units cabinet (2 from depot X and 1 from depot Y). Move to the next column and deliver 3 units to customer B (demand is 3 units). Now depot Y has 2 units left, move to the right to the next column and deliver 2 units to customer. Move down to the third row in the same column.

3. Since customer C requires 4 units, deliver 2 units from depot Z (2 units is from depot Y). At this stage there are 5 units available, deliver all 5 units to customer D (demand is 5 units).

Now when the initial feasible solution have been determine, we can calculate the initial transportation cost.



Stepping-Stone Method

• This is a method that moves an initial feasible solution to an optimal solution.

• Before we applied this method to the problem, we must make sure that the transportation routes from our table must have : the number of occupied routes (or squares) must always be equal to tone less than the sum of the number of rows plus the number of columns.

• Occupied transportation routes (squares) = number rows + number columns – 1



Testing The Solution For Possible Improvement

The test for each square is as follows:

a. Select an unused square to be evaluated

b. Beginning at this square, trace a closed path back to the original square via squares that are currently being used, moving with only horizontal and vertical only.

c. Beginning with a plus (+) sign at the unused square, place alternate minus (-) signs and plus signs on each corner square of the closed path.

d. Calculate an improvement index by adding together the unit cost found in each square containing a minus sign.

e. Repeat steps (a) to (d) until an improvement index has been calculated for all unused squares.

f. If all indices computed are greater than or equal to zero an optimal solution has been reached. If not, it is possible to improve the current solution and decrease the total transportation costs.











To Improve the Solution

• Every negative amount index computed means that the total transportation costs could be decreased by the amount if 1 unit of product used the route.

• If there were more than one negative index, choose the route with the largest negative index.

• Now to determine the maximum quantity that can be transport using this route we do the following:

 Refer to the closed path of plus and minus sign on this route.

 Select the smallest number found in those squares containing minus signs.

 Add this number to all squares with plus signs on the closed path and subtract from all squares on the path with minus signs.























































To



From

Customer

A



Customer

B



Customer

C



Customer

D



Depot

Available



Depot X



2 13

11



15



20



2



Depot Y



1 17



3 14



2 12



13



6



Depot Z



18



18



2 15



5 12



7

Customer

Requirements

3

3

4

5

15



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



Depot X - Customer B = +11 -14 +17 -13

= +1

Depot X - Customer C = +15 -12 +17 -13

= +7

Depot X - Customer D = +20 -12 +15 -12 +17 -13

= +15

Depot Y - Customer D = +13 -12 +15 -12

= +4

Depot Z - Customer A = +18 -15 +12 -17

= -2

Depot Z - Customer B = +18 -15 +12 -14

= +1



Total Transportation Cost = (2 X 13) + (17X 1) + (14 X 3) + (2 X 12) + (15 X 2) + (12 X 5)

= 26 + 17 +42 +24 +30 +60

= RM199





To



From

Customer

A



Customer

B



Customer

C



Customer

D



Depot

Available



Depot X



2 13

11



15



20



2



Depot Y



17



3 14



3 12



13



6



Depot Z



1 18



18



1 15



5 12



7

Customer

Requirements

3

3

4

5

15



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



Depot X - Customer B = +11 -13 +18 -15 +12 -14

= -1

Depot X - Customer C = +15 -13 +18 -15

= +5

Depot X - Customer D = +20 -12 +18 -13

= +13

Depot Y - Customer A = +17 -12 +15 -18

= +2

Depot Y - Customer D = +13 -12 +15 -12

= +4

Depot Z - Customer B = +18 -15 +12 -14

= +1



Total Transportation Cost = (2 X 13) + (14 X 3) + (3 X 12) + (18 X 1) + (15 X 1) + (12 X 5)

= 26 + 42 + 36 + 18 + 15 + 60

= RM197





To



From

Customer

A



Customer

B



Customer

C



Customer

D



Depot

Available



Depot X



1 13

1 11



15



20



2



Depot Y



17



2 14



4 12



13



6



Depot Z



2 18



18



15



5 12



7

Customer

Requirements

3

3

4

5

15



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



Depot X - Customer c = +15 -12 +14 -11

= +6

Depot X - Customer D = +20 -12 +18 -13

= +13

Depot Y - Customer A = +17 -14 +11 -13

= +1

Depot Y - Customer D = +13 -12 +18 -13 +11 -14

= +3

Depot Z - Customer B = +18 -18 +13 -11

= +2

Depot Z - Customer B = +15 -18 +13 -11 +14 -12

= +1



Total Transportation Cost = (1 X 13) + (11 X 1) + (2 X 14) + (12 X 4) + (18 X 2) + (12 X 5)

= 13 + 11 + 28 + 28 + 36 + 60

= RM176





UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

EXERCISE – CHAPTER 8





Question



The Giant Farmer Company process cement, which it stocks in various warehouse. The market research study provided the following data:





Table 1 : Warehouse



Warehouse Capacities (tones)

W 250

X 400

Y 300

Z 200



Table 2 : Demand



A 400

B 150

C 600



Table 3 : Shipping cost per tones



From W to A : 67 From Y to A : 29

From W to B : 42 From Y to B : 47

From W to C : 51 From Y to C : 60

From X to A : 61 From Z to A : 43

From X to B : 24 From Z to B : 31

From X to C : 39 From Z to C : 42





Use the Northwest – Corner Approach Method to find the shipping that minimize shipping cost



To



From

Customer

A



Customer

B



Customer

C



Warehouse



W



250 67

42



51



250



X



150 61



150 24



100 39



400



Y



29



47



300 60



300



Z



43



31



200 42



200

Customer

Requirements

400

150

600

1150





































Occupied transportation routes = M + N - 1

= 4 + 3 – 1

= 6

Index

Depot W - Customer B = +42 -24 +61 -67

= +12

Depot W- Customer C = +51 -39 +61 -67

= +6

Depot Y - Customer A = +29 -60 +39 -61

= -53

Depot Y - Customer B = +47 -60 +39 -24

= +2

Depot Z - Customer A = +43 -42 +39 -61

= -21

Depot Z - Customer B = +31 -42 +39 -24

= +4

Total Transportation Cost = (250 X 67) + (150X 61) + (150 X 24) + (100 X 39) + (300 X 60) + (200 X 42)

