1. Explain the structure of Mathematical model in OR.

Answer:

Many industrial and business situations are concerned with planning activities. In each case of planning, there are limited sources, such as men, machines, material and capital at the disposal of the planner. One has to make decision regarding these resources in order to either maximize production, or minimize the cost of production or maximize the profit etc. These problems are referred to as the problems of constrained optimization. Linear programming is a technique for determining an optimal schedule of interdependent activities, for the given resources. Programming thus means planning and refers to the process of decision-making regarding particular plan of action amongst several available alternatives.Any business activity of production activity to be formulated as a mathematical model can best be discussed through its constituents; they are:

Decision Variables, Objective function, Constraints.

Decision variables and parametersThe decision variables are the unknowns to be determined

from the solution of the model. The parameters represent the controlled variables of the system.

Objective functionsThis defines the measure of effectiveness of the system as a

mathematical function of its decision variables. The optimal solution to the model is obtained when the corresponding values of the decision variable yield the best value of the objective function while satisfying all constraints. Thus the objective function acts as an indicator for the achievement of the optimal solution.While formulating a problem the desire of the decision-maker is expressed as a function of ‘n’ decision variables. This function is essentially a linear programming problem (i.e., each of its item will have only one variable raise to power one). Some of the Objective functions in practice are:

Maximization of contribution or profit Minimization of cost Maximization of production rate or minimization of

production time Minimization of labour turnover Minimization of overtime Maximization of resource utilization Minimization of risk to environment or factory etc.

ConstraintsTo account for the physical limitations of the system, the

model must include constraints, which limit the decision variables to their feasible range or permissible values. These are expressed in the form of constraining mathematical functions.For example, in chemical industries, restrictions come from the government about throwing gases in the environment. Restrictions from sales department about the marketability of

some products are also treated as constraints. A linear programming problem then has a set of constraints in practice.The mathematical models in OR may be viewed generally as determining the values of the decision variables X J, J = 1, 2, 3,- - - -n, which will optimize Z=f (x1 , x2 ,−−−xn ) . Subject to the constraints:gi (x1 , x2 ,−−−xn ) bi, i = 1, 2, - - - mand X J≥0 j = 1, 2, 3, - - - - n where is ≤ , ≥ or =.The function f is called the objective function, where gi bi, represent the ith constraint for i = 1, 2, 3 - - - m where b i is a known constant. The constraints X J≥0 are called the non-negativity condition, which restrict the variables to zero or positive values only.

2. Explain briefly the graphical method of analysing a linear programming problem. Can a LPP have unbounded solution?

Answer: Graphical Method to Solve the Linear Programming Problems

A LPP with 2 decision variables x1 and x2 can be solved easily by graphical method. We consider the x1 x2 – plane where we plot the solution space, which is the space enclosed by the constraints. Usually the solution space is a convex set which is bounded by a polygon; since a linear function attains extreme (maximum or minimum) values only on boundary of the region, it is sufficient to consider the vertices of the polygon and find the value of the objective function in these vertices. By comparing the vertices of the objective function at these vertices, we obtain the optimal solution of the problem.

An Algorithm for solving a linear programming problem by Graphical Method:(This algorithm can be applied only for problems with two variables).

Step – I: Formulate the linear programming problem with two variables (if the given problem has more than two variables, then we cannot solve it by graphical method).

Step – II: Consider a given inequality. Suppose it is in the forma1 x1+a2 x2≤b(¿a1 x1+a2 x2≥b) . Then consider the relation a1 x1+a2 x2=b. Find two distinct points (k, l), (c, d) that lie on the

straight line a1 x1+a2 x2=b. This can be found easily: If x1=0, then x2=ba2

.

If x2=0, then x1=ba1

. Therefore

(k, l) = (0,ba2

 ) and (c, d) = (ba1

, 0) are two points on the straight

linea1 x1+a2 x2=b.

Step – III: Represent these two points (k, l), (c, d) on the graph which denotes X–Y-axis plane. Join these two points and extend this line to get the straight line which represents a1 x1+a2 x2=b

.Step – IV: a1 x1+a2 x2=b divides the whole plane into two half

planes, which are a1 x1+a2 x2≤b (one side) and a1 x1+a2 x2≥b (another side). Find the half plane that is related to the given inequality.

