數(shù)據(jù)、模型與決策(運籌學(xué))課后習(xí)題和案例答案002

1、Chapter 2linear programming: Basic conceptsReview Questions2.1-1Ponderosa Industrial uses linear programming monthly to guide the product-mix decision.2.1-2Overall profitability has increased by 20%. Better utilization of raw material, capital equipment, and personnel also resulted.2.1-3The goal was

2、 to improve the utilization of reservation personnel by matching work schedules to customer needs.2.1-4United Airlines saved more than $6 million annually in direct salary and benefit costs. Customer service also improved and workloads were reduced for support staff.2.1-5The SDM system is used to co

3、ordinate the supply, distribution and marketing of each of Citgos major products throughout the United States.2.1-6Citgo saved about $14 million annually in interest expenses. Improvements in coordination, pricing, and purchasing decisions added at least $2.5 million more to annual profits.2.2-1They

4、 provide the highest quality available in the industry for the most discriminating buyers.2.2-21)Should the company launch the two new products?2)What should be the product mix for the two new products?2.2-3The group was asked to analyze product mix.2.2-4Which combination of production rates for the

5、 two new products would maximize the total profit from both of them.2.2-51)available production capacity in each of the plants2)how much of the production capacity in each plant would be needed by each product3)profitability of each product2.3-11)What are the decisions to be made?2)What are the cons

6、traints on these decisions?3)What is the overall measure of performance for these decisions?2.3-2When formulating a linear programming model on a spreadsheet, the cells showing the data for the problem are called the data cells. The changing cells are the cells that contain the decisions to be made.

7、 The output cells are the cells that provide output that depends on the changing cells. The target cell is a special kind of output cell that shows the overall measure of performance of the decision to be made.2.3-3The Excel equation for each output cell can be expressed as a SUMPRODUCT function, wh

8、ere each term in the sum is the product of a data cell and a changing cell.2.4-11)Gather the relevant data.2)Identify the decisions to be made.3)Identify the constraints on these decisions.4)Identify the overall measure of performance for these decisions.5)Convert the verbal description of the const

9、raints and measure of performance into quantitative expressions in terms of the data and decisions2.4-2Algebraic symbols need to be introduced to represents the measure of performance and the decisions.2.4-3A decision variable is an algebraic variable that represents a decision regarding the level o

10、f a particular activity. The objective function is the part of a linear programming model that expresses what needs to be either maximized or minimized, depending on the objective for the problem. A nonnegativity constraint is a constraint that express the restriction that a particular decision vari

11、able must be greater than or equal to zero. All constraints that are not nonnegativity constraints are referred to as functional constraints.2.4-4A feasible solution is one that satisfies all the constraints of the problem. The best feasible solution is called the optimal solution.2.5-1Two.2.5-2The

12、axes represent production rates for product 1 and product 2.2.5-3The line forming the boundary of what is permitted by a constraint is called a constraint boundary line. Its equation is called a constraint boundary equation.2.5-4The coefficient of x1 gives the slope of the constraint boundary line.

13、The constant term gives the value where the line intercepts the x2-axis.2.5-5The easiest way to determine which side of the line is permitted is to check whether the origin (0,0) satisfies the constraint. If it does, then the permissible region lies on the side of the constraint where the origin is.

14、 Otherwise it lies on the other side.2.6-1The Solver dialogue box.2.6-2The Add Constraint dialogue box.2.6-3The Assume Linear Model option and the Assume Non-Negative option.2.7-1Cleaning products for home use.2.7-2Television and print media.2.7-3Determine how much to advertise in each medium to mee

15、t the market share goals at a minimum total cost.2.7-4The changing cells are in the column for the corresponding advertising medium.2.7-5The objective is to minimize total cost rather than maximize profit. The functional constraints contain rather than .2.7-6No.2.7-7Closer to the origin.2.8-1No.2.8-

16、2The graphical method helps a manager develop a good intuitive feeling for the linear programming is.2.8-31)where linear programming is applicable2)where it should not be applied3)distinguish between competent and shoddy studies using linear programming.4)how to interpret the results of a linear pro

17、gramming study.Problems2.1a)The two factors that often hinder the use of optimization models by managers are cultural differences and response time. Cultural differences cause managers and model developers to often have a hard time understanding each other. Response time is often slow due to the tim

