Transportation Method

Example 1

A mining company extracts gravel, the basic product it sells, from three mines, L1, L2 and L3. The
weekly production of each mine is 75, 150 and 75 tones of gravel respectively. The gravel has to
be transported to five main consumers, K1, K2, K3, K4 and K5 requiring for their needs 100, 60,
40, 75 and 25 tones of gravel per week respectively. The problem that concerns the company's
management is the minimization of the required cost for the transportation of the product to the
consumers. For this purpose a detailed cost analysis was carried out which gave the results of the
following table

Cost table of gravel transportation Consumers

K1 K2 K3 K4 K5

L1 3 2 3 4 1

L2 4 1 2 4 2

L3 1 0 5 3 2

Answer:

Table 1

K1 K2 K3 K4 K5

L1 3 2 3 4 1 75
75
L2 4 1 2 4 2 150
25 60 40 25
L3 1 0 5 3 2 75
50 25
100 60 40 75 25
Cost:

75 x 3 =225 60 x 1 = 60 25 x 4 = 100 25 x 2 = 50
25 x 4 =100 40 x 2 = 80 50 x 3 = 150 K/min = 765

C.I.I:

L3-K1: 1 - 3 + 4 – 4 = -37 L3-K2: 0 – 1 + 4 – 3 = 0 L3-K3: 5 – 2 + 4 – 3 = 4

L1-K2: 2 – 1 + 4 – 3 = 2 L1-K3: 3 – 2 + 4 – 3 = 2 L1-K4: 4 – 3 + 4 – 4 = 1

L1-K5: 1 – 2 + 3 – 4 + 4 – 3 = -1 L2-K5: 2 – 2 + 3 – 4 = -1

Table 2

K1 K2 K3 K4 K5

L1 3 2 3 4 1 75
50 25
L2 4 1 2 4 2 150
35 40 75
L3 1 0 5 3 2 75
15 60
100 60 40 75 25

Cost:

L1-K1:50 x 3 = 150 L3-K1:15 x 1 = 15 L2-K3:40 x 2 = 80 L1-K5:25 x 1 = 25

L1-K2:35 x 4 = 140 L3-K2:60 x 0 = 0 L2-K4:75 x 4 =300 K/min= 710

C.I.I:

L1-K2: 2 – 0 + 1 – 3 = 0 L1-K3: 3 – 2 + 4 – 3 = 2 L1-K4: 4 – 4 + 4 – 3 = 1

L2-K2: 1 – 0 + 1 – 4 = 2 L3-K3: 5 – 1 + 4 – 2 = 6 L3-K4: 3 – 1 + 4 – 2 = 4

L2-K5: 2 – 1 + 3 - 4 = 0 L3-K5: 2 – 1 + 3 – 1 = 3

Decision:
Transport from customer 1 to produce Project 1

Transport from customer 1 to produce Project 2

Transport from customer 3 to produce Project 1

Transport from customer 3 to produce Project 2

Transport from customer 2 to produce Project 3

Transport from customer 2 to produce Project 4

Transport from customer 1 to produce Project 5

Total Kilometer is 710/min

Example 2

A company transports desk top computers from three warehouses to three retail stores. The number of
computers available (supply) in the three warehouses are 40 50 and 30 while the number of computers
required at the three retailers (demand) are 20, 35 and 65 respectively. There is a cost of transporting
from the warehouses to the retail outlets. The cost transport one unit from a given warehouse to a
given retailer is given, How do we transport the computers such that we incur minimum total cost of the
transportation?

A B C Supply

1 9 6 10 40

2 10 8 9 50

3 5 4 7 30

Demand 20 35 65

Answer: Table 1
 A B C Supply

1 9 6 10 40
20 20
2 10 8 9 50
35 15
3 5 4 7 30
30
Demand 20 35 65

Cost:

20 x 9 = 180 20 x 10 = 200 30 x 7 = 210

35 x 8 = 280 15 x 9 = 135 Total Cost = 1005

C.I.I.:

A2: 10 – 9 + 10 – 9 = 2 B1: 6 – 8 + 9 – 10 = -3

A3: 5 – 9 + 10 – 7 = -1 B3: 4 – 8 + 9 – 7 = -2

Table 2

A B C Supply

1 9 6 10 40
20 20
2 10 8 9 50
15 35
3 5 4 7 30
30
Demand 20 35 65

Cost:
20 x 9 = 180 15 x 8 = 120 30 x 7 = 210

20 x 6 = 120 35 x 9 = 315 Total Cost = 945

C.I.I:

2A: 10 – 8 + 6 – 9 = -1 3B: 4 – 8 + 9 – 7 = -2

3A: 5 – 7 + 9 – 8 + 6 – 9 = -41C: 10 – 9 + 8 – 6 = 3

Table 3

A B C Supply

1 9 6 10 40
20 5 15
2 10 8 9 50
50
3 5 4 7 30
30
Demand 20 35 65

Cost:

20 x 9 = 180 30 x 4 = 120 50 x 9 = 450

5 x 6 = 30 15 x 10 = 150 Total Cost = 930

C.I.I. :

2A: 10 – 9 + 10 - 9 = 2 2B: 8 – 6 + 10 – 9 = 3

3A: 5 – 4 + 6 – 9 = -2 3C: 7 – 10 + 6 – 4 = -1

Table 4
 A B C Supply

1 9 6 10 40
25 15
2 10 8 9 50
50
3 5 4 7 30
20 10
Demand 20 35 65

Cost:

A3: 20 x 5 = 100 B3:10 x 4 = 40 C2:50 x 9 = 450

B1:25 x 6 = 150 C1:15 x 10 = 150 Total Cost = 890

Decision:

Transport from retailers 3 to produce Project A

Transport from retailers 1 to produce Project B

Transport from retailers 3 to produce Project B

Transport from retailers 1 to produce Project C

Transport from retailers 2 to produce Project C

Minimum Cost: 890.00

Assignment Method
 Example 1
Tony Company need to produce 4 iron (J1, J2, J3, and J4) need to be executed by four workers (W1,
W2, W3, and W4), one job per worker. The matrix below shows the cost of assigning a certain worker
to a certain job. The objective is to minimize the total cost of the assignment.

