operation research models and methods

Write the constraints of the linear programming problem whose feasible region is shown in the accompanying diagram.. Show that the objective function value obtained is thesame as that pr

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Chapter 1 Problem Solving in Operations Research

1 Why is the object labeled situation in Fig 1.1 depicted with ambiguous borders?

The problem associated with situation may be unknown in a variety of respects

2 In Fig 1.1, why is the object representing the model drawn with straight lines?

The model is an abstraction, eliminating a lot of complexity

3 In Fig 1.1, why is the object representing the procedure drawn as an oval with a box

around it?

The procedure is a packaged solution ready for the user

4 What are the assumptions associated with a model? Why is it necessary to makeassumptions?

Assumptions are the simplifications (or abstractions) made to simplify the

problem for analysis They are necessary to make the problem tractable

5 Why may it be necessary to go back and change the model after solving it?

On testing the model it may be found to be invalid (perhaps it does not addressthe original problem) The model found may be impossible to solve (intractible)

6 What is the meaning of the term systems approach?

A systems approach considers aspects of the situation broader than the

immediate problem

7 What is meant by the term optimal solution?

The optimal solution maximizes or minimizes some measure of effectiveness

8 Why is implementation of an operations research solution sometimes difficult?

Solutions to a problem usually involve change to some persons in the

organization Change is difficult for many

9 In the title of the book, what is meant by the term methods?

A method is a mathematical procedure for solving a particular model

10 How would you define an organization in the context of problem solving? Who areits members?

The organization is the society in which the problem arises or for which the

solution is important The organization may be the citizens of a governmentalentity, a branch of government, a corporation, a department, or perhaps even ahousehold or individual

11 What is an abstraction? Why does one make abstractions in the modeling process?

An abstraction is an assumption about the situation Alternatively, an abstraction

is a simplification of the situation They are necessary to make the problemtractable (or capable of solution)

12 What must happen if the solution to a model turns out to violate some importantconstraints not previously stated by the decision maker?

The process must return to the problem statement step (or modeling step) toincorporate the violated constraint

13 Why do operations research studies often involve a team rather than a single person?

A study involves a team so the solution is not be limited by past experience ortalents of a single individual A team also provides the collection of specializedskills that are rarely found in single individuals

14 Explain how the conflicting goals of tractability and validity may cause the

modeling process to fail

It may impossible to construct a model that both can be solved (that is tractable)and is valid (represent the situation)

15 How is a solution different than a model?

A solution specifies values for the variables of the model, while the model is arepresentation of the situation and does not specify a solution

16 In what two activities of the problem solving process should the decision maker play

a large role?

The decision maker plays a large role in the problem formulation and solutionimplementation

17 Why would an analyst make assumptions that he or she does not necessarily believewhen formulating a model?

The assumptions are necessary to obtain a model that is tractable

18 What is the purpose of a control procedure?

The control procedure implements a solution in a situation where the problemarises on a regular basis The control procedure solves repeated instances of aproblem as they arise The procedure should also recognize when a situation haschanged and requires new analyses

19 How should the value a model be determined?

A model has two important values For the decision makers and analyst themodel can add to the understanding of the situation Just building the modeloften describes features of the model in more objective terms The problem canthen be addressed more intelligently, possibly without further recourse to themodel Some models can be solved to suggest decisions that can be

implemented In these cases the value of the model can be measured by thesuccess of the decisions in addressing the original problem

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Copyright 2003 – All rights reserved

Chapter 2 Linear Programming Models

1 Write the constraints of the

linear programming problem

whose feasible region is

shown in the accompanying

diagram The axes intercepts

of the lines are:

Grid size = 1

To do this problem consider the equation of a line ax1 + bx2 = 1 Any line notpassing through the origin can have this form by assigning the correct values of a

and b Given any two points on the line, one can solve for a and b For instance

let the intercept on the x1 axis be at point (j, 0) and the intercept on the x2 axis be(0, k) Since both intercepts lie on the line

a*j =1 and b*k = 1 or a = 1/j and b = 1/k.

