QUME507_M6

QUME 507 Module 6: Project Management and Waiting Time Models

Introduction On Topic 6.1 we will study the quantitative methods with the purpose of monitoring, controling, ad planning project management. We will discuss the quantitative methods directed to reduction of project costs and time. We will study probabilistic models (such as PERT) and deterministic models (such as CPM). We will also discuss the concept of project crashing and its applications. Students will be taught to use computer programs in the solution of project management problems. Topic 6.1 Project Management In many situations, managers are responsible of planning, programming, and controling projects which consist of numerous tasks or independent tasks performed by a variety of departments and individuals. Frequently, these projects are so big or complex that the manager cannot remember all the information related to the plan, timeline, and process o the project. In these situations, the program evaluation and review technique (PERT) and the critical path method (CPM) have demonstrated to be extremely valuable (Anderson et al. 2018). PERT is a probabilistic technique. On the other hand, CPM is a deterministic model, since the times are supposed to be known with certainty. Even though the differences are still seen, both techniques are so similar that the term PERT/CPM is frequently used to describe the general approach (Render et al., 2018). Six Steps of PERT/CPM 1. Define the project and all its significant activities or tasks. 2. Develop the relationship between activities. Decide which activites must precede others. 3. Draw the network that connects all the activities. 4. Assign time and/or cost estimates to aeach activity. 5. Calculate the trajectory with the longest time through the network; it is known as critical path. 6. Use the network to help plan, program, supervise, and control the project.

Questions to Consider 1. When will the entire project be finished? 2. Which are the critical activites or tasks in the project, those which would delay the project if they are delayed? 3. Which are the non-critical activities, those which may be delayed without delaying the completion of the entire project? 4. If there are three time estimates, what are the probabilities the project will be completed at a specific date? 5. At a given date, is the project on time, delayed, or advanced? 6. At a specific date, is the money spent equal to, less than or more than the budgeted amount? 7. Are there sufficient available resources to complete the project on time? General Foundry Example General Foundry, Inc., a metal welding plant in Milwaukee, has been trying to avoid the cost of installing equipment for the control of atmospheric pollution. The area’s environmental protection group has just given the plant 16 weeks to install a complex air filtration system in its principle chimney. General Foundry was warned it will be forced to close unless the device is installed within the stipulated period. Lester Harky, the managing partner, wishes to make sure the filter system installation progesses without problems and on time. First and Second Step As we saw in the first step the project and all its significant activities and tasks have to be defined. The second step is to develop the relationship between activities. That is to say, decide which activities must precede others. Table 1 shows the activities and their descriptions. The activities which precede others have also been identified.

Table 1. Activities and immediate predecessors for General Foundry, Inc. ACTIVITY DESCRIPTION IMMEDIATE PREDECESSORS A Build internal components — B Modify roof and floor — C Build collection shaft A D Place the concrete and install the frame B E Build high-temperature burner C F Install control system C G Install anti-pollution device D, E H Inspect and test F, G Third Step The third step is related to drawing the network. There are two common techniques to draw PERT networks. 1. Activity on the node (AON) in which nodes represent activities. 2. Activity on arc (AOA) in which the arcs are used to represent activities. The AON approach is easier and is most commonly found in software packages. It is important to consider that one node represents the start of the project, one node for the end of the project, and nodes for each activity. Arcs are used to show each activity’s predecessors. Figure 1 shows the network for General Foundry Inc.

Figure 1. Network for General Foundry, Inc. Adapted from Render, B., Stair, R. M., Hanna, M. E. & Hale, T. S. (2018). Quantitative Analysis for Management. (13 ed). Pearson. Fourth Step The fourth step is to assign times to the activities. Nevertheless, for special projects or new tasks, providing time estimates is not always simple. Without solid historical data, managers many times are uncertain about activity times. Because of this, PERT developers used a probability distribution based on three estimates for each activity. PERT frequently assumes that time estimates follow a beta probability distribution. Time estimates with PERT are: Optimistic Time (a) Time an activity would take if everything results as well as possible. There must only be a small probability (let’s say, 1/100) of this happening. Pessimistic Time (b) Time an activity would take if there are unfavorable conditions. There must only be a small probability that the activity takes so much time.. Most Probable Time (m) Most realistic time estimate for the completion of the activity. To find the expected activity time (t), the beta distribution evaluates the estimates in the following manner: In order to calculate the dispersion or variance of activity completion time, we use the formula:

