operation research models and methods

Operations Research Models and Methods by Paul A Jensen and Jonathan F Bard John Wiley and Sons, Inc Copyright 2003 – All rights reserved Chapter Problem Solving in Operations Research 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 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 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 What are the assumptions associated with a model? Why is it necessary to make assumptions? Assumptions are the simplifications (or abstractions) made to simplify the problem for analysis They are necessary to make the problem tractable 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 address the original problem) The model found may be impossible to solve (intractible) Problem Solving in Operations Research What is the meaning of the term systems approach? A systems approach considers aspects of the situation broader than the immediate problem What is meant by the term optimal solution? The optimal solution maximizes or minimizes some measure of effectiveness 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 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 are its 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 governmental entity, a branch of government, a corporation, a department, or perhaps even a household 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 problem tractable (or capable of solution) Problem Solving in Operations Research 12 What must happen if the solution to a model turns out to violate some important constraints not previously stated by the decision maker? The process must return to the problem statement step (or modeling step) to incorporate the violated constraint 13 Why 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 or talents of a single individual A team also provides the collection of specialized skills 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 a representation 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 solution implementation Problem Solving in Operations Research 17 Why would an analyst make assumptions that he or she does not necessarily believe when 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 problem arises on a regular basis The control procedure solves repeated instances of a problem as they arise The procedure should also recognize when a situation has changed 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 the model can add to the understanding of the situation Just building the model often describes features of the model in more objective terms The problem can then be addressed more intelligently, possibly without further recourse to the model Some models can be solved to suggest decisions that can be implemented In these cases the value of the model can be measured by the success of the decisions in addressing the original problem Operations Research Models and Methods by Paul A Jensen and Jonathan F Bard John Wiley and Sons, Inc Copyright 2003 – All rights reserved Chapter Linear Programming Models Write the constraints of the linear programming problem whose feasible region is shown in the accompanying diagram The axes intercepts of the lines are: x2 Grid size = Line 1: (0, 1) and (2, 0) Line 2: (0, 2.5) and (-5, 0) Line 3: (0, -5) and (4, 0) Line 4: (0, 8) and (6, 0) x1 To this problem consider the equation of a line ax1 + bx2 = Any line not passing 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 = or a = 1/j and b = 1/k Using line in the problem the intercepts are at j = and k = The equation for line is therefore 1/2x1 + x2 = To write the constraint we must determine if it is a £ or ≥ constraint The easiest way to this is to see in the picture whether the origin (0, 0) is feasible for the constraint and set the sense of the constraint to accomplish this For line 1, the origin is not feasible, so the constraint must be 1/2x1 + x2 ≥ In like manner, we can write the entire set of constraints Constraint C1 C2 x1 intercept (j) -5 x2 intercept (k) 2.5 (0, 0) feasible? No Yes Constraint 0.5x1 + x2 ≥ -0.2x1 + 0.4x2 £ Linear Programming Models C3 C4 -5 Yes Yes 0.25x1 - 0.2x2 £ 0.167x1 + 0.125x2 £ Linear Programming Models The figure below shows the graphical model of a linear program The large numbers on the right 1, 2, and indicate the constraints The feasible region is shown in white and the infeasible region is cross hatched The small numbers indicate four feasible corner points: 0, 1, 2, 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 In each case, specify the location of the optimal solution If more than one optimum exists, characterize all of them a maximize A b maximize B c maximize C d minimize A e minimize B f minimize C g Drop x1 ≥ 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 a Point c The solution is unbounded b Point and all points above the point on line There are