Using line 1 in the problem the intercepts are at j = 2 and k = 1 The

equation for line 1 is therefore

1/2x1 + x2 = 1

To write the constraint we must determine if it is a £ or ≥ constraint The easiestway to do this is to see in the picture whether the origin (0, 0) is feasible for theconstraint and set the sense of the constraint to accomplish this For line 1, theorigin is not feasible, so the constraint must be

1/2x1 + x2 ≥ 1

In like manner, we can write the entire set of constraints

Constraint x1 intercept (j) x2 intercept (k) (0, 0) feasible? Constraint

C3 4 -5 Yes 0.25x1 - 0.2x2£ 1

2 The figure below shows the graphical model of a linear program The large numbers

on the right 1, 2, and 3 indicate the constraints The feasible region is shown inwhite and the infeasible region is cross hatched The small numbers indicate fourfeasible corner points: 0, 1, 2, 3 Three objective functions are under consideration, as

indicated by the three lines labeled A, B, and C The arrows represent the direction

of increasing objective function Objective B is parallel with constraint 3 In each

case, specify the location of the optimal solution If more than one optimum exists,characterize all of them

g Drop x1≥ 0 and minimize C

Note that this problem assumes the feasible region is defined by the three

constraints and nonnegativity conditions on x1 and x2 Thus the region is

unbounded in the x2 direction

point on line 3 There are aninfinite number of alternativeoptima

c The solution is unbounded d The solution is unbounded

e The solution is unbounded f Point 0

g The solution is unbounded

3 Consider the linear programming model shown below.

a Graph the feasible region using a

2-dimensional grid Show an

isovalue contour for the objective

function and indicate the direction

of decrease Identify the optimal

solution on the graph

b Graphically perform a sensitivity

analysis for each of the objective

function coefficients and each of the

Sensitivity Analysis

Num Name Value Status Reduced Cost

Objective Coefficient

Range Lower Limit

Range Upper Limit

Constraint Limit

Range Lower Limit

Range Upper Limit

b1 = -4

x1

x2

(1)(2)

b2 = -10

x1

x2

(1)(2)

1010

For b2 = •, constraint 2 moves down

and off the picture

For b2 = •, constraint 2 moves down

and off the picture

As b4 increases the optimumincreases along the constraint (1) line.There is no limit to the increase, sothe upper limit is •

x1

x2

(1)

(2)(3)

(4)

1010

Exercises 4 – 10 refer to the material in Section 2.6.

4 For the resource allocation problem, show that the dual variables can be used toestimate the effects of changes in the right-hand sides of the constraints Add onemore machine of type 3 with a corresponding increase in availability of 40 hours.Solve the resulting linear program using the software that accompanies the book orany LP code available to you Show that the objective function value obtained is thesame as that predicted from the dual variables associated with the optimal solutiongiven in the text

From the information given in the text one finds the dual variable for M3 is 8.96.The upper limit on the range of b for this constraint is 237.5 Increasing the rightside from the current value of 120 to 160, corresponding to an additional

machine, results in a right side within the range specified for M3 Since the value

of profit is 2988.73, and the change in b is 40, the predicted profit after the

change is

2988.73 + (8.96)(40) = 3347.13Solving the revised problem with the computer results in below

Optimal Solution

The difference between the predicted profit and the computed Z is due to

the numerical accuracy of the computations Note that the basic variables (theones with nonzero values) have the same identity but different values for the newsolution

5 For the blending problem, show that the ranges given in the objective function

analysis suggest how the solution will change as the price of corn changes Solve theproblem for the price equal to 24, 25, 90 and 91 Compare the four solutions with thesolution given in the text with the price of corn equal to 30.5

The following compares the solution to the results given in the text

∑ Cost 24: The basis changes The diet has minimum calcium rather thanmaximum calcium (L = 0.0176, C = 0.6614, S = 0.3209, Z = 44.94)

∑ Cost 25 and 90: The same solution as the text (L = 0.0282, C = 0.6486,

S = 0.3233, Z = 45.59 (for cost = 25), Z = 87.75 (for cost = 90)

∑ Cost 91: The basis changes The diet has maximum calcium and

maximum fiber The minimum protein and minimum calcium constraintsare loose (L = 0.0277, C = 0.4631, S = 0.5092, Z = 88.25)

6 Change the car rental problem to add a Monday-Wednesday-Friday plan that costs

$105 for the three days and a Tuesday-Thursday plan that costs 64 for the two days.What is the new optimum rental plan? If one of these new plans is not used, fromsensitivity analysis determine how much the cost of the plan should be reduced tohave the plan be adopted

Answer: Let the new variables be y6 for the MWF option and y7 for the TTh.option The solution has a cost of 1930 and the following assignments

Var x1 x2 x3 x4 x5 x6 x7 y3 y4 y5 we wd w y6 y7

From the sensitivity analysis we learn that if the cost of the TTh option werereduced by more than 4, the basis will change

7 Use an approach similar to the car rental problem to solve this workforce schedulingproblem A bus company is scheduling drivers for its buses The requirement fordrivers varies by time of day as in the table below These requirements repeat everyday of the week.

