Sample Chapter. Basic Hard Systems. Engineering: Part I

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2 Basic Hard Systems 10 Engineering: Part I 10.1 Introduction: Hard Systems Analysis In Chapter 1 we dealt briefly with the rudiments of the systems approach to problem solving. In this chapter, we begin by elaborating the meaning of some specific terms associated with the systems approach. This approach offers systemic (holistic rather than piecemeal) and/or systematic (step-by-step rather than intuitive) guidelines to problem solving. Both systemic and systematic methodologies and techniques are used by engineers. Techniques, in general, are precise, specific programs of action that will produce a standard result. Methodology, on the other hand, lacks the precision of a technique, but is a more definitive guide to action when compared to a philosophy. Nearly all important real-world problems that we face on a day-to-day basis are systemic problems, such as sustainability and environmental problems, homelessness and poverty problems, social and economic problems, and so forth. These complex problems are truly systemic in the sense that they cannot be attacked on a piecemeal basis, partly because of their interconnectedness. From the early 190s, systems analysis (which is the economic appraisal of different means of meeting a defined end) and systems engineering (which involves the design of complex, technical systems to insure that all components operate in an integrated, efficient way) have been widely used in problem solving all over the world. When systems analysis and systems engineering are put to use for solving problems concerning natural and physical systems, we describe this approach as the hard systems approach. Essentially, the hard systems approach defines the objectives to be achieved and then engineers the system to achieve these objectives. However, when dealing with problems involving human activity systems, one notices that the problem is usually ill-defined. Such cases are defined as soft systems. In contrast to hard systems engineering, soft systems methodology does not seek to mechanically design a solution as much as it orchestrates a process of learning (Checkland, 1981; Khisty, 199). Operations research (OR) and management science can also be classified as a hard systems methodology comprising a range of techniques that are typical of the means-end approach. It is well known that OR emerged as a means of tackling the vast logistical problems that were encountered during World War II. Later, a variety of formal quantitative techniques based on the principles of OR were developed for use in every conceivable area, including manufacturing, production, transportation, and construction management (CM).

3 6 Basic Hard Systems Engineering: Part I Chapter 10 This chapter deals with several techniques that form the basis of hard systems methodology. First, we deal with methods based on calculus. Next, three of the best known methods of network analysis (Critical Path Method [CPM], Program Evaluation and Review Technique [PERT], and Line-of-Balance [LOB]) used extensively in CM are presented, followed by three other methods of network analysis: shortest path, minimal spanning tree and maximal flow. Lastly, the basic ideas of linear programming (LP) are described. LP is a quantitative method of analysis used extensively in business and engineering. 10. Methods Based on Calculus The classical methods of calculus provide elegant and powerful solutions to a relatively large number of problems encountered in engineering and economics. At the same time, one of the principal assumptions on which it rests is that the variables that describe a problem must be continuous along all points. This assumption limits its use for practical problem solving, that is, network systems in transportation or choosing between discrete projects in CM. Many of the tools described in this chapter are predominantly linear and involve the solution of sets of linear equations. A large number of managerial problems consist of one or more nonlinear relationships, where traditional linear solution methods are not applicable. Fortunately, you have already been exposed to the basic principles of microeconomics in Chapter, and some of the examples worked out with respect to demand, supply, and elasticity are revisited in this section. The optimization of a nonlinear objective function may be constrained or unconstrained. The former may be solved by the method of substitution or by the use of Lagrange multipliers. The best way to get familiar with these techniques is to work through the examples given in this chapter Production Function Characteristics A production function is a basic representation for the conversion of resources to products. A production function could be represented as: Z = k(x 1, x,... x n ) For example, Z could be the maximum number of houses provided by a city, where x 1 represented the land provided, x the labor supplied, and so on. The shape of the production function has important implications regarding where to search for an optimum solution. The following example illustrates the use of calculus.

4 10. Methods Based on Calculus 7 Example 10.1 If the total cost (TC) of providing labor for the repair of motors is given by: 10.. Relationship among Total, Marginal, and Average Cost Concepts and Elasticity You were introduced to the price elasticity and cost functions in Chapter. We make use of these concepts in this section. Remember that the price elasticity (e) of demand is: e = which is frequently expressed as: dq/ dp e = = q/ p The next example uses this relationship. TC = 0 + X -. X + 1 X dq dp # p q where X = number of labor involved in repairs and TC = total cost, find the relative minimum and maximum labor force required for this cost function. What is your recommendation? Solution TC = 0 + X -. X + 1 X The necessary and sufficient condition for a maximum or minimum are: Taking the second derivative: d( TC ) = - 11X + X = 0 dx ` X = 8 or X = d ( TC) = X dx At X = 8, 11 + ()(8) = > 0 At X =, 11 + ()() = < 0 Thus, at X = 8, TC = 0 + (8).(8) + (8) = $0.67 (minimum) and at X =, TC = 0 + ().( ) + () / = $71.0 (maximum). It is recommended that the labor force be kept at 8 laborers to minimize the cost function. marginal cost average cost

