CH-9 Discrete Control and Programmable Logic – Automation, Production Systems, and Computer-Integrated Manufacturing : Book Guide

This document contains Chapters 9 to 12 Chapter 9 DISCRETE CONTROL AND PROGRAMMABLE LOGIC CONTROLLERS REVIEW QUESTIONS 9.1 Briefly define the two categories of discrete process control? Answer: Discrete process control can be divided into two categories: (1) logic control, which is concerned with event-driven changes in the system; and (2) sequence control, which is concerned with time-driven changes in the system. 9.2 What is an AND gate? How does it operate on two binary inputs? Answer: An AND gate outputs a value of 1 if all of the inputs are 1, and 0 otherwise. 9.3 What is an OR gate? How does it operate on two binary inputs? Answer: An OR gate outputs a value of 1 if either of the inputs has a value of 1, and 0 otherwise. 9.4 What is Boolean algebra? What was its original purpose? Answer: Boolean algebra is a special form of algebra based on the logic elements (AND, OR, and NOT) that was developed around 1847 by George Boole. Its original purpose was to provide a symbolic means of testing whether complex statements of logic were TRUE or FALSE. 9.5 What is the difference between a delay-off timer and a delay-on timer? Answer: A delay-off timer switches power on immediately in response to a start signal, and then switches power off after a specified time delay, whereas a delay-on timer waits a specified length of time before switching power on when it receives a start signal. 9.6 What is the difference between an up counter and a down counter? Answer: An up counter starts at zero and increments its contents (the count total) by one in response to each pulse. When a preset value has been reached, the up counter can be reset to zero. A down counter starts with a preset value and decrements the total by one for each pulse received. 9.7 What is a ladder logic diagram? Answer: A ladder logic diagram shows the various logic components along horizontal lines or rungs connected on either end to two vertical rails. The diagram has the general configuration of a ladder, hence its name. The components are contacts (representing logical inputs) and loads, also known as coils (representing outputs). 9.8 The two types of components in a ladder logic diagram are contacts and coils. Give two examples of each type. Answer: Contacts include switches and relay contacts, and coils include motors, lamps, and alarms. 9.9 What is a programmable logic controller? Answer: A programmable logic controller is a microcomputer-based controller that uses stored instructions in programmable memory to implement logic, sequencing, timing, counting, and arithmetic functions through digital or analog input/output (I/O) modules, for controlling machines and processes. 9.10 What are the advantages of using a PLC rather than conventional relays, timers, counters, and other hard-wired control components? Answer: The advantages listed in the text are (1) programming the PLC is easier than wiring the relay control panel; (2) the PLC can be reprogrammed, whereas conventional controls must be rewired and are often scrapped instead; (3) PLCs take less floor space than relay control panels; (4) reliability is greater, and maintenance is easier; (5) the PLC can be connected to computer systems more easily than relays; and (6) PLCs can perform a greater variety of control functions than relay controls can. 9.11 What are the four basic components of a PLC? Answer: The four basic components of a PLC are the following: (1) processor, (2) memory unit, (3) power supply, and (4) I/O module. In addition, a programming device is required to enter the control program to the PLC. In most cases, this can be disconnected from the PLC when not being used. 9.12 The typical operating cycle of the PLC, called a scan, consists of three parts: (1) input scan, (2) program scan, and (3) output scan. Briefly describe what is accomplished in each part. Answer: During the input scan, the inputs to the PLC are read by the processor and the status of each input is stored in memory. Next, the control program is executed during the program scan. The input values stored in memory are used in the control logic calculations to determine the values of the outputs. Finally, during the output scan, the outputs are updated to agree with the calculated values. 9.13 Name the five PLC programming methods identified in the International Standard for Programmable Controllers (IEC 61131–3). Answer: The standard specifies three graphical languages and two text-based languages for programming PLCs, respectively: (1) ladder logic diagrams, (2) function block diagrams, (3) sequential functions charts, (4) instruction list, and (5) structured text. 9.14 What are the reasons and factors that explain why personal computers are being used with greater and greater frequency for industrial control applications? Answer: Some of the reasons given in the text are the following: (1) The technological evolution of PLCs has not kept pace with the development of PCs. (2) New generations of PCs are introduced with much greater frequency than PLCs. (3) There is much more proprietary software and architecture in PLCs than in PCs, making it difficult to mix and match components from different vendors. (4) PCs are now available in more-sturdy enclosures for the dirty and noisy plant environment. (5) PCs can be equipped with membrane-type keyboards for protection against factory moisture, oil, and dirt. (6) PCs can be ordered with I/O cards and related hardware to provide the necessary devices to connect to the plant’s equipment and processes. (7) Operating systems designed to implement realtime control applications can be installed in addition to traditional office software. 9.15 Name the two basic approaches used in PC-based control systems. Answer: The two approaches are soft logic and hard real-time control. In the soft logic configuration, the PC’s operating system is Windows, and control algorithms are installed as high-priority programs under the operating system. However, it is possible to interrupt the control tasks in order to service certain system functions in Windows, such as network communications and disk access. When this happens, the control function is delayed, with possible negative consequences to the process. By contrast, in a hard real-time control system, the PC’s operating system is the real-time operating system, and the control software takes priority over all other software. Windows tasks are executed at a lower priority under the real-time operating system. Windows cannot interrupt the execution of the real-time controller. If Windows locks up, it does not affect the controller operation. Also, the real-time operating system resides in the PC’s active memory, so a failure of the hard disk has no effect in a hard real-time control system. 9.16 Modern industrial control applications have evolved to include requirements in addition to logic and sequence control. What are these additional requirements? Answer: The additional requirements of modern industrial control applications include the following: (1) analog control to cope with continuous process variables like temperature, (2) motion and servomotor control, (3) arithmetic functions to permit application of complex control algorithms, (4) matrix functions, (5) data processing and reporting, and (6) network connectivity and enterprise data integration. 9.17 What is a programmable automation controller? Answer: A programmable automation controller (PAC) is an industrial process controller designed to interface with the sensors and actuators whose signals are typically binary, like a conventional PLC, but it also has capabilities usually associated with personal computer applications, such as simulated analog control, data processing, advanced mathematical functions, network connectivity, and enterprise data integration. In effect, a PAC combines the input/output and discrete control capabilities of a PLC with the advanced computational, data processing, and enterprise integration capabilities of a PC. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. 9.1 Construct the truth table for the robot interlock system in Example 9.1. Answer: Truth table, given the symbols defined in Example 9.1. X1 X2 X3 Y 0 0 0 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 1 1 0 1 1 1 1 9.2 Construct the ladder logic diagram for the robot interlock system in Example 9.1. Answer: Ladder logic diagram, given the symbols defined in Example 9.1. 9.3 (A) Write the Boolean logic expressions for the pushbutton switch of Example 9.2 using the following symbols: X1 = START, X2 = STOP, P = POWER-TO-MOTOR, and M = MOTOR. Hint: See Figure 9.6(b). Answer: Given X1 = start, X2 = stop, P = power-to-motor, and M = motor, Boolean logic expressions: 9.4 In the circuit of Figure 9.1, suppose a photodetector were used to determine whether the lamp worked. If the lamp does not light when both switches are closed, the photodetector causes a buzzer to sound. Construct the truth table for this system. Answer: Using the following symbols: X1 = first switch, X2 = second switch, Y = lamp (Y = 1 when on), X3 = photodetector (X3 = 0 when light is detected, X3 = 1 when no light is detected), and B = buzzer (B = 1 means buzzer is sounding), the truth table is as follows: Line X1 X2 Y X3 B 1 0 0 0 1 0 2 0 1 0 1 0 3 1 0 0 1 0 4 1 1 1 0 0 5 1 1 0 1 1 Comment: In line 5, Y = 0 indicates that the lamp is not lighting, thus photodetector X3 = 1. Because X1 and X2 are also closed, the buzzer is energized to sound. 9.5 Construct the ladder logic diagram for the preceding problem. Answer: Using the symbols and their values from the preceding problem, the ladder logic diagram is as follows: 9.6 In the circuit of Figure 9.2, suppose a photodetector were used to determine whether the lamp worked. If the lamp does not light when either X1 or X2 switch is closed, the photodetector causes a buzzer to sound. Construct the truth table for this system. Answer: Using the following symbols: X1 = first switch, X2 = second switch, Y = lamp (Y = 1 when on), X3 = photodetector (X3 = 0 when light is detected, X3 = 1 when no light is detected), and B = buzzer (B = 1 means buzzer is sounding), the truth table is as follows: Line X1 X2 Y X3 B 1 0 0 0 1 0 2 0 1 1 0 0 3 1 0 1 0 0 4 1 1 1 0 0 5 0 0 0 0 0 6 0 1 0 1 1 7 1 0 0 1 1 8 1 1 0 1 1 9.9 Comment: In lines 6, 7, and 8, Y = 0 indicates that the lamp is not lighting, thus photodetector X3 = 1. Because either X1 or X2 is also closed, or both X1 and X2 are closed, the buzzer is energized to sound. 