= 16750 + 9150 + 9750 + 3900 + 18000 + 8400

= RM65950

To



From

Customer

A



Customer

B



Customer

C



Warehouse



W



250 67

42



51



250



X



61



150 24



250 39



400



Y



150 29



47



150 60



300



Z



43



31



200 42



200

Customer

Requirements

150

150

600

1150





































Occupied transportation routes = M + N - 1

= 4 + 3 – 1

= 6

Index

Depot W - Customer B = +42 -67 +29 -60 +39 -24

= -41

Depot W - Customer C = +51 -67 +29 -60

= -47

Depot XY - Customer A = +61 -39 +60 -29

= +53

Depot Y - Customer B = +47 -60 +39 -24

= +2

Depot Z - Customer A = +43 -42 +60 -29

= +32

Depot Z - Customer B = +31 -42 +39 -24

= +4

Total Transportation Cost = (250 X 67) + (150X 24) + (250 X 39) + (150 X 29) + (150 X 60 + (200 X 42)

= 16750 + 3600 + 9750 + 4350 + 9000 + 8400

= RM51850

To



From

Customer

A



Customer

B



Customer

C



Warehouse



W



100 67

42



150 51



250



X



61



150 24



250 39



400



Y



300 29



47



60



300



Z



43



31



200 42



200

Customer

Requirements

150

150

600

1150





































Occupied transportation routes = M + N - 1

= 4 + 3 – 1

= 6

Index

Depot W - Customer B = +42 -51 +39 -24

= +6

Depot X - Customer A = +61 -39 +51 -67

= +6

Depot Y - Customer B = +47 -24 +39 -51 +67 -29

= +49

Depot Y - Customer B = +60 -51 +67 -29

= +47

Depot Z - Customer A = +43 -42 +51 -67

= -15

Depot Z - Customer B = +31 -42 +39 -24

= +4

Total Transportation Cost = (100 X 67) + (150X 51) + (150 X 24) + (250 X 39) + (300 X 29) + (200 X 42)

= 6700 + 7650 + 3600 + 9750 + 8700 + 8400

= RM44800

To



From

Customer

A



Customer

B



Customer

C



Warehouse



W



67

42



250 51



250



X



61



150 24



250 39



400



Y



300 29



47



60



300



Z



100 43



31



100 42



200

Customer

Requirements

150

150

600

1150





































Occupied transportation routes = M + N - 1

= 4 + 3 – 1

= 6

Index

Depot W - Customer A = +67 -51 +42 -43

= +15

Depot W - Customer B = +42 -51 +39 -24

= +6

Depot X - Customer A = +61 -39 +51 -67

= +6

Depot Y - Customer B = +47 -29 +67 -51 +39 -24

= +49

Depot Y - Customer B = +60 -51 +67 -29

= +47

Depot Z - Customer B = +31 -42 +39 -24

= +4

Total Transportation Cost = (250 X 51) + (150X 24) + (250 X 39) + (300 X 29) + (100 X 43) + (100 X 42)

= 12750 + 3600 + 2250 + 9750 + 8700 + 4300 + 4200

= RM43300

UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

TUTORIAL – CHAPTER 8



Question 1

The Giant Farmer Company process cement, which it stocks in various warehouse. The market research study provided the following data:



Table 1 : Warehouse Table 2 : Demand

Warehouse Quantities (kg)

X 10000 A 12000

Y 16000 B 12000

Z 10000 C 12000



Table 3 : Shipping cost per kg (RM)

From / To I II III

X 10 6 12

Y 15 20 16

Z 8 18 22

Use the Northwest – Corner Approach Method to find the shipping that minimize shipping cost



Question 2

To



From

Sentosa

Sriwangi

Budi

C



Warehouse



Factory A



15



20



30



4000



Factory B



15



10



40



3500



Factory C



25



15



35



2500

Customer

Requirements

3500

3000

3500

10000































Use the Northwest – Corner Approach Method to find the shipping that minimize shipping cost

Answers



Question 1



To



From

I

II





III





Warehouse



X



10000 10

6



12



10000



Y



2000 15



12000 20



2000 16



16000



Z



8



18



10000 22



10000

Customer

Requirements

12000

12000

12000

26000

































Total transportation cost = 10000(10) + 2000(15) + 12000(20) + 2000(16) + 10000(22)

= 10000 + 30000 + 240000 + 32000 + 220000

= RM622000



X – II : The index is +6 -20 +15 -15 = -9

X – III : The index is +12 -16 +15 -10 = +1

Z – I : The index is +8 -22 +16 -15 = -13

Z – II : The index is +18 -20 +16 -22 = -8



















To



From

I

II





III





Warehouse



X



10000 10

6



12



10000



Y



15



12000 20



4000 16



16000



Z



2000 8



18



8000 22



10000

Customer

Requirements

12000

12000

12000

26000

































Total transportation cost = 10000(10) + 12000(20) + 4000(16) + 2000 (8) + 8000(22)

= 10000 + 240000 + 64000 + 16000+ 176000

= RM622000



X – II : The index is +6 -20 +16 -22 +8 -10 = -22

X – III : The index is +12 -22 +8 -10 = -12

Y – I : The index is +15 -16 +22 -8 = +13

Z – I : The index is +18 -20 +16 -22 = -8



























To



From

I

II





III





Warehouse



X



2000 10

8000 6



12



10000



Y



15



4000 20



12000 16



16000



Z



10000 8



18



22



10000

Customer

Requirements

12000

12000

12000

26000

































Total transportation cost = 2000(10) + 8000(6) + 4000(20) + 12000 (16) + 10000(8)

= 2000 + 48000 + 80000 + 192000+ 80000

= RM420000



X – III : The index is +12 -16 +20 -6 = +10

Y – I : The index is +15 -20 +6 -10 = -9

Z – II : The index is +18 -6 +10 -8 = +14

Z – III : The index is +22 -16 +20 -6 +10 -8 = +22



























To



From

I

II





III





Warehouse



X



10

10000 6



12



10000



Y



2000 15



2000 20



12000 16



16000



Z



10000 8



18



22



10000

Customer

Requirements

12000

12000

12000

26000

































Total transportation cost = 10000(6) + 2000(15) + 2000(20) + 12000 (16) + 10000(8)