18、e to translate, formulate, and solve the mangers problem using optimization systems.b)The company shifted from an emphasis on the manufacture of thicker plywoods to thinner plywoods.c)Ponderosa plans to use optimization in the use of timber for other products also. In addition, optimization may be u

19、sed for raw material and inventory management and for financial planning.2.2a)The shift schedules at airports and reservation offices were done by hand prior to this study.b)The project requirements were:(i)to determine the needs for increased manpower,(ii)to identify excess manpower for reallocatio

20、n,(iii)to reduce the time required for preparing schedules,(iv)to make manpower allocation more day- and time-sensitive, and(v)to quantify the cost associated with scheduling.c)Flexibility, such as the number of start times, the preferred shift lengths, the length of breaks, the preferred days-off c

21、ombinations, etc. were considered. This versatility was necessary to satisfy the group culture at each office, which was necessary to gather field support.d)Benefits included:(i)significant labor cost savings,(ii)improved customer service,(iii)improved employee schedules,(iv)quantified manpower plan

22、ning and evaluation.2.3a)During the years preceding this study, the price of crude oil increased tenfold and short-term interest rates more than tripled.b)Citgos distribution network of pipelines, tankers, and barges spanned the eastern two-thirds of the United States. They market their products in

23、all of the 48 contiguous states.c)An 11-week planning horizon, partitioned into six one-week periods and one five-week period, was used.d)Citgo used an IBM 4381. Typical run times for model generation, solution, and reports were two minutes, half a minute, and seven minutes, respectively.e)The four

24、types of model users were the product managers, the pricing manager, the product traders, and the budget manager. Product managers compared the model recommendations to the actual operational decisions to determine the existence and cause of discrepancies. They also used the models what-if capabilit

25、ies to generate economically viable alternatives to current and forecasted operations. The pricing manager used the model to set ranges for terminal prices for each product and to help set prices and recommend volumes for bulk sales made to reduce excess inventories. Product traders used the model t

26、o determine which side of the trading board they should be on for each product. They also used the models what-if capabilities to determine the sensitivity of spot prices to the required purchases or sales volumes as prices fluctuated during the week. The budget manager used the Financial Summary Re

27、port to generate various components of the monthly and quarterly budgets.f)The major reports generated by the SDM system are:(i)Infeasibility report,(ii)In-transit, Terminal, Exchange, Inventory reports,(iii)Spot recommendation report,(iv)Purchases, Sales, Trades reports,(v)Wholesale report,(vi)Volu

28、me summary report,(vii)Financial summary report.g)The education of the users was a challenge in addition to the collection, validation, and correction of input data for the model. Another challenge concerned the forecasting sales volumes and wholesales prices. Citgo forecasted for monthly and quarte

29、rly budgets, while SDM systems needed weekly forecasts.h)Direct benefits were:(i)the reduction in Citgos product inventory with no drop in service levels, and(ii)operational decision making improved.Indirect benefits were:(i)the establishment of a corporate database, which provided common, up-to-dat

30、e, on-line, operational information for current decision, support(ii)the utilization of a single forecast throughout the different departments,which kept the entire organization, focused,(iii)the closed-loop planning process fostered by the continual feedbackprovided by the project manager, when com

31、paring actual decision tomodel recommended decision,(iv)increased interdepartmental communication, and(v)the insight gained from the modeling process itself.2.4a)b)Maximize P = $600D + $300W,subject toD 42W 123D + 2W 18andD 0, W 0.c)Optimal Solution = (D, W) = (x1, x2) = (4, 3). P = $3300.2.5a)Optim

32、al Solution: (D, W) = (x1, x2) = (1.67, 6.50). P = $3750.b)Optimal Solution: (D, W) = (x1, x2) = (1.33, 7.00). P = $3900.c)Optimal Solution: (D, W) = (x1, x2) = (1.00, 7.50). P = $4050.d)Each additional hour per week would increase total profit by $150.2.6a)b)c)d)Each additional hour per week would

33、increase total profit by $150.2.7a)b)Let A = units of product A producedB = units of product B producedMaximize P = $3,000A + $2,000B,subject to2A + B 2A + 2B 23A + 3B 4andA 0, B 0.c)Optimal Solution = (A, B) = (x1, x2) = (0.667, 0.667). P = $33a)As in the Wyndor Glass Co. problem, we want