9.7 Construct the ladder logic diagram for the preceding problem. Answer: Using the symbols and their values from the preceding problem, the ladder logic diagram is as follows: 9.8 Construct the ladder logic diagrams for (a) the NAND gate and (b) the NOR gate. Answer: Ladder logic diagrams, using the following symbols: X1 = first switch, X2 = second switch, Y1 = intermediate output load, and Y = output load of interest. Draw the ladder logic diagram for the Boolean logic equation: Y = (X1 + X2)⋅X3. Answer: Draw the ladder logic diagram for the Boolean logic equation: Y = (X1 + X2)⋅(X3 + X4). Answer: Draw the ladder logic diagrams for the Boolean logic equation: Y = (X1⋅X2) + X3. Answer: (A) Write the language statements for Problem 9.9 using the instruction set in Table 9.10. Answer:: STR X1 OR X2 AND X3 OUT Y Write the language statements for Problem 9.10 using the instruction set in Table 9.10. Answer: STR X1 OR X2 AND X3 OR X4 OUT Y 9.10 Write the language statements for Problem 9.10 using the instruction set in Table 9.10. Answer: STR X1 OR X2 AND X3 OR X4 OUT Y Write the language statements for Problem 9.11 using the instruction set in Table 9.10. Answer: STR X1 AND X2 OR X3 OUT Y Write the low level language statements for the robot interlock system in Example 9.1 using the instruction set in Table 9.10. Answer: STR X1 AND X2 AND X3 OUT Y Write the low level language statements for the lamp and photodetector system in Problem 9.4 using the instruction set in Table 9.10. Answer: STR X1 AND X2 OUT Y STR X1 AND X2 AND X3 OUT B An industrial furnace is controlled as follows: The contacts of a bimetallic strip inside the furnace close if the temperature falls below the set point and open when the temperature is above the set point. The contacts regulate a control relay which turns on and off the heating elements of the furnace. If the door to the furnace is open, the heating elements are turned off until the door is closed. Let X1 = contacts of bimetallic strip, X2 = door interlock, C1 = relay contacts, and Y1 = heating elements. Construct the ladder logic diagram for the system. Answer: Let X1 = closed if temperature falls below set point, X2 = closed if door is closed, C1 = relay contacts, and Y1 = heating elements on. Ladder logic diagram: (A) For the previous problem, write the low level language statements for the system using the PLC instruction set in Table 9.10. Answer: Let X1 = closed if temperature falls below set point, X2 = closed if door is closed, C1 = relay contacts, and Y1 = heating elements on. Low level language: STR X1 AND X2 OUT C1 STR C1 OUT Y1 In the manual operation of a sheet metal stamping press, a two button safety interlock system is used to prevent the operator from inadvertently actuating the press while his hand is in the die. Both buttons must be depressed to actuate the stamping cycle. In this system, one press button is located on one side of the press while the other button is located on the opposite side. During the work cycle the operator inserts the part into the die and depresses both pushbuttons, using both hands. (a) Write the truth table for this interlock system. (b) Write the Boolean logic expression for the system. (c) Construct the ladder logic diagram for the system. Answer: Let X1 = button one, X2 = button 2, and Y = safety interlock 9.20 An emergency stop system is to be designed for a certain automatic production machine. A single "start" button is used to turn on the power to the machine at the beginning of the day. In addition, there are two "stop" buttons located at two locations on the machine, either of which can be pressed to immediately turn off power to the machine. Let X1 = start button (normally open), X2 = stop button 1 (normally closed), X3 = stop button 2 (normally closed), and Y = power-to-machine. (a) Construct the truth table for this system. (b) Write the Boolean logic expression for the system. (c) Construct the ladder logic diagram for the system. Answer: 9.21 An industrial robot performs a machine loading and unloading operation. A PLC is used as the cell controller. The cell operates as follows: (1) a human worker places a part into a nest, (2) the robot reaches over and picks up the part and places it into an induction heating coil, (3) a time of 10 sec is allowed for the heating operation, and (4) the robot reaches into the coil, retrieves the part, and places it on an outgoing conveyor. A limit switch X1 (normally open) is used to indicate that the part is in the nest in step (1). This energizes output contact Y1 to signal the robot to execute step (2) of the work cycle (this is an output contact for the PLC, but an input interlock signal for the robot controller). A photocell X2 is used to indicate that the part has been placed into the induction heating coil C1. Timer T1 is used to provide the 10-sec heating cycle in step (3), at the end of which, output contact Y2 is used to signal the robot to execute step (4). Construct the ladder logic diagram for the system. Answer: (a) Ladder logic diagram: 9.22 For the previous problem, write the low level language statements for the system using the PLC instruction set in Table 9.10. Answer: Low level language statements: STR X1 OUT Y1 STR X2 OUT C1 STR X2 OUT TMR T1 STR NOT T1 OUT C1 STR T1 OUT Y2 9.23 Write the low level language statements for the fluid filling operation in Example 9.6 using the instruction set in Table 9.10. Hint: See Figure 9.5. Answer: STR X1 OR C1 AND NOT FS OUT C1 STR C1 OUT S1 STR NOT FS OUT S1 STR FS OUT T1 120 STR T1 OUT C2 STR C2 (120 specifies timer delay in sec) OUT T2 90 STR C2 OR S2 AND NOT T2 OUT S2 (90 specifies timer delay in sec) 9.24 In the fluid filling operation of Example 9.6, suppose a sensor (e.g., a submerged float switch) is used to determine whether the contents of the tank have been evacuated, rather than rely on timer T2 to empty the tank. Construct the ladder logic diagram for this revised system. Answer: (a) Ladder logic diagram. Assume FS2 (new float switch) is open when tank is empty. FS1 is the tank full float switch. 9.25 For the previous problem, write the low level language statements for the system using the PLC instruction set in Table 9.10. Answer: Low level language statements: STR X1 OR C1 AND NOT FS1 OUT C1 STR C1 OUT S1 STR NOT FS1 OUT S1 STR FS1 OUT T1 (120 s) STR T1 OUT C2 STR C2 OR S2 AND NOT FS2 OUT S2 Chapter 10 MATERIAL TRANSPORT SYSTEMS REVIEW QUESTIONS 10.1 Provide a definition of material handling. Answer: Material handling is defined by the Material Handling Industry of America (MHIA) as “the movement, protection, storage and control of materials and products throughout the process of manufacture and distribution, consumption and disposal.” 10.2 How does material handling fit within the scope of logistics? Answer: Material handling is concerned with internal logistics – the movement and storage of material in a facility. By contrast, external logistics is concerned with the transportation of materials between facilities by rail, truck, seaway, air transport, and/or pipelines. 10.3 Name the four major categories of material handling equipment. Answer: As identified in the text, the four categories of material handling equipment are (1) transport equipment, (2) positioning equipment, (3) unit load formation equipment, (4) storage equipment, and (5) identification and control equipment. 10.4 What is included within the term unitizing equipment? Answer: As defined in the text, unitizing equipment refers to the containers used to hold individual items during handling and the equipment used to load and package the containers. 10.5 What is the unit load principle? Answer: The unit load principle recommends that the unit load should be designed to be as large as is practical for the material handling system that will move or store it, subject to considerations of safety, convenience, and access to the materials making up the unit load. 10.6 What are the five categories of material transport equipment commonly used to move parts and materials inside a facility? Answer: The five categories identified in the text are (1) industrial trucks, manual and powered, (2) automated guided vehicles, (3) monorails and other rail guided vehicles, (4) conveyors, and (5) cranes and hoists. 10.7 Give some examples of industrial trucks used in material handling. Answer: Examples given in the text include nonpowered trucks such as dollies and pallet trucks, and powered trucks such as walkie trucks, forklift rider trucks, and towing tractors. 10.8 What is an automated guided vehicle system (AGVS)? Answer: As defined in the text, an automated guided vehicle system is a material handling system that uses independently operated, self-propelled vehicles guided along defined pathways. The vehicles are powered by on-board batteries. 10.9 Name three categories of automated guided vehicles. Answer: The three categories are (1) driverless trains, consisting of a towing vehicle that pulls one or more trailers, (2) pallet trucks, used to move palletized loads, and (3) unit load carriers, which move unit loads. 10.10 What features distinguish laser-guided vehicles from conventional AGVs? Answer: Conventional AGVs use either imbedded guide wires in the floor or paint strips on the floor surface as the guidance technology, whereas laser-guided vehicles (LGVs) operate without continuously defined pathways. Instead, they use a combination of dead reckoning and reflective beacons located throughout the plant that can be identified by on-board laser scanners. Dead reckoning refers to the capability of the vehicle to follow a given route by counting its own wheel rotations along a specified trajectory. 10.11 What is forward sensing in AGVS terminology? Answer: Forward sensing involves the use of one or more sensors on each vehicle to detect the presence of other vehicles and obstacles ahead on the guide path. Sensor technologies include optical and ultrasonic devices. When the on-board sensor detects an obstacle in front of it, the vehicle is programmed to stop. 10.12 What are some of the differences between rail-guided vehicles and automated guided vehicles? Answer: Rail-guided vehicles ride on tracks on the floor or overhead, whereas AGVs ride on the building floor without rails. Rail-guided vehicles are guided by the tracks, whereas AGVs are guided by imbedded wires in the floor that emit a magnetic field, or by paint strips, or other means that does not rely on rails. Rail-guided vehicles obtain their electrical power from a “third rail”, whereas AGVs carry batteries as their electrical power source. 10.13 What is a conveyor? Answer: As defined in the text, a conveyor is a mechanical apparatus for moving items or bulk materials, usually inside a facility. Conveyors are used when material must be moved in relatively large quantities between specific locations over a fixed path, which may be inthe-floor, above-the-floor, or overhead. Conveyors divide into two basic categories: (1) powered and (2) non-powered. 10.14 Name some of the different types of conveyors used in industry. Answer: The conveyor types listed in the text include roller, skate wheel, belt, chain, in-floor towline, overhead trolley, and cart-on-track. 10.15 What is a recirculating conveyor? Answer: A recirculating conveyor is a closed-loop conveyor that allow parts or loads to remain on the return loop for one or more revolutions. 