Managing Change The Case of FastLink Logistics Company

Introduction Contextual Information Cut-throat competition, technological advances, geopolitical realignments and other related pressures are combining with concerns for security, shifting customer demands, expansion urges and organizational governance to generate momentous pressure for organizational change (Howard, 1994). Current literature as demonstrated by Kotter (2007) reveals that organizations need to continually change and reinvent their strategies, structures, processes and culture to remain relevant as well as maintain competitive advantage.Advertising We will write a custom essay sample on Managing Change: The Case of FastLink Logistics Company specifically for you for only $16.05 $11/page Learn More But as observed by McGahan (2004), the change process itself need to be planned beforehand and aligned to the business strategy and objectives if it is to achieve success. The capacity of an organization to plan and implement the change process is therefore fundamental for success. By presenting a case of the current challenges facing FastLink Logistics Company, this paper will focus on providing an outline through which the underlying challenges can be adequately addressed by initiating a change process. Headquartered in Boston, MA, FastLink Logistics has been in operation for the last 20 years, and has consistently maintained a leadership position in the industry judging from its financial returns and competitiveness. An interview with the company’s operations manager revealed that the company operates a fleet of 3000 trucks and 10 cargo airplanes for purposes of transporting goods to various destinations across the U.S. and internationally. From the beginning of the current financial year, however, the company’s management has been receiving sustained customer complaints regarding delayed deliveries of their orders, especially from truck drivers. Indeed, some of FastLink’s traditional customers have opted to look for other logistics companies who keep their word on deliveries, implying that profits for FastLink Logistics have been on a downward trend. The management, having studied all the facts, identified massive paperwork in cargo clearing departments and irresponsible behaviour on the part of truck drivers, who engage in personal business while performing official duties. Upon holding several consultative meetings, the management decided to undertake a change process that will automate the clearing departments and install remote sensors on their trucks to monitor their movements and timeliness. The intended change process, however, is likely to experience some challenges, especially from employees who may oppose the intended automation of duties for fear of being rendered redundant. Still, truck drivers may feel that the intended introduction of remote sensors on their trucks may impinge on their privacy, hence perceive the new system negatively. Overall, the change proces s may be resisted by employees since it requires dramatic shifts in business processes and employee roles (Suntano et al., 2008). The management, therefore, must come up with strategies that ensure employee participation, commitment and identification to the change process so as to achieve success. In line with the above, FastLink’s management decides to enrol the services of a consultant to prepare a project feasibility study that could be effectively used to manage the change process.Advertising Looking for essay on business economics? Let's see if we can help you! Get your first paper with 15% OFF Learn More Project Feasibility Study for FastLink Logistics Company The Project Scope Greer Conradi (2009) observes that â€Å"†¦the feasibility study usually refers to an assessment of the product/project against technical, operational, financial and social/political criteria† (p. 357). Academics and practitioners are in agreement that recognizing the challenges facing an organization and justifying the need for change must form the initial step of a system change effort (Suntano et al., 2008). Section A of this report demonstrates that FastLink is currently being faced with challenges arising from slower business processes and inability of some employees, particularly truck drivers, to identify with the values of the company. These challenges have caused the company to lose its traditional customers to competitors due to late delivery of orders, implying that the company’s profitability and competitive advantage in the market have been negatively affected. This particular feasibility study, hereafter identified as PFS, will therefore aim to come up with ways to balance business processes and technology constraints facing the firm so as to select a cost-effective solution. More importantly, the PFS will also involve matching the new system to organizational objectives in terms of deliverables, ability to meet customersâ⠂¬â„¢ requirements, efficient utilization of resources, budgetary allocations and time-frames. The PFS will be presented in the form of a short report. The Current Analysis FastLink logistics has for a very long time employed a manual process in its operations across all the offices located in the U.S. and abroad. This implies that the work environment has been largely paper-based and personnel-intensive. It is also imperative to note that FastLink’s working environment has been largely distributed over expansive geographical areas due to the nature of the transportation business. The advantages derived from the above working methodology may in fact be minimal as opposed to the disadvantages. In terms of advantages of the ‘status-quo’, it is important to note that manual processes of doing business in a way motivates employees to offer more, not mentioning that they are beneficial in aligning members of staff to organizational objectives and strategies (McGahan, 2004). A paper-based working environment also implies that information can be retrieved from the cabinets anytime without undue regard to other externalities such as the availability of power to run the