Step – V: Do step-II to step-IV for all the inequalities given in the problem. The intersection of the half-planes related to all the inequalities and x1≥0, x2≥0 , is called the feasible region ( or feasible solution space). Now find this feasible region.

Step – VI: The feasible region is a multisided figure with corner points A, B, C, … (say). Find the co-ordinates for all these corner points. These corner points are called as extreme points.

Step – VII: Find the values of the objective function at all these corner/ extreme points.

Step – VIII: If the problem is a maximization (minimization) problem, then the maximum (minimum) value of z among the values of z at the corner/extreme points of the feasible region is the optimal value of z. If the optimal value exists at the corner/extreme point, say A(u, v), then we say that the solution x1≥u and x2≥v is an optimal feasible solution.

Step – IX : Write the conclusion (that include the optimum value of z, and the co-ordinates of the corner point at which the optimum value of z exists).

Example 1: A grain merchant deals with two items: rice and wheat. He could invest only Rs.15000, and had a room for at most 80 bags. A bag of rice costs of Rs.250, and a bag of wheat costs Rs.150. He gains Rs.15 on a bag of rice, and Rs.12 on a bag of wheat. Assuming that he sells all the quantity he purchases, find how he should invest the money in order to get the maximum profit?

Solution: Let the number of bags of rice be x and the number of bags of wheat be y. Since both are non-negative, we have x≥0 , y ≥0.He has space to store only 80 bags. So x+ y≤80.He could invest only Rs.15000. Cost of x bags of rice is Rs.250x; and cost of y bags of wheat is Rs.150y. Therefore 250 x+150 y≤15000. This inequality can be written as 5 x+3 y≤300.He makes a profit of Rs.15 on one bag of rice, and Rs.12 on a bag of wheat; and so the total profit he makes is Rs.(15x + 12y). The profit is to be maximized. So we have Max z = 15x + 12y. So the mathematical model isMax z = 15x + 12y.Subject tox+ y≤80 5 x+3 y≤300 x≥0 , y ≥0. Now, graph the inequalities and find out the solution set. (Since all quantities involved are positive it is enough if we have the 1st quadrant).

Some points on x + y = 80 are (80, 0); (40, 40); (0, 80), and some points on 5x + 3y = 300 are (60, 0); (30, 50), (0, 100).

The solution set/space is the closed polygonal region OABC.Every point in this region satisfies the given constraints and is a feasible solution. So the region OABC is also called as the feasible region.From this region we have to find the point at which the value of the objective functionz = 15x + 12y is maximum.Now we have to find the values of z at the various vertices/extreme points of the polygonal region and decide the maximum.Value of z at 0(0, 0) = 0 + 0 = 0.Value of z at A(60, 0) = 15 x 60 + 0 = 900.Value of z at B(30, 50) = (15 x 30 + 12 x 50) =1050.Value of z at C(0, 80) = (15 x 0 + 12 x 80) = 960.Maximum profit = Rs.1050 at x = 30, and y = 50.

Conclusion: If grain merchant invests Rs.7500 for 30 bags of rice, and Rs.7500 for 50 bags of wheat, then he gets a maximum profit of Rs.1050.

Example 2 : 1 Kg. of food–I contains 3 units of vitamin A and 2 units of vitamin C, while 1 Kg. of food-II contains 2 units of vitamin A and 3 units of vitamin C. They cost Rs.10 per Kg. and Rs.8 per Kg. respectively. A dietician wants to mix the two kinds and prepare a mixture so that it contains at least 14 units of vitamin A and 16 units of vitamin C. Find how the dietician should mix them so that the cost to him may be minimum.

Solution: Let the mixture contain x Kg of food-I and y Kg. of food–II. So x≥0 , y ≥0. x Kg. of food-I contains 3x units of vitamin A, and y Kg of food-II contains 2y units of vitamin A. Therefore the mixture contains 3x + 2y units of vitamin A. Therefore 3 x+2 y ≥14.Similarly the mixture contains 2x + 3y units of vitamin C. Therefore2 x+3 y ≥16.