Table 2 shows weekly time Varianza = Table 2. Time estimates for estimates for General Foundry Inc. General Foundry, Inc Fifth Step In the fifth step we must calculate the trajectory with the longest time throughout the ACTIVITY OPTIMISTIC MOST PESSIMISTIC EXPECTED VARIANCE a PROBABLE b TIME [(b – a)/6]2 A1 m 3 t = [(a + 4m 4/36 B2 4 + b)/6] 4/36 C1 2 3 2 4/36 D2 3 6 3 16/36 E1 2 7 2 36/36 F1 4 9 4 64/36 G3 4 11 4 64/36 2 3 3 4/36 1 4 5 2 2 25 network; identified as critical path. It contains a series of activities which have zero clearance. It is the trajectory with the longest time in the network. A delay in any activity found in the critical path will delay the completion of the entire project. Clearance time is that which can delay an activity without delaying the entire project. Clearance is equal to the furthest starting time minus the closest starting time, or, the furthest completion time minus the closest completion time. To find the critical path, we need to determine the following amounts for each activity on the network: 1. Closest Initial Time (IC): the earliest an activity may start without contravening the requirements of immediate predecedence. 2. Closest Termination Time (TC): the earliest an activity may be completed.

3. Furthest Initial Time (IL): the latest an activity may start without delaying the entire project. 4. Furthest Termination Time (TL): the latest an activity may be completed without delaying the entire project. Excel QM Application Step 1 When you open the Program, on the left Menu, choose Project Management/Predecesor List as shown on image 1. Image 1 Step 2 Enter the number of activities and predecessors and mark Graph as shown on image 2. Image 2

Check here Step 3 Identify the predecessors by vertical lines Enter the data on the template. As the data is being entered, Excel QM calculates the expected times and the variances for every activity, and, automatically displays a table with the closest, furthest, and clearance times for the activities. A Gantt graphic is displayed which shows the critical path and the clearance time for the activities. The results are shown on images 3 and 4. Image 3

Image 4 expected termination date Interpretation of PERT Results 1. The expected termination date for the project is 15 weeks. 2. There is a 71.6% probability of the equipment being installed within the delivery time of 16 weeks. PERT may easily find the probability of completing the project within any period Harky wishes. 3. Five activites (A, C, E, G, H) are in the critical path. If one of them is delayed for any reason, the entire project will be delayed. 4. Three activites (B, D, F) are not critical since they have a certain time clearance. This means Harky may borrow some resources, if necessary, to maybe speed up the entire project. 5. A detailed program of initial and termination activity times may be obtained. To watch a demonstrative video of the General Foundry exercise on QM for Windows you may access the following video https://youtu.be/uCAzQkUppKg Yuang, C. (10 de septiembre de2012). QM for Windows Demo - Project Management. [Video]. Recuperado de QM for Windows https://youtu.be/uCAzQkUppKg

Project Acceleration: Crashing Projects sometimes have time limits which are impossible to meet using normal procedures. By using exceptional methods, it is possible to complete the project in less time than what it would normally require at a higher cost. Reducing the termination time of a project is called crashing. Project crashing starts with the use of normal time to create the critical path. Normal cost is the cost to complete the activity using normal procedures. If the project does not comply with the required deadline, extraordinary measures must be taken. Then, crashing time must be identified. This is the shortest possible activity time and it will require additional resources. The cost of crashing is the price of completing the activity at an earlier time than normal. Steps for Project Crashing 1. Find the normal critical path and identify the critical activities. 2. Calculate the cost of crashing by week (or any other period) for all activities in the network. This process uses the following formula: 3. Select the activity on the critical path with the least crashing cost by week. Accelerate this activity as much as possible or to the point where the desired deadline is achieved. 4. Verify the critical trajectory that is accelerated still be critical. Frequently, a reduction in time of an activity on the critical path causes another activity or other non-critical trajectories to become critical. If the critical path is still the longest path on the network, return to step 3. If not, find the new critical path and return to step 3. Table 3 shows the normal and crash data for General Foundry, Inc. Table 3. Data for General Foundry

ACTIVITY TIME (weeks) COST ($) CRASH CRASH Critical NORMAL CRASH NORMAL COST Path? A 23,000 by B 21 22,000 34,000 week Yes C 31 30,000 27,000 ($) No D 21 26,000 49,000 1,000 Yes E 43 48,000 58,000 2,000 No F 42 56,000 30,500 1,000 Yes G 32 30,000 86,000 1,000 No H 52 80,000 19,000 1,000 Yes 21 16,000 500 Yes 2,000 3,000 QM for Windows Application Step 1. On the left Menu Project Management/Crashing. Click on SOLVE. When the template appears, enter the data as shown on image 5 Image 5 Enter the data choose the method Step 2. The results are shown on image 6. The crashing cost is 18,000.