= 60000 + 30000 + 40000 + 192000+ 80000

= RM402000



X – I : The index is +10 -15 +20 -6 = +9

X – III : The index is +12 -16 +20 -6 = +10

Z – II : The index is +18 -6 +15 -8 = +5

Z – III : The index is +22 -16 +15 -8 = +13





























Question 2

To



From

Sentosa

Sriwangi

Budi

C



Warehouse



Factory A



3500 15



500 20



30



4000



Factory B



15



2500 10



1000 40



3500



Factory C



25



15



2500 35



2500

Customer

Requirements

3500

3000

3500

10000































Total transportation cost = 3500(15) + 500(20) + 2500(10) + 1000(40) + 2500(35)

= 52500 + 10000 + 25000 + 40000 +87500

= RM215000



Factory A – Budi : The index is +30 -40 +10 -20 = -20

Factory B – Sentosa : The index is +15 -10 +20 -15 = +10

Factory C – Sentosa : The index is +25 -35 +40 -10 +20 -15 = +25

Factory C – Sriwangi : The index is +15 -35 +40 -10 = +10

























To



From

Sentosa

Sriwangi

Budi

C



Warehouse



Factory A



2500 15



20



500 30



4000



Factory B



15



3000 10



500 40



3500



Factory C



25



15



2500 35



2500

Customer

Requirements

3500

3000

3500

10000































Total transportation cost = 2500(15) + 3000(10) + 500(30) + 500(40) + 2500(35)

= 52500 + 15000 + 30000 + 20000 + 87500

= RM205000



Factory A – Sriwangi : The index is +20 -10 +40 -30 = +20

Factory B – Sentosa : The index is +15 -40 +30 -15 = -10

Factory C – Sentosa : The index is +25 -35 +30 -15 = +25

Factory C – Sriwangi : The index is +15 -35 +40 -10 = +10





























To



From

Sentosa

Sriwangi

Budi

C



Warehouse



Factory A



3000 15



20



1000 30



4000



Factory B



500 15



3000 10



40



3500



Factory C



25



15



2500 35



2500

Customer

Requirements

3500

3000

3500

10000































Total transportation cost = 3000(15) + 1000(30) + 500(15) + 3000(10) + 2500(35)

= 45000 + 30000 + 7500 + 30000 + 87500

= RM200000



Factory A – Sriwangi : The index is +20 -10 +15 -15 = +10

Factory B – Budi : The index is +25 -35 +30 -15 = +5

Factory C – Sentosa : The index is +15 -35 +30 -15 +15 -10 = 0

Factory C – Sriwangi : The index is +40 -30 +15 -15 = +10

































Question

ABC Office Supplier has three depots located in various towns. It receives order for a total of 850 special filing cabinets from four customers.



Customer Order

A 300

B 200

C 200





In total the available filing cabinets in three depots are

Depot Available

X 250

Y 300

Z 300







The management wishes to minimize delivery costs by dispatching the filing cabinets from appropriate depot for each customer. The transportation costs from each depot to each customer per filing cabinet are given in Table 1.





Table 1 - Transportation Costs per Cabinet for ABC Office Supplier

To Requirements (customer)

From A B C

Depot X RM5 RM4 RM3

Depot Y 8 4 3

Depot Z 9 7 5





Use the Northwest – Corner Approach Method to find the shipping that minimize shipping cost

(25M)





To



From

Customer

A



Customer

B



Customer

C



Dummy





Depot

Available



Depot X



250 5

4



3



0



250



Depot Y



50 8



200 4



50 3



0



300



Depot Z



9



7



150 5



150 0



300

Customer

Requirements

300

200

200

150

850



Depot X – Customer A = +4 -4 +8 -5

= +3



Depot X – Customer C = +3 -3 +8 -5

= +3



Dummy = +0 -0 +5 -3 +8 -5

= +5



Dummy = +0 -0 +5 -3

= +2



Depot Z – Customer A = +9 -5 +3 -8

= -1



Depot Z – Customer B = +7 -5 +3 -4

= +1



Total transportation cost = 250(5) + 200(4) + 100(3) + 50(9) + 200(5) + 150(0)

= 1250 + 800 +300 +450 +1000 +0

=



To



From

Customer

A



Customer

B



Customer

C



Dummy





Depot

Available



Depot X



250 5

4



3



0



250



Depot Y



8



200 4



100 3



0



300



Depot Z



50 9



7



200 5



150 0



300

Customer

Requirements

300

200

200

150

850



Depot X – Customer B = +4 -4 +3 -5 +9 -5

= +2



Depot X – Customer C = +3 -5 +9 -3

= +4



Dummy = +0 -0 +9 -5

= +4



Depot Y – Customer A = +8 -3 +5 -9

= +1



Dummy = +0 -0 +5 -3

= +2



Depot Z – Customer B = +7 -5 +3 -4

= +1



Total transportation cost = 250(5) + 50(8) +200(4) + 50(3) +150(5) + 150(0)

= 1250 + 400 + 800 +150 +750 + 0

= RM33350











Question 1



The Giant Farmer Company process cement, which it stocks in various warehouse. The market research study provided the following data:





Table 1 : Warehouse



Warehouse Capacities (tones)

A 100

B 60

C 140





Table 2 : Demand



I 60

II 120

III 40

IV 80





Table 3 : Shipping cost per tones



I II III IV

A 30 36 38 26

B 40 28 32 34

C 50 24 38 44





Use the Northwest – Corner Approach Method to find the shipping that minimize shipping cost

(40M)





To



From

I

II

III

IV

Supply



A



60 30

40 36



38



26



100



B



40



60 28



32



34



60



C



50



20 24



40 38



80 44



140



Demand

60

120

40

80

300



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



A – III = +38 -38 +24 -36

= -12

A – IV = +26 -44 +24 -36

= -30

B – I = +40 -28 +36 -30

= +18

B – III = +32 -38 +24 -28

= -10

B – IV = +34 -44 +24 -28

= -14

C – I = +50 -24 +36 -30

= +32



Total Transportation Cost = (60 X 30) + (40X 36) + (60 X 28) + (20 X 24) + (40 X 38) + (80 X 44)