machines, computer viruses, and end-user capabilities. In terms of the disadvantages arising from the current business processes, it is evident that FastLink has been losing, productively, due to delay occasioned by the tedious paper-based working environment and centralized information architecture.Advertising We will write a custom essay sample on Managing Change: The Case of FastLink Logistics Company specifically for you for only $16.05 $11/page Learn More Preliminary observations reveals that the distributed information architecture coupled with lax rules are to blame for the truck drivers’ unbecoming behaviour by virtue of the fact that critical information aimed at necessitating effective order deliveries to customers is either delayed due to the manual operational processes, or the information does not get to the intended recipients at all due to lax operational rules. Another problematic area for the company is the fact that it has to employ a large number of employees to do the paperwork needed to necessitate order deliveries, implying that the company’s wage bill may exceed profitable limits. These disadvantages, according to preliminary observations, are to blame for the company’s continued dip in profitability and competitive advantage within the logistics industry. According to Davies (1994), these challenges are functional in nature owing to the fact that they hinder effective work-flow processes for the company, thus they demand functional solutions. Benefits Pitfalls of the Proposed System The next stage of designing a project feasibility study, according to Greer Conradi (2009), is to develop specific objectives and limits of the proposed system. Here, it is proposed that the company redesig n and automate its database management system for the obvious advantages of faster and better customer service as well as ensuring the distributed use and processing of order information. Automated work systems, according to Howard (1994), not only simplify and streamline work processes, but they also enhance efficiency by curtailing excessive use of an organization’s scarce resources. The proposed remote sensors to be installed on trucks will go a long way to ensure that truck drivers follow their routes in the process of delivering orders to customers, and the orders are delivered on a timely basis. However, financial resources must be channelled to purchase the needed automation hardware and software as well as the remote sensors. The decisions to automate the offices and install remote sensors on trucks will most probably be met with resistance from members of staff who may fear loosing their jobs or who may argue that their privacy is being impinged on. Here, Suntano et al (2008) advises that the company should take an all-inclusive approach to enable employees participate in the change process; hence offer their views, fears and expectations as concerns the intended change. In addition, the management should exercise an open-door policy, whereby employees should be encouraged to raise any underlying issues and seek clarifications to better understand the proposed change.Advertising Looking for essay on business economics? Let's see if we can help you! Get your first paper with 15% OFF Learn More According to Davies (1994), involving employees and effectively answering their concerns in any change process is advantageous since it minimizes resistance during the implementation process, not mentioning that it facilitates employees to identify more with the change process once it is operationalized. Alternatives Companies in the 21st century are increasingly faced with the dilemma of whether to automate their operations or to continue with paper-based working environment (Alkuwaiti, 2010). In the present case, one of the viable alternatives would be to recruit more human resources to clear the backlog, hence necessitate timely delivery of orders to customers. Another alternative would be to enlist the services of security firms to escort trucks so that drivers do not engage in personal business during official working hours. The company may also consider investing in telecommunications to facilitate a more distributed information infrastructure. Assessment The assessment of any viable change process or alternatives must be evaluated against a backdrop of risk management controls, ability to meet fundamental objectives of the project’s attention, program needs and costs, intensity of effort, time-scale, security features, and flexibility for future modification and expansion, among others (Kotter, 2007; Bryce, 2011). It is indeed true that the proposed automated process may in the short-term be costly for the company in terms of purchasing the needed hardware and software as well as training staff members to operate the automated systems. In the long-term, however, the company will largely benefit from enhanced customer satisfaction, improved sales and reduced overhead costs. The alternative of employing more members of staff to clear the backlog may not be tenable in these circumstances, especially when the company is working on new modalities that will enable it to return to profitability and maintain competitive advantage. The time-scale needed f or implementing automated systems and installing the remote sensors may indeed be minimal when it is compared to the time-scale and resources needed to advertise for positions, recruit, orient and train new members of staff. According to Howard (1994), companies must always be on the lookout for proposals that enhance value creation as well as offer viable solutions to the problems that triggered the change process. In this perspective, the proposed system change should be recommended over the other stated alternatives. Automation of business process and introduction of remote sensors for trucks seem the only way to go for FastLink Logistics Company if it has to overcome its present challenges. Lastly, it is indeed true that automated systems can be expanded or modified to meet future needs and expectations (Suntano et al., 2008). The same cannot be said of other alternatives such as utilizing security firms to make the company drivers toe the line. List of References Alkuwaiti, A. J. (2010). The Project Book: A Simple and Direct Approach to Project Management. New York, NY: Create Space. Bryce, T. (2011). The Elements of a Good Feasibility Study. Web. Davis, S. M. (1994). Attempting Major Change? 