Image 6 When you click on the Edit Data button and choose Cost Budgeting from the upper menu, and you click on Graph; the program also shows the graph of budget ranges for General Foundry. The blue line represents the budget using the closest initial times. The red line represents the budget using the furthest initial times. Budgets as shown on image 7 tend to be elaborated before beginning the project. So, as the project is done, the budget spent is supervised and controlled. Image 7

Below, youll find article and study reference which may help you delve into the topic of project management and are found in the folder of complementary material. Kholil, M., Alfa, B. N. & Hariad. M. (2018). Scheduling of House Development Projects with CPM and PERT Method for Time Efficiency. IOP Conference Series: Earth and Environmental Science, 140. IOP Publishing. Langley, M. (2017). Sucess Rates Rise: Transforming the High Cost of Low Performance. 9th Global Management Global Survey. Recuperado de http://www.pmi.org/-/media/pmi/documents/public/pdf/learning/thought- leadership/pulse/pulse-of-the-profession-2017.pdf Lermen, F. H., Morais, M. F., Matos, C., Rórder, R. & Róder, C. (2016). Optimization of Times and Costs of Project of Horizontal Production Using PERT/CPM Technical. Independent Journal of Management & Production (IJM&P), 7(3), 833-853. Tamrakar, P. (2013). Analysis and Improvement by the Application of Network Analysis (Pert/CPM). The International Journal of Engineering and Science (IJES), 2(2), 154-159. ------------------------------------------------------------------------------- Introduction to Topic 6.2 On Topic 6.1 we discussed the PERT/CPM techniques of project management and project crashing. On Topic 6.2 we will study the pertinence of waiting time models. These models look to minimize the waiting time for clients subject to company restrictions. Likewise, the exchange rates between the cost of waiting for a service and the cost of waiting in line. At the same time, we will study the queuing system applications in business. Lastly, we will use computer programs for the solution of waiting time models. 6.2 Waiting Time Models The queuing theory is the study of waiting lines. It is one the oldest and most used quantitative analysis theories. The three basic components of a queuing process are: 1. Arrivals 2. Service installations 3. Real waiting queues

Waiting line analytical models may help managers assess the cost and effectiveness of service systems. In fact, the majority of waiting line problems are centered in finding the ideal level of service which a company should offer. Service systems are generally classified in terms of number of channels, or number of servers, and the number of phases or number of service stops, which must be done. As shown on figure 1, a single-phased system means the client receives service in one station before abandoning the system. On the other hand, a multi-phased system implies two or more stops before exiting the system. Figure 2 shows the flow on a single-phased system and figure 3 shows the flow of a multi-phased system model. Figure 1. Single-phased and multi-phased channel Adapted from Heizer, J. Render, B. & Munson C. (2017). Operations Management: Sustainability and Supply Chain. (12 ed.). Person Education. Figure 2. Multi-channel single-phased system

Adapted from Heizer, J. Render, B. & Munson C. (2017). Operations Management: Sustainability and Supply Chain. (12 ed.). Person Education. Figure 3. Multi-channel Multi-phased System Adapted from Heizer, J. Render, B. & Munson C. (2017). Operations Management: Sustainability and Supply Chain. (12 ed.). Person Education. D. G. Kendall developed a notation for queuing models which specify the arrival pattern, the distribution of service time, and the amount of channels.