an infinite number of alternative optima d The solution is unbounded e The solution is unbounded f Point g The solution is unbounded Linear Programming Models 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 Minimize z = 2x1 + x2 subject to –x1 + 2x2 £ 10 (1) x1 – 2x2 £ (2) x1 + x2 ≥ (3) x1 + 2x2 ≥ 20 (4) x1 ≥ 0, x2 ≥ (5) b Graphically perform a sensitivity analysis for each of the objective function coefficients and each of the right-hand-side constants x2 unbounded feasible region Grid size = (1) 10 (2) (4) (3) 10 x1 Sensitivity Analysis Variable Analysis Num Name X1 X2 Constraint Num Name Con1 Con2 Con3 Con4 Objective Value: Objective Value Status Reduced Cost Coefficient Basic Basic 0 17.5 Range Lower Range Upper Limit Limit 0.5 -4 Analysis Value Status 10 12.5 20 Upper Basic Basic Lower Shadow Price -0.75 0 1.25 Constraint Limit 10 20 Range Lower Range Upper Limit Limit -4 -10 20 - 14 - - 12.5 Linear Programming Models c1 = 0.5 c1 = • x2 10 x2 10 (1) (1) (2) (3) (2) (4) (3) (4) 10 x1 10 x1 c2 = -4 c2 = x2 10 x2 10 (1) (1) (2) (3) (2) (3) (4) (4) 10 x1 10 x1 b1 = -4 b1 = 20 (1) x2 10 x2 10 (1) (2) (2) (3) (4) 10 x1 (3) (4) 10 Linear Programming Models b2 = -10 b3 = 12.5 x2 10 x2 10 (1) (2) (1) (2) (3) (3) (4) (4) 10 x1 10 x1 For b2 = •, constraint moves down and off the picture b4 = 14 x2 10 (1) (2) (4) (3) For b3 = -•, constraint moves down and off the picture As b4 increases the optimum increases along the constraint (1) line There is no limit to the increase, so the upper limit is • x2 10 (1) (4) 10 x1 For b2 = •, constraint moves down and off the picture (2) (3) 10 x1 Simulation 45 34 Using the random number table, generate 10 random variates for the following probability mass function Value, x pmf, p(x) 0.156 0.234 0.208 0.161 0.095 0.064 0.036 0.028 0.018 rn 0.005 0.192 0.941 0.72 0.596 0.206 0.233 0.711 0.909 0.994 Simulated x 1 45 Simulation 46 35 A firm wishes to investigate the profitability of a new product Consumer tests on samples of the product made at a small test plant have given favorable results From these tests and the other market surveys, it is estimated that the potential sales should range from 40 to 100 units in about out of 10 quarters within years of its introduction One of the alternatives considered involves the construction of a plant with a capacity of 100 units per quarter at a cost of $400,000 Such a plant would be in operation within 1year of the decision to go ahead with the new product Since the product can only be stored for very short periods, production would follow actual sales very closely Hence, any potential sales above 100 units per quarter would be lost The fixed production cost per quarter is estimated at $30,000, while the variable production costs per unit are highly nonlinear, as follows: for £ q £ 50, $(100 - q) per unit 50 £ q £ 80, $50 per ton 80 £ q £ 100, $0.625 q per unit Sales predictions are as follows: Operating year from on Sales range Sales price/unit 20 to 60 unit per quarter 30 to 90 unit per quarter 40 to 100 unit per quarter $880 $890 $900 All ranges are quoted with odds of out of 10, and actual sales are assumed to be approximately normally distributed (Note that sales cannot be negative.) Management would like to determine by simulation the distribution of the cumulative net cash flow over the first years of operating the plant (after its construction) Round sales figures to nearest unit a Using the random number table starting in row 1, simulate the cash flow quarter by quarter over a 5-year period What is the net cash flow? We constructed an Excel simulation using the Simulation add-in Here we are using the first two rows of the random numbers appearing in thiis Exercise section The total net income over the year period is $59,799 46 Simulation Quarter 10 11 12 13 14 15 16 17 18 19 20 RN Year 0.005 0.192 0.941 0.72 0.596 0.206 0.233 0.711 0.909 0.994 0.624 0.684 0.748 0.187 0.069 0.399 0.527 0.291 0.388 0.965 Demand 1 1 2 2 3 3 4 4 5 5 8.6801 29.415 59.007 47.087 64.432 45.037 46.704 70.146 94.342 115.82 75.763 78.735 82.187 53.786 42.947 65.332 71.235 59.96 64.81 103.05 Prod 29 59 47 64 45 47 70 94 100 76 79 82 54 43 65 71 60 65 100 47 Unit Cost Total Cost 91 71 50 53 50 55 53 50 58.75 62.5 50 50 51.25 50 57 50 50 50 50 62.5 30819 32059 32950 32491 33200 32475 32491 33500 35523 36250 33800 33950 34203 32700 32451 33250 33550 33000 33250 36250 Unit Rev Total Rev Net Rev 0 880 7920 -22899 880 25520 -6539 880 51920 18970 880 41360 8869 890 56960 23760 890 40050 7575 890 41830 9339 890 62300 28800 900 84600 49078 900 90000 53750 900 68400 34600 900 71100 37150 900 73800 39598 900 48600 15900 900 38700 6249 900 58500 25250 900 63900 30350 900 54000 21000 900 58500 25250 900 90000 53750 Total Net Rev 9 b (More computations) Run the simulation for 10 runs, continuing with the random number table where you left