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Transportation Problems:MOVING TOWARDS OPTIMALITY

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MOVING TOWARDS OPTIMALITY
The rationale behind optimality
Given an initial non-degenerate basic feasible solution, a question may arise as to how we can find a
successively better basic feasible solution. This means that we have to select an entering variable, a leaving
basic variable and identify the corresponding solution. In the transportation model, selecting an entering variable
means selecting a new cell in which to make a non-zero allocation in the transportation matrix. The principle of
selecting the variable which improves the value of the objective function (minimizes the total transportation
cost), at the fastest rate is still valid. So we select an unoccupied cell and make a unit allotment, so that the cost
of transportation will decrease by the greatest amount.
To illustrate, suppose that in the example 4.1 we choose arbitrarily the cell (B, P) which is unoccupied
and make + 1 allotment in this cell. This new allocation automatically violates the supply situation and demand
requirements of the origin B and destination P. So changes are required in other cells to restore feasibility. This
necessitates a reallocation in the cells to restore feasibility. This necessitates a reallocation in the cells (B, P),
(A, P), (A, Q) and (B, Q). The row total and column total will be unaltered. This idea is illustrated in tables 17 to
19.
Table 17
P
Q
R
A
5
7
8
B
4
4
6
C
6
7
7
Unit cost of transportation
Table 18
P
Q
R
A
65
5
70
B
30
30
C
7
43
50
65
42
43
Table 19
P
Q
R
A
65-1
5+1
70
B
1
30-1
30
C
7
43
50
65
42
43
Changed allotment
The unit costs in the cells (B, P), (A, P), (A, Q) and (B, Q) are 4, 5, 7, 4 respectively (refer table 17).
The total change in the total cost will be 4 - 5 + 7 - 4 = + 2. We see that this arbitrary unit allotment to cell (B,
P) increases the transportation cost by rupees 2. Hence an allocation to the cell (B, P) is not recommended.
This procedure is repeated by choosing some other unoccupied cell for allocation. This requires
corresponding changes in the allocation in the occupied cells to restore feasibility and calculation of the
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marginal cost. If, by this procedure, we get a +ve value, it indicates that the total cost is not decreasing but
increasing.
Consider the objective function to minimize,
mn
Z=∑∑
c
x
ij
ij
i =1 j =1
and the restrictions as
n
0 = ai - ∑ xij  for i = 1,2, ... , m
j =1
m
0 = b - ∑ xij  for j = 1,2, ... , n
j
i =1
Multiply each of these equations by a number and add this multiple to the objective function. This resultant can
be used to eliminate the basic variable. Let us represent the multiples by ui (i = 1, 2, ..., m) and vj = (j = 1, 2, ...,
n) respectively for rows and columns. These multipliers are called the simplex multipliers.
Therefore,
⎡
⎤
m
m
n
n
+ ∑ ui ⎢ ai - ∑ xij ⎥
Z = ∑  ∑ cij
x
i =1  ⎢
⎥
ij
i =1 j =1
⎣
j =1  ⎦
⎡
⎤
n
m
+ ∑ v j ⎢b  j - ∑ xij ⎥
⎢
i =1  ⎥
⎣
⎦
j =1
m
m
n
n
(
)
= ∑ ∑ cij -ui -v j x + ∑ ui ai + ∑ v j b  j
ij
i =1 j =1
i =1
i =1
Thus in order to have a zero coefficient (for a basic variable) it is necessary that
crs = ur + vs
for each basic variable xrs (i.e.) for each occupied cell (r, s). There are (m + n - 1) occupied cells and therefore
(m + n - 1) equations. Since we have (m + n) unknown (ui's and vj's) one of these variables can be assigned a
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value arbitrarily and rest of them can then be solved algebraically. This leads to the solution of a set of (m +
n - 1) equations simultaneously.
Having determined the values of ui and vj, the entering basic variable (if any) can be identified readily
by calculating cij - (ui + vj) for each of the unoccupied cells. If cij - (ui + vj) > 0 in every case, then the solution
must be optimal since no non-basic variable can decrease Z. On the other hand if cij - (ui + vj) yields a negative
result, the cell with the largest  negative (smallest) value of cij - (ui + vj) is selected to be the cell for new
allocation since it means that this non-basic variable decreases Z at the fastest rate. This is the basis for
conducting the optimality test, which is explained in the next section.