5 8 Basic Hard Systems Engineering: Part I Chapter 10 Example 10. If the demand for bus travel tickets between two cities is q = 800 p p, where p, the price of the ticket, is $10, and q is the number of tickets sold, what is the price elasticity of demand? Solution q = 800 p p dq =- - p dp Substituting the value of p = 10 in the equation: dq =- - # 10 =- dp Next, find the number of tickets sold when p = $10: Substituting these values: Hence, inelastic. q = 800 (10) (10) = 60 dq p e = # = (- ) 10 e o = dp q 60 Example 10. If a company s demand function for machines is p = 0.q and its average cost function is: AC = q - 8q q find the level of output which (a) maximizes total revenue, (b) minimizes marginal cost (MC), and (c) maximizes profits. Solution (a) Demand function is p = 0.q Total revenue (TR) is (p)(q) = ( 0.q)q = q 0.q To maximize q, Testing the second-order condition: d( TR ) = - ( 0. )( ) q and equating this to zero: dq q = d ( TR) = dq Continues

6 10. Methods Based on Calculus 9 Example 10.: Continued Thus, at q =, TR is a maximum. (b) From the average cost function AC = q 8q /q Marginal cost: Total cost, TC = (AC)(q) = eq MC is minimized when: Testing the second-order condition (c) Profit = TR TC: - 8 q o q q = q - 8 q + 7 q + d( TR) MC = = q - 16q + 7 dq d( MC ) = 6q - 16 = 0, and q = dq d ( MC) = 6 0 dq ` at q = MC is at a minimum = ( q - 0. q ) - ( q - 8q + 7q + ) =- q + 7. q - 1q - for maximizing profit (Pr): testing the second-order conditions: At q = 1; this results in 9 > 0 At q = ; this results in 9 < 0 ` Profits are maximized at q = and Profit = () + 7.() 1() = 6 d( Pr) =- q + 1q - 1 = 0 dq ` q = 1 or q = d ( Pr) =- 6q + 1 dq 10.. The Method of Lagrange Multipliers The method of Lagrange multipliers can be used for solving constrained optimization problems consisting of a nonlinear objective function and one or more linear or nonlinear constraint equations. The constraints, as multiples of a Lagrange multiplier, λ, are subtracted from the

7 0 Basic Hard Systems Engineering: Part I Chapter 10 objective function resulting from a unit change in the quantity value of the constraint equation. This characteristic will be obvious when the following example is worked through. Example 10. The cost function of a firm selling two products A and B is C = 8A AB + 1B. However, the firm is required by contract to produce a minimum quantity of A and B totaling. What are the values of A and B, and what is your interpretation of the value of λ? Solution Set the constraint to 0, multiply it by λ and form the Lagrange function: Take the first-order partials: C = 8A AB + 1B + λ(a + B ) C A = 16A B + λ = 0 C B = A + B + λ = 0 C λ = A + B = 0 (A = ; B = 17; and λ = 8 which means that a one-unit increase in the production quota will lead to an increase in cost by approximately $ Critical PATH Method One of the popular uses of network analysis is for the planning and monitoring of projects before and during execution. Such analysis is vital in order to finish a project within the budget allotment and prescribed time limit. CPM and PERT are the two most popular network analysis techniques used for project planning. Developed in the late 190s to aid in the planning and scheduling of large projects, today, CPM and PERT are used worldwide. Both techniques have many characteristics in common, although CPM is deterministic whereas PERT is probabilistic. Both involve the identification and proper sequencing of specific tasks or activities to complete projects in time. Also, the relationship between specific tasks and the logic of precedence is important, as is their duration and quantification. Coupled with these qualities is the classification and quantity of labor, along with their periods of time and wages. The planning of cash flows and financial assistance is also a crucial part of CPM and PERT. Project planning involves the identification and sequencing of specific tasks, their duration, and their relationships. This process is represented by a network, not necessarily drawn to scale. Two types of networks are currently in use; an Activity-on-Arrow (AOA) and an Activityon-Node (AON). We describe only the AOA network. The AOA network consists of arrows (branches) and nodes. The arrows represent activities (or tasks) while the nodes represent the

8 10. Critical Path Method 1 Start ES ES 0 1 A 0 8 LS A A EF FF IF ES EF LF TF 8 B 9 C LS FF 1 17 D 1 17 G 18 0 Dummy E task 1 1 H 1 6 F (a) B ES (b) 0 7 Finish 0 beginning and end of activities referred to as events. Since a number of terms are used in CPM, it is best to begin describing them with the help of a typical diagram (see Figure 10.1) Key Concepts 1. CPM is a linear graph consisting of nodes and arrows as shown in Figure Two methods of diagramming can be used: AON or AOA. We use the AOA method.. Dummy activities have zero duration. For example, Activity (-6) is a dummy activity which means that Activity (6-7) cannot start before Activity (-) is completed.. The forward pass gives the early start (ES) and the early finish (EF) time of an activity. The forward pass establishes the earliest time for each event.. The backward pass gives the late start (LS) and the late finish (LF) time of an activity. The backward pass is simply a reversal of that for calculating the earliest event time. 6. When ES = LS for an activity, it lies on the critical path. 7. The critical path (CP) is the set of activities that cannot be delayed if the project is to be completed on time. 8. Total float (TF) is the amount of time that an activity may be delayed without delaying the completion of the project. TF = LF EF = LS ES; also, TF = FF + IF. Free float (FF) is the time that the finish of an activity can be delayed without delaying the ES B EF EF LF ES = Early start EF = Early finish LS = Late start LF = Late finish TF = Total float FF = Free float IF = Interfering float Figure 10.1 (a) CPM network showing nodes and arrows and (b) total, free, and interfering float