= 1800 + 1440 + 1680 + 480 + 1520 + 3520

= RM10440





To



From

I

II

III

IV

Supply



A



60 30

36



38



40 26



100



B



40



60 28



32



34



60



C



50



60 24



40 38



40 44



140



Demand

60

120

40

80

300



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



A – II = +36 -24 +44 -26

= +30

A – III = +38 -38 +44 -26

= +18

B – I = +40 -28 +24 -44 + 26 -30

= -12

B – III = +32 -38 +24 -28

= -10

B – IV = +34 -44 +24 -28

= -14

C – I = +50 -44 +26 -30

= +2



Total Transportation Cost = (60 X 30) + (60X 28) + (60 X 24) + (40 X 38) + (40 X 26) + (40 X 44)

= 1800 + 1680 + 1440 + 1520 + 1040 + 1360

= RM9240





To



From

I

II

III

IV

Supply



A



60 30

36



38



40 26



100



B



40



20 28



32



40 34



60



C



50



100 24



40 38



44



140



Demand

60

120

40

80

300



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



A – II = +36 -26 +34 -28

= +16

A – III = +38 -38 +24 -28 +34 -26

= +4

B – I = +40 -34 + 26 -30

= +2

B – III = +32 -38 +24 -28

= -10

C – I = +50 -24 +28 -34 +26 -30

= +16

C – IV = +44 -34 +28 -24

= +14



Total Transportation Cost = (60X30) + (20X28) + (100 X 24) + (40 X 38) + (40 X 26) + (40 X 34)

= 1800 + 560 + 2400 + 1520 + 1040 + 1360

= RM8680





To



From

I

II

III

IV

Supply



A



60 30

36



38



40 26



100



B



40



28



20 32



40 34



60



C



50



120 24



20 38



44



140



Demand

60

120

40

80

300



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



A – II = +36 -24 +38 -32 +34 -26

= +26

A – III = +38 -32 +34 -26

= +14

B – I = +40 -34 + 26 -30

= +2

B – III = +28 -24 +38 -32

= +10

C – I = +50 -38 +32 -34 +26 -30

= +6

C – IV = +44 -34 +32 -38

= +4



Total Transportation Cost = (60 X 30) + (120X24) + (20 X 32) + (20 X 38) + (40 X 26) + (40 X 34)

= 1800 + 2880 + 640 + 760 + 1040 + 1360

= RM8480





Least – Cost Solution

• This is a method that moves from an initial feasible solution to an optimal solution.



Transportation Table for ABC Office Supplier

To



From

Customer

A



Customer

B



Customer

C



Customer

D



Depot

Available



Depot X



13

2 11



15



20



2



Depot Y



1 17



1 14



4 12



13



6



Depot Z



2 18



18



15



5 12



7

Customer

Requirements

3

3

4

5

15

Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



Depot X - Customer A = +13 -17 +14 -11

= -11

Depot X - Customer C = +15 -11 +14 -12

= +6

Depot X - Customer D = +20 -11 +14 -17 +18 -12

= +12

Depot Y - Customer D = +13 -17 +18 -12

= +2

Depot Z - Customer B = +18 -14 +17 -18

= +3

Depot Z - Customer C = +15 -12 +17 -18

= +2



Total Transportation Cost = (11 X 12) + (17 X 1) + (1 X 14) + (12 X 4) + (18 X 2) + (12 X 5)

= RM197

To



From

Customer

A



Customer

B



Customer

C



Customer

D



Depot

Available



Depot X



1 13

1 11



15



20



2



Depot Y



17



2 14



4 12



13



6



Depot Z



2 18



18



15



5 12



7

Customer

Requirements

3

3

4

5

15



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



Depot X - Customer A = +15 -12 +14 -11

= +6

Depot X - Customer C = +20 -12 +18 -13

= +13

Depot X - Customer D = +17 -14 +11 -13

= +1

Depot Y - Customer D = +18 -11 +13 -18

= +12

Depot Z - Customer B = +15 -12 +14 -11 +13 -18

= +1

Depot Z - Customer B = +15 -12 +17 -18

= +2



Total Transportation Cost = (1 X 13) + (11 X 1) + (2 X 14) + (12 X 4) + (18 X 2) + (12 X 5)

= RM196







UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

TUTORIAL – CHAPTER 8



Question 1

The Giant Farmer Company process cement, which it stocks in various warehouse. The market research study provided the following data:



Table 1 : Warehouse Table 2 : Demand

Warehouse Quantities (kg)

X 10000 A 12000

Y 16000 B 12000

Z 10000 C 12000



Table 3 : Shipping cost per kg (RM)

From / To I II III

X 10 6 12

Y 15 20 16

Z 8 18 22

Use the Least – Cost Method to find the shipping that minimize shipping cost



Question 2

To



From

Sentosa

Sriwangi

Budi

C



Warehouse



Factory A



15



20



30



4000



Factory B



15



10



40



3500



Factory C



25



15



35



2500

Customer

Requirements

3500

3000

3500

10000































Use the Least – Cost Method to find the shipping that minimize shipping cost

Answers



Question 1



To



From

I

II





III





Warehouse



X



10

10000 6



12



10000



Y



2000 15



2000 20



12000 16



16000



Z



10000 8



18



22



10000

Customer

Requirements

12000

12000

12000

26000

































Total transportation cost = 10000(6) + 2000(15) + 2000(20) + 12000 (16) + 10000(8)

= 60000 + 30000 + 40000 + 192000+ 80000

= RM402000



X – I : The index is +10 -15 +20 -6 = +9

X – III : The index is +12 -16 +20 -6 = +10

Z – II : The index is +18 -6 +15 -8 = +5

Z – III : The index is +22 -16 +15 -8 = +13



















Question 2



To



From

Sentosa

Sriwangi

Budi

C



Warehouse



Factory A



2500 15



20



500 30



4000



Factory B



15



3000 10



500 40



3500



Factory C



25



15



2500 35



2500

Customer

Requirements

3500

3000

3500

10000































Total transportation cost = 2500(15) + 3000(10) + 500(30) + 500(40) + 2500(35)

= 52500 + 15000 + 30000 + 20000 + 87500

= RM205000



Factory A – Sriwangi : The index is +20 -10 +40 -30 = +20

Factory B – Sentosa : The index is +15 -40 +30 -15 = -10

Factory C – Sentosa : The index is +25 -35 +30 -15 = +25

Factory C – Sriwangi : The index is +15 -35 +40 -10 = +10



























To



From

Sentosa

Sriwangi

Budi

C



Warehouse



Factory A



3000 15



20



1000 30



4000



Factory B



500 15



3000 10



40



3500



Factory C



25



15



2500 35



2500

Customer

Requirements

3500

3000

3500

10000































Total transportation cost = 3000(15) + 1000(30) + 500(15) + 3000(10) + 2500(35)