Optimality test:
For conducting the optimality test to any feasible solution of a (mxn) transportation problem, the
following two conditions must be satisfied.
1.
It consists of exactly (m + n - 1) individual cells being allocated.
2.
These allocations are in independent positions.
The first condition is normally satisfied in many problems. If the first condition is not satisfied then it results in
a state known as degeneracy, in the transportation mode. The discussion of degeneracy and its resolution is
explained in a later section.
A set of allocations comprising a feasible solution is said to be in 'independent positions' if it is
impossible to increase or decrease any individual allocation without either changing the positions of allocations
or violating the row or column restrictions. In other words, a simple criterion for independence is that, it is
impossible to travel from any allocation back to itself, by a series of alternating horizontal and vertical jumps
from one occupied cell to another, without a direct reversal of route.
For example, if the occupied cells form a closed loop as in table 20 they are not in independent
positions
Table 20
*
*
*
*
Applying the procedure outlined above to the initial basic feasible solution, (found by any method) we have the
following steps in conducting an optimality test.
STEP 1:Write the matrix of allotment found by NWC rule or by Vogel's Approximation Method. Check for
conditions for optimality (i.e.) there must be (m + n - 1) alloted cells and all of them must be in independent
positions. If these two conditions are satisfied, proceed to step 2.
STEP 2:Write the matrix of the cost of the alloted cells.
STEP 3:Determine a set of (m + n) numbers,
ui, i = 1, 2, ..., m
vj, j = 1, 2, ..., n
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called simplex multipliers such that for each occupied cell (r, s).
crs = ur + vs
where crs is the individual cost of an allocation from origin r to destinations s.
This can be done by arbitrarily assigning, any one of the ui's or vj's, any value (say 0) and satisfying the above
equation for all the occupied cells.
STEP 4:Write the matrix of cost elements for all unoccupied cells.
STEP 5:Find (ur + vs) for each unoccupied cell and form the matrix of (ui + vj) for each of the unoccupied cell.
STEP 6:Calculate cij - (ui + vj) for each of the unoccupied cells. This is the matrix of cell evaluation or square
evaluation. If the value of cell evaluation is positive, it indicates that the solution already found is optimal. If it
is negative there is a scope for relocation of items so that the total cost of transportation can be made minimum.
If there are many negative entries in the matrix of cell evaluation, select the cell yielding the most negative entry
as the entering variable. This requires reallotment of the allocation matrix. By including this cell for occupation,
the allocations of the other occupied cells require a revision to fulfill the row and column restrictions. If the
value of cell evaluation is zero, then this indicates that there exists another solution to the transportation problem
without reducing the total cost.
Consider the example whose cost matrix is as in table 21
Table 21
Origin
Destination
Supply
P
Q
R
A
5
7
8
70
B
4
4
6
30
Cost Matrix
C
6
7
7
50
Demand
65
42
43
We have the allocation as per NWC rule as in table 22
Table 22
Origin
Destination
Supply
P
Q
R
A
65
5
70
B
30
30
Allotment Matrix
C
7
43
50
Next we write the matrix of the cost of alloted cells and using explained below and shown in table 23
Table 23
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ui
Origin
Destination
P
Q
R
A
5
7
-
0
u1
B
-
4
-
-3
Finding ui, vj using the
u2
C
0
cost of the occupied cells
-
7
7
u3
vj
v1
v2
v3
5
7
7
To determine ui's and vj's we arbitrarily assign one of the multipliers say u3 the value 0.
Then it follows that v2 = 7 because we have the equation
u3 + v2 = 7
Since
u3 + v3 = 7
we have
v3 = 7
Similarly using the cost elements of the occupied cells and with the equation crs = ur + vs, we can find
the values of ui's and vj's.
The unoccupied (un-allotted) cells are indicated by placing dots in the cells.
Next we write the matrix of the cost elements cij of the unoccupied cells as in table 24.
