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P a g e | 1
Q: A firm has normally distributed forecast of usage with MAD=50 units. It desires a service level, which
limits the stock, out to one order cycle per year. Determine Standard Deviation (SD), if the order
quantity is normally a week`s supply.
Solution:
(50)2
3.14(50)
2
1.57 (50)
(1.25) (50)
62.64
SD
SD
SD
SD
SD
Q: A company centre has got four experts programmers. The centre needs four application
programmers to be developed. The head of the computer centre, after studying carefully the
programmer’s to be developed, estimate the computer time in minutes required by the respective
experts to develop the application programmers as follows.
Programmers A B C D
1 120 100 80 90
2 80 90 110 70
3 110 140 120 100
4 90 90 80 90
Solution:
Programmers A B C D
1 30 10 0 10
2 0 10 40 0
3 0 30 20 0
4 0 0 0 10
Programmers A B C D
1 30 10 0 10
2 0 10 40 0
3 0x 30 20 0x
4 0x 0 0x 10
Q: the cost of a new machine is Rs. 5000. The maintenance cost during the nth year is given by Mn =
Rs.500 (n-1), where n=1, 2, 3… If the discount rate per year is 0.05, determine discount factor (vn-1) for
each year.
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Solution:
M = 5000
Mn = 500 (n-1)
V = 0.05
(vn-1) = ?
Nth year Mn (vn-1)
1 0 1.00
2 500 0.05
3 1000 0.0025
4 1500 0.000125
Or
Nth year (vn-1)
1 1.00
2 0.05
3 0.0025
4 0.000125
Q: Determine whether the following Transportation model has initial feasible solution?
D1 D2 D3 D4 Supple
Q1 x11 x12 x13 x14 6
Q2 X21 X22 X23 X24 8
Q3 X31 X32 X33 X34 10
Demand 4 6 8 6
Solution:
The transportation problems can be represented mathematically as a linear programming model. The
Objective function in this problem is to minimize the total transportation cost given by
Z = c11x11+ c12x12+ ... + cmnxmn
Subject to the restrictions:
Row restrictions:
x11 + x12 + x13 + x14 = 6
x21 + x22 + x23 + x24 = 8
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x31 + x32 + x33 + x34 = 10
Column restrictions:
x11 + x21 + x31 + x41 = 4
x12 + x22 + x32 + x42 = 6
x13 + x23 + x33 + x43 = 8
x14 + x24 + x34 + x44 = 6
And
x11 + x12 + x13 + x14 ≥ 0
It should be noted that the model has feasible solutions only if
a1 + a2 + a3 + a4 = 4+6+8+6
Or
∑ ai 𝑚𝑖=0 = ∑ bj
𝑛
𝑗=1
Q:
Salesman Region
1 2 3 4 5 6
A 5 0 25 0 15 0
B 0 50 0 25 30 45
C 35 15 55 0 40 75
D 10 30 5 30 15 0
E 25 0 30 55 20 30
F 35 35 20 10 0 55
Do next step by applying Hungarian method?
Solution:
Salesman Region
1 2 3 4 5 6
A 5 0x 25 0x 15 0x
B 0 50 0x 25 30 45
C 35 15 55 0 40 75
D 10 30 5 30 15 0
E 25 0 30 55 20 30
F 35 35 20 10 0 55
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Here we have only three assignments. But we must have four assignments. With this maximal
assignment we have to draw the minimum number of lines to cover all the zeros.
Q: An oil company has 8 unit of money available for exploration of three sites. If oil is present at a site,
the probability of finding it depends upon the amount allocated for exploiting the site as given below.
0 1 2 3 4 5 6 7 8
Site I 0.0 0.0 0.1 0.2 0.3 0.5 0.7 0.9 1.0
Site II 0.0 0.1 0.2 0.3 0.4 0.6 0.7 0.8 1.0
Site III 0.0 0.1 0.1 0.2 0.3 0.5 0.8 0.9 1.0
The probability that the oil exits at sites I, II and III is 0.4, 0.3 and 0.2 respectively; we have to find the
optimal allocating of money. Stage I is given below, only do stage it.