= 45000 + 30000 + 7500 + 30000 + 87500

= RM200000



Factory A – Sriwangi : The index is +20 -10 +15 -15 = +10

Factory B – Budi : The index is +25 -35 +30 -15 = +5

Factory C – Sentosa : The index is +15 -35 +30 -15 +15 -10 = 0

Factory C – Sriwangi : The index is +40 -30 +15 -15 = +10





























UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

EXERCISE – CHAPTER 8





Question



The Giant Farmer Company process cement, which it stocks in various warehouse. The market research study provided the following data:





Table 1 : Warehouse



Warehouse Capacities (tones)

W 250

X 400

Y 300

Z 200



Table 2 : Demand



A 400

B 150

C 600



Table 3 : Shipping cost per tones



From W to A : 67 From Y to A : 29

From W to B : 42 From Y to B : 47

From W to C : 51 From Y to C : 60

From X to A : 61 From Z to A : 43

From X to B : 24 From Z to B : 31

From X to C : 39 From Z to C : 42





Use the Least – Cost Method to find the shipping that minimize shipping cost



To



From

Customer

A



Customer

B



Customer

C



Warehouse



W



100 67

42



150 51



250



X



61



150 24



250 39



400



Y



300 29



47



60



300



Z



43



31



200 42



200

Customer

Requirements

150

150

600

1150





































Occupied transportation routes = M + N - 1

= 4 + 3 – 1

= 6

Index

Depot W - Customer B = +42 -51 +39 -24

= +6

Depot X - Customer A = +61 -39 +51 -67

= +6

Depot Y - Customer B = +47 -24 +39 -51 +67 -29

= +49

Depot Y - Customer B = +60 -51 +67 -29

= +47

Depot Z - Customer A = +43 -42 +51 -67

= -15

Depot Z - Customer B = +31 -42 +39 -24

= +4

Total Transportation Cost = (100 X 67) + (150X 51) + (150 X 24) + (250 X 39) + (300 X 29) + (200 X 42)

= 6700 + 7650 + 3600 + 9750 + 8700 + 8400

= RM44800

To



From

Customer

A



Customer

B



Customer

C



Warehouse



W



67

42



250 51



250



X



61



150 24



250 39



400



Y



300 29



47



60



300



Z



100 43



31



100 42



200

Customer

Requirements

150

150

600

1150





































Occupied transportation routes = M + N - 1

= 4 + 3 – 1

= 6

Index

Depot W - Customer A = +67 -51 +42 -43

= +15

Depot W - Customer B = +42 -51 +39 -24

= +6

Depot X - Customer A = +61 -39 +51 -67

= +6

Depot Y - Customer B = +47 -29 +67 -51 +39 -24

= +49

Depot Y - Customer B = +60 -51 +67 -29

= +47

Depot Z - Customer B = +31 -42 +39 -24

= +4

Total Transportation Cost = (250 X 51) + (150X 24) + (250 X 39) + (300 X 29) + (100 X 43) + (100 X 42)

= 12750 + 3600 + 2250 + 9750 + 8700 + 4300 + 4200

= RM43300

UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

QUIZ – CHAPTER 8



Question 1



The Giant Farmer Company process cement, which it stocks in various warehouse. The market research study provided the following data:





Table 1 : Warehouse



Warehouse Capacities (tones)

A 300

B 150

C 250





Table 2 : Demand



I 350

II 125

III 100

IV 125





Table 3 : Shipping cost per tones



I II III IV

A 10 8 12 14

B 6 7 10 11

C 4 9 7 5





Use the Least – Cost Method to find the shipping that minimize shipping cost

(25M)





To



From

I

II

III

IV

Supply



A



175 10

125 8



12



14



300



B



150 6



7



10



11



150



C



25 4



9



100 7



125 5



250



Demand

350

125

100

125

700



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



A – III = +12 -7 +4 -10

= -1

A – IV = +14 -5 +4 -10

= +3

B – II = +7 -6 +10 -8

= +3

B – III = +10 -7 +4 -6

= +1

B – IV = +11 -5 +4 -6

= +4

C – II = +9 -8 +10 -4

= +7



Total Transportation Cost = 175 (10) + 150 (6) + 125 (8) + 25 (4) + 100 (7) + 125 (5)

= RM5075









To



From

I

II

III

IV

Supply



A



75 10

125 8



100 12



14



300



B



150 6



7



10



11



150



C



125 4



9



7



125 5



250



Demand

350

125

100

125

700



Occupied transportation routes = M + N - 1

= 3 + 4 – 1

= 6



A – IV = +14 -5 +4 -10

= +3

B – II = +7 -6 +10 -8

= +3

B – III = +10 -6 +10 –12

= +2

B – IV = +11 -5 +4 -6

= +4

C – II = +9 -8 +10 -4

= +7

C – III = +7 -12 +10 -4

= +1





Total Transportation Cost = 75 (10) + 100 (12) + 125 (8) + 150 (6) + 125 (4) + 125 (5)

= RM4975







Question







The Giant Farmer Company process cement, which it stocks in various warehouse. The market research study provided the following data:

Use the Northwest – Corner Approach Method to find the shipping that minimize shipping cost





To



From

A

B

C

Factory



W



4



3



3



35



Y



6



7



6



50





Z



8



2



5



50



Demand



30

65

40

135



(25m)

























UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

QUIZ – CHAPTER 8 (ANSWERS)

CLASS : DRM3



To



From

A

B

C

Factory



W



30 / 4



5 / 3



3



35



Y



6



50 / 7



6



50





Z



8



10 / 2



40 / 5



50



Demand



30

65

40

135



Each / ========= 1m





Indeks



W – C = +3 -5 +2 -3

= -3 (0.5m)



Y – A = +6 -7 +3 -4

= -2 (0.5m)



Y – C = +6 -7 +2 -5

= -4 (0.5m)



Z – A = +8 -2 +3 -4

= +5 (0.5m)





Total transportation cost = (30 X4) + (3X5) + (50X7) + (10X2) + (40X5) (1m)

= RM705 (1m)













To



From

A

B

C

Kilang



W



30 / 4



5 / 3



3



35



Y



6



10 / 7



40 / 6



50





Z



8



50 / 2



5



50



Permintaan



30

65

40

135



Each / ========= 1m





Indeks



W – C = +3 -3 +7 -6

= +1 (0.5m)