off in part a Once you reach the end of the table go to the upper right and read the numbers from right to left Find the average net cash flow and the standard deviation of the net cash flow Here we used the RAND() function to generate seeds and rounded them to four digits The 10 observations are shown below Run Seed 0.2536 0.3659 0.1165 0.0061 Total Net Income - 7 Grand Average Std Deviation 10 0.396 0.2371 0.4333 0.6991 0.6529 0.4355 58707 -65160 10693 65231 48275 141714 42768 70237 c (Computer implementation) Write a computer program to perform this simulation Make 100 runs, each over a 10-year period, discounting the quarterly cash flows at a rate of percent per quarter Construct a frequency histogram of the net discounted cash flows containing about 10 class intervals 47 Simulation 48 (If you need a refresher on discounting, see the material at the end of Section 6.2.) What conclusion you reach about the profitability of the project, assuming that 10 years is the productive life of the plant? We simulated 100 observations of the 10-year simulation The results are shown below with a histogram There was never a NPW less than and the average NPW was well above This investment should yield the minimum acceptable rate of return (3% a quarter) The values are in 1000’s of dollars Sample Size Mean Stand Dev Minimum Maximum Median Mid Range 31.79 44.6455 70.3565 96.0675 121.7785 147.4895 173.2005 198.9115 224.6225 250.3335 276.0445 288.9 100 140.8403 50.8193 31.7947 288.9116 150.5611 Frequency 0.09 0.08 0.09 0.15 0.17 0.25 0.14 0.01 0.01 0.01 48 Simulation 36 49 Canal A is connected to the lower level Canal B by a lock Boats enter from Canal A into the lock, the gates are closed, the water level is lowered to the level of Canal B, the gates toward Canal B are opened, and the boats leave At this point any boats in Canal B wanting to be raised to Canal A enter the lock, the gates are closed, the water level is raised to the level of Canal A, the gates are opened, the boats leave, and so on The lock had a capacity of boats, and it takes minutes to lower or raise the water level Only one boat may enter the lock at a time, and a boat takes minute to be moored However, all boats leave the lock one after the other taking a total of minutes (regardless of the number of boats) Boats arrive at the lock in Canal A at a rate of 12 per hour, and in Canal B at a rate of 15 per hour Arrivals in both canals follow a Poisson distribution The current mode of operation during the busy hours, for which the above arrival rates hold, is to fill or empty the lock whenever boats have been moored inside the lock At 11 AM, the system is in the following state: The lock gates are open to traffic from Canal A Three boats are already moored in the lock, and the next arrival from A is scheduled at 11:02 Five boats are currently waiting in Canal B to go through the lock to A The next boat arrival is scheduled at 11:04 Simulate this system until 11:30 AM Keep enough detail to determine the number of boats passing through the lock in each direction, the average time boats spend at the lock between arrival and departure in each direction, and the number of times the lock has been raised and lowered Use the following sequence of 2-digit random numbers to generate the arrivals: 56 03 09 78 38 47 01 98 03 16 14 56 17 11 98 82 51 97 93 04 We must sample from an exponential distribution with rates lA = 0.2/min and lB = 0.25/min to determine the times between arrivals at the locks in canals A and B, respectively Because entities and 10 are already scheduled, the first random number on the list, 56, is used to simulate the arrival of a boat to the lock ln(u) , where u = 56/100, we get x = 2.89 at canal A Using the formula x = -1 l Rounding to the nearest integer, the next arrival to canal A is 11:02 + = 11:06 The same logic was used to simulate the remaining arrivals The results for all operations are shown in the table 49 Simulation Time Entity no Event 11.00 11.02 11.03 11.04 11.04 11.04 11.06 11.07 Simulation starts arrival at A lock lowered 10 arrival at B 12 arrival at B 13 arrival at B 11 arrival at A – lock down 11.08 11.09 11.10 11.10 11.11 11.11 11.12 11.13 11.15 11.17 11.18 11.19 11.19 11.21 11.25 11.26 11.29 11.30 15 arrival at A – lock empty 14 arrival at B 17 arrival at B 16 arrival at A 19 arrival at A 20 arrival at A 21 arrival at A – lock raised 22 arrival at A 23 arrival at A 24 arrival at A – lock up – lock empty – lock lowered 18 arrival at B – lock down simulation ends Waiting at A B 50 In lock 10 6 Next scheduled Event arrival at A arrival at B arrival at A lock lowered lock down arrival at B arrival at B arrival at B 11.02 11.04 11.06 11.03 11.07 11.04 11.04 11.10 arrival at A lock empty arrival at A lock raised arrival at B arrival at B arrival at A arrival at A arrival at A arrival at A lock up arrival