10 Pitfalls to Avoid – And How. Human Resource Planning, Vol. 7, Issue 4, pp 175-188. Greer, D., Conradi, R. (2009) Software Project Initiation and Planning – An Empirical Study. IET Software, Vol. 3, Issue 5, pp 356-368. Howard, A. (1994). Diagnosis for Organizational Change: Methods and Models. New York, NY: The Guilford Press. Kotter, J. P. (2007). Leading Change. Harvard Business Review, Vol. 85, Issue 1, pp 96-103. McGahan, A. M. (2004). How Industries Change. Harvard Business Review, Vol. 82, Issue 10, pp 86-94. Suntano, J., Kankanhalli, A., Tay, J., Raman, K. S., Tan, B. C. Y. (2008). Change Management in Interorganizational Systems for the Public. Journal of Management Information Systems, Vol. 25, Issue 3, pp 133-175. Wischnevsky, J. D. (2007 ). Change as the Winds Change: The Impact of Organizational Transformation on Firm Survival in a Shifting Environment. Organizational Analysis, Vol. 12, Issue. 4 pp 361-377. This essay on Managing Change: The Case of FastLink Logistics Company was written and submitted by user Ian U. to help you with your own studies. You are free to use it for research and reference purposes in order to write your own paper; however, you must cite it accordingly. You can donate your paper here.

Linear Solution Essays

Linear Solution Essays Linear Solution Essay Linear Solution Essay CHAPTER 8 Linear Programming Applications Teaching Suggestions Teaching Suggestion 8. 1: Importance of Formulating Large LP Problems. Since computers are used to solve virtually all business LP problems, the most important thing a student can do is to get experience in formulating a wide variety of problems. This chapter provides such a variety. Teaching Suggestion 8. 2: Note on Production Scheduling Problems. The Greenberg Motor example in this chapter is largest large problem in terms of the number of constraints, so it provides a good practice environment. An interesting feature to point out is that LP constraints are capable of tying one production period to the next. Teaching Suggestion 8. 3: Labor Planning Problem- Hong Kong Bank of Commerce. This example is a good practice tool and lead-in for the Chase Manhattan Bank case at the end of the chapter. Without this example, the case would probably overpower most students. Teaching Suggestion 8. 4: Ingredient Blending Applications. Three points can be made about the two blending examples in this chapter. First, both the diet and fuel blending problems presented here are tiny compared to huge real-world blending problems. But they do provide some sense of the issues to be faced. Second, diet problems that are missing the constraints that force variety into the diet can be terribly embarrassing. It has been said that a hospital in New Orleans ended up with an LP solution to feed each patient only castor oil for dinner because analysts neglected to add constraints forcing a well-rounded diet. Alternative Examples Alternative Example 8. 1:  Ã‚  Natural Furniture Company manufactures three outdoor products, chairs, benches, and tables. Each product must pass through the following departments before it is shipped: sawing, sanding, assembly, and painting. The time requirements (in hours) are summarized in the tables below. The production time available in each department each week and the minimum weekly production requirement to fulfill contracts are as follows: | | |Minimum | | |Capacity | |Production | |Department |(In Hours) |Product |Level | |Sawing |450 |Chairs |100 | |Sanding |400 |Benches |50 | |Assembly |625 |Tables |50 | |Painting |550 | | | |Hours Required |Unit | |Product |Sawing |Sanding |Assembly |Painting |Profit | |Chairs |1. 5 |1. 0 |2. 0 |1. 5 |$15 | |Benches |1. 5 |1. 5 |2. 0 |2. 0 |$10 | |Tables |2. 0 |2. 0 |2. 5 |2. 0 |$20 | The production manager has the responsibility of specifying production levels for each product for the coming week. Formulate as a linear programming problem to maximize profit. Let X1= Number of chairs produced X2= Number of benches produced X3= Number of tables produced The objective function is Maximize profit = 15X1 + 10X2 + 20X3 Constraints 1. 5X1 + 1. 5X2 + 2. 0X3( 450 hours of sawing available 1. 0X1 + 1. 5X2 + 2. 0X3( 400 hours of sanding available 2. 0X1 + 2. 0X2 + 2. 5X3( 625 hours of assembly available 1. 5X1 + 2. 0X2 + 2. 0X3( 550 hours of painting available X1+ 2. 0X2 + 2. 0X3( 100 chairs X2 + 2. 0X3( 50 benches X3( 50 tables X1, X2, X3( 0 What mix of products would yield maximum profit? Solving with computer software we get: X1= 100 chairs; X2 = 50 benches; X3 = 112. 5 tables; profit = 4250. Alternative Example 8. 2:  Ã‚  A phosphate manufacturer produces three grades of phosphate, A, B, and C, which yield profit of $40, $50, and $60 per kilogram, respectively. The products require the labor and materials per batch that are shown in the table. Each batch of Grade A phosphate yields 800kg of phosphate; each batch of Grade B phosphate yields 700kg of phosphate; and each batch of Grade C phosphate yields 800 kg. |Grade |Grade |Grade |Available | | |A |B |C |Resources | |Labor hours |4 |4 |5 |80 hr | |Raw material #1 |200 |300 |300 |6,000 kg | |Raw material #2 |600 |400 |500 |5,000 kg | Formulate as an LP problem to maximize profit. Objective function Maximize profit = 40(800)A + 50(700)B + 60(800)C Constraints Labor:4A +4B +5C( 80 Raw material #1200A + 300B +300C( 6,000 Raw material #2600A + 400B +500C( 5,000 What mix of products would yield maximize profit? Solutions To Problems 8-1. Since the decision centers about the production of the two different cabinet models, we let X1= number of French Provincial cabinets produced each day X2= number of Danish Modern cabinets produced each day Objective: maximize revenue = $28X1 + $25X2 subject to 3X1 + 2X2( 360 hours  Ã‚  (carpentry department) [pic] X1 + 1X2( 200 hours  Ã‚  (painting department) [pic]X1 + [pic] X2( 125 hours  Ã‚  (finishing department) X1( 60 units  Ã‚  (contract requirement) X2( 60 units  Ã‚  (contract requirement) X1, X2( 0 Problem 8-1 solved by computer: Produce 60 French Provincial cabinets (X1) per day Produce 90 Danish Modern cabinets (X2) per day Revenue = $3,930 8-2. Let X1= dollars invested in Los Angeles municipal bonds X2= dollars invested in Thompson Electronics X3= dollars invested in United Aerospace X4= dollars invested in Palmer Drugs X5= dollars invested in Happy Days Nursing Homes Maximize return = 0. 