Y – A = +6 -7 +3 -4

= -2 (0.5m)



Z – A = +8 -2 +3 -4

= +5 (0.5m)



Z – C = +5 -6 +7 -2

= +4 (0.5m)





Total transportation cost = (30 X4) + (3X5) + (10X7) + (50X2) + (40X6) (1m)

= RM545 (1m)





















To



From

A

B

C

Factory



W



20 / 4



15 / 3



3



35



Y



10 / 6



7



40 / 6



50





Z



8



50 / 2



5



50



Demand

30

65

40

135



Each / ========= 1m





Indeks



W – C = +3 -4 +6 -6

= -1 (0.5m)



Y – B = +7 -3 +6 -4

= +6 (0.5m)



Z – A = +8 -2 +3 -4

= +5 (0.5m)



Z – C = +5 -6 +6 -4 +3 -2

= +2 (0.5m)





Total transportation cost = (20 X4) + (15X3) + (10X6) + (50X2) + (40X6)(1m)

= RM525 (1m)





















To



From

A

B

C

Factory



W



4



15 / 3



20 / 3



35



Y



30 / 6



7



20 / 6



50





Z



8



50 / 2



5



50



Demand



30

65

40

135



Each / ========= 1m





Indeks



W – A = +4 -6 +6 -3

= +1 (0.5m)



Y – B = +7 -6 +3 -3

= +1 (0.5m)



Z – A = +8 -2 +3 -3 +6 -6

= +6 (0.5m)



Z – C = +5 -3 +3 -2

= +3 (0.5m)





Total transportation cost = (30 X6) + (15X3) + (50X2) + (20X3) + (20X6)(1m)

= RM505 (1m)

chapter 8 assignment method

CHAPTER 8






ASSIGNMENT METHOD





• Involves assigning tasks or jobs resources.

• Eg: assigning jobs to machines, contracts to bidders, people to projects and salespeople to territories

• Objective – minimize total costs or time required to perform the tasks.

• One important characteristics of assignment problem is that only one job (or worker) to one machine (or project).

• Assignment method is a special form of transportation problem in which there is only one item at each of the various sources and only one item is required at each of the various destinations.

• Each assignment problem is usually arranged in either a table or a matrix.

• Generally, the rows contain the objects, people or sources to assign and column contain the tasks or destinations on things to assign on.

• Assignment service engineers to requests so as that the total costs are minimized.





































Step 1



Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.



Step 2

Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.



Step 3

Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



Step 4

a. Find the smallest number NOT covered by a line, called X.

b. Subtract X from every the uncovered number in the table.

c. Add X to every number lying at the intersection of horizontal and vertical lines.



Step 5

When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)













UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

EXERCISES – CHAPTER 8



Question 1



The operations manager of King Manufacturing must assign 3 tasks to 3 machines. Cost data are presented below.



A B C

C1

11 12 17

C2

7 11 20

C3

5 8 16



Use the assignment algorithm to solve this problem.









Answer





Step 1



Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 11 (row 1), 7 (row 2) and 5 (row 3).



A B C

C1

0 1 6

C2

0 4 13

C3

0 2 11





Step 2



Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 0 (column 1), 1 (column 2) and 6 (column 3).



A B C

C1

0 0 0

C2

0 3 7

C3

0 1 5







Step 3



Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



A B C

C1

0

0 0

C2

0 3 7

C3

0 1 5











Step 4

a. Find the smallest number NOT covered by a line, called X. The smallest number is 1.

b. Subtract X from every the uncovered number in the table.

(3 – 1 = 2), (7 - 1 = 6), (1 – 1 = 0) and (5 – 1 = 4)

c. Add X to every number lying at the intersection of horizontal and vertical lines.

(0 + 1 = 1)





A B C

C1

0 0 0

C2

0 3 7

C3

0 1 5





Step 5

When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)





A B C

C1

3 0 [0]

C2

[0] 1 5

C3

0 [0] 3





Total assignment method



C1 - C = 17

C2 - A = 7

C3 - B = 8

32









































































Question 2





The operations manager of King Manufacturing must assign 4 tasks to 4 machines. Cost data are presented below.



Kangar Tampin Alor Gajah Mersing

Irdina

40 50 44 56

Nadiatul

30 36 46 34

Nurin

38 34 42 48

Nomie

50 46 48 48



Use the assignment algorithm to solve this problem.









Answer





Step 1





Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 40 (row 1), 30 (row 2), 34 (row 3) and 46 (row 4).



Kangar Tampin Alor Gajah Mersing

Irdina

0 10 4 16

Nadiatul

0 6 16 4

Nurin

4 0 8 14

Nomie

4 0 2 2







Step 2



Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 0 (column 1), 0 (column 2), 2 (column 3) and 2 (column 4).



Kangar Tampin Alor Gajah Mersing

Irdina

0 10 2 14

Nadiatul

0 6 14 2

Nurin

4 0 6 12

Nomie

4 0 0 0







Step 3



Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



Kangar Tampin Alor Gajah Mersing

Irdina

0

10 2 14

Nadiatul

0 6 14 2

Nurin

4

0 6 12

Nomie

4

0 0 0







Step 4

a. Find the smallest number NOT covered by a line, called X. The smallest number is 2.

b. Subtract X from every the uncovered number in the table.

(10 – 2 = 8), (2 – 2 = 0), (14 – 2 = 20), (6 – 2 = 4), (14 – 2 = 12) and (2 – 2 = 0)

c. Add X to every number lying at the intersection of horizontal and vertical lines.

(4 + 2 = 6) and (4 + 2 = 6)





Kangar Tampin Alor Gajah Mersing

Irdina

0

10 2 14

Nadiatul

0 6 14 2

Nurin

4

0 6 12

Nomie

4

0 0 0





Step 5



When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)



Kangar Tampin Alor Gajah Mersing

Irdina

[0] 8 0 12

Nadiatul

0 4 12 [0]

Nurin

6 [0] 6 12

Nomie

6 0 [0] 0

Total assignment method

Irdina - Kangar = 40

Nadiatul - Mersing = 34

Nurin - Tampin = 34

Nomie - Alor Gajah = 48

156





































































Question 3





The operations manager of King Manufacturing must assign 4 tasks to 4 machines. Cost data are presented below.