Table 24
Origin
Destination
P
Q
R
A
8
B
4
6
cost of
C
6
unoccupied cells
Next we write the matrix of ui + vj for unoccupied cells as in
Table 25
Origin
Destination
P
Q
R
A
7
B
2
4
(ui + vj) for
C
5
unoccupied cells
Table 26
Origin
Destination
P
Q
R
A
1
B
2
2
Cell Evaluation
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C
1
Next we write the matrix of ui + vj for unoccupied cells as in table 25. Next we find cij - (ui + vj) for all
the unoccupied cells. This matrix is obtained by subtracting the elements in the (ui + vj) matrix from the
corresponding elements of cij in the matrix of the cost of unoccupied cells. This is the matrix of cell evaluation
or square evaluation as shown in table 26.
We see that all the elements of the matrix of cell evaluation are positive (> 0). This indicates that our
initial basic solution is optimal and the cost of transportation is Rs. 830. If the elements of matrix of cell
evaluation are negative, it indicates that further reduction in cost is possible.
Example
Mam enterprise has three factories located at A, B and C and supplies to three warehouses located at D,
E and F. Monthly factory capacities are 10, 80 and 15 units respectively. Monthly warehouse requirements are
45, 20 and 40 units respectively. Unit transportation costs in rupees are given below.
Warehouse
Factory
D
E
F
A
5
1
7
B
6
4
6
C
3
2
5
Starting with NWC rule, find the optimal allotment.
Solution:
Using NWC rule we obtain the initial basic feasible solution as in table 27.
Table 27
Warehouse
Factory
D
E
F
Supply
A
10
10
0
B
35
20
25
80
45
25
0
C
15
15
0
45
20
40
35
15
0
0
0
There are five allotments. As per the optimality test conditions there must be (3 + 3 - 1) cells occupied
and we see that this condition is satisfied. Secondly all the cells should be in independent positions.
To test the cells whether they are independent or not, we take an occupied cell and alter the allotment in
that cell by one unit. This requires the change in other occupied cells without violating the constraints. Or if we
start from one occupied cell and we do come back to the starting cell, then the cells are in dependent positions.
In the above problem all the cells are independent.
Having satisfied the two conditions we proceed to the optimality test as described above and are carried
in tables 28 to 33
Table 28
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Factory
Warehouse
D
E
F
A
5
1
7
B
6
4
6
C
3
2
5
Cost Matrix
Table 29
Factory
Warehouse
D
E
F
A
10
B
35
20
25
C
15
Allotment Matrix
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Table 30
Factory
Warehouse
ui
D
E
F
A
5
-
-
0
u1
B
6
4
6
1
u2
C
-
-
5
0
u3
vj
v1
v2
v3
5
3
5
Evaluation ui and vj
Table 31
Factory
Warehouse
D
E
F
A
1
7
B
C
3
2
Cost of un-allotted cells
Table 32
Factory
Warehouse
D
E
F
A
3
5
B
C
5
3
(u1 + vj) for un-allotted cells
Table 33
Factory
Warehouse
D
E
F
A
-2
2
B
C
-2
-1
Cell Evaluation
In the matrix of cell evaluation, there are three cells (A, E), (C, D), and (C, E) having negative entries
indicating the scope of minimizing the cost by bringing any of these cells for occupation. But the most negative
element is found in two cells (A, E) and (C, D). Since there is a tie, the same is broken arbitrarily. We select the
cell (C, D) for reallotment (as marked in table 34). With the initial feasible solution,
Table 34
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Warehouse
Factory
D
E
F
A
10
B
35
20
25
C
15
we have to allocate as much as possible in the empty cell (with the mark) with the most negative evaluation to
arrive at the minimum cost as quickly as possible. The reallocation is performed by identifying the loop joining
the empty cell to the occupied cells by horizontal and vertical jumps as indicated in table 35.
Table 35
Factory
Warehouse
D
E
F
A
10
B
35
20
25
C
15
It can be seen that we can allocate 15 units to cell (C, D) and still satisfy the row and column total and
thus no allocation becomes negative. The effect of allotting 15 units to empty cell leads to the matrix of
reallotment as shown in table 36.