10.16 What is the difference between a hoist and a crane? Answer: A hoist is a mechanical device for lifting and lowering loads vertically, whereas a crane is a mechanical apparatus for horizontal movement of loads. A crane invariably includes one or more hoists. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Analysis of Vehicle-based Systems 10.1 (A) An automated guided vehicle system has an average travel distance per delivery = 220 m and an average empty travel distance = 160 m. Load and unload times are each 24 sec and the speed of the AGV = 1 m/sec. Traffic factor = 0.9, and availability = 0.94. How many vehicles are needed to satisfy a delivery requirement of 35 deliveries/hr? Answer: Tc = 24 + 220/1 + 24 + 160/1 = 428 sec = 7.13 min Rdv = 60(0.90)(0.94)/7.13 = 7.12 deliveries/hr per vehicle nc = 35/7.12 = 4.92 → 5 vehicles 10.2 In Example 10.2 in the text, suppose that the vehicles operate according to the following scheduling rules: (1) vehicles delivering raw work parts from station 1 to stations 2, 3, and 4 must return empty to station 5; and (2) vehicles picking up finished parts at stations 2, 3, and 4 for delivery to station 5 must travel empty from station 1. (a) Determine the empty travel distances associated with each delivery and develop a from-to chart in the format of Table 10.5 in the text. (b) The AGVs travel at a speed of 50 m/min, and the traffic factor = 0.90. Assume reliability = 100%. From Example 10.2, the delivery distance Ld = 103.8 m. Determine the value of Le. (c) How many automated guided vehicles will be required to operate the system? Answer: (a) Enumeration of empty trips: Deliveries Associated empty trips Frequency Empty distance 1 to 2 2 to 1 9 110 2 to 5 1 to 2 and 5 to 1 9 50 + 30 1 to 3 3 to 1 5 200 3 to 5 1 to 3 and 5 to 1 3 120 + 30 1 to 4 4 to 1 6 115 3 to 4 1 to 3 and 4 to 1 2 120 + 115 4 to 5 1 to 4 and 5 to 1 8 205 + 30 From-To chart: To: 1 2 3 4 5 From: 1 0/0 9/50 5/120 8/205 - 2 9/110 0/0 - - - 3 5/200 - 0/0 - - 4 8/115 - - 0/0 - 5 20/30 - - - 0/0 (b) Le = 9(50) + 5(120) + 8(205) + 9(110) + 5(200) + 8(115) + 20(30)/42 = 6200/42 = 147.6 m Loaded travel distance from Example 10.2 Ld = 103.8 m (c) Tc = 1.0 + 103.8/50 + 147.6/50 = 6.03 min Rdv = 60(0.90)/6.03 = 8.96 deliveries/hr per vehicle Number of vehicles nc = 42/8.96 = 4.7 → 5 vehicles 10.3 In Example 10.2 in the text, suppose that the vehicles operate according to the following scheduling rule in order to minimize the distances the vehicles travel empty: vehicles delivering raw work parts from station 1 to stations 2, 3, and 4 must pick up finished parts at these respective stations for delivery to station 5. (a) Determine the empty travel distances associated with each delivery and develop a from-to chart in the format of Table 10.5 in the text. (b) The AGVs travel at a speed of 50 m/min, and the traffic factor = 0.90. Assume reliability = 100%. From Example 10.2, the delivery distance Ld = 103.8 m. Determine the value of Le. (c) How many automated guided vehicles will be required to operate the system? Answer: (a) Enumeration of empty trips: Deliveries Associated empty trips Frequency Empty distance 1 to 2 none 2 to 5 5 to 1 9 30 1 to 3 none 3 to 5 5 to 1 3 30 1 to 4 none 3 to 4 none 4 to 5 5 to 1 8 30 From-To chart: To: 1 2 3 4 5 From: 1 0/0 - - - - 2 - 0/0 - - - 3 - - 0/0 - - 4 - - - 0/0 - 5 20/30 - - - 0/0 Le = 20(30)/42 = 14.3 m Tc = 1.0 + 103.8/50 + 14.3/50 = 3.362 min Rdv = 60(0.90)/3.362 = 16.06 deliveries/hr per vehicle Number of vehicles nc = 42/16.06 = 2.625 → 3 vehicles Comment: The results of Problems 10.2 and 10.3 demonstrate the significance of vehicle scheduling in AGVS operations. Problem 10.2 represents the worst possible scheduling of vehicles, while Problem 10.3 represents almost perfect scheduling. 10.4 A planned manufacturing system will have the layout pictured in Figure P10.4 and will use an automated guided vehicle system to move parts between stations in the layout. All work parts are loaded into the system at station 1, moved to one of three processing stations (2, 3, or 4), and then brought back to station 1 for unloading. Once loaded onto its AGV, each work part stays onboard the vehicle throughout its time in the manufacturing system. Load and unload times at station 1 are each 0.5 min. Processing times at the processing stations are 6.5 min at station 2, 8.0 min at station 3, and 9.5 min at station 4. Vehicle speed = 50 m/min. Assume that the traffic factor = 1.0, and vehicle availability = 100%. (a) Construct the fromto chart for distances. (b) Determine the maximum hourly production rate for each of the three processing stations, assuming that 15 s will be lost between successive vehicles at each station; this is the time for the vehicle presently at the station to move out and the next vehicle to move into the station for processing. (c) Find the total number of AGVs that will be needed to achieve these production rates. Answer: (a) First develop the distances from the FMS layout. Distance from 1 to 2: 10 + 10 + 10 = 30 m Distance from 2 to 1: 5 + 10 + 5 = 20 m Distance from 1 to 3: 10 + 20 + 10 = 40 m Distance from 3 to 1: 5 + 20 + 5 = 30 m Distance from 1 to 4: 10 + 30 + 10 = 50 m Distance from 4 to 1: 5 + 30 + 5 = 40 m From-To chart for distances: To: 1 2 3 4 From: 1 0 30 40 50 2 20 0 - - 3 30 - 0 - 4 40 - - 0 To determine the maximum production rates for the three stations, first determine the time required for processing plus the 15 s lost time at each processing station: Station 2: Rp = 60/(6.5 + 0.25) = 8.89 pc/hr Station 3: Rp = 60/(8.0 + 0.25) = 7.27 pc/hr Station 4: Rp = 60/(9.5 + 0.25) = 6.15 pc/hr To determine the number of vehicles required to achieve these production rates, find the vehicle delivery cycle time for each processing station. Station 2: Tc = 0.5 + (30 + 20)/50 + 6.5 + 0.5 = 8.5 min Station 3: Tc = 0.5 + (40 + 30)/50 + 8.0 + 0.5 = 10.4 min Station 4: Tc = 0.5 + (50 + 40)/50 + 9.5 + 0.5 = 12.3 min Total workload WL = 8.89(8.5) + 7.27(10.4) + 6.15(12.3) = 226.8 min/hr Available time per vehicle AT = 60(1.0)(1.0) = 60 min/hr per vehicle Number of vehicles required nc = 226.8/60 = 3.78 rounded up to 4 vehicles 10.5 In the previous problem, it is unrealistic to assume that the traffic factor will be 1.0 and that vehicle availability will be 100%. It is also unrealistic to believe that the processing stations will operate at 100% reliability. Solve the problem except the traffic factor = 90%, and availability of the vehicles and processing workstations = 95%. Answer: (a) The distances and from-to chart are the same as in the previous problem. Distance from 1 to 2: 10 + 10 + 10 = 30 m Distance from 2 to 1: 5 + 10 + 5 = 20 m Distance from 1 to 3: 10 + 20 + 10 = 40 m Distance from 3 to 1: 5 + 20 + 5 = 30 m Distance from 1 to 4: 10 + 30 + 10 = 50 m Distance from 4 to 1: 5 + 30 + 5 = 40 m To: 1 2 3 4 From: 1 0 30 40 50 2 20 0 - - 3 30 - 0 - 4 40 - - 0 To determine the maximum production rates for the three stations, the time required for processing plus the 15 s lost time at each processing station is divided into 60 min: Station 2: Rp = 60/(6.5 + 0.25) = 8.89 pc/hr at 100% reliability However, with A = 95%, Rp = 8.89(0.95) = 8.45 pc/hr Station 3: Rp = 60/(8.0 + 0.25) = 7.27 pc/hr at 100% reliability However, with A = 95%, Rp = 7.27(0.95) = 6.91 pc/hr Station 4: Rp = 60/(9.5 + 0.25) = 6.15 pc/hr at 100% reliability However, with A = 95%, Rp = 6.15(0.95) = 5.84 pc/hr To determine the number of vehicles required to achieve these production rates, find the vehicle delivery cycle time for each processing station. Station 2: Tc = 0.5 + (30 + 20)/50 + 6.5 + 0.5 = 8.5 min Station 3: Tc = 0.5 + (40 + 30)/50 + 8.0 + 0.5 = 10.4 min Station 4: Tc = 0.5 + (50 + 40)/50 + 9.5 + 0.5 = 12.3 min Total workload WL = 8.45(8.5) + 6.91(10.4) + 5.84(12.3) = 215.5 min/hr Available time per vehicle AT = 60AFt = 60(0.95)(0.90) = 51.3 min/hr per vehicle Number of vehicles required nc = 215.5/51.3 = 4.20 rounded up to 5 vehicles 10.6 (A) A fleet of forklift trucks is being planned for a new warehouse. The average travel distance per delivery will be 500 ft loaded and the average empty travel distance will be 400 ft. The fleet must make a total of 50 deliveries/hr. Load and unload times are each 0.75 min and the speed of the vehicles = 350 ft/min. Assume the traffic factor for the system = 0.85, availability = 0.95, and worker efficiency = 90%. Determine (a) ideal cycle time per delivery, (b) the resulting average number of deliveries/hr that a forklift truck can make, and (c) how many trucks are required to accomplish the 50 deliveries/hr. Answer: (a) Tc = 0.75 + 500/350 + 0.75 + 400/350 = 4.07 min/delivery (b) Ideally, Rdv = 60/4.07 = 14.74 deliveries/hr per truck Accounting for traffic factor, availability, and worker efficiency, Rdv = 14.74(0.85)(0.95)(0.90) = 10.71 deliveries/hr per truck (c) nc = 50/10.71 = 4.67 → 5 forklift trucks 10.7 Three forklift trucks are used to deliver pallet loads of parts between work cells in a factory. Average travel distance loaded is 350 ft and the travel distance empty is estimated to be the same. The trucks are driven at an average speed of 3 miles/hr when loaded and 4 miles/hr when empty. Terminal time per delivery averages 1.0 min (load = 0.5 min and unload = 0.5 min). If the traffic factor is assumed to be 0.90, availability = 100%, and worker efficiency = 0.95, what is the maximum hourly delivery rate of the three trucks? Answer: When loaded, vc = (3 miles/hr)(5280/60) = 264 ft/min When empty, vc = (4 miles/hr)(5280/60) = 352 ft/min Tc = 1.0 + 350/264 + 350/352 = 3.32 min/delivery Rdv = 60(1.0)(0.90)(0.95)/3.32 = 15.45 deliveries/hr per vehicle With three trucks, Rd = 3(15.45) = 46.4 deliveries/hr 10.8 An AGVS has an average loaded travel distance per delivery = 300 ft. The average empty travel distance is not known. Required number of deliveries/hr = 50. Load and unload times are each 0.5 min and the AGV speed = 200 ft/min. Anticipated traffic factor = 0.85 and availability = 0.95. Develop an equation that relates the number of vehicles required to operate the system as a function of the average empty travel distance Le. Answer: Tc = 0.5 + 300/200 + 0.5 + Le/200 = 2.5 + 0.005Le AT = 60(0.95)(0.85)(1.0) = 48.45 min/hr per vehicle WL = 50(2.5 + 0.005Le) = 125 + 0.25 Le nc = WL/AT = (125 + 0.25Le)/48.45 nc = 2.58 + 0.00516 Le 10.9 A rail-guided vehicle system is being planned as part of an assembly cell consisting of two parallel lines, as in Figure P10.9. In operation, a base part is loaded at station 1 and delivered to either station 2 or 4, where components are added to the base part. The RGV then goes to either station 3 or 5, respectively, where further assembly of components is accomplished. From stations 3 or 5, the completed product moves to station 6 for removal from the system. Vehicles remain with the products as they move through the station sequence; thus, there is no loading and unloading of parts at stations 2, 3, 4, and 5. After unloading parts at station 6, the vehicles then travel empty back to station 1 for reloading. The hourly moves (parts/hr, above the slash) and distances (ft, below the slash) are listed in the table below. Moves indicated by "L" are trips in which the vehicle is loaded, while "E" indicates moves in which the vehicle is empty. RGV speed = 150 ft/min. Assembly cycle times at stations 2 and 3 = 4.0 min each, and at stations 4 and 5 = 6.0 min each. Load and unload times at stations 1 and 6 respectively are each 0.75 min. Traffic factor = 1.0 and availability = 1.0. How many vehicles are required to operate the system? To: 1 2 3 4 5 6 From: 1 0/0 13L/100 - 9L/80 - - 2 - 0/0 13L/30 - - - 3 - - 0/0 - - 13L/50 4 - - - 0/0 9L/30 - 5 - - - - 0/0 9L/70 6 22E/300 - - - - 0/0 Answer: Assembly through stations 2 and 3: Tc = 0.75 + 100/150 + 4.0 + 30/150 + 4.0 + 50/150 + 0.75 + 300/150 = 12.7 min Rdv = 60/12.7 = 4.72 deliveries/hr per vehicle nc = 13/4.72 = 2.75 vehicles Assembly through stations 4 and 5: Tc = 0.75 + 80/150 + 6.0 + 30/150 + 6.0 + 70/150 + 0.75 + 300/150 = 16.7 min Rdv = 60/16.7 = 3.59 deliveries/hr per vehicle nc = 9/3.59 = 2.51 vehicles Total vehicles nc = 2.75 + 2.51 = 5.26 → 6 vehicles Comment: Note that rounding up is done after the fractional values are summed. 