at A arrival at A arrival at A lock empty lock lowered lock down arrival at B lock empty 11.08 11.09 11.11 11.15 11.10 11.26 11.11 11.12 11.13 11.17 11.19 11.18 11.19 11.39 11.21 11.25 11.29 11.33 11.31 Entity no 10 11 12 13 14 15 16 17 18 19 20 21 22 – 23 24 25 – – – Number of lock movements: Number of boats moved: down, up Number of arrivals at A: 10, at B: Waiting time (including boats already there) at A: 189 min, at B: 236 50 Simulation 51 37 A job shop has three work centers X, Y, Z Each job has to go through some or all centers in a prescribed unique sequence A work center can only process one job at a time but the next job can enter immediately after the previous job departs The current state of the system is as follows: Jobs currently in the system Sequence of centers left to be entered Z Z X XY XYZ 10 XZ YXZ Y YZ Y Work center X is currently processing job 1; the scheduled release time from the center is at simulated time 124 minutes Work center Y is currently processing job with a scheduled release time at 180 minutes Jobs are processed by each center on a first-come-first served basis The current order at center X is job 1, 3, 4, 5, 6, and at center Y, 2, 7, 8, 9, 10 Note that center Z is currently idle Processing times in minutes at each work center are normally distributed as follows: Statistical measure Mean (min) Standard deviation (min) Work center X Y Z 120 40 200 40 180 60 New jobs enter the system at the beginning of each day, at which time they are added to the file of jobs waiting for processing at each center Current simulated time is t = 100 minutes a Identify the components of this system in terms of entities, attributes of the entities, activities and their associated events, and the calendar Entities (attributes) Jobs (processing sequence) Work centers (processing time distributions) Calendar ([arrival at a work center, time], [completion at a work center, time]; must be specified for each job) b Draw a flow diagram similar to the one in the text for the inventory system that depicts the logic of the work center operations 51 Simulation 52 In the flow diagram, FSE is the “file of scheduled events” or calendar and “file i” refers to jobs waiting to be processed at work center i.” REMOVE FIRST EVENT SCHEDULED ON OR AFTER TIME t FROM FSE NEXT EVENT FSE UPDATE TIME t WHAT TYPE EVENT WORK CENTER Y COMPLETED WORK CENTER X COMPLETED WORK CENTER Z COMPLETED i YES FILE i NO FILE i EMPTY WORK CENTER i IDLE START WORK ON HIGHEST PRIORITY JOB FILE i GENERATE JOB COMPLETION TIME AND ENTER IN FSE YES FSE NO LAST PROCESS FOR JOB ? WHAT PROCESS IS NEXT ? CANCEL JOB, RECORD STATISTICS X Y Z i YES CENTER i IDLE? STATUS: CENTER i BUSY GENERATE JOB COMPLETION TIME AND ENTER IN FSE NO ENTER JOB IN FILE i FILE i FSE 52 Simulation 53 c Simulate the system with the data given until simulated time t = 480 minutes Use the random number table starting at the upper left and go down the columns Time Completion job center center (Status at 100) 124 X 124 Status† Calendar Files‡ job time X Y Z X Y X 124 B B I in in Y 180 in in in in in 10 in Z 318 X 171 B out B out 171 X X 321 180 Y Y 375 318 Z Z 533 321 X X 457 375 Y Y 594 457 X X 645 480 end Completions Z job time 171 318 B B out B B B in out out in in out out in † B = busy, I = idle ‡ Indicates the status of the job at the specified point in time: in = waiting, out = finished 53 Simulation 54 38 Draw two flow diagrams representing the activities and logic for the problem in Exercise 31, one for the movement of ships and one for the unloading and loading operations The latter should contain the details associated with the load and unload activities of the former Solution not available 54 Simulation 55 39 Bonzi Air operates a 24-hour parcel pick-up service for air freight delivery to East Coast cities Customers make pick-up requests by phone to a dispatcher, who fills in an order form and then assigns the request to the first available driver After completing a pick-up, the driver fills in a pick-up report, which he or she files with the dispatcher, and then waits for a new assignment A dispatcher will interrupt assigning a job to a driver to take customer phone calls, and only continues with the job assignment after having prepared the pick-up order form There are two dispatchers who alternate taking customer calls except when a call comes in while a dispatcher is still engaged with the previous customer The number of incoming phone lines is sufficiently large so that no customer ever gets a busy signal All pick-up orders are available to both dispatchers for assignment to drivers Past records for a particular office show that ∑ The average rate of pick-up requests is 24/hr and follows a Poisson process ∑ The time required to receive a pick-up request and fill in the order form is normal with mean 120 sec and standard deviation