Table 36
Factory
Warehouse
D
E
F
A
10
B
20
20
40
C
15
Reallotment among cells of loop
We conduct optimality test to see whether this solution is optimal. The computations are carried in
tables 37 to 42. From the table 42 we see that the unoccupied cell (A, E) has a negative evaluation and hence
there exists another feasible solution as shown in table 4
We have to conduct the optimality test to find whether the third feasible solution is optimal. By
carrying out the iterations (not shown), we observe that the third feasible solution (Table 43) is optimal and the
corresponding cost is
10 × 1 + 30 × 6 + 10 × 4 + 40 × 6 + 15 × 3 = Rs. 515.
Table 37
Factory
Warehouse
D
E
F
A
5
1
7
B
6
4
6
C
3
2
5
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Cost Matrix
Table 38
Factory
Warehouse
D
E
F
A
10
B
20
20
40
C
15
Allotment Matrix
Table 39
Factory
Warehouse
ui
D
E
F
A
5
-1
B
6
4
6
0
C
3
-3
6
4
6
vj
Cost of alloted cells & Evaluation of ui and vj
Table 40
Factory
Warehouse
D
E
F
A
1
7
B
C
2
5
Cost of unalloted cells
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Table 41
Factory
Warehouse
D
E
F
A
3
5
B
C
1
3
(ui + vj) for unalloted cells
Table 42
Factory
Warehouse
D
E
F
A
-2
2
B
C
1
2
Cell Evaluation
Table 43
Factory
Warehouse
D
E
F
A
10
B
30
10
40
C
15
Reallotment
Vogel's Approximation Method for maximization problem
The VAM method can be used in a maximization problem making a slight modification. The required
modification is to multiply all the elements in the (profit) matrix by -1, with the concept that minimizing the
negative of a function maximizes the original function.
Consider the example given below.
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Example:
Solve the following transportation problem to maximize profit and give criteria for
optimality.
Profits (Rs) / Unit
Destination
Origin
1
2
3
4
Supply
A
40
25
22
33
100
B
44
35
30
30
30
C
38
38
28
30
70
Demand
40
20
60
30
We first convert the elements of profit matrix by multiplying by -1 and then we adopt the
Solution:
method of minimization. This is done in table 44
Table 44
Destination
Origin
1
2
3
4
Supply
A
-40
-25
-22
-33
100
B
-44
-35
-30
-30
30
C
-38
-38
-28
-30
70
Demand
40
20
60
30
In the above problem total supply is not equal to total demand. (supply demand) and hence it is an
unbalanced transportation model. To balance, we introduce a dummy column 5. to find the initial feasible
solution by VAM. We put the elements in the dummy column 90 as in table 45
Table 45
Destination
Origin
1
2
3
4
Supply (Penalty
A
-40
-25
-22
-33
100
(7)
B
-44
-35
-30
-30
30
(9)
C
-38
-38
-28
-30
70
(0)
Demand
40
20
60
30
Penalty
(4)
(3)
(2)
(3)
Allot in the row B having maximum penalty and in the cell (B, I) with least cost.
The origin B is deleted for further analysis as it is exhausted. We have the reduced matrix given in table
46.
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Table 46
Destination
Origin
1
2
3
4
5
Supply
Penalty
A
-40
-25
100
(7)
20
-28
-30
0
70
(0)
C
-38
-38
Demand
10
20
60
30
50
Penalty
(2)
(13)
(6)
(3)
(0)
Allot in the column (2), which has a maximum penalty (13) and in the cell (C, 2) with least cost. Column 2 is
deleted as the supply is satisfied and we have the reduced matrix as in table 46
Table 47
Origin
Destination
Supply
Penalty
1
3
4
5
A
-40
-22
-33
0
C
10
-28
-30
0
100
(7)
-38
Demand
10
60
30
0
50
(8)
Penalty
(2)
(6)
(3)
(0)
Allot (C, 1) with 10 units and delete column 1. Then we have table 48
Table 48
Origin
Destination
Supply
Penalty
3
4
5
30
0
100
(11)
A
-22
-33
-28
-30
0
40
(2)
C
Demand
60
30
50
Penalty
(6)
(3)
(0)
Allot 30 to cell (A, 4) and delete column 4. Then we have table 49
Table 49
Origin
Destination
Supply
Penalty
3
5
A
-22
0
70
(22)
40
-28
C
0
40
(28)
Demand
60
50
Penalty
(6)
(0)
Deleting row C we have the table 50 as we allot 40 to cell (C, 3).