10.10 An AGVS will be used to satisfy the material flows in the from-to chart below, which shows delivery rates between stations (pc/hr, above the slash) and distances between stations (m, below the slash). Moves indicated by "L" are trips in which the vehicle is loaded, while "E" indicates moves in which the vehicle is empty. It is assumed that availability = 0.90, traffic factor = 0.85, and efficiency = 1.0. Speed of an AGV = 0.9 m/sec. If load handling time per delivery cycle = 1.0 min (load = 0.5 min and unload = 0.5 min), how many vehicles are needed to satisfy the indicated deliveries per hour? To: 1 2 3 4 From: 1 0/0 9L/90 7L/120 5L/75 2 5E/90 0/0 - 4L/80 3 7E/120 - 0/0 - 4 9E/75 - - 0/0 Answer: vc = 0.9 m/s = 54 m/min Route 1 → 2 → 1: Tc = 1.0 + (90 + 90)/54 = 4.33 min, 5 deliveries Route 1 → 3 → 1: Tc = 1.0 + (120 + 120)/54 = 5.44 min, 7 deliveries Route 1 → 4 → 1: Tc = 1.0 + (75 + 75)/54 = 3.78 min, 5 deliveries Route 2 → 4 → 1*: Tc = 1.0 + (80 + 75)/54 = 3.87 min, 4 deliveries Route 1 → 2*: Tc = 1.0 + 90/54 = 2.67 min, 4 deliveries * Assumes vehicles on route 1 → 2 are used to make deliveries on route 2 → 4 → 1 Average Tc = 5(4.33) + 7(5.44) + 5(3.78) + 4(3.87) + 4(2.67)/25 = 4.192 min/delivery cycle Rdv = 60(0.85)/4.192 = 12.166 deliveries/hr per vehicle Including effect of availability factor, Rdv = 12.166(0.90) = 10.95 deliveries/hr per vehicle nc = 25/10.95 = 2.28 → 3 vehicles Alternative solution: Total workload to make all deliveries, neglecting traffic factor: WL = (9 + 7 + 5 + 4)(1.0) + (9(90) + 7(120) + 5(3.78) + 4(80)/25) + (5(90 + 7(120) + 9(75)/54) WL = 104.8 min Time available per vehicle/hr, AT = 60(.90)(.85)(1.0) = 45.9 min nc = 104.8/45.9 = 2.28 → 3 vehicles 10.11 An automated guided vehicle system is being proposed to deliver parts between 40 workstations in a factory. Loads must be moved from each station about once every hour; thus, the delivery rate = 40 loads/hr. Average travel distance loaded is estimated to be 250 ft and travel distance empty is estimated to be 30 ft. Vehicles move at a speed = 200 ft/min. Total handling time per delivery = 1.5 min (load = 0.75 min and unload = 0.75 min). Traffic factor Ft becomes increasingly significant as the number of vehicles nc increases; this can be modeled as Ft = 1.0 - 0.05(nc −1), for nc = Integer > 0. Determine the minimum number of vehicles needed in the factory to meet the flow rate requirement. Assume that availability = 1.0 and worker efficiency = 1.0. Answer: 10.12 A driverless train AGVS is being planned for a warehouse complex. Each train will consist of a towing vehicle plus four carts. Speed of the trains = 160 ft/min. Only the pulled carts carry loads. Average loaded travel distance per delivery cycle is 2000 ft and empty travel distance is the same. Assume the travel factor = 0.95, and availability = 1.0. The load handling time per train per delivery is expected to be 10 min. If the requirements on the AGVS are 25 cart loads/hr, determine the number of trains required. Answer: Tc = 10 + 2000 + 2000/160 = 35.0 min/delivery cycle Rdv = 60(0.95)/35.0 = 1.629 deliveries/hr per vehicle Expressing cart loads as train loads, Rf = 25 cart loads/hr = 25/4 = 6.25 train loads/hr Number of train loads/hr = 6.25/1.629 = 3.84 → Use 4 AGV trains 10.13 The from-to chart below indicates the number of loads moved per 8-hr day (above the slash) and the distances in ft (below the slash) between departments in a particular factory. Fork lift trucks are used to transport the materials. They move at an average speed = 275 ft/min (loaded) and 350 ft/min (empty). Load handling time (loading plus unloading) per delivery is 1.5 min, and anticipated traffic factor = 0.9. Availability = 95% and worker efficiency = 110%. Determine the number of trucks required under each of the following assumptions: (a) the trucks never travel empty; and (b) the trucks travel empty a distance equal to their loaded distance. To Dept. A B C D E From Dept A 0/0 62/500 51/450 45/350 - B - 0/0 - 22/400 - C - - 0/0 - 76/200 D - - - 0/0 65/150 E - - - - 0/0 Answer: 10.14 A warehouse consists of five aisles of racks (racks on both sides of each aisle) and a loading dock. The rack system is four levels high. Forklift trucks are used to transport loads between the loading dock and the storage compartments of the rack system in each aisle. The trucks move at an average speed = 120 m/min (loaded) and 150 m/min (empty). Load handling time (loading plus unloading) per delivery totals 1.0 min per storage/retrieval delivery on average, and the anticipated traffic factor = 0.90. Worker efficiency = 100% and vehicle availability = 95%. The average distance between the loading dock and the centers of aisles 1 through 5 are 150 m, 250 m, 350 m, 450 m, and 550 m, respectively. These values are to be used to compute travel times. The required rate of storage/retrieval deliveries is 75/hr, distributed evenly among the five aisles, and the trucks perform either storage or retrieval deliveries, but not both in one delivery cycle. Determine the number of forklift trucks required to achieve the 75 deliveries per hour. Answer: Ld = (150 + 250 + 350 + 450 + 550)/5 = 350 m, Le = Ld = 350 m Tc = 1.0 + 350/120 + 350/150 = 6.25 min Rdv = 60/6.25 = 9.6 deliveries/hr per truck nc = 75/(9.6 × 0.90 × 0.96 × 1.0) = 75/8.21 = 9.14 rounded to 10 trucks 10.15 Suppose the warehouse in the preceding problem were organized according to a class-based dedicated storage strategy based on activity level of the pallet loads in storage, so that aisles 1 and 2 accounted for 70% of the deliveries (class A) and aisles 3, 4, and 5 accounted for the remaining 30% (class B). Assume that deliveries in class A are evenly divided between aisles 1 and 2, and that deliveries in class B are evenly divided between aisles 3, 4, and 5. How many forklift trucks would be required to achieve 75 storage/retrieval deliveries per hour? Answer: Class A: Ld = (150 + 250)/2 = 200 m, Le = Ld = 200 m Class B: Ld = (350 + 450 + 550)/3 = 450 m, Le = Ld = 450 m Weighted average Ld = Le = 0.70(200) + 0.30(450) = 275 m Tc = 1.0 + 275/120 + 275/150 = 5.125 min Rdv = 60/5.125 = 11.71 deliveries/hr per truck nc = 75/(11.71 × 0.90 × 0.95 × 1.0) = 75/10.01 = 7.49 rounded to 8 trucks 10.16 Major appliances are assembled on a production line at the rate of 50/hr. The products are moved along the line on work pallets (one product per pallet). At the final workstation the finished products are removed from the pallets. The pallets are then removed from the line and delivered back to the front of the line for reuse. Automated guided vehicles are used to transport the pallets to the front of the line, a distance of 500 ft. Return trip distance (empty) to the end of the line is also 500 ft. Each AGV carries three pallets and travels at a speed of 200 ft/min (loaded or empty). The pallets form queues at each end of the line, so that neither the production line nor the AGVs are ever starved for pallets. Time required to load each pallet onto an AGV = 15 sec; time to release a loaded AGV and move an empty AGV into position for loading at the end of the line = 12 sec. The same times apply for pallet handling and release/positioning at the unload station located at the front of the production line. Assume availability = 100% and traffic factor = 1.0 because the route is a simple loop. How many vehicles are needed to operate the AGV system? Answer: (a) TL = Tu = 12 sec + 3(15 sec) = 57 sec = 0.95 min Tc = 0.95 + 500/200 + 0.95 + 500/200 = 6.9 min/delivery cycle Rdv = 60/6.9 = 8.7 delivery cycles/hr per vehicle Each delivery means 3 pallets, so Rdv = 3(8.7) = 26.09 pallets/hr per vehicle nc = 50/26.09 = 1.92 → Use nc = 2 vehicles 10.17 For the production line in the previous problem, assume that a single AGV train consisting of a tractor and multiple trailers is used to make deliveries rather than separate vehicles. Time required to load a pallet onto a trailer = 15 sec; and the time to release a loaded train and move an empty train into position for loading at the end of the production line = 30 sec. The same times apply for pallet handling and release/positioning at the unload station located at the front of the production line. The velocity of the AGV train = 175 ft/min (loaded or empty). Assume availability = 100% and traffic factor = 1.0 because there is only one train. If each trailer carries three pallets, how many trailers should be included in the train? Answer: 10.18 (A) An AGVS will be installed to deliver loads between four workstations: A, B, C, and D. Hourly flow rates (loads/hr, above the slash) and distances (m, below the slash) within the system are given in the table below (travel loaded denoted by “L” and travel empty denoted by “E”). Load and unload times are each 0.45 min, and travel speed of each vehicle is 1.4 m/s. A total of 43 loads enter the system at station A, and 30 loads exit the system at station A. In addition, during each hour, six loads exit the system from station B and seven loads exit the system from station D. This is why there are a total of 13 empty trips made by the vehicles within the AGVS. How many vehicles are required to satisfy these delivery requirements, assuming the traffic factor = 0.85 and availability = 95%? To A B C D From A - 18L/95 10L/80 15L/150 B 6E/95 - 12L/65 C - 22L/80 D 30L/150 7E/150 - Answer: Analysis of Conveyor Systems 10.19 (A) An overhead trolley conveyor is configured as a closed loop. The delivery loop has a length of 100 m and the return loop is 60 m. All parts loaded at the load station are unloaded at the unload station. Each hook on the conveyor can hold one part and the hooks are separated by 2 m. Conveyor speed = 0.5 m/sec. Determine (a) number of parts in the conveyor system under normal operations, (b) parts flow rate; and (c) maximum loading and unloading times that are compatible with the operation of the conveyor system? Answer: (a) Number of parts on the conveyor = (100 m)/(2 m/pc) = 50 pc Rf = npvc/sc = (1 pc/hook)(0.5 m/sec)/(2 m/hook) = 0.25 pc/sec = 900 pc/hr TL = Tu = 1/Rf = 1/0.25 = 4 sec Comment: The time of 4 sec is the time it takes for the hook to move 2 m at the load and unload stations. The actual loading and unloading of parts would have to be accomplished in less than 4 sec. 10.20 A 400-ft long roller conveyor operates at a velocity = 50 ft/min and is used to move parts in containers between load and unload stations. Each container holds 15 parts. One worker at the load station is able to load parts into containers and place the containers onto the conveyor in 45 sec. It takes 30 sec to unload at the unload station. Determine (a) center-to-center distance between containers, (b) number of containers on the conveyor at one time, and (c) hourly flow rate of parts. (d) By how much must conveyor speed be increased in order to increase flow rate to 1500 parts/hr? Answer: (a) sc = TL vc = (45/60 min)(50 ft/min) = 37.5 ft/container Number containers on conveyor = Ld/sc = (400 ft)/(37.5 ft/container) = 10.667 Comment: Of course, the number of containers must be an integer, so this answer must be interpreted to mean that there are 10 containers on the conveyor 33.33% of the time, and 11 containers on the conveyor 66.67% of the time. Rf = npvc/sc = (15 parts/container)(50 ft/min)/(37.5 ft/container) = 20 parts/min Rf = 1200 parts/hr Increasing vc would have no effect on flow rate Rf in this problem. The loading rate, set by TL = 45 sec, is what limits the flow rate in this system. If vc were increased, it would increase the distance between containers and cancel the increase in conveyor velocity. 