53X1 + 0. 068X2 + 0. 049X3 + 0. 084X4 + 0. 118X5 subject to X1 + X2 + X3 + X4 + X5 ( $250,000 (funds) X1( . 2 (X1 + X2 + X3 + X4 + X5) (bonds) or ( . 8X1 – . 2X2 – . 2X3 – . 2X4 – . 2X5 ( 0 X2 + X3 + X4( . 4 (X1 + X2 + X3 + X4 + X5) (combination of electronics, aerospace, and drugs) or –0. 4X1 + 0. 6X2 + 0. 6X3 + 0. 6X4 – 0. 4X5 ( 0 (X5 ( 0. 5X1) rewritten as –0. 5X1 + X5 ( 0 (nursing home as percent of bonds) X1, X2, X3, X4, X5 ( 0 Problem 8-2 solved by computer: $50,000invested in Los Angeles municipal bonds (X1) $0invested in Tho mpson Electronics (X2) $0invested in United Aerospace (X3) 175,000invested in Palmer Drugs (X4) $25,000invested in Happy Days (X5) This produces an annual return on investment of $20,300. 8-3. Minimize staff size = X1 + X2 + X3 + X4 + X5 + X6 where Xi = number of workers reporting for start of work at period i (with i = 1, 2, 3, 4, 5, or 6) X1 + X2 ( 12 X2 + X3 ( 16 X3 + X4 (   9 X4 + X5 ( 11 X5 + X6 (   4 X1 + X6 ( 3 All variables ( 0 The computer solution is to hire 30 workers: 16 begin at 7 a. m. 9 begin at 3 p. m. 2 begin at 7 p. m. 3 begin at 11 p. m. An alternative optimum is 3 begin at 3 a. m. 9 begin at 7 a. m. 7 begin at 11 a. m. begin at 3 p. m. 9 begin at 7 p. m. 0 begin at 11 p. m. 8-4. Let X1= number of pounds of oat product per horse each day X2= number of pounds of enriched grain per horse each day X3= number of pounds of mineral product per horse each day Minimize cost = 0. 09X1 + 0. 14X2 + 0. 17X3 subject to 2X1+ 3X2 + 1X3 ( 6 (ingredient A) [pic]X1+ 1X2 + [pic]X3 ( 2 (ingredient B) 3X1+ 5X2 + 6X3 ( 9 (ingredient C) 1X1+ 1[pic]X2 + 2X3 ( 8 (ingredient D) [pic]X1+ [pic]X2 + 1[pic]X3 ( 5 (ingredient E) X1+ X2 +X3 ( 6 (maximum feed/day) All variables ( 0 Solution: X1= 1[pic] X2= 0 X3= 3[pic] cost= 0. 87 8-5. Let E1, E2, and E3 represent the ending inventory for the three months respectively. Let RT1, RT2, and RT3 represent the reguar production for the three months and OT1, OT2, an d OT3 represent the overtime production quantities during the three months respectively. Then the formulation is: Minimize cost: 300RT1 + 300RT2 + 300RT3 + 325OT1 + 325OT2 + 325OT3 + 20E1 + 20E2 + 20E3 subject to RT1 lt; 200 June regular production RT2 lt; 200 July regular production RT3 lt; 200 August regular production OT1 lt; 15 June overtime production OT2 lt; 15 July overtime production OT3 lt; 15 August overtime production (E1 + RT1 + OT1= 195 Ending inventory from first month (E2 + E1 + RT2 + OT2= 215 Ending inventory from second month (E3 + E2 + RT3 + OT3= 205 Ending inventory from third month {All variables}? 0 Non-negativity constraints The optimal production schedule is to produce 200 each month during regular production and to use overtime to produce 10 units in July and 5 in August for a total cost of $184,975. 8-6. Let T = number of TV ads R = number of radio ads B = number of billboard ads N = number of newspaper ads Maximize total audience = 30,000T + 22,000R + 24,000B + 8,000N Subject to 800T + 400R + 500B + 100N ( 15,000 ? ( 10 R (10 ? (10 ? (10 ? + R ( 6 500B + 100N ( 800T ?, R, ? , ? ( 0 Solution: T = 6. 875; R = 10; B = 9; N = 10; Audience reached = 722,250. If integer solutions are necessary, integer programming (see Chapter 11) could be used. 8-7. Let_X1= number of newspaper ads placed X2= number of TV spots purchased Minimize cost =$925X1 + $2,000X2 subject to0. 04X1 + 0. 05X2 ( 0. 40 (city exposure) 0. 03X1 + 0. 03X2 ( 0. 60 (exposure in northwest suburbs) X1, X2 ( 0 Note that the problem is not limited to unduplicated exposure (e. g. one person seeing the Sunday newspaper three weeks in a row counts for three exposures). Problem 8-7 solved by computer: Buy 20 Sunday newspaper ads (X1) Buy 0 TV ads (X2) This has a cost of $18,500. Perhaps the paint store should consider a blend of TV and newspaper, not just the latter. 8-8. Let Xij = number of new leases in month i for j-months, i = 1, . . . , 6; j = 3, 4, 5 Minimize cost =1260X13 + 1260X23 + 1260X 33 + 1260X43 + 840X53 + 420X63 + 1600X14 + 1600X24 + 1600X34 + 1200X44 + 800X54 + 400X64+ 1850X15 + 1850X25 + 1480X35 + 1110X45 + 740X55 + 370X65 subject to:X13 + X14 + X15 ( 420 – 390 X13 + X14 + X15 + X23 + X24 + X25 ( 400 – 270 X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 ( 430 – 130 X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 ( 460 X15 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 ( 470 X25 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 + X63 + X64 + X65 ( 440 X15 + X25 + X35 + X45 + X55 + X65 ( 0. 0(X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 + X63 + X64 + X65) All variables ( 0 Solving this on the computer results in the following solution: X15 = 305-month leases in March X25 = 1005-month leases in April X35 = 1705-month leases in May X45 = 1605-month leases in June X55 = 105-month leases in July All other variables equal 0. Total cost = $677,100. As a result of this, there are 440 cars remaining at the end of August. 8-9. The linear program has the same constraints as in problem 8-8. The objective function changes and is now: Minimize cost =1260(X13 + X23 + X33 + X43 + X53 + X63) + 1600(X14 + X24 + X34 + X44 + X54 + X64) + 1850(X15 + X25 + X35 + X45 + X55 + X65) Solving this on the computer results in the following solution: X15 = 305-month leases in March X25 = 1005-month leases in April X34 = 654-month leases in May X35 = 1055-month leases in May X43 = 1603-month leases in June X53 = 103-month leases in July All other variables equal 0. Total cost = $752,950. 