Hours midnight –

4am

4am –8am

8am –noon

noon –4pm

4pm –8pm

8pm –midnight

Drivers are hired for eight hour shifts that start at midnight, 4a.m., 8a.m., 12 noon,4pm, or 8pm That is, a driver starting at midnight will work until 8a.m Drivers thatstart at 8pm work until 4pm the next morning Find a schedule of drivers that willminimize the number of drivers necessary to meet the daily requirements Notethat some drivers may be idle for a part of their shifts

The matrix representation of this model when the workers have 8 hour shifts isshown below

The optimal solution uses 26 drivers There are a number of alternative

schedules, but one is shown below

x0 x4 x8 x12 x16 x20 Z

8 Solve the workforce scheduling problem with each of the work rules below Use thesame requirements Comment on the integrality of the solution each case.

a Each driver works a 12 hour shift

b Drivers brought in at a particular time work for four hours, break for four hours,and then work for another four hours

a The matrix representation of this model when the workers have 12 hour shifts

9 How must the model for the aggregate planning problem be modified to account forcosts incurred when production levels change? In particular, if production goes upfrom one month to the next, there is an additional cost of $2 per unit of change Ifproduction goes down from one month to the next, there is a cost of $1 per unit ofchange The current production level is 1000.

Hint: Introduce variables E t and F t in the model to represent respectively the

increase or decrease in production in month t.

For each period add the appropriate objective function terms and a constraintthat relates the variables Et and Ft to the production in adjacent months; i.e.,

Pt -1 – Pt + Et – Ft = 0 for t = 1, ,6

Modifying the model along these lines and solving gives the following solution.Because of the added costs of changing production levels, the solution has amore uniform production plan

10 For the power distribution problem, assume that a generator failure occurs at station

B so that it can no longer serve any outlying areas and falls 25 MW short of servingits own area Set up and solve the linear programming model to find the optimaldistribution of power in this circumstance

The network model when generator B has no capacity and requires 25 extraunits is shown below

To modify the linear programming model, change the station B

conservation constraint as below and set the upper bound on PB to zero

11 Consider the following linear program.

Sketch the feasible region and several isovalue contours for the objective function in

the (x1, x2)-space Show the optimal solution on the graph

Z = 36

Z = 60

Note that the objective function line isparallel to the second constraint Allfeasible points on that constraint lineare alternative optima

12 Ten jobs are to be completed by three men during the next week Each man works a40-hour week The times for the men to complete the jobs are shown in the tablebelow The values in the cells assume that each job is completed by a single person;however, jobs can be shared with completion times being determined proportionally.

If no entry exists in a particular cell, it means that the corresponding job cannot beperformed by the corresponding person

Set up and solve a linear programming model that will determine the optimalassignment of men to jobs The goal is to minimize the total time required to

complete all the jobs

xij : Proportion of job j performed by man i, for i = A, B, C, and j = 1, , 10.

Variables should be defined for only the man–job combinations not eliminatedwith an X in the table Thus there are 20 variables in this problem

CONSTRAINTS

Each man can only work 40 hours

MANA: + 7xA2 + 3xA3 +18xA6 +13xA7 + 6xA8 + 9xA10 £ 40

NONNEGATIVITY AND SIMPLE UPPER BOUNDS

0 < xij < 1 for all i and j for which variables are defined.

OBJECTIVE FUNCTION

MINIMIZE: Total time required for the jobs

Min Z = 7xA2 + 3xA3 +18xA6 +13xA7 + 6xA8 + 9xA10

+12xB1 + 5xB2 +12xB4 + 4xB5 +22xB6 +17xB8 +13xB9

+18xC1 + 6xC3 + 8xC4 +10xC5 +19xC7 + 8xC9 +15xC10

Assignment: Value 88

13 A company has two manufacturing plants (A and B) and three sales outlets (I, II,and III) Shipping costs from the plants to the outlets are as follows.