10.21 (A) A roller conveyor moves tote pans in one direction at 200 ft/min between a load station and an unload station, a distance of 350 ft. With one worker, the time to load parts into a tote pan at the load station is 3 sec per part. Each tote pan holds 10 parts. In addition, it takes 15 sec to load a tote pan of parts onto the conveyor. Determine (a) spacing between tote pan centers flowing in the conveyor system and (b) flow rate of parts on the conveyor system. (c) Consider the effect of the unit load principle. Suppose the tote pans were smaller and could hold only one part instead of 10. Determine the flow rate of parts in this case if it takes 7 sec to load a tote pan onto the conveyor (instead of 15 sec for the larger tote pan), and it takes the same 3 sec to load the part into the tote pan. Answer: (a) TL = 15 + 3(10) = 45 sec = 0.75 min/pan sp = (200 ft/min)(0.75 min/pan) = 150 ft/pan Rf = npvc/sc = (10 pc/pan)(200 ft/min)/(150 ft/pan) = 13.33 pc/min = 800 pc/hr TL = 7 + 3(1) = 10 sec = 0.167 min/pan sp = (200 ft/min)(0.167 min/pan) = 33.33 ft/pan Rf = npvc/sc = (1 pc/pan)(200 ft/min)/(33.33 ft/pan) = 6.0 pc/min = 360 pc/hr Comment: The merits of the unit load principle are demonstrated by comparing answers in (b) and (c). 10.22 A closed loop overhead conveyor must be designed to deliver parts from one load station to one unload station. The specified flow rate of parts that must be delivered between the two stations is 300 parts/hr. The conveyor has carriers spaced at a center-to-center distance that is to be determined. Each carrier holds one part. Forward and return loops will each be 90 m long. Conveyor speed = 0.5 m/sec. Times to load and unload parts at the respective stations are each = 12 sec. Is the system feasible and if so, what is the appropriate number of carriers and spacing between carriers that will achieve the specified flow rate? Answer: 10.23 Consider the previous problem, only the carriers are larger and capable of holding up to four parts (np = 2, 3, or 4). The loading time TL = 9 + 3np, where TL is in seconds. With other parameters defined in the previous problem, determine which of the three values of np are feasible. For those values that are feasible, specify the appropriate design parameters for (a) spacing between carriers and (b) number of carriers that will achieve this flow rate. Answer: 10.24 A recirculating conveyor has a total length of 700 ft and a speed of 90 ft/min. Spacing of part carriers = 14 ft. Each carrier holds one part. Automated machines load and unload the conveyor at the load and unload stations. Time to load a part is 0.10 min and unload time is the same. To satisfy production requirements, the loading and unloading rates are each 2.0 parts per min. Evaluate the conveyor system design with respect to the three principles developed by Kwo. Answer: 10.25 A recirculating conveyor has a total length of 200 m and a speed of 50 m/min. Spacing of part carriers = 5 m. Each carrier holds two parts. Time needed to load a part carrier = 0.15 min. Unloading time is the same. The required loading and unloading rates are 6 parts per min. Evaluate the conveyor system design with respect to the three Kwo principles. Answer: 10.26 There is a plan to install a continuous loop conveyor system with a total length of 1000 ft and a speed of 50 ft/min. The conveyor will have carriers separated by 25 ft. Each carrier can hold one part. A load station and an unload station are to be located 500 ft apart along the conveyor loop. Each day, the conveyor system is planned to operate as follows, starting empty at the beginning of the day. The load station will load parts at the rate of one part every 30 sec, continuing this loading operation for 10 min, then resting for 10 min during which no loading occurs. It will repeat this 20 min cycle of loading and then resting throughout the 8-hr shift. The unload station will wait until loaded carriers begin to arrive, then will unload parts at the rate of one part every minute during the eight hours, continuing until all carriers are empty. Will the planned conveyor system work? Present calculations and arguments to justify your answer. Answer: Time for a carrier to complete one loop = L/Vc = (1000 ft)/(50 ft/min) = 20 min vc Analysis: The time for one loading cycle = 20 min (10 min loading + 10 min rest). On average, loading and unloading schedules seem balanced (parts loaded in 20 min = 10 min × 2 parts/min + 10(0) = 20 parts total or 60 parts/hr, and parts unloaded in 20 min = 20 min × 1 part/min = 20 parts total or 60 parts/hr). However, due to the loading schedule, the 1000 ft conveyor length will be divided into two sections: the first 500 ft will be fully loaded when it leaves the loading station, but the second 500 ft will always be empty. When the first 500 ft flows past the unload station, only half the parts will be unloaded, leaving every other carrier loaded. When the second 500 ft flows past the unload station, all carriers are empty so there are no parts to unload. When the first 500 ft comes back around to the load station, the loading rate will be only half the intended rate since every other carrier already contains a part. Thus, the intended loading and unloading rates will be only half their planned values (10 parts in 20 min or 30 parts/hr). Conclusion: The system is not feasible as the operation is planned. Chapter 11 STORAGE SYSTEMS REVIEW QUESTIONS 11.1 Materials stored in manufacturing include a variety of types. Name six of the categories listed in Table 11.1. Answer: The material types listed in Table 11.1 are (1) raw materials, (2) purchase parts, (3) work-in-process, (4) finished product, (5) rework and scrap, (6) refuse, (7) tooling, (8) spare parts, (9) office supplies, and (10) plant records. 11.2 Name and briefly describe the six measures used to assess the performance of a storage system? Answer: The six performance measures discussed in the text are the following: (1) storage capacity, which is defined and measured either as the total volumetric space available or as the total number of storage compartments in the system available for items or loads; (2) storage density, defined as the volumetric space available for actual storage relative to the total volumetric space in the storage facility; (3) accessibility, which refers to the capability to access any desired item or load stored in the system; (4) system throughput, defined as the hourly rate at which the storage system receives and puts loads into storage and/or retrieves and delivers loads to the output station; (5) utilization, which is defined as the proportion of time that the system is actually being used for performing storage and retrieval operations compared with the time it is available, and (6) availability, defined as the proportion of time that the system is capable of operating (not broken down) compared with the normally scheduled shift hours. 11.3 Briefly describe the two basic storage location strategies. Answer: The two basic strategies are (1) randomized storage and (2) dedicated storage. In randomized storage, items are stored in any available location in the storage system. In the usual implementation of randomized storage, incoming items are placed into storage in the nearest available open location. When an order is received for a given SKU, the stock is retrieved from storage according to a first-in-first-out policy so that the items held in storage the longest are used to make up the order. In dedicated storage, SKUs are assigned to specific locations in the storage facility. This means that enough locations are reserved for all SKUs stored in the system, and so the number of storage locations for each SKU must be sufficient to accommodate its maximum inventory level. 11.4 What is a class-based dedicated storage strategy? Answer: A class-based dedicated storage allocation is a compromise between a randomized storage strategy and a dedicated storage strategy. The storage system is divided into several classes according to activity level, and a randomized storage strategy is used within each class. The classes containing more-active SKUs are located closer to the input/output point of the storage system for increased throughput, and the randomized locations within the classes reduce the total number of storage compartments required. 11.5 Name the four traditional (non-automated) methods for storing materials. Answer: The four traditional methods for storing materials are (1) bulk storage, (2) rack systems, (3) shelving and bins, and (4) drawer storage. 11.6 Which of the four traditional storage methods is capable of the highest storage density? Answer: Bulk storage is capable of the highest storage density, given that materials can be stacked on top of each other and very little allowance is made for aisle space. 11.7 What are some of the objectives and reasons behind company decisions to automate their storage operations? Name six of the ten objectives and reasons listed in Table 11.3. Answer: The ten objectives and reasons given in Table 11.3 are to (1) increase storage capacity, (2) increase storage density, (3) recover factory floor space presently used for storing work-in-process, (4) improve security and reduce pilferage, (5) improve safety in the storage function, (6) reduce labor cost and/or increase labor productivity in storage operations, (7) improve control over inventories, (8) improve stock rotation, (9) improve customer service, and (10) increase throughput. 