8-10. Let Xij = number of students bused from sector i to school j Objective: minimize total travel miles = 5XAB+ 8XAC + 6XAE 0XBB+ 4XBC + 12XBE + 4XCB+ 0XCC + 7XCE + 7XDB+ 2XDC + 5XDE + 12XEB+ 7XEC + 0XEE subject to XAB + XAC + XAE= 700 (number of students in sector A) XBB + XBC + XBE= 500 (number of students in sector B) XCB + XCC + XCE= 100 (number of students in sector C) XDB + XDC + XDE= 800 (number of students in sector D) XEB + XEC + XEE= 400 (number students in sector E) XAB + XBB + XCB + XDB + XEB ( 900 (school B capacity) XAC + XBC + XCC + XDC + XEC ( 900 (school C capacity) XAE + XBE + XCE + XDE + XEE ( 900 (school E capacity) All variables ( 0 Solution: XAB= 400 XAE= 300 XBB= 500 XCC= 100 XDC= 800 XEE= 400 Distance = 5,400 â€Å"student miles† 8-11. Maximize number of rolls of Supertrex sold = 20X1 + 6. 8X2 + 12X3 – 65,000X4 whereX1= dollars spent on advertising X2= dollars spent on store displays X3= dollars in inventory X4= percent markup subject to X1 + X2 + X3 ( $17,000 (budgeted) X1( $3,000 (advertising constraint) X2( 0. 05X3 (or X2 – 0. 05X3 ( 0) (ratio of displays to inventory) [pic] (markup ranges) X1, X2, X3, X4 ( 0 Problem 8-11 solved by computer: Spend $17,000 on advertising (X1). Spend nothing on in-store displays or on-hand inventory (X2 and X3). Take a 20% markup. The store will sell 327,000 rolls of Supertrex. This solution implies that no on-hand inventory or displays are needed to sell the product, probably due to an oversight on Mr. Kruger’s part. Perhaps a constraint indicating that X3 ( $3,000 of inventory should be held might be needed. 8-12. Minimize total cost = $0. 60X1 + 2. 35X2 + 1. 15X3 + 2. 25X4 + 0. 58X5 + 1. 17X6 + 0. 33X7 subject to 295X1 + 1,216X2 + 394X3 +358X4 + 128X5+ 118X6 + 279X7 ( 1,500 295X1 + 1,216X2 + 394X3 +358X4 + 128X5+ 118X6 + 279X7 ( 900 .2X1 + 121. 2X2 + . 4. 3X3 + 3. 2X4 + 3. 2X5+ 14. 1X6 + 2. 2X7 ( 4 16X1 +1,296X2 + . 4. 9X3 + 0. 5X4 + 0. 8X5+ 1. 4X6 + 0. 5X7 ( 50 6X1 + 81X2 + 74X3 + 83X4 + 7X5+ 14X6 +  8X7 ( 26 22X1 + 28X5 + 19X6 + 63X7( 50 All Xi ( 0 Problem 8-12 solved by computer: The meal plan for the evening is No milk (X1 = 0) 0. 499 pound of ground meat (X2) 0. 173 pound of chicken (X3) No fish (X4 = 0) No beans (X5 = 0) 0. 105 pound of spinach (X6) 0. 762 pound of white potatoes (X7) Each meal has a cost of $1. 75. The meal is fairly well -balanced (two meats, a green vegetable, and a potato). The weight of each item is realistic. This problem is very sensitive to changing food prices. Sensitivity analysis when prices change: Milk increases 10 cents/lb: no change in price or diet Milk decreases 10 cents/lb: no change in price or diet Milk decreases 30 cents/lb (to 30 cents): potatoes drop out and milk enters, price = $1. 42/meal Ground meat increases from $2. 35 to $2. 75: price = $1. 93 and spinach leaves the optimal solution Ground meat increases to $5. 25/lb: price = $2. 07 and meat leaves; milk, chicken, and potatoes in solution Fish decreases from $2. 25 to $2. 00/lb: no change Chicken increases to $3. 00/lb: price = $1. 91 and meat, fish, spinach, and potatoes in solution If meat and fish are omitted from the problem, the solution is chicken= 0. 774 lb milk= 1. 891 lb potatoes= 0. 33 lb If chicken and meat are omitted; fish= 0. 679 lb spinach= 0. 0988 lb milk= 2. 188 lb 8-13. a. Let X1= no. of units of internal modems produced per week X2= no. of units of external modems produced per week X3= no. of units of circuit boards produced per week X4= no. of units of floppy disk drives produced per week X5= no. of units of hard drives produced per week X6= no. of units of memory boards produced per week Objective function analysis: First find the time used on each test device: hours on test device 1 [pic] hours on test device 2 [pic] hours on test device 3 [pic] Thus, the objective function is aximize profit = (revenue) – (material cost) – )test cost) = (200X1 + 120X2 + 180X3 + 130X4 + 430X5 + 260X6 – 35X1 – 25X2 – 40X3 – 45X4 – 170X5 – 60X6)[pic] [pic] [pic] This can be rewritten as maximize profit =$161. 35X1 + 92. 95X2 + 135. 50X3 + 82. 50X4 + 249. 80X5 + 191. 75X6 subject to 7X1 + 3X2 + 12X3 + 6X4 + 18X5 + 17X6 lt; 120(60) Minutes on test device 1 2X1 + 5X2 + 3X3 + 2X4 + 15X5 + 17X6 lt; 120(60) Minutes on test device 2 5X1 + 1X2 + 3X3 + 2X4 + 9X5 + 2X6 lt; 100(60) Minutes on test device 3 All variables ( 0 b. The solution is X1= 496. 55 internal modems X2= 1,241. 38 external modems X3 through X6= 0 profit= $195,504. 80 c. The shadow prices, as explained in Chapter 7 and Module 7, for additional time on the three test devices are $21. 41, $5. 75, and $0, respectively, per minute. 8-14. a. Let Xi = no. of trained technicians available at start of month i Yi = no. of trainees beginning in month i Minimize total salaries paid = $2,000X1 + 2,000X2 + 2,000X3 + 2,000X4 + 2,000X5 + 900Y1 + 900Y2 + 900Y3 + 900Y4 + 900Y5 subject to 130X1 – 90Y1( 40,000 (Aug. need, hours) 130X2 – 90Y2( 45,000 (Sept. need) 130X3 – 90Y3( 35,000 (Oct. need) 130X4 – 90Y4( 50,000 (Nov. need) 130X5 – 90Y5( 45,000 (Dec. eed) X1= 350 (starting staff on Aug. 1) X2= X1 + Y1 – 0. 05X1 (staff on Sept. 1) X3= X2 + Y2 – 0. 05X2 (staff on Oct. 1) X4= X3 + Y3 – 0. 05X3 (staff on Nov. 1) X5= X4 + Y4 – 0. 05X4 (staff on Dec. 1) All Xi, Yi ( 0 b. The computer-generated results are: | |Trained | | | |Technicians |Trainees | |Month |Available |Beg inning | |Aug. 350 |13. 7 (actually 14) | |Sept. |346. 2 |0 | |Oct. |328. 8 |72. 2 (actually 72) | |Nov. |384. 6 |0 | |Dec. |365. 4 |0 | Total salaries paid over the five-month period = $3,627,279. 8-15. a. Let Xij = acres of crop i planted on parcel j wherei = 1 for wheat, 2 for alfalfa, 3 for barley = 1 to 5 for SE, N, NW, W, and SW parcels Irrigation limits: 1. 6X11 + 2. 9X21 + 3. 5X31( 3,200 acre-feet in SE 1. 6X12 + 2. 9X22 + 3. 5X32( 3,400 acre-feet in N 1. 6X13 + 2. 9X23 + 3. 5X33( 800 acre-feet in NW 1. 6X14 + 2. 9X24 + 3. 5X34( 500 acre-feet in W 1. 6X15 + 2. 9X25 + 3. 