Table 50
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Destination
Supply
Origin
3
5
50
20
0
-22
70
A
Demand
20
50
Thus we have the initial feasible solution given in table 51
Table 51
Origin
Destination
1
2
3
4
5
A
20
30
50
B
C
10
20
40
Allotment Matrix
The optimality test is conducted table 52 to 55
Table 52
Origin
Destination
1
2
3
4
5
ui
A
-
-
-22
-30
0
6
B
-44
-
-
-
-
-6
C
-38
-38
-28
-
-
0
-38
-38
-28
-36
-6
vj
Evaluation of ui's and vj's
Table 53
Origin
Destination
1
2
3
4
5
A
-40
-25
B
-35
-30
-30
0
C
-30
0
Cost of un-allotted cells
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Table 54
Origin
Destination
1
2
3
4
5
A
-32
-32
B
-44
-34
-42
-12
C
-36
-6
(ui + vj) for unalloted cells
From table 55 see that the cell (A, 1) has a negative value indicating that there is scope for optimality.
This is a potential box to be alloted. Therefore we go back to the allotment matrix in table 51. Allotting to cell
(A, 1) we have to remove from cell (A, 3), add to cell (C, 3) and remove from cell (C, 1) so that the loop is
complete indicating that reallotment is possible. Now we have to decide the amount to be alloted and this is
decided with the idea that in the process of reallotment no cell can have negative allotment. Hence we have the
new allotment as per the following table 56
Table 55
Origin
Destination
1
2
3
4
5
A
-8
-7
B
9
4
12
12
C
6
6
Cell Evaluation
Table 56
Origin
Destination
1
2
3
4
5
A
10
10
30
50
B
30
C
20
50
Reallotment
Table 57
Origin
Destination
ui
1
2
3
4
5
A
-40
-
-22
-33
0
0
B
-44
-
-
-
-
-4
C
-38
-28
-
-
-6
-40
-32
-22
-33
0
vj
Evaluation of ui and vj
Table 58
Origin
Destination
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1
2
3
4
5
A
-25
B
-35
-30
-30
0
C
-38
-30
0
Cost of unalloted cells
Table 59
Origin
Destination
1
2
3
4
5
A
-32
B
-36
-26
-37
-4
C
-46
-39
-6
(ui + vj) for unalloted cells
Table 60
Origin
Destination
1
2
3
4
5
A
7
B
1
-4
7
4
C
8
9
6
Cell Evaluation
The matrix of cell evaluation of cell evaluation in table 60 shows a negative entry in the cell (B, 3)
indicating that this is a potential cell for allotment so that we get a better result. Hence we make a reallotment as
shown by dotted lines in table 61.
Table 61
Origin
Destination
1
2
3
4
5
A
20
30
50
B
20
10
C
20
50
Reallotment
We can conduct once again the optimality test. One more iteration would reveal that the allotments in
the table 61 are optimal. This iteration is left to the reader. Hence the profit
= (20 × 40) + (30 × 33) + (20 × 44) + (50 × 0) + (10 × 30) + (20 × 38) + (50 × 20) = Rs.5130.
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REVIEW QUESTIONS
1.
Solve the following transportation problem for minimization (Unit costs are given in rupees).
Destination
Origin
D1
D2
D3
D4
D5
Supply
01
4
3
1
2
6
40
02
5
2
3
4
5
30
03
3
5
6
3
2
20
04
2
4
4
5
3
10
Demand
30
30
15
20
5
2.
A manufacturer has distribution centres at X, Y and Z. These centres have availability of 40, 20 and 40
units of the product, His retail outlets at A, B, C, D and E require 25, 10, 20, 30 and 15 units
respectively. The transport cost per unit between each centre and each outlet is given below.
Retail Outlets
Distribution
A
B
C
D
E
Centre
X
55
30
40
50
50
Y
35
30
100
45
60
Z
40
60
95
35
30
Determine the optimal distribution to minimize the cost of transportation.
