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q30 = [( 2 350 200 ) /(0.02 8.75)] = 894 units.
From the above q30 = 894 units is within the given range. Hence we have to calculate the total
cost of C8940.
C8940 = 200 8.75 + 350 (200 / 984) + 0.02 8.75 (894 / 2) = Rs.1750 + Rs.
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Figure 8.21
Problem 8.61.
The average monthly consumption for an item is 300 units and the normal lead-time is one month.
If the maximum consumption has been up to 370 units per month and maximum lead-time is 1
months, what should
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per year. Storage cost Rs. 1000/- per year for 50,000 units. Calculate the economic order quantity,
Total inventory carrying cost and optimal replacement period.
Solution
Data: = 50,000 units per year.
C3 = Rs. 3/- + Rs. 12/- = Rs.
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Problem 8.17
(a) Compute the EOQ and the total variable cost for the data given below:
Annual demand = = 25 units, Unit price = p = Rs. 2.50, Cost per order = Rs. 4/-, Storage
rate = 1% Interest rate = 12%, Obsolescence rate = 7%.
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Hence total savings = Rs. 360 + Rs. 54 = Rs. 414/Increase in the holding cost : (1000/2) 0.20 0.90 Rs. 6/- = Rs. 480/- As the savings is less
than the increase in the total cost the discount offer of 10% can not be accepted.
Proble
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Operations Research
i.e., a + 1/2 t b + 1/3 t c 2500
.(3)
Now the market demand constraints are: a 500, b 500 and c 375
.(4)
As the ratio of production must be 3 : 2 : 5,
A = 3k. b = 2k and c = 5k which gives the equations:
1/3 a = 1/2/ b and 1/2 b = 1
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The solution is infeasible. As net evaluation row elements are negative the solution remains optimal
and as basic variables are negative, it is infeasible.
Now both rows of S1 and S3 are having 1 in capacity column, any one of them
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Table II: x = 0, y = 11.11, p = 0, q = 0, A1 = 1.32, A2 = 0 , Z = Rs. 11.11 x 10 + 1.32M =
111.1 +1.32 M
Program
Cost in
Rs.
Cj =
requirement
8
x
10
y
0
p
0
q
M
A1
M Replacement
A2
ratio
y
10
11.1
0.33
1
0.11
0
0.11
0
33.6
A2
M
106
374
S.No
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Selective control technique
Basis of classification
Main Use.
1
ABC
Annual consumption value
Controlling raw material
components and work in
process inventory.
2
VED
Criticality of item.
Determining the inventory
levels of spa
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9. The curve used to interpret machine life cycle is
(a) Bath tub curve
(b) Time curve
(c) Product life cycle
(d) Ogive curve.
( )
10. Decreasing failure rate is usually observed in . stage of the machine
(a) Infant
(b) Youth
(c) O
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Operations Research
In figure 8.2 q1, q2 and q3 are the different quantities to be ordered at period to depending on the
demand rate r1, r2 and r3. This system is not recommended for A class items but it is very useful in
controlling the inventory of
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24. Distinguish between deterministic and stochastic models of inventory.
25. What function does inventory perform? State the two basic inventory decisions management
must make as they attempt to accomplish the functions of invento
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cfw_( 2C3 r (C1 + C 2 ) / C1C 2 ) (Attention is to be given to see that the EOQ model is
multiplied by a factor (C1 + C2) / C2
q0 = r t0 =
OR q0 =
(C1 + C 2 ) / C2
t0 = q0 /r =
cfw_2C3 (C1 + C 2 ) / C1 r C2
OR t0 = q0 / =
C0 =
( 2
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Operations Ressearch
4. All players act rationally and intelligently.
5. Each player is interested in maximizing his gains or minimizing his losses. The winner,
i.e. the player on the left side of the matrix always tries to maximize his gains and is k
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There are 1000 bulbs in use, and it costs Rs.2/- to replace an individual bulb, which is burnt out.
If all bulbs were replaced simultaneously it would cost 50 paise per bulb. It is proposed to replace all
bulbs at fixed intervals o
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Operations Research
(i) Find the number of staff members to be recruited every year.
(ii) If there are seven posts of Head of Departments for which length of service is the only
criterion of promotion, what will be average length of service after whic
CHAPTER 9
Waiting Line Theory or Queuing Model
9.1. INTRODUCTION
Before going to waiting line theory or queuing theory, one has to understand two things in clear. They
are service and customer or element. Here customer or element represents a person or ma
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Operations Research
arrivals that have already occurred prior to the beginning of time interval. Figures 9.5 and
9.6 shows the Poisson distribution and negative exponential distribution curves.
Figure 9.5. Poisson Distribution
Figure 9.6. Negative exp
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Operations Research
w: The waiting time for any customer before it is taken into service.
v: Time spent by the customer in the system.
n: Number of customers in the system, that is customers in the waiting line at any time,
including the number of cus
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production schedules and satisfy the market demands of different segments of the market. As the
profits of all alternate solutions are equal, the manager can select the solution, which is more needed by
him. Now let us workout the a
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Operations Research
Table: II. X = 0, y = 40, S = 10, A1 = 0, A2 = 60, p = 0, q = 0 and Z = Rs. 120 60 M.
Problem
variable
Profit
Profit:
Rs.
capacity units
4
x
3
y
0
S
0
p
0
q
M
A1
M Replace
A 2 ment ratio
S
0
10
0.5
0
1
0.5
0
0.5
0
20
y
3
40
0.5
1
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1x + 1y + 0p 1q + 0A1 + 1A2 = 1 and
x, y, p, q, A1 and A2 all 0
3a + 1b + 0S1 + 1S2 = 1 and
a, b, S1, S2 all 0
Problem 3.33: Write the dual of the primal problem given and solve the both and interpret the
results.
Primal Problem:
S
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similar to slack variable in maximization problem). But the artificial surplus variable has to be purchased
by paying a very high price for it. In character it is very much similar to surplus variable p because
one unit of A1 consis
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Operations Research
Figure 2.5 shows the feasible area for all the three machines combined. This is the fact because
a products X and Y are complete when they are processed on machine A, B, and C. The area covered
by all the three lines PQ. RS, and TU
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Random
Number
Service
Time
Arrival
Time
Bearer
One
Service
Begins
Bearer
One
Service
Ends
0.0
Arrival
Number
Random
Number
Inter
Arrival
Time
1
31
3
0
2
48
1.0
46
3
1.0
3
51
1.0
24
2
2.0
4
06
0.5
54
3
2.5
5
22
0.5
63
3
3.0
6
80
2.
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In general, for item i fi (xi) = Max. [av + fi 1 (xi ai wi) is the recursive equation.
In the given problem the recursive equations are:
1. x1v1
2. 5x2 + f 1 (x2 3x2)
3. 7x3 + f 2 (x3 4x3 )
4. 11x4 + f3 (x4 6x4)
Substituting the va