Kangar Tampin Alor Gajah Mersing

Irdina

25 18 23 14

Nadiatul

38 15 53 23

Nurin

15 17 41 30

Nomie

26 28 36 29



Use the assignment algorithm to solve this problem.









Answer





Step 1





Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 14 (row 1), 15 (row 2), 15 (row 3) and 26 (row 4).



Kangar Tampin Alor Gajah Mersing

Irdina

11 4 9 0

Nadiatul

23 0 38 8

Nurin

0 2 26 15

Nomie

0 2 10 3







Step 2



Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 0 (column 1), 0 (column 2), 9 (column 3) and 0 (column 4).





Kangar Tampin Alor Gajah Mersing

Irdina

11 4 0 0

Nadiatul

23 0 29 8

Nurin

0 2 17 15

Nomie

0 2 1 3







Step 3



Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



Kangar Tampin Alor Gajah Mersing

Irdina

11

4 0 0

Nadiatul

23

0 29 8

Nurin

0 2 17 15

Nomie

0 2 1 3





Step 4

a. Find the smallest number NOT covered by a line, called X. The smallest number is 2.

b. Subtract X from every the uncovered number in the table.

(10 – 2 = 8), (2 – 2 = 0), (14 – 2 = 20), (6 – 2 = 4), (14 – 2 = 12) and (2 – 2 = 0)

c. Add X to every number lying at the intersection of horizontal and vertical lines.

(4 + 2 = 6) and (4 + 2 = 6)





Kangar Tampin Alor Gajah Mersing

Irdina

12 4 0 0

Nadiatul

24 0 29 8

Nurin

0 1 16 14

Nomie

0 1 0 2





Step 5



When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)



Kangar Tampin Alor Gajah Mersing

Irdina

12 4 0 [0]

Nadiatul

24 [0] 29 8

Nurin

[0] 1 16 14

Nomie

0 1 [0] 2



Total assignment method

Irdina - Kangar = 14

Nadiatul - Mersing = 15

Nurin - Tampin = 15

Nomie - Alor Gajah = 36

80

































































Question 4





The operations manager of King Manufacturing must assign 4 tasks to 4 machines. Cost data are presented below.



A B C D

Z1

5 3 1 5

Z2

6 6 2 7

Z3

5 5 3 8

Z4

8 2 4 3



Use the assignment algorithm to solve this problem.











Answer







Step 1





Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 1 (row 1), 2 (row 2), 3 (row 3) and 3 (row 4).



A B C D

Z1

4 2 0 4

Z2

4 4 0 5

Z3

2 2 0 5

Z4

6 0 2 1





Step 2



Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 2 (column 1), 0 (column 2), 0 (column 3) and 1 (column 4).



A B C D

Z1

2 2 0 3

Z2

2 4 0 4

Z3

0 2 0 4

Z4

4 0 2 0





Step 3



Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



A B C D

Z1

2

2 0

3

Z2

2 4 0 4

Z3

0

2 0 4

Z4

4

0 2 0









Step 4

a. Find the smallest number NOT covered by a line, called X. The smallest number is 2.

b. Subtract X from every the uncovered number in the table.

(2 – 2 = 0), (2 – 2 = 0), (3 – 2 = 1), (2 – 0 = 2) and (4 – 2 = 2)

c. Add X to every number lying at the intersection of horizontal and vertical lines.

(0 + 2 = 2) and (2 + 2 = 4)



A B C D

Z1

0 0 0 1

Z2

0 2 0 2

Z3

0 2 2 4

Z4

4 0 4 0





Step 5



When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)



A B C D

Z1

0 [0] 0 1

Z2

0 2 [0] 2

Z3

[0] 2 2 4

Z4

4 0 4 [0]



Total assignment method

Z1 - B = 3

Z2 - C = 2

Z3 - A = 5

Z4 - D = 3

13





































































UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

ASSIGNMENT – CHAPTER 8





Question 1



The operations manager of King Manufacturing must assign 5 tasks to 5 machines. Cost data are presented below.



I II III IV V

A

3 5 4 8 2

B

9 4 3 6 2

C

11 6 8 10 9

D

6 10 4 12 5

E

3 5 6 4 9



Use the assignment algorithm to solve this problem.



Answer



Step 1



Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 2 (row 1), 2 (row 2), 6 (row 3), 4 (row 4) and 3 (row 5)



I II III IV V

A

1 3 2 6 0

B

7 2 1 4 0

C

5 0 2 4 3

D

2 6 0 8 1

E

0 2 3 1 6



Step 2

Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 0 (column 1), 0 (column 2), 0 (column 3), 1 (column 4) and 0 (column 5)



I II III IV V

A

1 3 2 5 0

B

7 2 1 3 0

C

5 0 2 3 3

D

2 6 0 7 1

E

0 2 3 0 6





Step 3

Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



I II III IV V

A

1 3 2 5 0


B

7 2 1 3 0

C

5

0 2 3 3

D

2

6 0 7 1

E

0

2 3 0 6







Step 4

a. Find the smallest number NOT covered by a line, called X. The smallest number is 1.

b. Subtract X from every the uncovered number in the table.

(1 – 1 = 0), (3 - 1 = 2), (2 – 1 = 1), (5 – 1 = 4), (7 – 1 = 6), (2 -1 = 1), (1 – 1 = 0) and (3 – 1 = 2)

c. Add X to every number lying at the intersection of horizontal and vertical lines.

(3 + 1 = 4), (1 + 1 = 2) and (6 + 1 = 7)



I II III IV V

A

0 2 1 4 0

B

6 1 0 2 0

C

5 0 2 3 4

D

2 6 0 7 2

E

0 2 3 0 7



Step 5

When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)



I II III IV V

A

[0] 2 1 4 0

B

6 1 0 2 [0]

C

5 [0] 2 3 4

D

2 6 [0] 7 2

E

0 2 3 [0] 7



Total assignment method

A - I = 3

B - V = 2

C - II = 6

D - III = 4

E - IV = 4

19



































































Question 2



The operations manager of King Manufacturing must assign 6 tasks to 6 machines. Cost data are presented below.



I II III IV V VI

A

60 22 34 42 30 60

B

22 52 16 32 18 48

C

29 16 58 28 22 55

D

42 32 28 46 15 30

E

30 18 25 15 45 42

F

50 48 27 30 44 60



Use the assignment algorithm to solve this problem.