5X35( 600 acre-feet in SW [pic] water acre-feet total Sales limits: X11 + X12 + X13 + X14 + X15 ( 2,200 wheat in acres (= 110,000 bushels) X21 + X22 + X23 + X24 + X25 ( 1,200 alfalfa in acres (= 1,800 tons) X31 + X32 + X33 + X34 + X35 ( 1,000 barley in acres (= 2,200 tons) Acreage availability: X11 + X21 + X31( 2,000 acres in SE parcel X12 + X22 + X32( 2,300 acres in N parcel X13 + X23 + X33( 600 acres in NW parcel X14 + X24 + X34( 1,100 acres in W parcel X15 + X25 + X35( 500 acres in SW parcel Objective function: maximize profit [pic] b. The solution is to plant X12= 1,250 acres of wheat in N parcel X13= 500 acres of wheat in NW parcel X14= 312[pic] acres of wheat in W parcel X15= 137[pic] acres of wheat in SW parcel X25= 131 acres of alfalfa in SW parcel X31= 600 acres of barley in SE parcel X32= 400 acres of barley in N parcel Profit will be $337,862. 10. Multiple optimal solutions exist. c. Yes, need only 500 more water-feet. 8-16. Amalgamated’s blending problem will have eight variables and 11 constraints. The eight variables correspond to the eight materials available (three alloys, two irons, three carbides) that can be selected for the blend. Six of the constraints deal with maximum and minimum quality limits, one deals with the 2,000 pound total weight restriction, and four deal with the weight availability limits for alloy 2 (300 lb), carbide 1 (50 lb), carbide 2 (200 lb), and carbide 3 (100 lb). Let X1 through X8 represent pounds of alloy 1 through pounds of carbide 3 to be used in the blend. Minimize cost = 0. 12X1 + 0. 13X2 + 0. 15X3 + 0. 09X4 + 0. 07X5 + 0. 10X6 + 0. 12X7 + 0. 09X8 subject to manganese quality: 1  Ã‚  0. 70X1 + 0. 55X2 + 0. 12X3 + 0. 01X4 + 0. 05X5 ( 42 (2. 1% of 2,000) 2  Ã‚  0. 70X1 + 0. 55X2 + 0. 12X3 + 0. 01X4 + 0. 05X5 ( 46 (2. 3% of 2,000) silicon quality: 3  Ã‚  0. 15X1 + 0. 30X2 + 0. 26X3 + 0. 10X4 + 0. 025X5 + 0. 24X6 + 0. 25X7 + 0. 23X8 ( 86 (4. 3% of 2,000) 4  Ã‚  0. 15X1 + 0. 30X2 + 0. 26X3 + 0. 10X4 + 0. 025X5 + 0. 24X6 + 0. 25X7 + 0. 23X8 ( 92 (4. 6% of 2,000) carbon quality: 5  Ã‚  0. 03X1 + 0. 01X2 + 0. 03X4 + 0. 18X6 + 0. 20X7 + 0. 25X8 ( 101 (5. 5% of 2,000) 6  Ã‚  0. 03X1 + 0. 01X2 + 0. 03X4 + 0. 18X6 + 0. 20X7 + 0. 25X8 ( 107 (5. 35% of 2,000) Availability by weight: 7  Ã‚  X2 ( 300 8  Ã‚  X6 ( 50 9  Ã‚  X7 ( 200 10  Ã‚  X8 ( 100 One-ton weight: 11  Ã‚  X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 2,000 The solution is infeasible. 8-17. This problem refers to Problem 8-16’s infeasibility. Some investigative w ork is needed to track down the issues. The two issues are: 1. Requiring at least 5. 05% carbon is not possible. 2. Producing 1 ton from the materials is not possible. If constraints 5 and 11 are relaxed (or removed), one solution is X2 = 83. lb (alloy 2), X6 = 50 lb (carbide 1), X7 = 83. 6 lb (carbide 2), and X8 = 100 lb (carbide 3). Cost = $34. 91. Each student may take a different approach and other recommendations may result. 8-18. X1= number of medical patients X2= number of surgical patients Maximize revenue = $2,280X1 + $1,515X2 subject to 8X1 + 5X2( 32,850 (patient-days available = 365 days ( 90 new beds) 3. 1X1 + 2. 6X2 ( 15,000 (lab tests) 1X1 + 2X2( 7,000 (x-rays) X2( 2,800 (operations/surgeries) X1, X2( 0 Problem 8-18 solved by computer: X1= 2,791 medical patients X2= 2,105 surgical patients revenue= $9,551,659 per year To convert X1 and X2 to number of medical versus surgical beds, find the total number of hospital days for each type of patient: medical= (2,791 patients)(8 days/patient) = 22,328 days surgical= (2,105 patients)(5 days/patient) = 10,525 days total= 32,853 days This represents 68% medical days and 32% surgical days, which yields 61 medical beds and 29 surgical beds. (Note that an alternative approach would be to formulate with X1, X2 as number of beds. ) See the printout on the next page for the solution and sensitivity analysis. 8-19. This problem, suggested by Professor C. Vertullo, is an excellent exercise in report writing. Here is a chance for students to present management science results in a management format. Basically, the following issues need to be addressed in any report: (a)  Ã‚  As seen in Problem 8-18, there should be 61 medical and 29 surgical beds, yielding $9,551,659 per year. (b)  Ã‚  Referring to the QM for Windows printout, there are no empty beds because the slack for constrain 1 has a value of 0.. (c)  Ã‚  There are 876 (the slack for constraint 2) lab tests of unused capacity. (d)  Ã‚  The x-ray is used to its maximum (slack for constraint 3 is 0) and has a $65. 5 dual price. The revenue would increase by this amount for each additional x-ray. (e)  Ã‚  The operating room still has 695 operations available (the slack for constraint 4). [pic] [pic] 8-20. For the Low Knock Oil Company example it was originally assumed that a one to one ratio of raw materials (crude oil) to finished goods (gasoline). In reality, that ratio is closer to 46%. Hence, the example problem needs to be modified with 0. 46 as the coefficient throughout the first two constraints as follows: Minimize 30X1 + 30X2 + 34. 80X3 + 34. 80X4 subject to: 0. 46X1 + 0. 46X3( 25000 0. 46X2+ 0. 46X4( 32000 0. 10 X1 + 0. 15X3 ( 0 0. 05X2 – 0. 25X4 ( 0 The rounded solution is X1 = 32609; X2 = 57971; X3 = 21739; X4 = 11594; Cost = 3877391 8-21. Minimize time = 12XA1 + 11XA2 + 8XA3 + 9XA4 + 6XA5 + 6XA6 + 6XG1 + 12XG2 + 7XG3 + 7XG4 + 5XG5 + 8XG6 + 8XS1 + 9XS2 + 6XS3 + 6XS4 + 7XS5 + 9XS6 subject to XA1+ XA2+ XA3+ XA4+ XA5+ XA6= 200 XG1+ XG2+ XG3+ XG4 + XG5+ XG6= 225 XS1+ XS2+ XS3+ XS4+ XS5+ XS6= 275 XA1+ XG1+ XS1= 80 XA2+ XG2+ XS2= 120 XA3+ XG3+ XS3= 150 XA4+ XG4+ XS4= 210 XA5+ XG5+ XS5= 60 XA6+ XG6+ XS6= 80 All variables ( 0 Solution: |Source |Destination |Number of |(Station) |(Wing) |Trays | |5A |5 |60 | |5A |6 |80 | |5A |3 |60 | |3G |1 |80 | |3G |3 |90 | |3G |4 |55 | |1S |4 |155 | |1S |2 |120 | Optimal cost = 4,825 minutes. Multiple op timal solutions exist. 8-22. Let Xi = proportion of investment invested in stock i for i = 1, 2, . . . , 5 Minimize beta = 1. 2X1 + 0. 85X2 + 0. 55X3 + 1. 40X4 + 1. 25X5 subject to X1 + X2 + X3 + X4 + X5 = 1total of the proportions must add to 1 0. 11X1 + 0. 09X2 + 0. 065X3 + 0. 15X4 + 0. 13X5 ( 0. 11return should be at least 11% X1 ( 0. 5no more than 35% in any single stock X2 ( 0. 35 X3 ( 0. 35 X4 ( 0. 