of 20 sec ∑ The time to make a job assignment is a constant 60 sec ∑ The time to make a pickup and file the report is normal with mean of 1500 sec and standard deviation of 400 sec ∑ There are 16 drivers a Identify entity classes, and activities, and list all entities jointly engaged in each activity Identify the calendar to be used for a next-event simulation Entities (attributes) Requests (arrival process and statistics) Dispatchers (rate at which requests are processed and job assignments made) Drivers (distribution of time to make pick-up and file the report) Most imminent events (entities jointly engaged in each activity) Arrival of pick-up request Dispatcher finishes taking pick-up order Dispatcher finishes assigning job to driver Driver returns after completing pick-u[ Conditions to be checked to see if other activities can be started or completed (entities jointly engaged in each activity) Dispatcher starts pick-up request Dispatcher starts assigning job to driver Driver starts pick-up assignment Calendar (Event, Entities associated with event, Scheduled time of event) 55 Simulation 56 Events Ci in means ‘request i has arrived’ (phone is ringing) Ci out means “dispatcher finishes with request i’ Asn i means ‘dispatcher assigns job i’ Reasn i means ‘assignment of job i is interrupted and a new time is set after the completion of the call’ Ji end means ‘driver completes job i’ Entities Vj refers to driver j Dk refers to dispatcher k b Draw a flow diagram similar to the one in the text for the inventory system depicting the logic of the dispatch operations 56 Simulation 57 Customer source Call arrives Create job Call waits Dispatchers idle Drivers idle Job file Start assignment Call taken - Assign job - Interupt call Pick up job & file report End assignment Calls are initiated from the customer source node and arrive at the company’s phone bank A job is created for every call and stored in a job file When the pick-up is made and the driver finishes the paperwork, the job file is updated with this information Calls wait on hold if both dispatchers are busy with other calls If at least one dispatcher is free or in the processes of assigning a job, he is interrupted to take the call This is shown in the diagram with the dashed lines An assignment is started as soon as at least one dispatcher and one driver is free After the assignment ends, the pick-up is made and the paperwork is filed The driver then returns to the driver idle node where he or she awaits another assignment 57 Simulation 58 c Assume that at 11:00 AM the system is in the following state: (1) no pending requests; (2) dispatcher is idle; (3) dispatcher is processing a pick-up request with 80 seconds to go; (4) the next pick-up request will be received at 100 seconds from now; (5) drivers 1, 2, 3, and are idle; (6) busy drivers and pick-up completion times (including filing a pick-up report) in seconds after 11:00 AM are: 240, 1250, 680, 560, 1650, 10 960, 11 180, 12 -1140, 13 1370, 14 510, 15 870, 16 1410 Simulate this system for hour Round all times to the nearest 10 seconds Collect statistics on the total idle time of dispatchers and drivers, and on the total time pick-up orders wait in the office until they are assigned to a driver Generate all activity times at the start of each activity only The following table can be used to organize the simulation Only the first few increments are shown Of course, the results depend on the random numbers generated, and the priority rules used for processing events that occur simultaneously 58 Simulation 59 File of scheduled Time Event Entities Event Entity Time X † Calls waiting Dispatch Status Order file File of idle drivers Status at time zero C1 C0 80 100 C0 end C1 in D2 D1 110 C2 in D2 180 200 220 240 270 J end C3 in C1 end C2 end Reasn V11 D1 D1 D2 D2/V1 in end jobs Asn C1 C2 C2 Reasn C3 end in end in D1 D2 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 D2/V1 D1 D2 D2 D2/V1 D1 100 80 240 1250 680 560 1650 960 180 1140 1370 510 870 1410 140 220 110 240 270 200 C4 C3 in end D2 D1 390 350 JO Asn end V1 D2/V2 1960 330 x x I B in in in in I B B B out B B B B B B B B B B B B x x x x x x x 11 in in out in in out out etc † X = event has been processed, I = idle, B = busy 59 ... 61.5 61.5 62 Operations Research Models and Methods by Paul A Jensen and Jonathan F Bard John Wiley and Sons, Inc Copyright 2003 – All rights reserved Chapter Linear Programming Methods Consider... problem Operations Research Models and Methods by Paul A Jensen and Jonathan F Bard John Wiley and Sons, Inc Copyright 2003 – All rights reserved Chapter Linear Programming Models Write the constraints... Programming Models Man A Man B Man C X Man A Man B Man C Hrs 40 32 16 Dual -.22 0 X X X 0.5 X 16 X X X 10 X Linear Programming Models 17 13 A company has two manufacturing plants (A and B) and three