11.8 What are the two basic categories of automated storage systems? Answer: The two basic categories of automated storage systems are (1) fixed-aisle automated storage/retrieval systems and (2) carousel storage systems. 11.9 What are the differences between the two basic types of automated storage systems? Answer: The fixed-aisle AS/RS consists of a rack structure for storing loads and a storage/retrieval machine whose motions are linear (x-y-z motions). By contrast, the carousel system uses storage baskets attached to a chain-driven conveyor that revolves around an oval track loop to deliver the baskets to a load/unload station. Both types include horizontal and vertical structures, with the horizontal configuration being much more common in both cases. 11.10 What is a vertical lift module? Answer: Most fixed-aisle automated storage/retrieval systems are designed around a horizontal aisle. The same principle of using a center aisle to access loads is used in a VLM except that the aisle is vertical. The structure consists of two columns of trays that are accessed by a S/R machine (also called an extractor) that delivers the trays one-by-one to a load/unload station at floor level. Vertical lift modules, some with heights of 10 m (30 ft) or more, are capable of holding large inventories while saving valuable floor space in the facility. 11.11 Identify the three application areas of fixed-aisle automated storage/retrieval systems. Answer: The three AS/RS application areas identified in the text are (1) unit load storage and handling, (2) order picking, and (3) work-in-process storage. 11.12 What are the four basic components of nearly all automated storage/retrieval systems? Answer: The four basic components of nearly all automated storage/retrieval systems are (1) storage structure, which is a rack system, (2) S/R machine, (3) storage modules (e.g., pallets for unit loads), and (4) one or more pickup-and-deposit stations. In addition, a control system is required to operate the AS/RS. 11.13 What is the advantage of a vertical storage carousel over a horizontal storage carousel? Answer: A vertical storage carousel requires less floor space. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Sizing the AS/RS Rack Structure 11.1 (A) Each aisle of a six-aisle Automated Storage/Retrieval System is to contain 50 storage compartments in the length direction and 8 compartments in the vertical direction. All storage compartments will be the same size to accommodate standard size pallets of dimensions: x = 36 in and y = 48 in. The height of a unit load z = 30 in. Using the allowances a = 6 in, b = 8 in, and c = 10 in, determine (a) how many unit loads can be stored in the AS/RS, and (b) the width, length, and height of the AS/RS. The rack structure will be built 18 in above floor level. Answer: (a) Capacity per aisle = 2(50(8) = 800 loads/aisle With six aisles, AS/RS capacity = 6(800) = 4800 loads (b) W = 3(x + a) = 3(36 + 6) = 126 in/aisle With 6 aisles, AS/RS width = 6(126) = 756 in = 63 ft L = ny (y + b) = 50(48 + 8) = 2800 in = 233.33 ft H = nz (z + c) = 8(30 + 10) = 320 in = 26.67 ft The rack structure is built 18 in above floor level: H = 320 + 18 = 338 in = 28.167 ft 11.2 A unit load AS/RS is being designed to store 1000 pallet loads in a distribution center located next to the factory. Pallet dimensions are: x = 1000 mm, y = 1200 mm; and the maximum height of a unit load = 1300 mm. The following is specified: (1) the AS/RS will consist of two aisles with one S/R machine per aisle, (2) length of the structure should be approximately five times its height, and (3) the rack structure will be built 500 mm above floor level. Using the allowances a = 150 mm, b = 200 mm, and c = 250 mm, determine the width, length, and height of the AS/RS rack structure. Answer: Assumption: the L/H ratio does not include the 500 mm foundation. 1000 pallets/ 2 aisles = 500 pallets/aisle. 500 pallets/aisle → 250 pallets per aisle side. Thus ny nz = 250 Eq. (1) L = ny (y + b) = ny (1200 + 200) = 1400 ny (mm) = 1.4 ny (m) H = nz (z + c) = nz (1300 + 250) = 1550 nz (mm) = 1.55 nz (m) Given the specification L/H = 5 1.40ny/1.55nz = 0.9032ny/nz = 5 0.9032 ny = 5.0 nz ny = 5.536 nz Eq. (2) Combining Eqs. (1) and (2): ny nz = (5.536 nz) nz = 250 nz2 = 250/5.536 = 45.161 nz = 6.72 → use nz = 7 ny = 250/nz = 250/7 = 35.71 → use nz = 36 W = 3(1000 + 150) = 3450 mm = 3.45 m/aisle. With 2 aisles, W = 2(3.45) = 6.9 m L = 1.4 ny = 1.4(36) = 50.4 m H = 1.55 nz = 1.55(7) = 10.85 m Given that the rack structure is built 500 mm above floor level, H = 10.85 + 0.5 = 11.35 m Check on specifications: Capacity = 2 x 2 x 36 x 7 = 1008 pallets L/H = 50.4/10.85 = 4.645 11.3 Given the rack structure dimensions computed in Problem 11.2. Assuming that 85% of the storage compartments are occupied on average, and that the average volume of a unit load per pallet in storage = 1.0 m3, compute the ratio of the total volume of unit loads in storage relative to the total volume occupied by the storage rack structure. Answer: Number of unit loads = 0.85(1008) = 856.8 unit loads on average Volume of loads at 0.75 m3 per load: Vu = 856.8(1.0) = 856.8 m3 Volume of rack structure, excluding elevation above floor level (using values computed in Problem 11.2): Vr = W × L × H = 6.9(50.4)(10.85) = 37773.2 m3 Ratio Vu/ Vr = 856.8/3773.2 = 0.227 = 22.7% 11.4 A unit load AS/RS for work-in-process storage in a factory must be designed to store 2000 pallet loads, with an allowance of no less than 20% additional storage compartments for peak periods and flexibility. The unit load pallet dimensions are: depth (x) = 36 in and width (y) = 48 in. Maximum height of a unit load = 42 in. It has been determined that the AS/RS will consist of four aisles with one S/R machine per aisle. The maximum ceiling height (interior) of the building permitted by local ordinance is 60 ft, so the AS/RS must fit within this height limitation. The rack structure will be built 2 ft above floor level, and the clearance between the rack structure and the ceiling of the building must be at least 18 in. Determine the dimensions (height, length, and width) of the rack structure. Answer: Total number of storage compartments = 2000(1 + 0.20) = 2400 compartments With 4 aisles, number of compartments per aisle = 2400/4 = 600 Number of compartments per aisle side = 600/2 = 300 compartments/side H = nz (z + c) = nz (42 + 10) = 52 nz Given that the rack structure is elevated 2 ft = 24 in above floor level and that the clearance above the rack structure = 18 in, then the AS/RS rack structure itself must have a height H less than 60(12) - 24 - 18 = 678 in. Thus, 52 nz ≤ 678 nz ≤ 678/52 = 13.04 → use nz = 13 H = 52 nz = 52(13) = 676 in = 56.33 ft (clearance above rack structure > 18 in by 2 in) ny nz = 300 ny = 300/13 = 23.07 → use ny = 24 compartments along aisle length L = ny (y + b) = 24(48 + 8) = 1344 in = 112 ft W = 3(x + a) = 3(36 + 6) = 126 in/aisle. With 4 aisles, W = 4(126) = 504 in = 42 ft Actual capacity = 4 × 2 × 24 × 13 = 2496, which slightly exceeds the specification of 2400. AS/RS Throughput Analysis 11.5 (A) The length of the storage aisle in an AS/RS = 240 ft and its height = 60 ft. Horizontal and vertical speeds of the S/R machine are 400 ft/min and 60 ft/min, respectively. The S/R machine requires 18 sec to accomplish a pick and deposit operation. Find (a) the single-command and dual-command cycle times per aisle, and (b) throughput for the aisle under the assumptions that storage system utilization = 85% and the number of single command and dual command cycles are equal. Answer: 11.6 Solve Problem 11.5 except that the ratio of single-command to dual-command cycles is 3:1 instead of 1:1. Answer: 11.7 An AS/RS is used for work-in-process storage in a manufacturing facility. The AS/RS has five aisles, each aisle being 120 ft long and 40 ft high. The horizontal and vertical speeds of the S/R machine are 400 ft/min and 50 ft/min, respectively. The S/R machine requires 12 sec to accomplish a pick and deposit operation. The number of single command cycles equals the number of dual command cycles. If the requirement is that the AS/RS must have a throughput rate of 200 S/R transactions/hr during periods of peak activity, will the AS/RS satisfy this requirement? If so, what is the utilization of the AS/RS during peak hours. Answer: 11.8 An automated storage/retrieval system installed in a warehouse has five aisles. The storage racks in each aisle are 30 ft high and 150 ft long. The S/R machine for each aisle travels at a horizontal speed of 350 ft/min and a vertical speed of 60 ft/min. The pick and deposit time = 0.25 min. Assume that the number of single command cycles/hr is equal to the number of dual command cycles/hr and that the system operates at 75% utilization. Determine the throughput rate (loads moved/hr) of the AS/RS. Answer: 11.9 A 10-aisle automated storage/retrieval system is located in an integrated factory-warehouse facility. The storage racks in each aisle are 18 m high and 95 m long. The S/R machine for each aisle travels at a horizontal speed of 2.5 m/sec and a vertical speed of 0.5 m/sec. Pick and deposit time = 20 sec. Assume that the number of single command cycles/hr is one-half the number of dual command cycles/hr and that the system operates at 80% utilization. Determine the throughput rate (loads moved/hr) of the AS/RS. Answer: 11.10 An automated storage/retrieval system for work-in-process has five aisles. The storage racks in each aisle are 10 m high and 50 m long. The S/R machine for each aisle travels at a horizontal speed of 2.0 m/sec and a vertical speed of 0.4 m/sec. Pick and deposit time = 15 sec. Assume that the number of single command cycles/hr is equal to three times the number of dual command cycles/hr and that the system operates at 90% utilization. Determine the throughput rate (loads moved/hr) of the AS/RS. Answer: 11.11 (A) The length of one aisle in an AS/RS is 100 m and its height is 20 m. Horizontal travel speed is 4.0 m/sec. The vertical speed is specified so that the storage system is "square in time," which means that L/vy = H/vz. The pick-and-deposit time is 12 sec. Determine the expected throughput rate (transactions/hr) for the aisle if the expected ratio of the number of transactions performed under single-command cycles to the number of transactions performed under dual-command cycles is 2:1. The system operates continuously during the hour. Answer: Tcs = 2(0.5(100)/4.0) + 2(12) = 49 sec/cycle = 0.817 min/cycle Tcd = 2(0.75(100)/4.0)+ 4(12) = 85.5 sec/cycle = 1.425 min/cycle Tts = Tcs = 0.817 min since one transaction is accomplished each single command cycle Ttd = Tcd/2 = 0.7125 min since two transactions are accomplished each dual command cycle Rts Tts + Rtd Ttd = 60 0.817 Rts + 0.7125 Rtd = 60 Given that Rts = 2 Rtd, 0.817 (2 Rtd) + 0.7125 Ttd = 2.3465 Rtd = 60 Rtd = 25.57 trans/hr Rts = 2 Rtd = 2(25.57) = 51.14 trans/hr Rt = Rts + Rtd = 51.14 + 25.57 = 76.71 