50 and the inventory cost is $1 per unit Find the solution that maximizes profit

Variables

PA1, PB1, PA2, PB2: production at A and B in the two periods

S11, S21, S31: sales in period 1 at three locations

S12, S22, S42: sales in period 2 at three locations

xA11, xA21, etc: shipments in period 1

xA12, xA22, etc: shipments in period 2

IA, IB: Inventory storage at A and B

Max profit: Z = 15S11 + 20S21 + … sales revenue

- 8PA1 - 7PB1 - … production costs

- 4xA11 - xA21 - … transportation costs

- 1IA - 1IB

Conservation constraints and upper limits on variables

Profit $4450.

P A P B S1 S2 S3 XA1 XA2 XA3 XB1 XB2 XB3 IA IB

Period 1 175 200 100 200 0 100 7 5 0 0 125 0 0 7 5 Period 2 150 170 150 9 5 150 150 0 0 0 9 5 150

14 A company is planning its aggregate production schedule for the next three months.Units may be produced on regular time or overtime The relevant costs and

capacities are shown in the table below The demand for each month is also shown.There are three ways of meeting this demand: inventory, current production, andbackorders Units produced in a particular month may be sold in that month or kept

in inventory for sale in a later month There is a $1 cost per unit for each month anitem is held in inventory Initially, there are 15 units in inventory Also, sales can

be backordered at a cost of $2/unit/mo Backorders represent production in futuremonths to satisfy demand in past months, and hence incur an additional cost

Capacity (units)

Production cost($/unit)Period Regular time Overtime Regular time Overtime Demand

Part a

Variables for part a

Rk = Regular production in month k

Ok = Overtime production in month k

Ik = Inventory stored in month k

Bk = items backordered from month k

1 R1 (Regular prod.)

O1 (Regular prod.)

(60) Demand

2 R2 (Regular prod.)

O2 (Regular prod.)

(80) Demand

3 R3 (Regular prod.)

15 Initial Inv.

Conservation of product

Limits on regular production R1 £ 100, R2 £ 100, R3 £ 60 O1 £ 20, O2 £ 10, O3 £ 20 Objective: Minimize cost

Z = 14R1 + 17R2 + 17R3 (regular prod + 18O1 + 22O2 + 22O3 (overtime prod.) + 1(I1 + I2) (inventory)

Inventory to Next Period

Backorder to Previous Period

Part b

Variables for part b

Inventory variables: We will use a variable for the amount of material put into

inventory in each period and separate variables for the amounts used in period

jthat were produced in period k, where j ≥ k.

Initial Inventory: I0 and I01, I02, I03

Period 1 Inventory: I1 and I12, I13

Period 2 Inventory: I2 and I23

Backorder variables: B k for k = 1…3.

Constraints defining Inventory:

I = 15; I = I + I + I I = I + I I = I

Production Variables:

Regular time and Over time Production: R k, O k, for k = 1…3.

Production used immediately for sales: P Sk for k = 1…3.

Constraints for Production

O1 (Regular prod.)

2_P R2 (Regular prod.)

O2 (Regular prod.)

3_P R3 (Regular prod.)

I03

This is the same solution as in part a

15 (Shuttle Bus Staffing) A shuttle bus system operates from 6 a.m to 12 midnight, a

period of 18 hours To meet student demands for service, the following schedule hasbeen determined for the number of bus drivers required during each hour of the day.Times are stated with respect to a 24-hour clock

Eight-hour work schedule

Work Work Break Work Work Break Work Work

The company would like to determine how many drivers to call in at eachhour of the day Drivers begin their shifts on the hour (at 6, 7, 8, …, 16) No driversare called after time 16 There must be at least enough drivers scheduled to coverthe hourly requirements under the condition that each follows the 8-hour patterngiven above The goal is to minimize the total number of drivers used Formulate alinear programming model that can be used to solve this problem

Let x i = Number of drivers called at time i, i = 6, 7, 8, …, 16.