The objective function is, say z, the cost of the mixture which is Rs.(10x + 8y). This has to be minimized. So we have min z = 10x + 8y.

The graphs of 3 x+2 y ≥14 and 2 x+3 y ≥16 are the half-planes not containing the origin. The shaded region in the graph represents the feasible region. Its vertices are A(8, 0), B(2, 4) and C(0, 7).

Value of z at A(8, 0) = Rs.80.Value of z at B(2, 4) = Rs.52.Value of z at C(0, 7) = Rs.56.

Therefore the minimum value of z exists at the point B(2, 4). So the dietician has to mix 2 Kg. of food-I, and 4Kg of food-II so that the cost is minimum.Note:

I. If the set of all values of the objective function at different feasible solutions is not bounded above, and if the problem is a maximization problem, then we say that the given problem has an unbounded solution.

II. If the set of all values of the objective function at different feasible solutions is not bounded below, and if the problem is a minimization problem, then we say that the given problem has an unbounded solution.

That means a LPP have unbounded solution.

3. Apply simplex procedure to solve the L.PP maximize z=3 x1+4 x2 subject to 5 x1+4 x2≤200 ;3 x1+5 x2≤150 ;5x1+4 x2≥100; 8x1+4 x2≥80 ; x1≥0; x2≥0.

Answer:

4. Solve the following transportation problem

Answer:

5. Explain Project Management (PERT).

Answer:Project Management – PERT

Probability and Cost Consideration in Project Scheduling:

The analysis in CPM does not take into the case where time estimates for the different activities are probabilistic. Also it does not consider explicitly the cost of schedules. Here we will consider both probability and cost aspects in project scheduling.

Probability considerations are incorporated in project scheduling by assuming that the time estimate for each activity is based on 3 different values. They are –

a = The optimistic time, which will be required if the execution of the project goes extremely well.b = The pessimistic time, which will be required if everything goes bad.m = The most likely time, which will be required if execution is normal.

The most likely estimate m need not coincide with mid-point a+b2  

of a and b. Then the expected duration of each activity D can be

obtained as the mean of a+b2

 and 2 m. i.e. D=

a+b2

+2m

3=a+b+4m

6.

This estimate can be used to study the single estimate D in the critical path calculation.The variance of each activity denoted by V is defined by variance 

V=( b−a6 )

2

.

The earliest expected times for the node i denoted by E (μi ) for each node i is obtained by taking the sum of expected times of all activities leading to the node i, when more than one activity leads to a node i, then greatest of all E (μi ) is chosen. Let μibe the earliest occurrence time of the event i, we can consider mi as a random variable. Assuming that all activities of the network are statistical independent, we can calculate the mean and the variance of the μi as follows E (μi )=ES l and Var (μi ) = k

∑ V k. Where K defines the activities along the largest path leading to i.For the latest expected time, we consider the last node. Now for each path move backwords, substituting the D eJ for each activity (ij). Thus we have E (LJ )=E (μa ) and E (μi )=L (LJ )−D iJ if only one path events from J to i or it is the minimum of E (LJ )−D iJ for all J for which the activities (i, j) is defined.

Note: The probability distribution of times for completing an event can be approximated by the normal distribution due to central limit theorem.Since μi represents the earliest occurrence time, event will meet a certain schedule time STi (specified by an analyst) with probability

Pr (μi≤ST i )=Pr( μi−E (μi )√V (μi )

≤ST i−E (μi )

√V (μ i ) )=Pr (Z ≤ K i )

where Z N(01) and K i=ST i−E (μ i )

√V (μi )  . It is common practice to

compute the probability that event i will occur no later than its LC e such probability will then represent the chance that the succeeding events will occur within the (ES e , LCe ) duration.

Example: A project is represented by the network shown below and has the following data.Task A B C D E F G H IOptimistic Time 5 18 26 16 15 6 7 7 3Pessimistic Time 10 22 40 20 25 12 12 9 5Most Likely Time 8 20 33 18 20 9 10 8 4Determine the following:a) Expected task time and their variance.b) The earliest and latest expected times to reach each event.c) The critical path.

d) The probability of an event occurring at the proposed completion data if the original contract time of completing the project is 41.5 weeks.e) The duration of the project that will have 96% channel of being completed.