Stage I
Max. Z=0.4P1(x1) + 0.3P2(x2)
Subject to: x1+x1+x1≤8
No. of boxes x1
0 1 2 3 4 5 6 7 8
f1(x1) 0 0 4 8 12 20 28 36 40
Q. a person wants to decide the constituents of a diet which will fulfill his daily requirements of protein,
fats and carbohydrates at the minimum cost. The choice is to be made from four different types of
foods. The yields per unit of these foods are given in the table below:
Food type
Yield per unit 3 Cost per unit (Rs.) Proteins Fats carbohydrates
1
2
3
4
3
4
8
6
2
2
7
5
6
4
7
4
45
40
85
65
Min
Requirement
800 200 700
Solution:
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Let x1, x2, x3 and x4 denote the number of units of food of type 1, 2, 3 & 4 respectively. Objective is to
minimize the cost i.e.
Minimize Z = 45x1+40x2+85x3+65x4
Constraints are on the fulfillment of the daily requirements of various constituents i.e.
Proteins - 3x1 + 4x2 + 8x3 + 6x4 ≥ 800
Fats - 2x1 + 2x2 + 7x3 + 5x4 ≥ 200,
Carbohydrates - 6x1 + 4x2 + 7x3 + 4x4 ≥ 700.
Where x1, x2, x3, x4 each ≥ 0
Fall 2012
Question No: 41 ( Marks: 2 )
A branch of Punjab National Bank has only one typist. Since the typing work varies in length
(number of pages to be typed), the typing rate is randomly distributed approximating a Poisson
distribution with mean service rate of 8 letters per hour. The letters arrive at a rate of 5 per hour
during the entire 8 – hour worki9ng day. If the typewriter is valued at Rs. 1.50 per hour,
Determine Average system time.
Answer:
Ws= 1/µ-λ = 1/8-5 =1/3hr=1/3*60=20 min
Question No: 42 ( Marks: 2 )
An oil company has 8 unit of money available for exploration of three sites. If oil is present at a
site, the probability of finding it depends upon the amount allocated for exploiting the site as
given below:
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The probability that the oil exits at sites I, II and III is 0.4, 0.3 and 0.2 respectively; we have to
find the optimal allocating of money.
Do stage I only.
Answer
Not Attempted
Question No: 43 ( Marks: 2 )
Write the relationship between the activities.
0 1 2 3 4 5 6 7 8
Site I 0.0 0.0 0.1 0.2 0.3 0.5 0.7 0.9 1.0
Site II 0.0 0.1 0.2 0.3 0.4 0.6 0.7 0.8 1.0
Site III 0.0 0.1 0.1 0.2 0.3 0.5 0.8 0.9 1.0
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Answer
X approches to Y
X also approches to Z
Y approches to Z
Whether A and b might have the values between the centre points
Question No: 44 ( Marks: 2 )
For the mathematical form of a Transportation problem (T.P)
1 1
min (1)j ni m
ij ij
i j
z c x
subject to
1
1
(2) , 1, 2, , (Sources)
(3) , 1, 2, , (Destinations)
j n
ij i
j
i m
ij j
i
x a i m
x b i n
Describe the practical significance of all the above equations(1), (2) and (3).
Answer :
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The above is a mathematical formulation of a transportation problem and we can adopt the linear
programming technique with equality constraints. Here the algebraic procedure of the simple method
may not be the best method to solve the problem and hence more efficient and simpler streamlined
procedures have been developed to solve transportation problems.
Question No: 45 ( Marks: 3 )
The milk plant at a city distributes its products by trucks, located at the loading dock. It has its
own fleet of trucks plus trucks of a private transport company. This transport company has
complained that sometimes its trucks have to wait in line and thus the company loses money
paid for a truck and driver that is only waiting. The company has asked the milk plant
management either to go in for a second loading dock or discount prices equivalent to the
waiting time, the following data available
3
4
Averagearrival rate per hour
Average service rate per hour
The transport company has provided 40%of the total number of trucks. Assuming that these
rates are random according to Poisson distribution, determine
a) The probability that a truck has to wait.
b) The waiting time of a truck that waits.
Answer
The probability that a truck has to wait.
The waiting time of a truck that waits.
4 44
4 3 1
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round about 40 minutes of each truck.
Question No: 46 ( Marks: 3 )
A company has a machine whose cost is Rs. 30,000. Its maintenance cost and resale value at
the end of different years are as given below:
Years. 1 2 3 4 5 6
Maintenance
Cost.