transactions/hr 11.12 An automated storage/retrieval system has four aisles. The storage racks in each aisle are 40 ft high and 200 ft long. The S/R machine for each aisle travels at a horizontal speed of 400 ft/min and a vertical speed of 60 ft/min. If the pick and deposit time = 0.3 min, determine the throughput rate (loads moved/hr) of the AS/RS, under the assumption that time spent each hour performing single command cycles is twice the time spent performing dual command cycles, and that the AS/RS operates at 90% utilization. Answer: Tcs = 2 Max(0.5(200)/400, 0.5(40)/60) + 2(0.3) = 1.267 min/cycle Tcd = 2 Max(0.75(200)/400, 0.75(40)/60) + 4(0.3) = 2.2 min/cycle 1.267 Rcs + 2.2 Rcd = 60(0.90) = 54.0 min Given Rcs Tcs = 2(Rcd Tcd), 1.267 Rcs = 2(2.2 Rcd ) = 4.4 Rcd Rcs = 3.473 Rcd 1.267 (3.473 Rcd) + 2.2 Rcd = 6.60 Rcd = 54 Rcd = 8.182 cycles/hr Rcs = 3.473(8.182) = 28.415 cycles/hr Rt = Rcs + Rcd = 28.415 + 2(8.182) = 44.78 transactions/hr Check: Time performing single command cycles = Rcs Tcs = 28.415(1.267) = 36.0 min Time performing dual command cycles = Rcd Tcd = 8.182(2.2) = 18.0 min. This is a 2:1 ratio. With 4 aisles, Rt = 4(44.78) = 179.12 transactions/hr 11.13 An AS/RS with one aisle is 300 ft long and 60 ft high. The S/R machine has a maximum speed of 300 ft/min in the horizontal direction. It accelerates from zero to 300 ft/min in a distance of 15 ft. On approaching its target position (where the S/R machine will transfer a load onto or off of its platform), it decelerates from 300 ft/min to a full stop in 15 ft. The maximum vertical speed is 60 ft/min, and the vertical acceleration and deceleration distances are each 3 ft. Assume simultaneous horizontal and vertical movement, and that the rates of acceleration and deceleration are constant in both directions. The pick and deposit time = 0.3 min. Using the general approach of the MHI method for computing cycle time but adding considerations for acceleration and deceleration, determine the single command and dual command cycle times. Answer: Because of the square-in-time feature, only one direction needs to be considered to calculate travel time, since both y and z travel times are equal 11.14 An AS/RS with four aisles is 80 m long and 18 m high. The S/R machine has a maximum speed of 1.6 m/sec in the horizontal direction. It accelerates from zero to 1.6 m/sec in a distance of 2.0 m. On approaching its target position (where the S/R machine will transfer a load onto or off of its platform), it decelerates from 1.6 m/sec to a full stop in 2.0 m. The maximum vertical speed is 0.5 m/sec, and the vertical acceleration and deceleration distances are each 0.3 m. Rates of acceleration and deceleration are constant in both directions. Pick and deposit time = 12 sec. Utilization of the AS/RS is assumed to be 90%, and the number of dual command cycles = the number of single command cycles. (a) Calculate the single command and dual command cycle times, including considerations for acceleration and deceleration. (b) Determine the throughput rate for the system. t 11.15 Your company is seeking proposals for an automated storage/retrieval system that will have a throughput rate of 300 storage/retrieval transactions/hr during the one 8-hr shift per day. The request for proposal indicates that the number of single command cycles is expected to be four times the number of dual command cycles. The first proposal received is from a vendor who specifies the following: ten aisles, each aisle 150 ft long and 50 ft high; horizontal and vertical speeds of the S/R machine = 200 ft/min and 66.67 ft/min, respectively; and pick and deposit time = 0.3 min. As the responsible engineer for the project, you must analyze the proposal and make recommendations accordingly. One of the difficulties you see in the proposed AS/RS is the large number of S/R machines that would be required - one for each of the 10 aisles. This makes the proposed system very expensive. Your recommendation is to reduce the number of aisles from 10 to 6 and to select a S/R machine with horizontal and vertical speeds of 300 ft/min and 100 ft/min, respectively. Although each high speed S/R machine is slightly more expensive than the slower model, reducing the number of machines from 10 to 6 will significantly reduce total cost. Also, fewer aisles will reduce the cost of the rack structure even though each aisle will be somewhat larger since total storage capacity must remain the same. The problem is that throughput rate will be adversely affected by the larger rack system. (a) Determine the throughput rate of the proposed 10-aisle AS/RS and calculate its utilization relative to the specified 300 transactions/hr. (b) Determine the length and height of a six-aisle AS/RS whose storage capacity would be the same as the proposed 10-aisle system. (c) Determine the throughput rate of the 6-aisle AS/RS and calculate its utilization relative to the specified 300 transactions/hr. (d) Given the dilemma now confronting you, what other alternatives would you analyze and recommendations would you make to improve the design of the system? Answer: (a) Tcs = 2(0.5(150)/200) + 2(0.3) = 1.35 min/cycle Tcd = 2(0.75(150)/200) + 4(0.3) = 2.325 min/cycle 1.35 Rcs + 2.325 Rcd = 60 Given Rcs = 4 Rcd, 1.35(4 Rcd) + 2.325 Rcd = 7.725 Rcd = 60 Rcd = 7.767 cycles/hr Rcs = 4(7.767) = 31.068 cycles/hr Rt = Rcs + 2 Rcd = 31.068 + 2(7.767) = 46.6 transactions/hr per aisle With 10 aisles, Rt = 10(46.6) = 466 transactions/hr Given the specification Rt = 300 transactions/hr, U = 300/466 = 0.644 = 64.4% This utilization is somewhat lower than desirable. Assume capacity is a function of (# aisles) L H. Let subscript 1 represent the 10 aisle AS/RS. Let subscript 2 represent the 6 aisle AS/RS. Then, 6 L2 H2 = 10 L1 H1 = 10(150)(50) = 75,000 To maintain square-in-time proportions for the 6-aisle system, L2/H2 = L1/H1 = 150/50 = 3.0 Thus, L2 = 3 H2. Thus, 6(3 H2)H2 = 18(H2)2 = 75,000 (H2)2 = 75,000/18 = 4166.67 H2 = 64.55 ft and L2 = 3 H2 = 3(64.55) = 193.65 ft Tcs = 2(0.5(193.65)/300)+ 2(0.3) = 1.246 min/cycle Tcd = 22(0.75(193.65)/300)+ 4(0.3) = 2.168 min/cycle 1.246 Rcs + 2.168 Rcd = 60 Given Rcs = 4 Rcd, 1.246(4 Rcd ) + 2.168 Rcd = 7.152 Rcd = 60 Rcd = 8.389 cycles/hr Rcs = 4(8.389) = 33.556 cycles/hr Rt = Rcs + 2 Rcd = 33.556 + 2(8.389) = 50.334 transactions/hr per aisle With 6 aisles, Rt = 6(50.334) = 302.0 transactions/hr Given the specification of 300 transactions/hr for the AS/RS, U = 300/302 = 0.993 = 99.3% This utilization is much higher than desirable. Other alternatives and recommendations: Investigate the possibility of an AS/RS with 7 or 8 aisles. The throughput performance should be between the two alternatives analyzed above. Improve scheduling of the AS/RS to achieve Rcs = Rcd rather than Rcs = 4 Rcd as given in problem statement. Operate a 9-hr shift rather than an 8-hr shift. Although hourly throughput will not be improved, the extra hour of AS/RS operation will improve daily throughput rates. AS/RS Throughput for Class-Based Dedicated Storage Strategy 11.16 (A) The aisles in the AS/RS of Example 11.3 in the text will be organized to follow a classbased dedicated storage strategy. There will be two classes, according to activity level. The more active stock is stored in the half of the rack system that is located closest to the input/output station, and the less active stock is stored in the other half of the rack system farther away from the input/output station. Within each half of the rack system, random storage is used. The more active stock accounts for 80% of the transactions, and the less active stock accounts for the remaining 20%. As in Example 11.3, assume that system utilization = 90%, and the number of single command cycles = thenumber of dual command cycles. Determine the throughput of the AS/RS, basing the computation of cycle times on the same kinds of assumptions used in the MHI method. Answer: With a total length of 280 ft, each half of the rack system will be 140 ft long and 46 ft high. Let the stock nearest the input/output station (accounting for 80% of the transactions) be identified as Class A, and the other half of the stock (accounting for 20% of the transactions) as Class B. The cycle times are computed as follows: For Class A stock: For four aisles, Rt = 4(37.605) = 150.42 transactions/hr, which is almost a 28% increase over the randomized storage strategy in Example 11.3 11.17 A unit load automated storage/retrieval system has five aisles. The storage racks are 60 ft high and 280 ft long. The S/R machine travels at a horizontal speed of 200 ft/min and a vertical speed of 80 ft/min. The pick and deposit time = 0.30 min. Assume that the number of single command cycles/hr is four times the number of dual command cycles/hr and that the system operates at 80% utilization. A class-based dedicated storage strategy is used for organizing the stock, in which unit loads are separated into two classes, according to activity level. The more active stock is stored in the half of the rack system located closest to the input/output station, and the less active stock is stored in the other half of the rack system (farther away from the input/output station). Within each half of the rack system, random storage is used. The more active stock accounts for 75% of the transactions, and the less active stock accounts for the remaining 25% of the transactions. Determine the throughput rate (loads moved/hr into and out of storage) of the AS/RS, basing the computation of cycle times on the same types of assumptions used in the MHI method. Assume that when dual command cycles are performed the two transactions per cycle are both in the same class. Answer: Class 1 stock (75%): Tcs1 = 2 Max (0.5(140)/200), 0.5(60)/60+ 2(0.30) = 1.35 min/cycle Tcd1 = 2 Max (0.75(140/200), 0.75(60)/80) + 4(0.30) = 2.325 min/cycle Class 2 stock (25%): Tcs2 = 2 Max(140 + 0.5(140)/200), 0.5(60)/80) + 2(0.30) = 2.7 min/cycle Tcd2 = 2 Max(140 + 0.75(140)/200) + 4(0.30) = 3.65 min/cycle 1.35 Rcs1 + 2.325 Rcd1 + 2.7 Rcs2 + 3.65 Rcd2 = 60(0.80) = 48.0 min Given: Rcs1 = 4 Rcd1 and Rcs2 = 4 Rcd2. Also, due to 75%: 25% ratio, Rcs1 = 3 Rcs2 and Rcd1 = 3 Rcd2 1.35 (4 Rcd1 ) + 2.325 Rcd1 + 2.7(4 Rcd2 ) + 3.65 Rcd2 = 48.0 (5.4 + 2.325) Rcd1 + (10.8 + 3.65) Rcd2 = 7.725 Rcd1 + 14.45 Rcd2 = 48.0 7.725(3 Rcd2 ) + 14.45 Rcd2 = 37.625 Rcd2 = 48.0 Rcd2 = 1.276 cycles/hr Rcd1 = 3(1.276) = 3.827 cycles/hr Rcs1 = 4(3.827) = 15.309 cycles/hr Rcs2 = 4(1.276) = 5.103 cycles/hr Throughput per aisle Rt = Rcs1 + Rcs2 + 2(Rcd1 + Rcd2 ) = (15.309 + 5.103) + 2(3.827 + 1.276) Rt = 30.62 transactions/hr per aisle AS/RS throughput Rt = 5(30.62) = 153.1 transactions/hr 11.18 The AS/RS aisle of Problem 11.5 will be organized following a class-based dedicated storage strategy. There will be two classes, according to activity level. The more active stock is stored in the half of the rack system that is located closest to the input/output station, and the less active stock is stored in the other half of the rack system farther away from the input/output station. Within each half