Solution:a) Using the formula we can calculate expected activity times and variance in the following table

D=16

(a+b+4m)V=( b−a6 )

2

Activity a B m v1-21-31-42-52-63-64-75-76-7

5182616156773

1022402025121295

8203318209

1084

7-820-0033-018-020-09-09-88-04-0

0.6960.4445.4290.4432.7801.0000.6940.1110.111

Forward Pass:E1 = 0 E2= 7.8 E3= 20 E4= 33 E5= 25-8 E6= 29 E7= 42Backward Pass:L7= 42.8 L6= 38.8 L5= 34.8 L4= 33.0 L3= 29.8 L2= 16.8 L1= 0

b) The E-values and L-values are shown in Fig.

c) The critical path is shown by thick line in fig. The critical path is 1-4-7 and the earliest completion time for the project is 42.8 weeks.d) The last event 7 will occur only after 42.8 weeks. For this we require only the duration of critical activities. This will help us in calculating the standard duration of the last event.Expected length of critical path = 33+9.8 = 42.8Variance of article path length = 5.429+0.694 = 6.123Probability of meeting the schedule time is given by Pi (Z ≤K i)=Pi (Z−0.52 )=0 .30(From normal distribution table)Thus the probability that the project can be completed in less than or equal to 41.5 weeks is 0.30. In other words probably that the project will get delayed beyond 41.5 weeks is 0.70.e) given that P (Z ≤ K i ) = 0.95. But Z0.9S = 1.6 u, from normal

distribution table. Then 1.6 u = ST i−E (μi )

√V (μi ) is 1.6 u=

ST i−42.82.47

Sji = 1.64´2.47+42.8 = 46.85 weeks.

Solution:

Activity Expected Times & Variances:

t=a+b+4m6

σ2=( (b−a )6 )

2

Activity Expected Times VariancesA 6 4/9B 4 4/9C 3 0D 5 1/9E 1 1/36F 4 1/9G 2 4/9H 6 1/9I 5 1J 3 1/9

K 5 4/9

Earliest/Latest Times and Slack:

Determining The Critical Path: A critical path is a path of activities, from the start node to

the finish node, with 0 slack times. Critical path: A - C - F - I - K The project completion time equals the maximum of the

activities’ earliest finish times. Project completion Time: 23 Hours

Normal Distribution of Project Time:

Probability the project will be completed within 24 hrss2 = s2A + s2C + s2F + s2H + s2K= 4/9 + 0 + 1/9 + 1 + 4/9= 2s = 1.414z = (24 – 23)/s = (24-23)/1.414 = 0.71From the Standard Normal Distribution table:P(z < 0.71) = 0.5 + 0.2612 = 0.7612

6. Explain the use of finite queuing tables

Answer:Finite Queuing Models:The models discussed so far relate to situations

involving infinite population of customers i.e. the queue can increase indefinitely. There will be cases, where the possible number of arrivals is limited and is relatively small. In a production shop, if the machines are considered as customers requiring service from repair crews or operators, the population is restricted to the total number of machines in the shop. In a hospital ward, the probability of the doctors or nurses being called for service is governed by the number of beds in the ward. Similarly, in an aircraft the number of seats is finite and the number of stewardesses provided by the airlines will be based on the consideration of the maximum number of passengers who can demand service. As in the case of a queuing system with infinite population, the efficiency of the system can be improved in terms of reducing the average length of queues, average waiting time and time spent by the customer in the system by increasing the number of service channels. However, such increases mean

additional cost and will have to be balanced with the benefits likely to accrue. If the queuing system in a machine shop is under study, the cost of providing additional maintenance crews or operators can be compared with the value of additional production possible due to reduced downtime of the machines. In cases where it is not possible to quantify the benefits, the management will have to base its decisions on the desired standards for customer service.

The queue discipline in a finite queuing process can be:i) First come-first servedii) Priority e.g.: Machines of high cost may be given priority for maintenance while others may be kept waiting even if they had broken down before.iii) Random e.g.: in a machine shop if a single operator is attending to several machines and several machines call for his attention at a time, he may attend first to the one nearest to him.