9 Basic Hard Systems Engineering: Part I Chapter 10 time of any activity that follows. FF = ES of the following activity minus EF of the activity in question. 9. The CP is the minimum time in which a project can be completed and is the duration of the longest path through the network CPM Scheduling To keep control over a CPM network while it is being prepared and worked out, the following steps are useful: (a) List all the activities sequentially and estimate their duration. There may be two or more activities that are performed simultaneously. (b) Pay special attention to which activity precedes (or follows) another activity, so that a proper logic of the project is maintained. (c) Draw an AOA network with the activities and events properly interconnected. If necessary, introduce dummy activities to maintain the logic and sequencing (in time) of all the activities in question. (d) Make a forward and backward pass through the network to establish ES, LS, EF, and LF times for all the activities. (e) Determine the CP and the corresponding critical activities. (f) Prepare a table with all the details as shown in Table Note that the first activity starts at zero and we add the duration to its ES to obtain its EF time. In this manner, you can progress through the network calculating ES and EF times for all activities, always choosing the preceding EF with largest time, at that node. Next, we can work backward from right to left, which is called the backward pass. On the last activity the EF time becomes the LF time, in order to finish the project as soon as possible. The LF time of the last activity is its LF time minus its duration. Working backward, the LF and LS times for preceding activities can be determined, noting always that the smaller value has to be taken into account. The CP is the longest interconnected path through the network. All activities on this path have the same ES and LS times (and similarly they have the same EF and LF times). Note that these activities have no float to their durations. Finally, all values of ES, LS, and LF times are put in a table (see Table 10.1) and the TF and FF are calculated as per definitions given before The Time-grid Diagram and Bar Charts CPM networks are not generally drawn to scale and, therefore, the lengths of the arrows do not represent the duration of tasks. However, the arrows in time-grid diagrams are drawn to scale in the horizontal direction (but not in the vertical scale). FTs are represented by broken horizontal lines whose lengths indicate time. (See Figure E10..) Project network activities can also be represented by bar charts (or Gantt charts) as shown in Figure E10..

10 10. Critical Path Method Table 10.1 Activities, times, and floats 0 1 A 0 8 Activity Duration ES EF LS LF TF FF IF *A *B C *D E 6 0 F *G H *Activities on the critical path 10.. Resource Scheduling 8 B 9 C 1 17 D 1 17 G 18 0 E 1 1 H 1 6 F Project managers generally want to proceed with projects under the condition that they will be executed efficiently at scheduled rates. They would like to verify that the resources, in the shape of man power, cash flows, machinery, and requirements are available to them. It is worthwhile for managers to draw up a resource allocation diagram, say with an ES timing, to see whether it could be improved. One can, of course, only tinker with those activities that are not on the critical path. This is exactly what has been done through the resource leveling procedure shown in Figure E10.. Notice how the fluctuations of labor have been leveled off. Example 10. An electric substation is to be installed and the following basic tasks are identified (Table E10.). The site will be cleared and leveled and the necessary materials for preparing the foundations and fencing will be procured and stored at site. Next, the foundation excavation and pouring of concrete will be done, a fence will be constructed, and the site will be cleared up and handed over to the authorities. Draw the activities network, perform the forward and backward passes, complete the activities, times, and floats table, draw the time-grid and Gantt chart, and finally sketch resource allocation diagrams, first with ES times followed by the procedure of leveling the labor resource. Continues

11 Basic Hard Systems Engineering: Part I Chapter 10 Example 10.: Continued Table E10.a Activities and tasks for electric substation Activity Task Duration Labor Logic and sequence A Select site A is independent B Clear site B and C can be done simultaneously C Procure materials 6 C follows A; F can follow C and D D Excavate foundation 10 E Fix fence 6 E can be done only after A and B are finished F Pour concrete and cure 8 8 F can be done only after D is finished G Fix gate to fence After E and F are completed H Dummy 0 0 Clean up cannot start till gate is fixed I Clean up and hand over 7 Solution Table E10.b presents activities and their corresponding events. Table E10.b Activities, times, and floats for electric substation Activity Event Duration ES EF LS LF TF FF IF A* B* C D* E F* G H I* *On the critical path Figures E10. shows (a) the plan of the substation, (b) the activities network, (c) the forward and backward passes, (d) a typical way of showing ES, EF, LS, and LF, (e) the time-grid diagram, (f) the Gantt chart, (g) resource allocation based on ES, and (h) resource leveling. Continues

12 10. Critical Path Method Example 10.: Continued Gate A 1 0 A ft X X X X X X 00 ft X X X X Foundation Base X X X X X X B B (a) Plan of electrical substation C D G (b) Activities network 100 ft X X X X X (c) Forward and backward pass E 6 10 ft (d) Typical way of showing ES, EF, LS, LF on activities Figure E10.a-d Critical path F 8 H 0 I G 19 8 D E H C F I ES EF LS 1 19 LF Continues