of the rack system, random storage is used. The more active stock accounts for 80% of the transactions, and the less active stock accounts for the remaining 20%. Assume that system utilization = 85% and the number of single command cycles = the number of dual command cycles in each half of the AS/RS. (a) Determine the throughput of the AS/RS, basing the computation of cycle times on the same kinds of assumptions used in the MHI method. (b) A class-based dedicated storage strategy is supposed to increase throughput. Why is throughput less here than in Problem 11.5? Answer: (b) Throughput is less than in Problem 11.5 because dividing the AS/RS length into two classes did not improve average cycle time. The reason for this lack of improvement is that the limiting travel time in nearly all cases is the vertical travel (z direction). In class 1 stock, the cycle times are the same as those in Problem 11.5 due to the vertical speed limitation. Without an improvement in class-1 cycle times, throughput cannot possibly be better than in a pure random storage strategy. Carousel Storage Systems 11.19 (A) A single carousel storage system is located in a factory making small assemblies. It is 18 m long and 1.0 m wide. The pick and deposit time is 0.20 min. The speed at which the carousel operates is 0.6 m/sec. The storage system has a 90% utilization. Determine the hourly throughput rate. Answer: C = 2(L - W) + πW = 2(18 - 1) + 1π = 37.14 m Tc = C/4vc+ Tpd = 37.14/4(0.6) + 0.20(60) = 27.48 sec = 0.458 min Rt = 60/0.458 = 131.0 transaction/hr With a 90% utilization, Rt = 0.90(131.0) = 117.9 transactions/hr 11.20 A storage system serving an electronics assembly plant has four storage carousels, each with its own manually-operated pick and deposit station. The pick and deposit time is 0.25 min. Each carousel is 50 ft long and 2.5 ft wide. The speed at which the system revolves is 100 ft/min. Determine the throughput rate of the storage system. Answer: C = 2(L - W) + πW = 2(50 - 2.5) + 2.5π = 102.85 ft Tc = C/4vc+ Tpd = 102.85/4(100) + 0.25 = 0.507 min Rt = 60/0.507 = 118.34 transaction/hr With 4 carousels, Rt = 4(118.34) = 473.4 transaction/hr 11.21 A single carousel storage system has an oval rail loop that is = 40 ft long and 3 ft wide. Fifty carriers are equally spaced around the oval. Suspended from each carrier are 5 bins. Each bin has a volumetric capacity = 0.95 ft3. Carousel speed = 100 ft/min. Average pick and deposit time for a retrieval = 20 sec. Determine (a) volumetric capacity of the storage system and (b) hourly retrieval rate of the storage system. Answer: (a) Total number of bins = nc nb = 50(5) = 250 bins Total volume capacity = 250(0.95) = 237.5 ft3 (b) C = 2(L - W) + π W = 2(40 - 3) + 3π = 83.42 ft Tc = C/4vc + Tpd = 83.42/4(100) + 20/60 = 0.567 min Rt = 60/0.567 = 105.9 transaction/hr 11.22 A carousel storage system is to be designed to serve a mechanical assembly plant. The specifications on the system are that it must have a total of 400 storage bins and a throughput of at least 125 storage and retrieval transactions/hr. Two alternative configurations are being considered: (1) a one-carousel system and (2) a two-carousel system. In either case, the width of the carousel is to be 4.0 ft and the spacing between carriers = 2.5 ft. One pickeroperator will be required for the one-carousel system and two picker-operators will be required for the two-carousel system. In either system, vc = 75 ft/min. For the convenience of the picker-operator, the height of the carousel will be limited to 5 bins. The standard time for a pick and deposit operation at the load/unload station = 0.4 min if one part is picked or stored per bin and 0.6 min if more than one part is picked or stored. Assume that 50% of the transactions will involve more than one part. Determine (a) the required length of the onecarousel system and (b) the corresponding throughput rate; (c) the required length of the twocarousel and (d) the corresponding throughput rate. (e) Which system better satisfies the design specifications? Answer: Total bins = 400 = nc nb. Given nb = 5 bins, thus nc = 400/5 = 80 carriers on the carousel. Circumference C = 80(2.5 ft/carrier) = 200 ft Given W = 4 ft, C = 2(L - W) + π W = 2(L - 4) + 4π = 200 ft 2L = 200 + 8 - 4π = 195.43 L = 97.7 ft (b) Average Tpd = 0.5(0.4) + 0.5(0.6) = 0.5 min Tc = 200/4(75) + 0.5 = 1.1667 min/transaction Rt = 60/1.1667 = 51.43 transactions/hr Total bins = 400. With 2 carousels, bins/carousel = 200/5 = 40 carriers on each carousel Circumference C = 40(2.5 ft/carrier) = 100 ft Given W = 4 ft, C = 2(L - W) + π W = 2(L - 4) + 4π = 100 ft 2L = 100 + 8 - 4π = 95.43 L = 47.7 ft Tc = 100/4(75) + 0.5 = 0.833 min/transaction Rt = 60/0.833 = 72.0 transactions/hr With 2 carousels, Rt = 2(72.0) = 144.0 transactions/hr Only the two-carousel system satisfies the throughput specification. 11.23 Given your answers to Problem 11.22, the costs of the two carousel systems are to be compared. The one-carousel system has an installed cost of $50,000, and the comparable cost of the two-carousel system is $75,000. Labor cost for a picker-operator is $20/hr, including fringe benefits and applicable overhead. The storage systems will be operated 250 days/yr for 7 hr/day, although the operators will be paid for 8 hours. Using a 3 year period in your analysis, and a 25% rate of return, determine (a) the equivalent annual cost for the two design alternatives, assuming no salvage value at the end of three years; and (b) the average cost per storage/retrieval transaction. Answer: (a) One-carousel system: EUAC = $50,000(A/P,25%,3) + 250(8)($15) = 25,615 + 30,000 = $55,615/yr Two-carousel system: EUAC = $75,000(A/P,25%,3) + 250(8)(2)($15) = 38,422 + 60,000 = $98,422/yr (b) One-carousel system: Annual Rt = 250(7)(51.43) = 90,002 transactions/yr Cost/transaction = 55,615/90,002 = $0.618/transaction Two-carousel system: Annual Rt = 250(7)(144.0) = 252,000 transactions/yr Cost/transaction = 98,422/252,000 = $0.391/transaction Chapter 12 AUTOMATIC IDENTIFICATION AND DATA CAPTURE REVIEW QUESTIONS 12.1 What is automatic identification and data capture? Answer: The definition given in the text is the following: Automatic identification and data capture (AIDC) refers to technologies that provide direct entry of data into a computer or other microprocessor controlled system without using a keyboard. 12.2 What are the drawbacks of manual collection and entry of data? Answer: Three drawbacks are identified in the text: (1) Errors. Errors occur in both data collection and keyboard entry of the data when accomplished manually. (2) Time factor. Manual methods are inherently more time consuming than automated methods. Also, when manual methods are used, there is a time delay between when the activities and events occur and when the data on status are entered into the computer. (3) Labor cost. The fulltime attention of human workers is required in manual data collection and entry, with the associated labor cost. 12.3 What are the three principal components in automatic identification technologies? Answer: The three components given in the text are (1) Data encoder. The data are translated into a machine-readable code. A label or tag containing the encoded data is attached to the item that is to be later identified. (2) Machine reader or scanner. This device reads the encoded data, converting them to alternative form, usually an electrical analog signal. (3) Data decoder. This component transforms the electrical signal into digital data and finally back into the original alphanumeric characters. 12.4 Name four of the six categories of AIDC technologies that are identified in the text. Answer: The six categories of AIDC technologies identified in the text are (1) optical, such as bar codes and optical character recognition, (2) electromagnetic, such as RFID, (3) magnetic, used in plastic credit cards, (4) smart cards, which are imbedded with microchips capable of containing large amounts of data, (5) touch techniques, such as touch screens, and (6) biometric, such as voice recognition and fingerprint analysis. 12.5 Name five common applications of AIDC technologies in production and distribution? Answer: The applications listed in the text are (1) receiving, (2) shipping, (3) order picking, (4) finished goods storage, (5) manufacturing processing, (6) work-in-process storage, (7) assembly, and (8) sortation. 12.6 There are two forms of linear bar codes. Name them, and indicate what the difference is. Answer: The two forms of linear bar codes are (a) width-modulated, in which the symbol consists of bars and spaces of varying width; and (b) height-modulated, in which the symbol consists of evenly spaced bars of varying height. 12.7 What was the major industry to first use the Universal Product Code (UPC)? Answer: The major industry to first use the Universal Product Code was the grocery industry, starting in 1973. 12.8 What are the two basic types of two-dimensional bar codes? Answer: The two basic types of two-dimensional bar codes are (1) stacked bar codes and (2) matrix symbologies. 12.9 What does RFID stand for? Answer: RFID stands for radio frequency identification. 12.10 What is a transponder in RFID? Answer: The RFID identification tag is a transponder. A transponder is defined as a device that emits a signal of its own when it receives a signal from an external source. 12.11 What is the difference between a passive tag and an active tag? Answer: A passive tag has no internal power source; it derives its electrical power for transmitting a signal from radio waves generated by the reader when in close proximity. An active tag includes its own battery power packs. 12.12 What are the relative advantages of RFID over bar codes? Answer: The relative advantages of RFID over bar codes are (1) read-write capability, (2) large storage capacity, (3) line-of sight reading is not required, (4) not susceptible to dirt or scratching that would destroy the label, and (5) the tags can be reused. 12.13 What are the relative advantages of bar codes over RFID? Answer: The relative advantages of bar codes over RFID are (1) lower cost and (2) the technology is widely available. 12.14 What are the reasons why magnetic stripes are not widely used in factory floor operations? Answer: Three reasons are given in the text: (1) The magnetic stripe must be in contact with the scanning equipment for reading to be accomplished, (2) unavailability of convenient shop floor encoding methods to write data into the stripe, and (3) the magnetic stripe labels are more expensive than bar code labels. 12.15 What is the advantage of optical character recognition technology over bar code technology? Answer: The advantage is that OCR symbols can be read by humans, whereas bar codes cannot. 12.16 What is the principal application of machine vision in industry? Answer: The principal application of machine vision is automated inspection. Solution Manual for Automation, Production Systems, and Computer-Integrated Manufacturing Mikell P. Groover 9780133499612, 9780134605463