Answer



Step 1



Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 22 (row 1), 16 (row 2), 16 (row 3), 15 (row 4), 15 (ROW 5) and 27 (row 65)



I II III IV V VI

A

38 0 12 20 8 38

B

6 36 0 16 2 32

C

13 0 42 12 6 39

D

27 17 13 31 0 15

E

15 3 10 0 30 27

F

23 21 0 3 17 33





Step 2

Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 6 (column 1), 0 (column 2), 0 (column 3), (column 4), 0 (column 5) and 15 (column 6)



I II III IV V VI

A

32 0 12 20 8 23

B

0 36 0 16 2 17

C

7 0 42 12 6 14

D

21 17 13 31 0 0

E

9 3 10 0 30 12

F

17 21 0 3 17 18





Step 3

Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



I II III IV V VI

A

32 0 12 20 8 23

B

0 36 0 16 2 17

C

7 0 42 12 6 14

D

21

17 13 31 0 0

E

9 3 10 0 30 12

F

17 21

0 3 17

8




Step 4



a. Find the smallest number NOT covered by a line, called X. The smallest number is 1.

b. Subtract X from every the uncovered number in the table.

(32 - 6 = 26), (12 - 6 = 6), (20 – 6 = 14), (8 – 6 = 2), (23 – 6 = 17), (7 - 6 = 1), (42 – 6 = 36), (12 – 6 = 6), (6 – 6 = 0) and (14 – 6 = 8)

c. Add X to every number lying at the intersection of horizontal and vertical lines.

(36 + 6 = 42), (17 + 6 = 23), (3 + 6 = 9) and (21 + 6 = 27)





I II III IV V VI

A

32 0 12 20 8 23

B

0 36 0 16 2 17

C

7 0 42 12 6 14

D

21 17 13 31 0 0

E

9 3 10 0 30 12

F

17 21 0 3 17 18



















Step 5



When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)





I II III IV V VI

A

26 0 6 14 2 17

B

0 42 0 16 2 17

C

1 0 36 6 0 8

D

21 23 13 31 0 0

E

9 9 10 0 30 12

F

17 27 0 3 17 8



Total assignment method

A - II = 22

B - I = 22

C - V = 22

D - VI = 30

E - IV = 15

F - III = 27

138





UNGKU OMAR POLYTECHNICS

P3724 – MATHEMATICS FOR RETAILING

QUIZ – CHAPTER 8



Question 1



The operations manager of King Manufacturing must assign 3 tasks to 3 machines. Cost data are presented below.



1 2 3

C3

800 1100 1200

C5

500 1600 1300

C8

500 1000 2300



Use the assignment algorithm to solve this problem.









Answer





Step 1



Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 800 (row 1), 500 (row 2) and 500 (row 3).



1 2 3

C3

0 300 400

C5

0 1100 800

C8

0 500 1800





Step 2



Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 0 (column 1), 300 (column 2) and 400 (column 3).



1 2 3

C3

0 0 0

C5

0 800 400

C8

0 200 1400







Step 3



Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



1 2 3

C3

0

0 0

C5

0 800 400

C8

0 200 1400











Step 4

a. Find the smallest number NOT covered by a line, called X. The smallest number is 1.

b. Subtract X from every the uncovered number in the table.

(800 – 200 = 600), (400 - 200 = 200), (200 – 200 = 0) and (1400 – 200 = 1200)

c. Add X to every number lying at the intersection of horizontal and vertical lines.

(0 + 1 = 1)





1 2 3

C3

0 0 0

C5

0 600 200

C8

0 0 1200





Step 5

When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row.

c. Assignments which is NOT in a) or b)





1 2 3

C3

0 0 [0]

C5

[0] 600 200

C8

0 [0] 1200





Total assignment method



C8 - 2 = 1200

C5 - 1 = 500

C3 - 3 = 1200

2700









































































Question 1



The operations manager of King Manufacturing must assign 5 tasks to 5 machines. Cost data are presented below.



I II III IV V

A

20 25 15 10 12

B

18 20 20 16 15

C

18 19 16 18 12

D

20 22 18 15 10

E

18 20 15 16 10



Use the assignment algorithm to solve this problem.



Answer



Step 1



Reduce each row by the SMALLEST number in the row. Subtract these values from each element in the row produces.

The smallest numbers are 10 (row 1), 15 (row 2), 12 (row 3), 10 (row 4) and 10 (row 5)



I II III IV V

A

10 15 5 0 2

B

3 5 5 1 0

C

6 7 4 6 0

D

10 12 8 5 0

E

8 10 5 6 0









Step 2

Reduce each column by the SMALLEST number in the column. Deduct each smallest number from every number in that column.

The smallest numbers are 3 (column 1), 5 (column 2), 4 (column 3), 0 (column 4) and 0 (column 5)



I II III IV V

A

7 10 1 0 2

B

0 0 1 1 0

C

3 2 0 6 0

D

7 7 4 5 0

E

5 5 1 6 0





Step 3

Draw the MINIMUM possible number of lines to cover all zeros. The lines may be horizontal or vertical.



I II III IV V

A

7

10 1 0 2


B

0

0 1 1 0

C

3

2 0 6 0

D

7 7 4 5 0

E

5 5 1 6 0







Step 4

a. Find the smallest number NOT covered by a line, called X. The smallest number is 1.

b. Subtract X from every the uncovered number in the table.

c. Add X to every number lying at the intersection of horizontal and vertical lines.





I II III IV V

A

7 10 1 0 3

B

0 0 1 1 1

C

3 2 0 6 1

D

6 6 3 4 0

E

4 4 0 5 0



Step 5

When the optimal assignment is found, the actual assignments can be made by assigning to:

a. any zero which is unique to BOTH a column and a row.

b. Any zero which is unique to column OR a row. [0]

c. Assignments which is NOT in a) or b)



I II III IV V

A

7

10 1

[0] 3


B

0

[0] 1 1 1

C

3 2 0 6 1

D

6 6 3 4 [0]

E

4 4 [0] 5 0









I II III IV V

A

7 10 3 [0] 5

B

[0] 0 3 2 3

C

1 [0] 0 4 1

D

4 4 3 2 [0]

E

2 2 [0] 3 0



Total assignment method

A - IV = 10

B - I = 18

C - II = 19

D - V = 10

E - III = 15

72