13 6 Basic Hard Systems Engineering: Part I Chapter 10 Example 10.: Continued 0. A B C D E F G H I A B C D G E F Time in days (e) Time-grid diagram H Time in days (f) Gantt chart of activities I Figure E10.e-h CPM example Number of workers Number of workers Days (g) Resource allocation with early start 10.. Time-cost Optimization CPM is effective in discovering the possibility of minimizing the cost of the project from its normal duration by reducing (or crashing) the duration of some of the individual activities in the network that lie on the critical path, by paying a higher cost (or crash cost). This extra cost is often offset by savings gained through lower overall indirect cost (or overhead cost). This technique of compressing certain tasks and thereby optimizing a project cost is often called project crashing. Typically, the crash cost per unit of time for an activity can be found as follows: Crash cost per unit time = Crash cost - Normal Cost Normal time - Crash time A A G 0 E 16 C B C B 1 C D D G E G E 1 1 F Days 8 G E 10 G F I 7 7 (h) Resource leveling I Labor requirements (Men) A B C 6 D 10 E F 8 G H 0 I 7

14 10. Critical Path Method 7 The first step in crashing a project is to identify the critical activity with the minimum crash cost per unit of time and crash it with the allowable limit, taking into consideration the amount of float available with respect to multiple critical paths. The second step is to revise the network to identify new or multiple critical paths until all of the activities available for crashing have been utilized. Finally, all the crashing steps together with the corresponding cost increases are compared with the savings derived from overhead costs. See the following example for clarification. Example 10.6 Determine the optimal completion time for a project whose logic is shown in Figure E10.6. Indirect cost is $10 per da. All other details are given in Table E10.6a. Table E10.6a Activities and times Activity Normal duration T N (days) Total cost (Normal) C N Crash duration T C (days) Total cost (Crash) C C A $600 $1000 B 6 $800 $100 C 8 $00 $100 D 7 $600 $100 E $00 $00 F 1 $100 1 $ B 6 A C 8 B A C 8 E D (a) Activities diagram E 1 1 F 6 I 7 19 F D 7 (b) Forward and backward passes Cost Figure E10.6a-b Activities and activity slopes Continues

15 8 Basic Hard Systems Engineering: Part I Chapter 10 Example 10.6: Continued Cost C C C N T C (b) Forward and backward passes T N (c) Slope calculations Activity C Slope = T A 00 B 00 C 10 D 10 E 0 Days F 0 Figure E10.6c Activities and activity slopes Solution The first task is to complete the activities table shown, after performing the forward and backward passes, as shown in Figure E10.6. Table E10.6b is completed followed by working out the time schedules to find the one that is the cheapest. These schedules are graphically shown in Figure E10.6a. The crash cost per unit time is the slope of the cost line and is computed for various activities as shown in Figure E10.6. Table E10.6b ES, EF, LS, LF and TF Activity Duration ES EF LS LF TF A B C D E F Schedule 1: Normal Schedule; Critical Path 1 6 = 0 da Cost = ( ) = $100 Schedule : Select activities on CP; choose D (smallest slope) Compress D (limit da; total float 11 da) Direct cost = (10 ) = $700; Time: 1 da TF: 11 = 6, CP: 1 6 = 1 da C C C T Schedule : Compress next cheapest activity Compress C; TF = 6 Crash Direct cost = (10 ) = $00 Time 9 da; TF: 6 = 1 CP: 1 6 = 10 da N N Continues

16 10. Critical Path Method 9 Example 10.6: Continued Schedule : Compress next cheapest activity Compress A; TF = 1 Crash = Compress A for only 1 Direct cost = $00 + (00 l) = $600 Time 9 da TF: 1 1 = 0 CP: 1 6 = 9 da and CP: 1 6 = 9 da Schedule : Any other compression will affect both CPs. Also, A can be compressed only 1 more day. Compress A and B, one day each = $00 Direct cost = $600 + $00 = $000 Cost $000; Time 8 da. The time-cost optimization is shown in Figure E10.6d; and a summary of costs for schedules is given in Table E10.6c. As indicated in the table, Schedule is the cheapest at $900. Cost 6,000,000,000 6,00,90,900,90 6,100,000 1,00,600 1,0,00 1,00,700, Days (d) Total, direct, and indirect cost Figure E10.6d Total, direct, and indirect cost Table E10.6c Schedule costs,100,000 Schedule 1 Days Direct cost ($) Indirect cost ($) Total cost Total cost Direct cost Indirect cost

17 0 Basic Hard Systems Engineering: Part I Chapter PROGRAM EVALUATION AND REVIEW TECHNIQUE and THE Line-of-Balance The PERT and the LOB techniques are both closely associated with CPM. PERT was developed to analyze projects in an environment of uncertainty, particularly with projects where the specific duration of activities could not be estimated with reliability. PERT uses two probability density functions: (1) the beta (β) distribution for each activity and () the normal distribution for estimating the completion time of the entire project (see Figure 10.). Other than the use of these probability functions, PERT is similar to CPM. PERT is described first in Section The LOB technique was developed by the U.S. Navy for controlling and managing production processes. It has since been used in the construction industry. LOB is described in Section Probability density Probability density t a t m t e Normal distribution Range β-distribution t b (a) Beta (β) distribution t a = Optimistic time t b = Pessimistic time t m = Most likely time t e = Expected mean time Time T P x = Expected total project time T s = Target time T s T P x Time (b) Normal distribution Figure 10. Key concepts (PERT): (a) beta distribution and (b) normal distribution