1. A single- channel model with Poisson arrivals and exponential service times would be represented by: M/ M/1 2. If a second channel is added, the notation would be: M/M/2 3. A system of three channels with Poisson arrivals and constant service time would be: M/D/3 4. A system of four channels with Poisson arrivals and normally distributed service times would be: M/G/4 Single-channel queuing model with Poisson arrivals and exponential service times (M/M/1) The single-channel single-phased model considered here is one of the simplest and broadly used queuing models. It implies supposing there are seven conditions: 1. Arrivals are assisted with on a basis of PEPS. 2. Each arrival waits for assistance egardless of the longitude of the line; that is to say, it isn’t eluded or refused. 3. Arrivals are independent from previous arrivals, but its average number (the arrival rate) does not change through time. 4. Arrivals are described with a Poison distribution of probability and come from an infinite or very large population. 5. Service times also vary from one client to the next and are independent between them, but its average rate is known. 6. Service times occur according to a negative exponential probability distribution. 7. The average service rate is greater than the average arrival rate. The queuing equations are the following:  = medium number of arrivals by period (for example, by hour)  = medium number of people or articles assisted by period 1. The average number of clients or units in the system, L, that is, the number on the queue plus the number that is being assisted:

2. The average time a client spends in the system, W, that is, the time the client spends on queue more than the time spent being assisted: 3. The average number of clients on queue, Lq: 4. The average time a client waits on the queue, Wq: 5. The system utilization factor, (the Greek letter rho), that is, the probability the service installation is being used: 6. Percentage of idle time P0, that is, the probability no one is in the system: 7. The probability that the number of clients in the system be greater than k, Pn>k: Arnold Silencer Workshop Example Arnold’s mechanic, Reid Blank, is able to install new silencers at an average rate of 3 per hour, or approximately 1 every 20 minutes. Clients who need this service arrive to the workshop at an average of 2 per hour. Larry Arnold, the workshop’s owner, studied queuing models at a Masters in Business Administration program and he feeld all seven conditions for the single server model are met.

Excel QM Application On the Excel QM program, choose Waiting lines/Single Server Model. Enter the arrival rate and the service rate on the template. The results will be displayed as they are shown on image 1. Image 1 Multiple channel queuing models with Poisson arrivals and exponential service times (M/M/m) The next logical step is to study a multiple channel queuing system, where two or more servers or channels are available to assist the clients who arrive. Clients are supposed to wait for the service at a single queue and then direct themselves to the first available server. Arnold Silencer Workshop Example In order to apply the multiple queu model, let’s return to the case of Arnold’s silencer workshop. Previously, Larry Arnold had examined two options. He could keep his

current mechanic, Reid Blank, at a total expected cost of $653 per day, or fire him and hire a more expensive, but faster mechanic named Jerry Smith. With Smith on board, the service system costs could be reduced to $360 per day. Now let’s explore a third option. Arnold finds that with a minimum cost after taxes he can open a second service area in the workshop, in which he could install silencers. Instead of hiring his first mechanic, Blank, he would hire a second worker. It could be expected that the second mechanic would install silencers at the same rate as Blankm approxiamtely  = 3 per hour. Clients, who woul continue to arrive at the rate of  = 2 per hour, would wait on a single queu until one of the new mechanics would be free. In order to know how this option compares to the previous single channel waiting line system, Arnold calculates various operation characteristics for the channel ssytem m = 2. Excel QM Application On the Excel QM program, choose Waiting lines/MMs Exponential Service Times. Enter the arrival rate and service rate on the template. The results will be displayed as shown on image 2. Image 2

Introduction of Model Costs Now that the queuing system characteristcs have been calculated, Arnold decides to make an economic analysis of its impact. The waiting line model was valuable to predict the waiting times, the longitude of queues, the potential idle times, among others. But, he did not identify the optimum decisions nor the cost factors. As states before, the solution to the queuing problem could require management to balance the increased costs of offering better services with the reduced costs of waiting that arise from offering the service. These are known as waiting cost and service cost, respectively. The total cost of the queuing system when based on the time in the system is: Total cost = Total service cost + Total waiting cost Total cost = mCs + WCw