35 X5 ( 0. 35 Xi ( 0 for i = 1, 2, . . . , 5 b. Solving this on the computer, we have X1 = 0 X2 = 0. 10625 X3 = 0. 35 X4 = 0. 35 X5 = 0. 19375 Minimum beta = 1. 015 Return = 0. 11(0) + 0. 09(0. 10625) + 0. 065(0. 35) + 0. 15(0. 35) + 0. 13(0. 19375) = 0. 11 8-23. Let A = 1,000 gallons of fuel to purchase in Atlanta L = 1,000 gallons of fuel to purchase in Los Angeles H = 1,000 gallons of fuel to purchase in Houston N = 1,000 gallons of fuel to purchase in New Orleans FA = fuel remaining when plane lands in Atlanta FL = fuel remaining when plane lands in Los Angeles FH = fuel remaini ng when plane lands in Houston FN = fuel remaining when plane lands in New Orleans Minimize cost = 4. 15A + 4. 25L + 4. 10H + 4. 18N subject to A + FA ( 24minimum amount of fuel on board when leaving Atlanta A + FA ( 36maximum amount of fuel on board when leaving Atlanta L + FL ( 15minimum amount of fuel on board when leaving Los Angeles L + FL ( 23maximum amount of fuel on board when leaving Los Angeles H + FH ( 9minimum amount of fuel on board when leaving Houston H + FH ( 17maximum amount of fuel on board when leaving Houston N + FN ( 11minimum amount of fuel on board when leaving New Orleans N + FN ( 20maximum amount of fuel on board when leaving New Orleans FL = A + FA – (12 + 0. 05(A + FA – 24)) This says that the fuel on board when the plane lands in Los Angeles will equal the amount on board at take-off minus the fuel consumed on that flight. The fuel consumed is 12 (thousand gallons) plus 5% of the excess above 24 (thousand gallons). This simplifies to: 0. 95A + 0. 95 FA – FL = 10. 8 Similarly, FH = L + FL – (7 + 0. 05(L + FL – 15)) becomes 0. 95L + 0. 95FL – FH = 6. 25 FN = H + FH – (3 + 0. 05(H + FH – 9)) becomes 0. 95H + 0. 95FH – FN = 2. 55 FA = N + FN – (5 + 0. 05(N + FN – 11)) becomes 0. 95N + 0. 95FN – FA = 4. 45 All variables ( 0 The optimal solution is A=18 (1,000 gallons of fuel to purchase in Atlanta) FA=6 (1,000 gallons of fuel remaining when plane lands in Atlanta) L=3 (1,000 gallons of fuel to purchase in Los Angeles) FL=12 (1,000 gallons of fuel remaining when plane lands in Los Angeles) H=1 (1,000 gallons of fuel to purchase in Houston) FH=8 (1,000 gallons of fuel remaining when plane lands in Houston) N=5 (1,000 gallons of fuel to purchase in New Orleans) FN=6 (1,000 gallons of fuel remaining when plane lands in New Orleans) Total cost = 112. 45 (( 1,000) Solutions to Internet Homework Problems 8-24. Let X1 = number of Chaunceys mixed X2= number of Sweet Italians mixed X3= number of bourbon on the rocks mixed X4= number of Russian martinis mixed Maximize total drinks = X1 + X2 + X3 + X4 subject to 1X1 +4X3 ( 52 oz (bourbon limit) 1X1 +1X2 ( 38 oz (brandy limit) 1X1 +2[pic]X4 ( 64 oz (vodka limit) X2 +1[pic]X4 ( 24 oz (dry vermouth limit) 1X1 +2X2 ( 36 oz (sweet vermouth limit) All variables ( 0 Because a Chauncey (X1) is [pic] sweet vermouth, it requires 1 oz of that resource (each drink totals 4 oz). Problem 8-27 solved by computer: Mix 25. 99 (or 26) Chaunceys (X1) Mix   5. 00 (or 5) Sweet Italians (X2) Mix   6. 50 (or 6[pic]) bourbon on the rocks (X3) Mix 14. 25 (or 14[pic]) Russian martinis (X4) This is a total of 51. 75 drinks (in five iterations). 8-25. Minimize 6X11 + 8X12 + 10X13 + 7X21 + 11X22 + 11X23 + 4X31 + 5X32 + 12X33 subject to X11 + X12 + X13( 150 X21 + X22 + X23( 175 X31 + X32 + X33( 275 X11 + X21 + X31= 200 X12 + X22 + X32= 100 X13 + X23 + X33= 300 All variables ( 0 The solution is: X11 = 25, X13 = 125, X23 = 175, X31 = 175, X32 = 100 Cost = $4,525. 8-26. Let Xi = number of BR54 produced in month i, for i = 1, 2, 3. Yi   = number of BR49 produced in month i, for i = 1, 2, 3. IXi = number of BR54 units in inventory at end of month i, for i = 0, 1, 2, 3. IYi = number of BR49 units in inventory at end of month i, for i = 0, 1, 2, 3. Minimize cost = 80(X1 + X2 + X3) + 95(Y1 + Y2 + Y3) + 0. 8(IX1 + IX2 + IX3) + 0. 95(IY1 + IY2 + IY3) Subject to: IX0 = 50initial inventory of BR54 IY0 = 50initial inventory of BR49 IX3 = 100ending inventory of BR54 IY3 = 150ending inventory of BR49 X1 + Y1 ( 1,100maximum production level in August X2 + Y2 ( 1,100maximum production level in September X3 + Y3 ( 1,100maximum production level in October X1 + IX0 = 320 + IX1  Ã‚  BR54 requirements for August X2 + IX1 = 740 + IX2  Ã‚  BR54 requirements for September X3 + IX2 = 500 + IX3  Ã‚  BR54 requirements for October Y1 + IY0 = 450 + IY1  Ã‚  BR49 requirements for August Y2 + IY1 = 420 + IY2  Ã‚  BR49 requirements for September Y3 + IY2 = 480 + IY3  Ã‚  BR49 requirements for October All variables ( 0 A computer solution to this results in IX0 = 50, IX1 = 190, IX2 = 130, IX3 = 100, IY0 = 50, IY3 = 150, X1 = 460, X2 = 680, X3 = 470, Y1 = 400, Y2 = 420, Y3 = 630. All other variables = 0. The total cost = $267,028. 50. Solution to Chase Manhattan Bank Case This very advanced and challenging scheduling problem can be solved most expeditiously using linear programming, preferably integer programming. Let F denote the number of full-time employees. Some number, F1, of them will work 1 hour of overtime between 5 p. m. and 6 p. m. each day and some number, F2, of the full-time employees will work overtime between 6 p. m. and 7 p. m. There will be seven sets of part-time employees; Pj will be the number of part-time employees who begin their workday at hour j, j = 1, 2, . . . , 7, with P1 being the number of workers beginning at 9 a. m. , P2 at 10 a. . , . . . , P7 at 3 p. m. Note that because part-time employees must work a minimum of 4 hours, none can start after 3 p. m. since the entire operation ends at 7 p. m. Similarly, some number of part-time employees, Qj, leave at the end of hour j, j = 4, 5, . . . , 9. The workforce requirements for the first two hours, 9 a. m. a nd 10 a. m. , are: F + P1( 14 F + P1 + P2 ( 25 At 11 a. m. half of the full-time employees go to lunch; the remaining half go at noon. For those hours: 0. 5F + P1 + P2 + P3( 26 0. 