4500 4700 5000 5500 6500 7500
Resale Value 27000 25300 24000 21000 18000 13000
Determine capital cost for each year.
Answer
Question No: 47 ( Marks: 3 )
A firm produced three products. These products are processed on three different machines. The time
required to manufacturer one unit of each of the three products and the daily capacities of the three
machines are given in the table:
Machines
Time per unit (minutes) Machine Capacity
(minutes / day) Product 1 Product 2 Product 3
M1 2 3 2 440
M2 4 --- 3 470
M3 2 3 --- 430
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It is required to determine the daily number of units to be manufactured for each product.
The profit per unit for product 1, 2 and 3 is Rs. 4, Rs. 3 and Rs. 6 respectively. It is assumed that all the
amounts produced are consumed in the market. Write the constraints of above Linear Programming
Problem.
Answer
Step 1
Find the key decision to be made. The key decision is to decide the extent of product 1,2&3 to
be produced as this can vary.
Step 2
Assume symbols for the extent of production. Let the extent of Product 1,2&3 be X1, X2 & X3.
Step 3
Express the feasible alternatives mathematically in terms of variables. Feasible alternatives are
those which are physically, economically and financially possible. In this example, feasible
alternatives are sets of values of x1, x2 & x3, where x1,x2 &x3 ≥ 0 since negative production
has no meaning and is not feasible.
Step 4
Mention the object quantitatively and express it as a linear function of variables. IN the present
example, objective is to maximize the profit.
i.e. Maximize Z = 4x1+3x2+6x3
Step 5
Express the constraints as linear equations/inequalities in terms of variables.
Here, constraints are o the machine capacities and can be mathematically expressed as
2x1 + 3x2 + 2x3 ≤ 440,
4x1 + 0x2 + 3x3 ≤ 470,
2x1 + 5x2 + 0x3 ≤ 430.
Question No: 48 ( Marks: 3 )
Express the following Transportation problem (T.P) table into algebraic form with proper
objective function and non-negative constraints
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Answer
Question No: 49 ( Marks: 5 )
1 2 3 Supply
1 1
2 6
7
7
2 0
10
4
2
12
3 3 1 5
1
11
Deman
d
10 10 10
Complete the above transportation Model by Vogel Approximation Method.
And also find the starting basic feasible solution.
Answer
2 1
1 1
1 2
4 5
6 3
8 10
x S D
S y D
S S z
D1 D2 D3 Supply
O1 c1
1
c1
2
c1
3
5
O2 c2
1
c2
2
c2
3
3
O3 c3
1
c3
2
c3
3
10
Deman
d
4 6 8
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1 2 3 Supply
1 1
2 6
7
7
2 0
10
4
2
12
3 3 1 5
1
11
Deman
d
10 10 10
Cost = 1 + 2 + 7(6) + 2 + 5(1) = 52
This is the initial basic solution consider u1 = 1 and v1 = 2 and v2 = 3
Question No: 50 ( Marks: 5 )
Check whether the given initial basic feasible solution is optimal or not.
1
2
6
7
7
0
10
4
2
2
12
3 1
10
5
1
11
10 10 10
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Answer
1 2 3 Supply
1 1
2 6
7
7
2 0
10
4
2
12
3 3 1 5
1
11
Deman
d
10 10 10
Cost = 1 + 2 + 7(6) + 2 + 5(1) = 52
This is the initial basic solution consider u1 = 1 and v1 = 2 and v2 = 3
it is a n optimal solution according to if we put formula
Question No: 51 ( Marks: 5 )
A company cost Rs. 500 operations and maintenance costs are zero for the first year and
increased by Rs. 100 every year. If money is worth 5% every year, calculate present worth
(P(r)) for each year. The resale value of the machine is negligibly small.
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Answer
for each year increase in money = 5%
means to say that company significes 105 ruppes every year the company capital must be increaing as
105 * 500
Question No: 52 ( Marks: 5 )
Express the following linear programming problem in standard form and also construct its initial
simplex table.
Max Z = 3x+2y
Subject to constraints:
x + y ≤4
x – y ≤2
x,y ≥0
Answer
Blank Data
Mth601 30 July 2013 final term paper:
2 marks qs
Q: ek bohat sari activities wali diagram di hui thi or qs ye tha
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Find EFT for each activity?