3.
(a) A company has three warehouses, W1, W2 and W3 and four retail stores S1, S2, S3 and S4. The
availability of a given commodity at these warehouses are as follows: W1 = 14, W2 = 16, W3 = 5. The
demands at these stores are: S1 = 6, S2 = 10, S3 = 15, S4 = 4. The costs of transporting one unit of the
commodity from warehouse i to store j, in rupees, are given in the following table.
S1
S2
S3
S4
6
4
1
5
W1
8
9
2
7
W2
4
3
6
2
W3
Determine the optimum transportation schedule, which minimizes the total transportation cost.
4.
A firm manufacturing a single product has three plants at locations X, Y and Z. The three plants have
produced 60, 35 and 40 units respectively during the week. The firm has made commitments to sell 22,
45, 20, 18 and 30 units of the product to customers A, B, C, D and E respectively. The net per unit cost
of transporting from the three plants to the five customers is given in the table below.
192
Operations Research (MTH601)
193
Customers
Plant Locations
A
B
C
D
E
X
4
1
3
4
4
Y
2
3
2
2
3
Z
3
5
2
4
4
Use Vogel's approximation method to determine the cost of shifting the product from plant locations to
the customers. Does your solution provide a least cost transportation schedule?
5.
A steel company has three furnaces and five rolling mills. Transportation costs (Rupees per quintal) for
sending steel from furnaces to rolling mills are given in the following table.
Rolling Mills
Furnaces
M1
M2
M3
M4
M5
Availability
A
4
2
3
2
6
8
B
5
4
5
2
1
12
C
6
5
4
7
3
14
Required
4
4
10
8
8
How should they meet the requirement?
6.
A company has three factories at A, B and C, which supply warehouses at D, E, F and G respectively.
Monthly product capacities of these factories are 250, 300 and 400 units respectively. The current
warehouse requirements are 200, 275 and 300 units respectively. Unit transportation costs from
factories to warehouses are given below.
To
D
E
F
G
A
11
13
17
14
B
16
18
14
10
C
21
24
13
10
Determine the optimum distribution to minimize cost.
7.
Solve the following transportation problem for minimization, the cost matrix is given as:
D1
D2
D3
D4
D5
ai
01
12
4
9
5
9
55
02
8
1
6
6
7
45
03
1
12
4
7
7
30
04
10
15
6
9
1
50
40
20
50
30
40
b1
Find all the alternate optimal solutions.
8.
Solve the following transportation problem for minimization.
To
I
II
III
IV
V
ai
20
19
14
23
16
40
A
193
Operations Research (MTH601)
194
15
20
13
19
16
60
B
18
15
18
20
100
70
C
30
40
50
40
60
bj
9.
Pir Iron and Steel Company (PISCO) has three open hearth furnaces and five rolling mills.
Transportation costs (Rs. per quintal) for shipping steel from furnaces to rolling mills are shown in the
following table.
Mills
Furnaces
M1
M2
M3
M4
M5
Capacities
4
2
3
2
6
8
F1
5
4
5
2
1
12
F2
6
5
4
7
7
14
F3
Required
4
4
10
8
8
What is an optimal shipping schedule for PISCO?
10.
A company having plants at P, Q and R supplies to the warehouses at W, X, Y and Z. Monthly plant
capacities are 75, 95, and 120 respectively. Monthly warehouse requirements are 55, 65, 75 and 100
respectively. Unit shipping costs are as follows.
W
X
Y
Z
P
18
21
15
12
Q
16
22
26
15
R
16
15
16
17
Determine the optimum distribution for this company to minimize the shipping costs using VAM for
initial solution.
11.
The Products of three plants F1, F2 and F3 are to be transported to 5 warehouses W1, W2, W3, W4 and
W5. The capacities of plants, the requirements of ware houses and the cost of transportation are
indicated in the following table.
W1
W2
W3
W4
W5
Plant
Capacity
74
56
54
62
68
400
F1
58
64
62
58
54
500
F2
66
70
52
60
60
600
F3
Demand
200
280
240
360
320
Find the optimum transportation schedule and the associated cost of transportation.
