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In Linear Programming, the objective function and also the constraints were linear in decision
variable. Though this linearity is justified in many real life situations, there do arise such problems in
business and industry that th
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Operations Research
Maximize Z = f (x1, x2, x3, . xn), subject to the constraints
G (x1, x2, x3,.xn) c and x1, x2, .xn 0 and c is a constant.
The constraints can be modified to the form h(x1, x2, .xn) 0 by introducing a function h (x1,
x2, .xn = g (x1
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Operations Ressearch
Solution
Decision tree is shown in figure 12.4. The cost associated with each outcome is written on the
decision tree.
EMV of node B = Rs. [0.2 0 + 0.8 18000] = Rs. 14,400/Therefore EMV of node 2 = Rs. 16,900/- lesser of the two v
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The company will select the third strategy, C, which yields highest utility.
Now let us consider the problem of making decision with multiple objectives.
Problem 12.2.
Consider a M/s XYZ company, which is developing its annual pla
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Operations Research
Problem 8.44.
For the following data, determine approximately the economic order quantities, when the total
value of average inventory level of the products is Rs. 1000/Costs.
Product 1.
Product 2.
Product 3.
Holding Cost C1 (%)
20
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Operations Research
9.2. HISTORICAL DEVELOPMENT OF THE THEORY
9.3. QUEUING SYSTEM OR PROCESS
Figure 9.3. Components of queuing system.
9.3.1. Input Process
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Waiting Line Theory or Queuing Model
Figure 9.4. Characteristics of Arrivals or input.
a
Si
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Solution
Strategies of party A are: A1: Send airplanes in two different regions and A2: Send both airplanes
in one region.
Strategies of B are: B1 To place three guns in three different regions.
B2To place one gun in one region, t
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7. To find the values of the game, we can fix the higher and lower limits of value. Take the
highest element in the last column i.e. 11 and lowest element, in the last row, i.e. 5. Divide
these two by 10, i.e. the number of times
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After drawing the graph, the lower bound is marked, and the highest point of the lower bound is
point Q, lies on the lines P1 and P2. Hence B plays the strategies II, and I so that he can minimize his
losses. Now the game is reduc
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Consider the 2 2 game given below:
B
y1
I
y2
II
x1
I
a 11
a 12
x2
II
a 21
a 22
A
Let x1 and x2 be the probability with which A plays his first and second strategies respectively.
Similarly B plays his first and second strategies w
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5. The rule of game refers to the expected outcome per play when both players follow
their best or optimal strategies. A game is known as fair game if its value is zero, and
unfair if its value is nonzero.
6. An Optimal strategy r
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Operations Ressearch
Before concluding this chapter, it is better to introduce the students to further developments or
advanced topics in network techniques.
1. Updating the network: In large project works, as the project progresses, we may come
acros
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Operations Research
Whenever, we solve the primal problem, may be maximization or minimization, we get the solution
for the dual automatically. That is, the solution of the dual can be read from the final table of the primal
and vice versa. Let us try
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Operations Research
Table I. a = 0, b = 0, c = 0, d = 15, e = 20, f = 10 and Z = Rs. 0.
Profit
Rs.
Cj Capacity,
units
1
a
2
b
3
c
1
d
d
1
15
1
2
3
1
0
0
5
e
M
20
2
1
5
0
1
0
4
f
M
10
1
2
1
0
0
1
10
0
0
0
Problem
variable
Net
2 + 3M 4 + 3M 4 + 8M
evalua
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Operations Research
Coils for each alloy are 400 feet long and weigh 5 tons. Set up objective function and restrictions
to set up matrix.
Solution: Let the company produce a units of alloy 1, b units of alloy 2 and c units of alloy 3.
Then the objectiv
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Operations Research
the isoprofit or isocost line coincides with a line, then the problem will have innumerable
number of solutions.
8. The different situation was found when the objective function could be made arbitrarily
large. Of course, no corner
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Operations Research
Step 5: Elements of Net evaluation row are obtained by:
Objective row element at the top of the row key column element profit column
element.
Step 6: Select the highest positive element in net evaluation row or highest opportunity c
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after certain number of iterations, the solution cannot be improved and we have to accept it
as the expected optimal solution.
(iii) Monte-Carlo Method: This method is based on random sampling of variable's values from a
distributio
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Operations Research
Table: I. x = 0, y = 0, S1 = 2, S2 = 0 and Z = Rs. 0
Profit
Rs.
Cj
Capacity units
6
x
2
y
0
S1
0
S2
Replacement
ratio
S1
0
2
2
1
1
0
2/2 = 1
S2
0
4
1
0
0
1
4/1 = 4
Net evaluation
6
2
0
0
Problem
variable
Table: II. x = 1, y = 0, S1
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Operations Research
Maximise Z 3x + 4y S.T.
Maximise Z = 3x + 4y S.T.
5x + 4y 200
3x + 5y 150
5x + 4y = 200
5x + 4y 100
8x + 4y 80
5x + 4y = 100
8x + 4y = 80
And both x and y are 0
And both x and y are 0
3x + 5y = 150
In the graph the line representing
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Operations Research
Since the demand is rectangular between 4000 and 5000, assuming it a continuous variate, the
density function is given by: f (r) = (1 / 1000) Therefore, (Note: f (x) = 1 / (b a) where a x b)
S
(1 / 1000) dr = 7/(1+7) = (7/8) or (1
QUIZ PAPERS
Unit II
Transprotation Model and
Assignment Model
1. Transportation problem is basically a
(a) Maximisation model,
(b) Minimisation model,
(c) Transshipment problem,
(d) Iconic model.
( )
2. The column, which is introduced in the matrix to bal
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Operations Research
38. The property of capital spares is:
(a) They have very low reliability;
(b) These can be purchased in large quantities, as the price is low,
(c) These spares have relatively higher purchase cost than the maintenance spares.
(d)
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Operations Research
As t0 is greater than the lead-time, and the safety stocks 400 units, the re-order level will be =
Safety stock + Normal lead-time consumption = 400 + 12 80 = 1360 units.
Average inventory = B + (q0 / 2) = 400 + 4000 / 2 = 2400 uni
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Operations Research
Problem 8.59.
A newspaper boy buys papers for 30 paise each and sells them for 70 paise each. He cannot
return unsold newspapers. Daily demand has the following distribution:
Number of
Customers.:
23
24
25
26
27
28
29
30
31
32
Prob