5F + P1 + P2 + P3 + P4 ( 38 Starting at 1 p. m. , some of the part-time employees begin to leave. For the remainder of the straight-time day: F + P1 + P2 + P3 + P4 + P5 – Q4( 55 F + P1 + P2 + P3 + P4 F + P1 + P2 + P5 + P6 – Q4 – Q5( 60 F + P1 + P2 + P3 + P4 + P5 F + P1 + P6 + P7 – Q4 – Q5 – Q6( 51 F + P1 + P2 + P3 + P4 + P5 + P6 F + P1 + P7 – Q4 – Q5 – Q6 – Q7( 29 For the two overtime hours: F1 + P1 + P2 + P3 + P4 + P5 + P6 F1 + P1 + P2 + P7 – Q4 – Q5 – Q6 – Q7 – Q8( 14 F2 + P1 + P2 + P3 + P4 + P5 + P6 + P7 F1 + P1 + P2 – Q4 – Q5 – Q6 – Q7 – Q8 – Q9( 9 If the left-hand sides of these 10 constraints are added, one finds that 7F hours of full-time labor are used in straight time (although 8F are paid for), F1 + F2 full-time labor hours are used and paid for at overtime rates, and the total number of part-time hours is 0P1 + 9P2 + 8P3 + 7P4 + 6P5 + 5P6 + 4P7 – 6Q4– 5Q5 – 4Q6 – 3Q7 – 2Q8 – Q9 ( 128. 4 which is 40% of the day’s total r equirement of 321 person-hours. This also leads to the objective function. The total daily labor cost which must be minimized is Z = 8(10. 11)F + 8. 08(F1 + F2) + 7. 82(10P1 + 9P2 + 8P3 + 7P4 + 6P5 + 5P6 + 4P7 – 6Q4 – 5Q5 – 4Q6 – 3Q7 – 2Q8 – Q9) Total overtime for a full-time employee is restricted to 5 hours or less, an average of 1 hour or less per day per employee. Thus the number of overtime hours worked per day cannot exceed the number of full-time employees: F1 + F2 ( F Since part-time employees must work at least 4 hours per day, Q4 ( P1 for those leaving at the end of the fourth hour. At the end of the fifth hour, those leaving must be drawn from the P1 – Q4 remaining plus the P2 that arrived at the start of the second hour: Q5 ( P1 + P2 – Q4 Similarly, for the remainder of the day, Q6( P1 + P2 + P3 – Q4 – Q5 Q7( P1 + P2 + P3 + P4 – Q4 – Q5 – Q6 Q8( P1 + P2 + P3 + P4 + P5 – Q4 – Q5 – Q6 – Q7 Q9( P1 + P2 + P3 + P4 + P5 + P6 – Q4 – Q5 – Q6 – Q7 – Q8 To ensure that all part-timers who began at 9 a. m. do not work more than 7 hours: Q4 + Q5 + Q6 + Q7 ( P1 Similarly, Q4 + Q5 + Q6+ Q7 + Q8 ( P1 + P2 Q4 + Q5 + Q+ Q7 + Q8 + Q9 ( P1 + P2 + P3 Finally, to ensure that all part-time employees leave at some time: P1 + P2 + P3 + P4 + P5 + P6 + P7 = Q4 + Q5 + Q6 + Q7 + Q8 + Q9 The resulting problem has 16 integer variables and 22 constraints. If integer programming software is not available, the linear programming problem can be solved and the solution rounded, making certain that none of the constraints have been violated. Note that the integer programming solution might also need to be adjusted- if F is an odd integer, 0. 5F will not be an integer and the requirement that â€Å"half† of the full-time employees go to lunch at 11 a. m. and the other half at noon will have to be altered by assigning the extra employee to the appropriate hour. 1. The least-cost solution requires 29 full-time employees, 9 of whom work two hours of overtime per day. In actuality, 18 of the full-time employees would work overtime on two different days and 9 would work overtime on one day. Fourteen of the full-time workers would take lunch at 11 a. m. and the other 15 would take it at noon. Eleven part-timers would begin at 11 a. m. , with 9 of them leaving at 3 p. m. and the other 2 at 4 p. m. Fifteen part-time employees would work from noon until 4 p. m. , and 5 would work from 2 p. m. until 6 p. m. The resulting cost of 232 hours of straight time, 18 hours of overtime, and 126 hours of part-time work is $3,476. 28 per day. This solution is not unique- other work assignments can be found that result in this same cost. 2. The same staffing would be used every day. In fact, one would expect different patterns to present themselves on different days; for example, Fridays are usually much busier bank days than the others. In addition, the person-hours required for each hour of the day are assumed to be deterministic. In a real situation, wide fluctuations will be experienced in a stochastic manner. The optimal solution results in a considerable amount of idle time, partly caused by the restriction that employees can start at the beginning of an hour and leave at the end. Eliminating this restriction might yield better results at the risk of increasing the problem size.

Pathways to entry into the practice of professional nursing is Essay

Pathways to entry into the practice of professional nursing is confusing - Essay Example I strongly believe that these numerous entry points are not confusing, but they offer a range of benefits to nursing candidates. When there are many entry points to the practice of professional nursing, candidates get a great exposure to several potential entry choices. The numerous entry points greatly assist students to choose the most appropriate one by giving specific focus to their long term career goals, opportunities for college entrance, and personal and work obligations. These options are also beneficial for candidates to choose the most affordable entry point and avoid excess financial burden. To illustrate, as Black (2013, p. 134) purports, typically the annual cost of diploma and associate degree programs is less than baccalaureate degree programs. Those candidates who choose to attend local schools and live at home may incur lesser costs than students attending schools at distance and living in hostels. Hence, they can thoughtfully identify the entry point that would best fit to their financial interests. Another potential benefit is that students can take the issues of accreditation and articula tion into account while considering entry choice. Similarly, the emerging entry points address various issues associated with traditional entry points, and therefore students can avoid unnecessary concerns associated with their entry into the nursing profession. To sum up, the numerous entry points do not cause any confusion to students who have a clear vision of their long term career goals and financial potentials. Maneval, R. E & Teeter, M. M. (2010). THE STUDENT PERSPECTIVE on RN-Plus-10 Legislation: A Survey of Associate Degree and Diploma Nursing Program Students. Nursing Education Perspectives 31 (6): 358-361. Raines, F & Taglaireni, M. E. (2008). Career Pathways in Nursing: Entry Points and Academic Progression. The Online Journal of Issues in Nursing. Retrieved from