The analysis of Finite Queuing Models is more complex than those with infinite population although the approach is similar. L.G. Peck and R. N. Hazelwood have provided solutions to such problems in their book “Finite Queuing Tables” (John Wiley & Sons Inc – 1958)

Notations:Notations used are different and are given below:N = Population (machines, customers etc.)M = Service channels (repairmen, telephone lines etc.)T = Average service time (repair time, length of conversation on a telephone etc.)W = Average waiting timeU = Average running time (of machines) or mean time between calls for service per unitH = Average number of units being servicedL = Average number of units waiting for serviceJ = Average number of units in operationF = Efficiency FactorX = Service FactorD = Probability that if a unit calls for service, it will have to wait.

Let us consider a machine shop with N machines. The inter breakdown time of these machines follows a negative exponential distribution with mean U. The number of breakdowns follows

Poisson distribution with mean λ=1U

It is assumed that machines are kept running (or in operation) except when they are under repairs or waiting for repair crew to attend.If M repair crews are available, the time taken by any crew follows a negative exponential distribution with mean T. Naturally, a machine which has broken down will have to wait for repairs if all the repair crews are busy.

Measures of System Efficiency:Efficiency of the Repair SystemFor a given set of machines, the efficiency of the repair system may be judged by the extent to which machines have to wait for repairs. If W is the average time for which a machine has to wait, the efficiency factor F is defined as

F= T+UT+U+W

At any point of time, a machine will either be running or under repair or waiting for repairs. Therefore, the total number of machines N = J + H + L.

HN,

JN

∧L

N Correspond to the probability that a machine is being 

repaired, running or waiting for repairs respectively.

In the finite queuing tables, service factor X is defined as X= TT +U  X

is an indicator of the utilisation of repair crew.The formulae for other properties of the system are given below:

H= NTT+W+U

=FN X

L= NWT+W +U

=N (1−F )

J= NUT+W+U

=NF (1−X )

F= H+JH+ L+J

= T +UT +W+U

Use of Finite Queuing Tables:The tables give the values of F and D for different values of N, M and X. They are arranged in the ascending order of the values of the population. For each N, the value of X increases from .001 to .950. For a given service factor X, several values of M can be found.For each value of X and M, values of D and F are tabulated.The steps in the use of Finite Queuing Tables may be summarized as follows: (i) Find mean service time T and mean running time U.(ii) Compute the service factor(iii) Select the table corresponding to the population N.(iv) For the given population, locate the service factor value.(v) Read off from tables, values of D and F for the number of service crews M. If necessary, these values may be interpolated between relevant values of X.(vi) Calculate the other measures L, W, H, and J from the formulae given.The overall efficiency F of the system will increase with the number of service channels (M) provided. As mentioned earlier, addition of service crews involves cost, which should be justified by the increase in the efficiency of the system i.e. additional running time of machines possible. However, it will be seen from the tables that as M increase the rate of increase in efficiency decreases. The practical significance is that beyond a certain value of M, it is not worthwhile increasing M as there would be no appreciable increase in the efficiency of the system.

Example:In a chemical plant there are five hoppers of the same size which give food to material to grinding mills. Due to changes in the requisite of material, there are variations in the time taken for emptying the hoppers. On the basis of past experience this time was found to follow negative exponential distribution with an average of 10 hours between getting emptied. Whenever a hopper gets empty it has to be filled by a pay loader. Although the capacity of the hoppers is the same, the time taken to fill the hoppers varies due to different locations from which the material is to be loaded.

The time for filling the hopper also was found to follow negative exponential distribution with an average of 2.5 hours. The company hires the pay loaders at a cost of Rs. 100 per hour irrespective of whether it is operated or not. If the mill has to be stopped due to its hopper getting empty it costs Rs. 1000 per hour in terms of loss of profits. Determine the number of pay loaders which the company should engage to minimize overall cost.

Solution:Since T = 2.5 and U = 10

X= TT +U

= 2.52.5+10

=0.200

For N = 5, X = 0.200 we have the following values from the tables:

We now prepare a table as below:Table:

As increasing the number of pay loaders beyond two reduces profits, the company should engage only two pay loaders.