18 10. Program Evaluation and Review Technique and the Line-of-Balance Key Concepts of PERT PERT introduces the concepts of uncertainty into time estimates as opposed to CPM, which is essentially deterministic. PERT uses expected mean time (t e ) with standard deviation, σ te or variance ν te. The expected mean time (t e ) of an individual task is an estimate having an approximate chance of 0 percent success. The value of t e is calculated from: t a the optimistic time of completion of an individual task; t m, the most likely time; and t b, the pessimistic time of completion of this task. This forces the planner to take an overall view of each task s duration. The beta (β) distribution uses t a, t b, and t m to estimate the expected mean time. The expected mean time t e is: t t t a m b t = + + e 6 with a standard deviation, and a variance, o te t t b v = - te 6 t = ( v ) = e te b a - ta o 6 t a and t b have small probabilities, around to 10 percent. Once t e and v te are found for each activity, the critical path is found in the same fashion as in CPM. Project duration = Expected mean duration T XP which is the expected mean time along the CP. Once the expected mean time for an event (T x ) and its standard deviation v T X are determined, calculate the event schedule time (T s ). This has a normal probability distribution with mean T x and v T X. The effect of adding a series of independent β-distribution gives a normal distribution. To determine the probability that the expected total project time T x will exceed some target time T s first calculate the value of Z, a dimensionless parameter expressing the horizontal axis of the standardized normal distribution function, where: ( T - T ) S X Z = v Refer to Table 10. to find the corresponding probability associated with the value of Z. TX

19 Basic Hard Systems Engineering: Part I Chapter 10 Table 10. Values of Z and probability Z Probability Probability Z Example 10.7 An activity network for a small house is shown in Figure E10.7. Table E10.7a shows the optimistic, most likely, and pessimistic times for various activities under Columns,, of the table. What is the probability of finishing this project in 110, 11, 117, 119, 1 da? Solution The expected mean time of all the activities are calculated using the formula t e = (t a + t m + t b )/6 and entered in Column of Table E10.7. These values are also shown against each activity on the network. One can now do a forward and backward pass to determine the CP (in a manner similar to what was explained in the section on CPM). Now that the CP is known, one can calculate the standard deviation σ te and variance ν te of the critical activities and enter these in Columns 6 and 7 respectively. The critical activities are marked with an asterisk and the total project duration works out to be 117 da with a corresponding variance of 7.1 da. Taking the square root of 7.1 da gives a standard deviation of 8.6 da. Notice that we cannot add values under Column 6 to get the standard deviation. We add the variance of critical activities and then take the square root of this total to get the standard deviation. While we have found that the expected mean duration of this project is 117 da with a standard deviation of 8.6 da, the probability of finishing the project at a target time of 110, 11, 119, or 1 da is needed. For finding these probabilities we must calculate the corresponding values of Z, as shown in Table E10.7b. Notice that the probability of completing the project in 117 da is merely 0%. Continues

20 10. Program Evaluation and Review Technique and the Line-of-Balance Example 10.7: Continued D i Figure E10.7 PERT network i i j indicates the critical path Table E10.7a Activities and their statistics Activity t a t m t b t e σ te ν te * * * * * * * * * *On the critical path. Note that 7. 1 = 8. 6 Continues

21 Basic Hard Systems Engineering: Part I Chapter 10 Example 10.7: Continued Table E10.7b Expected time and probabilities T T S X Expected time T Xe Standard deviation σ Target time Z = - Probability (%) v The LOB Technique The LOB technique is a management-oriented tool for collecting, measuring, and presenting information relating to the time and accomplishment of repetitive tasks during production. One of the major problems facing managers is obtaining information on the status of various operations soon enough to take effective action. LOB is particularly useful in repetitive construction work such as multi-house projects, road pavement construction, and the manufacturing of hundreds of identical units, such as small septic tanks, pylons, and beams. The LOB technique consists of four elements: (1) the objective chart, () the program chart, () the progress chart, and () the comparison. The objective chart is a graph showing the cumulative end product to be manufactured over a period of time, while the program chart is a flow process diagram showing sequenced tasks and their interrelationship with lead times. Lead time is the number of time periods by which each activity must precede the end event to meet the objective. The progress chart consists of vertical bars representing the cumulative progress of each monitoring point based on site visits to the production area indicating the actual performance. The comparison activity is derived from the objective, program, and progress charts to draw the LOB. When one draws the LOB on the program chart, it represents the number of completed units that should have passed through each control point at the time of the study in

22 10. Program Evaluation and Review Technique and the Line-of-Balance order to deliver the completed units according to the contract schedule. Many of these terms will become clear in working through the illustrative examples that follow Progress Charts and Buffers Suppose a company has been awarded a contract to erect ten steel pylons. The sequential operations involved are A = excavate; B = pour concrete; and C = erect pylon, as shown in Figure 10.a. This sequence needs to be repeated ten times to complete the work. However, to provide for a margin of error in the time taken to complete each operation, a time buffer is provided between two operations, as shown in Figure 10.b. If the work order to begin work is given on Day 1, and one pylon has to be handed over at the end of every fifth working day, then the first pylon 1 A B C 18 days 1 days 10 days (a) Activities A, B and C without buffers A Buffer B Buffer C 1 18 days days 1 days days 10 days = 0 days 1st (b) Activities A, B and C with buffers Pylons th 9th t 1 t (c) Completion schedule of 10 pylons = 0 days Day of delivery Figure 10. Line of balance: (a) activities A, B, and C without buffer; (b) activities A, B, and C with buffers; and (c) completion schedule of 10 pylons

