QT Assignment Prateek

8/8/2019 QT Assignment Prateek

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Quantitativetechniques

Assignment

Prateek Bhati

FB 1 Roll No 46



ANSWERS

A:1Linear programming is a quantitative technique for selecting an optimum plan. It is anefficient search procedure for finding the best solution to a problem containing many

interactive variables. The desired objective is to maximize some function e.g., contributionmargin, or to minimize some function, e.g., costs. Determination of the optimum objectiveis usually subject to various constraints or restrictions on possible alternatives. Theseconstraints describe availabilities, limitations, and relationships of resources to alternatives.

The key assumption is linearity, which prevails in two respects. First, the contributionmargin or cost associated with with one unit of product or activity is assumed to be thesame for all identical units. Second, resource inputs per unit of activity are assumed to bethe same for all units. Another assumption inherent in l inear programming is that all factorsand relationships are deterministic.

Accounting data such as the contribution margin or cost factors would be used indetermining the objective - to maximize the contribution margin or to minimize cost.Accounting data would also be used to establish the constraints. Such constraints mightinclude one or more of the following: machine capacity, labor force, quantity of outputdemanded, time, or capital.

Once the data are available, the linear programming model (equations) might be solvedgraphically, if no more than two variables are involved, or by the simplex method. When themodel contains many variables and constraints, the solution may require the use of acomputer.



A2:In linear programming problems, the unit cost refers to the directly traceable variable costrather than the total cost.

A3:(a = 0, b = 20); $3(0) + $4(20) = $80 CM(a = 20, b = 10); $3(20) + $4(10) = $100 CM - Maximum CM(a = 30, b = 0); $3(30) + $4(0) = $90 CM

A4:The Simplex method is an iterative process which approaches an optimum solution in such away that an objective function of maximization or minimization is fully reached. Eachiteration in this process shortens the distance (mathematically and graphically) from theobjective function.

A5:CM = 2a + 5b + 4cAssembling: 2a + 3b + 2c 30,000

Painting: 1a + 2b + 2c 38,000Finishing: 2a + 3b + 1c 28,000

A6:C = 25a + 10bSubject to: a +b = 50; a 20; b 40

A7:Components of a simplex tableau:

y The objective row contains the coefficients of the objective function. y The variable row contains the variables of the problem, including slack variables. y

The problem row contains the coefficient of the variables in the constraints with onerow for each constraint. Variable not included in a constraint are assigned zerocoefficients. New problem rows are computed with each iteration.

y The objective column represents the contribution margin per unit of the variables inthe solution, and receives different entries at each iteration.

y The variable column contains the variables used to find the solution. In the firsttableau, only slack and artificial variables are entered in this column, since a no-production situation is the starting point of the iteration process.

y The quantity column shows the constant values of the constraints in the first tableau,it shows the solution mix.

y The index row contains values which indicate whether an optimum solution has beenreached; if there are any negative numbers in it, an optimum solution has not yetbeen reached. The value in the objective column represents the total contributionmargin.

A8:The slack variable is a fictitious variable that takes up the slack in the inequalities.Mathematically speaking, slack variables are treated like other variables and their fictitiouscharacter disappears in the solution process. Slack variables enter the objective function butreceive a coefficient of zero and do not influence the final result.



A9:(a) The index row is computed as follows:

Step 1 and 2:

4(7) + 4(0) = 28

2 /3(7) + 4/3(0) = 14/3

1(7) + 0(0) = 7

1/3(7) + [-1/3(0)] = 7/3

0(7) + 1(0) = 0

Step 3:

28 - 0 = 28

14/3 - 6 = -4/3

7 -7 = 0

7/3 - 0 = 7/3

0 -0 = 0

(b) An optimum solution has not been reached because a negative figure, -4/3, still remainsin the index row. Hence, the contribution margin can be increased by introducing x intosolution

(c) Product y would be reduced by 1/3 of a unit.

A10:The artificial variable is a computational device allowing two types of restrictions to betreated: the equality type (=) and the greater-than-or-equal-to type ( ).

A11:

(a) $14 (1/2) = $7 $12 (-1/4) = -3 ------

$4 ====

(b) (1) (2) Product Units S2 S2

x 3 3 / 4 4* y 2 -1 / 2 -4

*Maximum decrease over which shadow price of S2 is valid

A12:

(a) 4 (b) 3 (c) 3 (d) 4 (e) 3 (f) 3 (g) 2



A13: Dynamic programming involves breaking a problem into a set of smaller problems andthen reassembling the result. It is best suited for decisions that must be in sequence and thatinfluence future decisions in sequence.

A14: Let us first write the sample space S of the experiment.

S = {1,2,3,4,5,6}

y Let E be the event "an even number is obtained" and write it down.

E = {2,4,6}

y We now use the formula of the classical probability.

P(E) = n(E) / n(S) = 3 / 6 = 1 / 2

A15:

The sample space S is given by.

S = {(H,T),(H,H),(T,H),(T,T)}

y Let E be the event "two heads are obtained".

E = {(H,H)}

y We use the formula of the classical probability.

P(E) = n(E) / n(S) = 1 / 4

A16::



A17:

a) The sample space S of two dice is shown below.

S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }

y Let E be the event "sum equal to 1". There are no outcomes which correspond toa sum equal to 1, hence

P(E) = n(E) / n(S) = 0 / 36 = 0y b) Three possible ouctcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.

P(E) = n(E) / n(S) = 3 / 36 = 1 / 12

y c) All possible ouctcomes, E = S, give a sum less than 13, hence.

P(E) = n(E) / n(S) = 36 / 36 = 1

A 18:

y The sample space S of the experiment described in question is as follows

S = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}

y Let E be the event "the die shows an odd number and the coin shows a head".Event E may be described as follows

E={(1,H),(3,H),(5,H)}

y The probability P(E) is given by

P(E) = n(E) / n(S) = 3 / 12 = 1 / 4



A 19:

a) x = 80 , z = (80 - 70)/10 = 1

Probablity for grade to be greater than 80 = 1 - 0.8413 = 0.1587

b) x = 50 , z = (50 - 70)/10 = -2

Probablity for grade to be less than 50 = 0.0228

c) The z-scores for x = 50 and x = 80 have already been calculated above.

Probablity for grade to be between 50 and 80 = 0.8413 - 0.0228 = 0.8185

d) 0.1587 * 1000 = 159 (rounded to the nearest unit)

QT Assignment Prateek

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