12.
Goods are to be transported from three warehouses to six customers. The availabilities at the
warehouses are 100, 120 and 150 units respectively. The demands of customers are 50, 40, 50, 90, 60
and 80 respectively. The unit transportation costs are given in the following table.
C1
C2
C3
C4
C5
C6
15
25
18
35
40
26
W1
22
36
40
60
50
38
W2
26
38
45
52
45
48
W3
194
Table of Contents:
  1. Introduction:OR APPROACH TO PROBLEM SOLVING, Observation
  2. Introduction:Model Solution, Implementation of Results
  3. Introduction:USES OF OPERATIONS RESEARCH, Marketing, Personnel
  4. PERT / CPM:CONCEPT OF NETWORK, RULES FOR CONSTRUCTION OF NETWORK
  5. PERT / CPM:DUMMY ACTIVITIES, TO FIND THE CRITICAL PATH
  6. PERT / CPM:ALGORITHM FOR CRITICAL PATH, Free Slack
  7. PERT / CPM:Expected length of a critical path, Expected time and Critical path
  8. PERT / CPM:Expected time and Critical path
  9. PERT / CPM:RESOURCE SCHEDULING IN NETWORK
  10. PERT / CPM:Exercises
  11. Inventory Control:INVENTORY COSTS, INVENTORY MODELS (E.O.Q. MODELS)
  12. Inventory Control:Purchasing model with shortages
  13. Inventory Control:Manufacturing model with no shortages
  14. Inventory Control:Manufacturing model with shortages
  15. Inventory Control:ORDER QUANTITY WITH PRICE-BREAK
  16. Inventory Control:SOME DEFINITIONS, Computation of Safety Stock
  17. Linear Programming:Formulation of the Linear Programming Problem
  18. Linear Programming:Formulation of the Linear Programming Problem, Decision Variables
  19. Linear Programming:Model Constraints, Ingredients Mixing
  20. Linear Programming:VITAMIN CONTRIBUTION, Decision Variables
  21. Linear Programming:LINEAR PROGRAMMING PROBLEM
  22. Linear Programming:LIMITATIONS OF LINEAR PROGRAMMING
  23. Linear Programming:SOLUTION TO LINEAR PROGRAMMING PROBLEMS
  24. Linear Programming:SIMPLEX METHOD, Simplex Procedure
  25. Linear Programming:PRESENTATION IN TABULAR FORM - (SIMPLEX TABLE)
  26. Linear Programming:ARTIFICIAL VARIABLE TECHNIQUE
  27. Linear Programming:The Two Phase Method, First Iteration
  28. Linear Programming:VARIANTS OF THE SIMPLEX METHOD
  29. Linear Programming:Tie for the Leaving Basic Variable (Degeneracy)
  30. Linear Programming:Multiple or Alternative optimal Solutions
  31. Transportation Problems:TRANSPORTATION MODEL, Distribution centers
  32. Transportation Problems:FINDING AN INITIAL BASIC FEASIBLE SOLUTION
  33. Transportation Problems:MOVING TOWARDS OPTIMALITY
  34. Transportation Problems:DEGENERACY, Destination
  35. Transportation Problems:REVIEW QUESTIONS
  36. Assignment Problems:MATHEMATICAL FORMULATION OF THE PROBLEM
  37. Assignment Problems:SOLUTION OF AN ASSIGNMENT PROBLEM
  38. Queuing Theory:DEFINITION OF TERMS IN QUEUEING MODEL
  39. Queuing Theory:SINGLE-CHANNEL INFINITE-POPULATION MODEL
  40. Replacement Models:REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION
  41. Replacement Models:ITEMS DETERIORATING WITH TIME VALUE OF MONEY
  42. Dynamic Programming:FEATURES CHARECTERIZING DYNAMIC PROGRAMMING PROBLEMS
  43. Dynamic Programming:Analysis of the Result, One Stage Problem
  44. Miscellaneous:SEQUENCING, PROCESSING n JOBS THROUGH TWO MACHINES
  45. Miscellaneous:METHODS OF INTEGER PROGRAMMING SOLUTION