23 6 Basic Hard Systems Engineering: Part I Chapter 10 will have to be handed over at the end of the fiftieth day, and subsequent pylons will be handed over every fifth day. Refer to Figure 10.c, if we use the straight-line equation Q = mt + c: Q 1 = 1; t 1 = 0; Q = 10; m = 1/; c = 1, and t =? then t = 10-1 e o + t = ( 9 ) + 0 = 9 da 1/ 1 # At the end of the ninety-fifth day, the tenth pylon will be completed. Notice that when we graph this problem, we must be mindful that on the fiftieth day we are ready to complete the first pylon, with nine more to go. Also, notice that we must begin work on the first pylon on the first day to get it ready by the fiftieth day. We will now consider another example to illustrate a more realistic case of LOB. Example 10.8 A septic tank prefabricator has received a contract to supply 1000 septic tanks, and to deliver 0 units each mo beginning on the first of the twentieth month. The major control points of the production scheme are shown in Figure E10.8a. Units A C E B Objective chart D (7 0)=80 F G Lead-time in months H 1 I 0 Program progress chart (a) LOB (0) (80) (0) (00) (60) (0) Figure E10.8 LOB example A Months (b) 8 0 B C D E F G H I Control points Continues

24 10. Program Evaluation and Review Technique and the Line-of-Balance 7 Example 10.8: Continued An LOB study performed on the first of the twenty-sixth month revealed that the number of completed units that actually passed through each control point is as follows (Table E10.8a): Table E10.8a Control points and units Control point Units A 0 B 0 C 0 D 0 E 00 F 0 G 10 H 8 I 0 Draw the objective chart, the program progress chart, and the LOB chart. Determine the deviation of units. Solution Table E10.8b presents the results of calculations. The table indicates that the performance of control Points A, C, E, H, and I are behind schedule and that corrective action is needed. Table E10.8b Control point performance results Control point Cumulative units to be delivered Units actually completed Deviation A B C D E F G H 0 8 I

25 8 Basic Hard Systems Engineering: Part I Chapter Resource and LOB Schedule Estimation of resources including labor requirements are an important feature of LOB schedules. In general such estimation is best done by: preparing a logic diagram of all the activities and tasks including sequenced and parallel (or simultaneous) activities; estimating the manhours required to complete each task; choosing realistic buffer times that reflect the risk entailed in not completing sequenced activities; calculating the required output target to meet a given project completion date; and finally putting all the information in the form of a convenient table as shown in Table 10.. Table 10. Calculation sheet for LOB schedule Activity Man power Men per Theoretical Actual gang Actual rate Time in days Time in per activity activity gang size size of output for 1 activity days Explanation of columns: 1: Major activities or tasks : Estimate of man-hours needed for each activity : The optimum number of laborers needed for each task (which is labor in each team) : The theoretical gang size needed to maintain the output rate (R) given by: R # ( Column ) Number of hr/wk : The actual gang size is chosen as a number that is a multiple of men required for one team 6: Actual rate of output = ( Actual gang size) # ( Target rate) Theoretical gang size 7: Time taken for one activity = Man-hours for activity ( Number of men in one team) # ( Number of hours in a working day) 8: the time in days from the start of the first section to the start of the last section is: ( Number of sections - 1) # ( Number of working days per week) Actual rate of build Table 10. helps us to draw the various activities which show the sections completed against time. Also, the slope of the activities is a function of several factors, such as the total units of time required to complete Q repetitive units; F, the resource unit factor, is the number of units

26 10. Program Evaluation and Review Technique and the Line-of-Balance 9 Units Units 1 a a a a 7 days = days Figure 10. Logic diagrams for teams a Days of the resource that are required to achieve the rate of working necessary to meet the handover program; d, the activity duration; and m, the rate of handover. The buffer time allowed plays an important role in the entire completion time. Lastly, the actual gang size in relation to the men per activity is important as it determines the slope of the activity as shown in Figure 10.. If a task takes 7 da, with one team shifting from one unit to the next, it would take da for the completion of units. If the same task is arranged with two teams (a and b), the units would be completed in 1 da. These alternatives are shown in Figure 10.. The following example explains the procedure of LOB. 1 a b a b 1 days Example 10.9 A contract has been awarded to erect 1 steel pylons at the rate of three per wk, (assuming da of 8 hr each) as per details given in Table E10.9a. Table E10.9a Tasks and man-hours Task Man-hours Optimum number of men per operation A Excavate pit B Concrete pit 0 C Fix pylon 00 D Adjust/finish 60 A minimum buffer of da is provided between tasks to take care of delays. Prepare a LOB schedule showing gang sizes and rates of build. Finally, draw a sketch showing the LOB schedule with buffers for all 1 pylons. Solution The calculation of gang sizes and rates of build, with a target rate of three per wk is shown in Tables E10.9b and c; and the LOB schedule for the four tasks is sketched in Figure E10.9. a Days Continues