And when it based on queuing time: Total cost = mCs + WqCw Arnold estimates the waiting time cost of clients, in terms of their unsatisfaction and the loss of good will, is of $50 per hour of the time they spend in the queu. Due to the fact that in average, a car waits 2/3 of an hour and approximately 16 cars are worked on per day (2) per hour multiplied by 8 work hours daily, the total number of hours clients spend waiting for their silencers to be installed each day is 16 hours. In this case: Total daily waiting cost = (8 hours per day) WqCw = (8)(2) (2/3) ($50) = $533.33 The only additonal cost that Larry Arnold may identify in this queuing situation is the compensation of Reid Blank, the mechanic. Blank (current mechanic) gets paid $15 per hour: Total daily service cost = (8 hours per day) mCs = 8(1) ($15) = $120 As currently configured, the total daily cost of the system is the total waiting cost and service cost, from which we obtain: Total daily queuing system cost $533.33 + $120 = $653.33 Now a decision needs to be made. Arnold finds out through his business connections that the competition at the other side of the city, Rusty Muffler, employs a mechanic named Jimmy Smith, who can efficiently install new silencers at a rhythm of 4 per hour. Larry Arnold calls Smith and asks him if he is interested in changing employment. Smith says he would consider leaving Rusty Muffler only if he gets paid a salary of $20 per hour. Arnold, who is a skilled businessman, decides it would probably be worth it to fire Blank and replace him with Smith, who is as fast as he is expensive. In this case, the problem is recalculated with a service rate of 4 silencers per hour. The calculations are made again: Total daily waiting cost = (8 hours per day) lWqCw = (8)(2) (1/4) ($50) = $200 per day Smith Service Cost = 8 hours/day x $20/hour = $160 per day Total expected cost = Waiting cost + Service cost = $200 + $160 = $360 per day.

The total daily costs of the system are less with the new mechanic, which translates to significant savings: $653.33 – $360 = $293.33 Constant Service Time Model (M/D/1) Some systems have constant service times instead of exponentially distributed times. When clients or equipments are processed according to a fixed cycle, as in the case of an automatic car wash or of a game at an amusement park, the constant service rates are appropriate. Constant service rates accelerate the process in comparison with exponentially distributed service times with the same value of m. García-Golding Recycling Example The García-Golding Recycling company collects and compacts aluminum cans and glass bottles in the city of New York. Truck drivers, who unload the material for recycling, wait approximately an average of 15 minutes before empyting their loads. The cost of the driver’s salary and the truck’s inactive time while on queue is valued at $60 per hour. A new automatic compactor, which would process the truck loads at a constant rate of 12 vehicles per hour (5 minutes per truck) may be bought. The trucks arrive in accordance to a Poisson distribution at an average rate of 8 per hour. If the new compactor is used, its cost would be redeemed at a rate of $3 per unloaded truck. QM for Windows Application On the QM for Windows program, choose Waiting Lines/M/D/1 Constant Services Line. Enter the arrival rate and service rate in the template. The results will be displayed as shown on image 3.

Image 3 Finite Population Model (M/M/1 with Finite Source) When there is a limited population of potential clients for a service installation, it is necessary to consider a different queuing model. This model would be used, for example, if equipment repairs would be considered in a factory with only five machines, if we were in charge of the maintenance of a fleet of 10 airplanes of intensive use, or if we were managing a hospital wing with 20 beds. The limited population model allows for the consideration of any number of people who work on repairs (servers). The reason why this model differs from the three previous queuing models is that there is a dependent relationship between the longitude of the queue and the arrival rate. This section describes a finite potential population model which is supported by the following suppositions: 1. There is only one server. 2. The population of units who search for service is finite. 3. Arrivals follow a Poisson distribution and service times are exponentially distributed. Department of Commerce Example Existing records indicate that each of the five flat high-speed printers on the United States Department of Commerce in Washington, D.C. need reparation after approximately 20 hours of use. It has been determined that the decompositions have a Poisson distribution. The only techician on call may service a printer at an average of

two hours, following an exponential distribution. Clients are assisted with the principle of first come, first served basis. In order to calculate the system’s operational characteristics, we first observe the average arrival rate is  = 1/20 = .05 printers/hour. The average service rate is  = 1/2 = 0.50 printers/hour. QM for Windows Application On the QM for Windows program, choose Waiting lines/M/M/s with a finite population. Enter the arrival rate and the service rate on the template. The results will be displayed as shown on image 3. Image 3 Below, we show article or study references which might help you delve into the topic of queuing systems, which are found in the folder of complementary material at the end of the modules. Al-Matar, N. (2017). Theories and Applications Related to Queuing Systems. International Journal of Advances in Electronics and Computer Science, 4(2), 14-17. Anbarasi, R. (2018). A study on M/M/3 queuing model on waiting time reduction in a local health care centre. International Journal of Statistics and Applied Mathematics, 3(2): 21-26. Rahman Chowdhury, M. R., Rahman, M. T. & Kabir, M. R. (2013). Solving of Waiting Lines Models in the Bank Using Queuing Theory Model the Practice Case: Islami Bank Bangladesh Limited, Chawkbazar Branch, Chittagong. IOSR Journal of Business and Management (IOSR- JBM), 10(1), 22-29.