J1 J2 J3 J4

W1 82 83 69 92

W2 77 37 49 92

W3 11 69 5 86

W4 8 9 98 23

Answer:

Table 1
J1 J2 J3 J4

W1 13 13 0 23 (-69)

W2 40 0 12 55 (-37)

W3 6 64 0 81 (-5)

W4 0 1 90 15 (-8)

J1 J2 J3 J4

Table 2 W1 13 13 0 23

W2 40 0 12 55

W3 6 64 0 81

W4 0 1 90 15
  Smallest uncovers number is 6

Table 3

J1 J2 J3 J4

W1 7 8 0 2

W2 40 0 18 40

W3 0 58 0 60

W4 0 1 96 0

Decision:
worker 1 should perform job 3 = 69

worker 2 job 2 = 37

worker 3 job 1 = 11

worker 4 should perform job 4 = 23

Total Cost = 140

Example 2
 An Accounts Officer has 4 suboridnates and 4 tasks. The subordinates differ in efficiency. The
tasks also differ in their intrinsic difficulty. His estimates of the time each would take to perform
each task is given in the matrix below. How should the tasks be allocated one to one man, so that
the total man hours are minimized?

A B C D

1 8 26 17 11

2 13 28 4 26

3 38 19 18 15

4 19 26 24 10

Answer:

Table 1

A B C D

1 0 18 9 3 (-8)

2 9 24 0 22 (-4)

3 24 4 3 0 (-15)

4 9 16 14 0 (-10)

A B C D

1 0 14 9 3

2 9 20 0 22
Table 2
3 24 0 3 0

4 9 12 14 0

(-0) (-4) (-0) (-0)

Table 3

A B C D

1 0 14 9 3

2 9 20 0 22

3 24 0 3 0

4 9 12 14 0

Decision:

Subordinates A should perform task 1 = 8

Subordinates C should perform task 2 = 4

Subordinates B should perform task 3 = 19

Subordinates D should perform task 4 = 10

Total hours = 41 hours

PERT Method
Example 1

 Allan Corporation assigned Hino Corporation as project manager of a power plant project and Hino
Corporation listed;
– All the Activities
– Predecessors
– Optimistic, Pessimistic and Most Likely Activity Durations

Activity Description predessors a b m Expected variance

time
0 Start milestone - 0 0 0 0 0
A Select technical staff 0 12 18 15 15 1
B Site Survey 0 6 12 9 9 1
C Select Equipment’s A 9 15 12 12 1
D Prepare Designs B 6 18 9 10 4
E Bring Utilities to the Site B 18 36 30 29 9
F Interview Applicants A 9 15 12 12 1
G Purchase the Equipment C 36 42 36 37 1
H Construct the Powerplant D 42 54 48 48 4
I Develop an Information A 6 18 12 12 4
System
J Install the Equpiment H,G,E 3 9 6 6 1
K Train the Staff to Run the F,J,I 3 9 6 6 1
System

Formula:
a+ 4 m+ b
Expected time: t= 6
2
b−a
Variance = ( )
6
 Activity ES EF LS LF Slack (LS – ES) Critical Path
A 0 15 3 18 3 NO
B 0 9 0 9 0 YES
C 15 27 61 73 46 NO
D 9 19 9 19 0 YES
E 9 38 38 67 31 NO
F 15 27 61 73 46 NO
G 27 64 30 67 3 NO
H 19 67 19 67 0 YES
I 15 27 61 73 46 NO
J 67 73 67 73 0 YES
K 73 79 73 79 0 YES

Example 2

Maxwell Corporation give to the Casey Corporation the following things need to do

• Determine the expected value and the variance of the completion time for each activity.

• (b) Use the expected times from (a) to find the critical path.

• (c) Assuming that the normal distribution applies, determine the probability that the critical path will
take between 18 and 26 days to complete.

• (d) How much time must be allowed to achieve a 90% probability of timely completion?

• (e)By using modified probability of completion method, what is the probability that all paths will take
before 18 weeks?
 Activity Activity Optimistic (a) Most Likely Pessimistic Expected Variance
Time
(m) (b)

a - 5 6 7 6 4/36

b - 4 5 18 7 196/36

c a 4 15 20 14 4/36

d b,c 3 4 5 4 4/36

e a 5 16 18 14.5 169/36

Formula:
a+ 4 m+ b
Expected time: t= 6
2
b−a
Variance =
6( ) FINISH

A
C D
START

B
 6 20.5
9.5 24
0 6 FINISH

0 6 6 20 20 24
START
6 20 20 24
0 7
13 20

Activity ES EF LS LF Slack Critical

(LS-ES) Path
A 0 6 0 6 0 YES

B 0 7 13 20 13 NO

C 6 20 6 20 0 YES

D 20 24 20 24 0 YES

E 6 20.5 9.5 24 3.5 NO