27 0 Basic Hard Systems Engineering: Part I Chapter 10 Example 10.9: Continued Table E10.9b Details of calculations for Example 10.9 Activity Man power per activity Man per activity Theoretical gang size Actual gang size Actual rate of output Time in days for 1 activity Actual time Time from start on first section to start on last section Actual time Buffer A B C D Note: Numbers in Column must be multiples of those in Column. TABLE E10.9c Details of calculations for Example 10.9 (continued) First unit Last unit Begin End Begin End Buffer A B C D Continues

28 10. Program Evaluation and Review Technique and the Line-of-Balance 1 Example 10.9: Continued Units Network Flow Analysis 1 A BUF Figure E10.9 LOB schedule showing buffers B C D BUF indicates buffers necessary Days 7 BUF Everybody is familiar with highway, telephone, and cable networks. They are all arrangements of paths and crossings for facilitating the flow of goods, people, and communication. In this section we deal with three types of network flow problems: (1) the minimum spanning tree, () the maximum flow, and () the shortest-path techniques Key Concepts Graph theory and network theory are branches of mathematics that have grown in the last 7 years. The building of large physical systems such as highways, railroads, and pipelines has created wide interest in network theory. We have already seen many problems of sequencing and scheduling such as CPM and PERT that can also be looked upon as problems in graph and network theory. A graph is formed when a number of points, nodes, or vertices are joined together by one or more lines, arcs, links, branches, or edges. Nodes may be joined by more than one arc and may BUF

29 Basic Hard Systems Engineering: Part I Chapter 10 be oriented by indicating a sense of direction for them, using an arrow. A loop is formed when the extremity nodes of a path through a graph are one and the same node (see Figure 10.). A network is a graph through which flows of money, traffic, commodities, and so forth may take place, and the direction of the arc represents the direction of flow. In some networks there are distinct nodes from which flows emanate, and there are other nodes to which all flows finally go. These are called sources and sinks respectively (see Figure 10.) Minimum Spanning Tree This type of problem involves finding the least length of links needed for connecting all the nodes in a network. For example, if it is required to find the shortest length of cable needed to connect all the nodes in a city network, this problem would fall under the category of finding the minimum spanning tree. Loop A 1 A Figure 10. Network details C A = Source B = Sink B D F E G Node 1 Cost or time A B E D C 8 H 6 I 0 Link 7 7 Loop J Bidirectional links B 9

30 10. Program Evaluation and Review Technique and the Line-of-Balance The procedure for finding the minimum spanning tree of a network is: 1. Select the shortest link in the network.. Identify an unconnected node with the shortest distance to the node in Step 1.. Connect these two nodes. In case of a tie, select one arbitrarily.. Continue connecting one node after another till all nodes are connected. Example A small village needs to be connected by cable to seven main nodes. What is the minimum length of cable needed to connect all the nodes? Distances are marked on the nodes. Solution See Figure E The shortest link in the network is EG = 11.. The next shortest link connected to either E or G is GF = 1. Join GF.. The next shortest link, either E or F, is ED = 1. Join ED.. The next link at D, F or G is DC = 6. Join DC and so on. A 0 F A F B B 7 G E 11 G E C D 6 C 1 11 D 6 Total length of minimum spanning tree = 107 units Figure E10.10 Minimum spanning tree Continues

31 Basic Hard Systems Engineering: Part I Chapter 10 Example 10.10: Continued The sequence of linking is EG, GF, ED, DC, CB, and BA, totaling 107 units, which is the minimum length of cable needed to join all the nodes in the network The Maximal Flow Problem There are countless instances in which one would like to know the maximum number of trucks or wagons that can flow on a railroad or highway network from a source (or origin) to a sink (or destination). For example, in a highway network you could have traffic flow on a one-way street or a two-way street, whereas in a pipeline network, oil could flow in both directions. The procedure for determining the maximum flow in a network is: 1. Identify the source node and the sink node of the network. Determine all the feasible paths from source to sink that would be able to handle the flow. Determine whether there are possibilities of reverse flows, depending on the information supplied. Sum up all the flows through each link of the network Example A small railroad network with the indicated link flow capacities is shown in Figure E Determine the maximum flow from Source Node 1 to Sink Node, and indicate the flow on each link Figure E10.11 Maximal flow problem Solution Source node = 1 Sink node = Continues

32 10. Network Flow Analysis Example 10.11: Continued Feasible paths with flows: A, Path 1 Flow units B, Path 1 Flow units C, Path 1 Flow units D, Path 1 Flow units E, Path 1 Flow units Total 11 units Table E10.11 summarizes the flows on various links. Table E10.11 Flows on various links Link Flow (Note that Link has flow in both directions) 10.. Shortest-path or Minimum-path Technique The shortest-path or minimum-path technique is based on the assumption that travelers want to use the minimum impedance route between two points. Efficient methods of determining minimum paths were developed, because manual determinations would be nearly impossible. In Figure 10.6, for example, 0 different paths must be tested to determine the minimum between A and B. You can imagine the problem of finding the shortest path in a network, such as a city, with thousands of links and nodes. Many years ago the work that was undertaken to determine the minimum paths for longdistance telephone calls provided the help that planners needed. Rather than simply testing each path, algorithms allowed planners to find minimum paths for complete networks. The algorithm used most commonly is Moore s algorithm. Using Moore s algorithm, minimum paths are developed by fanning out from their origin to all other nodes. Determining the minimum path from Node 1 to each of the other nodes results in a skimtree from Node 1 to all other nodes, as shown in Figure 10.6.