Q:
0
5
2 1
1
6
2 1
5
5
2
7
2 4 3
7
7
5 5 10
To find optimality condition we use UV multiplier process
Find
a) U2+V2
b) U3+V3
Q: 3 ya 5 mark ka tha ye qs
Contractor side wali values yad nai…
Building: Contractor:
A
B
C
D
Operate first step by optimizing row wise the above assignment model.
Q: state principal of optimality (optimal policy) for dynamic programming?
Q: fin EST and EFT for each activity.
A B 8 6 D
2 C
10 E
0 1
2
2
4
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3
Q: 5 marks
0
5
2 1
1
6
2 1
5
5
2
7
2 4 3
7
7
5 5 10
Find
a) P31
b) P32
Using Pij= Ui + Vj – Cij suppose U1 = 0 and U2 = 5
Q: ek statement thi us me se Average Queuing length find krna thi.
Q: Minimizing setup times, which are given? (5 marks)
Job ki values yad nai
Job 1 Job2 Job3 Job4
Machine 1 14 5 7
Machine 2
Machine 3
Machine 4
Q: ek 5 mark ka qs itna long tha k word pe paste krne se ek se zyada page ki just statement
thi….
Replacement Of Items with change in value and time
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It is assumed that the maintenance cost increases with time and each cost is to be paid just in the
start of the period. Let the money carry a rate of interest r per year. Thus a rupee invested now
will be worth (1 + r) after a year, (1+r)2after two years and so on.
Do first step?
Answer
Diff b/w pert n CPM??
Answer
PERT (Programme Evaluation & Review Technique) is event oriented whereas CPM
(Critical Path Method) is activity oriented.
In CPM based network analysis no allowance is made for the uncertainties in the
duration of time involved.
In CPM, times are related to costs
Q:…Make two steps, of rows and columns of the following table…
No. of stores 1 2 3
No. of boxes
0 0 0 0
1 4 2 6
2 6 4 8
3 7 6 8
4 7 8 8
5 7 9 8
7 7 10 8
No. of stores 1 2 3
No. of boxes
0 0 0 0
1 2 0 4
2 2 0 4
3 1 0 2
4 0 1 1
5 0 2 1
7 0 3 1
P a g e | 18
Answer:
Least ko sab me se Minus krna hay pehlay rows, then columns, to have atleast one
zero in all….
Markets /
salesmen
I II III IV
A 44 80 52 60
B 60 56 40 72
C 36 60 48 48
D 52 76 36 40
Markets /
salesmen
I II III IV
A 8 24 16 20
B 24 0 4 32
C 0 4 12 8
D 16 20 0 0
Complete the table By VOgha ‘s method:
P a g e | 19
Sol: Red is solved one
1 2 3 Supply
1 0
5
2
1
1
6
2 2 1
5
5
2
7
3 2 4 3
7
7
Demand 5 5 10
21 july 2013
1. a branch of bank has only one typist. typing rate is randomly distributed approximating a Poisson distribution with mean service
rate of 8/hour . Letter arrive rate is 5/hour during the entire 8 hour working day if writer is value 1.50 per hour determine the
equipment utilization? 2marks and same this question is appeared as 5 marks question
2. Scenario was given and we have to tell the objective function of linear programming.... 2 marks
3. Table was given n determine that transportation model has initial feasible solution....... 2 marks
4. State principle of optimality for dynamic programming ...... 2 marks
5. Transportation model was given and one block has x we have to find the value of x..... 3 marks
6. Values were given and we need to tell the capital cost for each year ....... 3 marks
7. In the context of pert and CPM summarize the project planning techniques
8. One stage problem is given find the two stage problem
s Fi*(s) Xi
8 25 10
9 35 10
9. Table was given and asked that check initial solution is feasible or not........... 5 marks
10. Question no 1 was again appeared as 5 marks question
11. The cost of the new machine is 5000. Maintenance during the nth year is given by Mn = 500 rps (n-1) when n = 1, 2, 3 ....if
discount rate per year is 0.05 calculate the present worth ...... 5 marks
12. Graph was given and question was construct the table relation show between events and activities. 5 mark s