Objective: Minimize the number of drivers called

A similar constraint is constructed for each hour The variables representing a

particular hour must only be included when the workers are available for that

hour thus x6 only appears in the constraints: 6-7, 7-8, 9-10, 10-11, 12-13, 13-14.All variables must be nonnegative

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Copyright 2003 – All rights reserved

Chapter 3 Linear Programming Methods

1 Consider the following linear program

a Show the equality form of the model

b Sketch the graph of the feasible region and identify the extreme point solutions.From this representation find the optimal solution

c Analytically determine all solutions that derive from the intersection of twoconstraints or nonnegativity restrictions Identify whether or not these solutionsare feasible, and indicate the corresponding objective function values Whichone is optimal?

d Let the slack variables for the first two constraints, x3 and x4, be the axes of thegraph, and sketch the geometric representation of the model Show an iso-objective line in these variables, and from it determine the optimal solution

b

0 1 2 3 4 5

d First we must solve the objective function and constraints in terms of the basicvariables z, x1 x2, x5 and x6 We obtain

2 You are given the following linear program.

iso-b Analytically determine the set of all solutions that are intersections of two

constraints or nonnegativity restrictions Identify whether or not the solutionsare feasible, and from the feasible subset select the optimum by evaluating theobjective function at each

a Solving in terms of x1 and x2 we obtain the representation of the problem in terms of

the variables x3 and x4

Corner Nonbasic Basic Solution Feasibility Objective Point Variables Variables (x1, ,x

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 3 For the linear program

Maximize z = 10x1 + 5x2 + 8x3 – 3x4

subject to -2x1 + x2 + 2x3 –3 x4 ≥ 12

x1 + x2 + x3 + x4 £ 20

x j ≥ 0, j = 1,…, 4

a Construct the equality form of the model by introducing slack variables x5 and x6

for the two constraints

b What is an upper bound on the number of basic solutions?

c Analytically determine the set of basic solutions by listing all possible selections

of the basic variables Identify which are feasible and which are infeasible.Compute the objective function values for the feasible solutions and select theoptimum

b Number of basic solutions £15

c Set of basic solutions:

Corner Nonbasic Basic Solution Feasibility Objective Point Variables Variables (x1, ,x6) (Yes, No) Value

#14 (2, 4, 5, 6) (1, 3) (7, 0, 13, 0, 0, 0) Yes 174

#15 (3, 4, 5, 6) (1, 2) (2.67, 017.33 0, 0, 0, 0) Yes 113.33––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– The solution is optimum for #14

4 Given the following linear program,

Set of basic solutions: x4 and x5 are slacks

Corner Nonbasic Basic Solution Feasibility Objective Point Variables Variables (x1, ,x5) (Yes, No) Value

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 5 The tableau below is not in the simplex form.

a Write the set of equations described by the tableau as it stands

b Using linear operations put the tableau in the simplex form by making x1 and x4

basic What is the solution corresponding to the new tableau

c From the new tableau predict the effects of increasing x5 by 1, by 0.5, and by 2

d From the new tableau predict the effects of increasing x3 by 1, by 0.5, and by 2

a

z + x1 - x2 - x3 +2x5 = 20

x1 + 5x2 - x3 + x4 +12x5 = 12

+ 8x2 + x3 + 2x4 +16x5 = 16

b The tableau for B(1)=1 and B(2) = 4 is as follows:

the basic solution is x1 = 4, x2 = 0, x3 = 0, x4 = 8, x5= 0

c Increasing x5 changes the objective value x1 and x4

d Increasing x3 changes the objective value x1 and x4.

6 Starting with the tableau found in part b of problem 5, consider the three cases listed

below Construct the new tableau in the simplex form, write the basic solution tained, and use the marginal information available from the tableau to comment onany characteristics the solution exhibits The parts of the problem all start from thesame basis definition

ob-a. Allow x2 to replace x4 as a basic variable

b. Allow x5 to replace x4 as a basic variable

c. Allow x5 to replace x1 as a basic variable

The tableau found in part b is

a Allow x2 to replace x4 as a basic variable The solution is optimal

7 For the example Section 3.5, start with the basic solution denoted by #1 in Fig 3.5.Sequentially change the basis by allowing variables to enter and leave so that thebasic solutions are #2, #3, #4 and #5 in this order Show the equations in the

simplex form for each of the four cases, and sketch the view of the feasible regiondescribed for each basis

Point #1: x2 and x5 are nonbasic

Feasible region in terms of x2 and x5

4 x2+ x3 – x5 = 20

3 x2 + x4+ 2 x5 = 35

x1 +2 x2 – x5 = 5

Point #2: x1 and x5 are nonbasic

-2x1 + x3 + x5 = 20

1.5x1 + x4+ 0.5x5= 35

0.5x1 + x2 – 0.5x5 = 5

Point #3; x1 and x3 are nonbasic

-2x1 + x3 + x5 = 20

2.5x1 -0.5x3 + x4 = 25

-0.5x1 + x2 + 0.5x3 = 15