33 6 Basic Hard Systems Engineering: Part I Chapter 10 A (a) Home node 10, 0 1,1, 9, Node 6 7 8,1, 6, 8,7, 7,6 Impedance (b) (c) 6 8 Figure 10.6 Minimum path technique: (a) small network links, 16 nodes; (b) minimum path through a network; and (c) skim tree from Node 1 to all other nodes Moore s algorithm is now applied to Figure 10.6b: 1. Start at Node 1 (the origin) and determine the shortest time (or distance) to get to a directly connected node. The two nodes directly connected to Node 1 are Nodes and, with the shortest time of 1 and units, respectively.. Since there is no doubt the shortest times to reach Nodes and are 1 and, respectively, we refer to these nodes as comprising the permanent set.. Repeat the foregoing steps by determining all the nodes directly connected to the nodes in the permanent set (Nodes 1,, and ). Now, Nodes and 6 are directly connected to Nodes and.. Two paths can be identified connected to Node 6 (-6 and -6), with times of and units respectively. Also, the path along 1--6 takes units of time, while Path 1--6 B

34 10. Network Flow Analysis 7 takes units. Therefore, the shortest path to Node 6 from Node 1 is And, this path is indicated in Figure 10.6c.. Nodes 1,,, and 6 are now part of the permanent set, and one can proceed to other nodes repeating the process described in Steps 1 through. 6. The shortest path from the starting (or home) node to all other nodes can be tabulated as follows (see Table 10.): Once the minimum paths are found, the trips between the zones are loaded onto the links making up the minimum path. This technique of assigning trips to the network is sometimes referred to as all-or-nothing, because all trips between a given origin and destination are loaded on links comprising the minimum path, and nothing is loaded on the other links. After all possible interchanges are considered, the result is an estimate of the volume on each link in the network. Moore s algorithm can be stated as follows: 1. Label the start node (or home node) as zero.. Calculate working values for each node that is directly connected to the node labeled zero, using working values indicated on the links. Select the minimum of the labeled values plus the distance from the labeled node. This establishes a permanent set.. Select an unlabeled node with the lowest working value and label it with that value.. Repeat Steps and until all nodes have been labeled.. Mark all links lying on the shortest path; this is the skim tree rooted in the start (or home) node. 6. Repeat Steps 1 through, selecting successive nodes as start nodes. Table 10. Shortest paths From To Path Duration Example 10.1 A highway network consisting of nodes and 10 links is shown in Figure E10.1. A trip table showing the number of vehicles wanting to use the network per hour from one node to another is also provided. Assign the trips to the network. Continues

35 8 Basic Hard Systems Engineering: Part I Chapter 10 Example 10.1: Continued Solution Refer to Figure E10.1. The origin-to-destination flows corresponding to each node of the trip table are assigned with the links that make up the minimum. The aggregate flow on each link is then shown. 100 Volume, (veh/hr) To node Zone Zone centroid (c) 100 node number Arc Home Assigned trips (a) Arc travel time (min) node 00 Home node 1 Minimum path tree (b) 1100 Travel time from home node (min) From node 0 10 (e) (d) Figure E10.1 Example of all-or-nothing traffic assignment: (a) network; (b) minimum-path trees; (c) origin-destination trip table; (d) assignment of trips to minimum-path trees; and (e) assigned traffic volume Assigned volume (veh/hr) 10.6 Linear Programming The objective of linear programing (LP) is to determine the optimal allocation of scarce resources among competing products or outputs. In most engineering and economic problems, one is frequently called upon to optimize a function that is subject to several constraints. When

36 10.6 Linear Programming 9 a single constraint is involved, the Lagrange method is used because of its simplicity. However, when more than one constraint is involved, LP is usually adapted. On the other hand, if the constraints are limited to only two variables (or at most three in some special cases), the graphical approach is the easiest to use. When a problem involves more than two variables, the best way of dealing with it is by using the simplex algorithm. We first deal with the graphical method and then follow up with the simplex The Graphical Method This method is generally used for solving maximization or minimization problems involving two variables. It is best described by means of an example. Example 10.1 A steel firm produces two products, small beams (X 1 ) and small poles (X ). Each beam requires. hr for cutting and welding, hr for finishing, and 1 hr for checking and testing. Each pole requires 1 hr for cutting and welding, hr for finishing, and hr for checking and testing. The firm is limited to no more than 0 hr for cutting and welding, 0 hr for finishing, and 16 hr for checking and testing. The firm makes a profit of $ per beam and $ per pole. How many beams and poles should the firm produce to maximize profit? Solution The firm s objective function is to maximize profit, and the total profit is the sum of the individual profit gained from each of the two products. First, express the information given in the form of equations or inequalities. The objective function can be written: Z = X 1 + X, subject to constraints. X 1 + X 0 X 1 + X 0 X 1 + X 16 X 1, X 0 The first three inequalities are technical constraints dictated by the availability of time, while the last constraint is imposed on all such problems to avoid negative values from the solution. The three inequality constraints are treated as: (1) X = 0. X 1 () X = 10 X 1 () X = 8 0.X 1 The graph of these three equations is shown in Figure E10.1. The feasibility area satisfying the equations above and the inequalities originally derived are shown by the area included in OABCD. Continues