34、to find the optimal levels of two activities that compete for limited resources.Let x1 be the fraction purchased of the partnership in the first friends venture.Let x2 be the fraction purchased of the partnership in the second friends venture.The following table gives the data for the problem:Resour

35、ce Usageper Unit of ActivityAmount ofResource12Resource AvailableFraction of partnership in first friends venture101Fraction of partnership in second friends venture011Money$5000$4000$6000Summer Work Hours400500600Unit Profit$4500$4500b)The decisions to be made are how much, if any, to participate i

36、n each venture. The constraints on the decisions are that you cant become more than a full partner in either venture, that your money is limited to $6,000, and time is limited to 600 hours. In addition, negative involvement is not possible. The overall measure of performance for the decisions is the

37、 profit to be made.c)First venture:(fraction of 1st) 1Second venture:(fraction of 2nd) 1Money:5000 (fraction of 1st) + 4000 (fraction of 2nd) 6000Hours:400 (fraction of 1st) + 500 (fraction of 2nd) 600Nonnegativity:(fraction of 1st) 0, (fraction of 2nd) 0Profit = $4500 (fraction of 1st) + $4500 (fra

38、ction of 2nd)d)Data cells:B2:C2, B5:C6, F5:F6, and B11:C11Changing cells:B9:C9Target cell:F9Output cells:D5:D6e)This is a linear programming model because the decisions are represented by changing cells that can have any value that satisfy the constraints. Each constraint has an output cell on the l

39、eft, a mathematical sign in the middle, and a data cell on the right. The overall level of performance is represented by the target cell and the objective is to maximize that cell. Also, the Excel equation for each output cell is expressed as a SUMPRODUCT function where each term in the sum is the p

40、roduct of a data cell and a changing cell.f)Letx1 = share taken in first friends venturex2 = share taken in second friends ventureMaximize P = $4,500 x1 + $4,500 x2,subject tox1 1x2 1$5,000 x1 + $4,000 x2 $6,000400 x1 + 500 x2 600 hoursandx1 0, x2 0.g)Algebraic Versiondecision variables:x1, x2 funct

41、ional constraints:x1 1x2 1$5,000 x1 + $4,000 x2 $6,000400 x1 + 500 x2 600 hoursobjective function:Maximize P = $4,500 x1 + $4,500 x2,parameters:all of the numbers in the above algebraic modelnonnegativity constraints:x1 0, x2 0Spreadsheet Versiondecision variables:B9:C9functional constraints:D4:F7ob

42、jective function:F9parameters:B2:C2, B5:C6, F5:F6, and B11:C11nonnegativity constraints:“Assume nonnegativity in the Options of the Solverh)Optimal solution = (x1, x2) = (0.667, 0.667). P = $6000.2.9a)objective functionZ = x1 + 2x2functional constraintsx1 + x2 5x1 + 3x2 9nonnegativity constraintsx1

43、0, x2 0b & e)c)Yes.d)No.f)Optimal Solution: (x1, x2) = (3, 2) and Z = 7. 2.10a)objective functionZ = 3x1 + 2x2functional constraints3x1 + x2 9x1 + 2x2 8nonnegativity constraintsx1 0, x2 0b & f)c)Yes.d)Yes.e)No.g & h)Optimal Solution: (x1, x2) = (2, 3) and Z = 12. 2.11a)As in the Wyndor Glass Co. pro

44、blem, we want to find the optimal levels of two activities that compete for limited resources. We want to find the optimal mix of the two activities.Let W be the number of wood-framed windows to produce.Let A be the number of aluminum-framed windows to produce.The following table gives the data for

45、the problem:Resource Usage per Unit of ActivityAmount ofResourceWood-framedAluminum-framedResource AvailableGlass6848Aluminum014Wood106Unit Profit$60$30b)The decisions to be made are how many windows of each type to produce. The constraints on the decisions are the amounts of glass, aluminum and woo

46、d available. In addition, negative production levels are not possible. The overall measure of performance for the decisions is the profit to be made.c)glass:6 (#wood-framed) + 8 (# aluminum-framed) 48aluminum:1 (# aluminum-framed) 4wood:1 (#wood-framed) 6Nonnegativity:(#wood-framed) 0, (# aluminum-f

47、ramed) 0Profit = $60 (#wood-framed) + $30 (# aluminum-framed)d)Data cells:B2:C2, B5:C5, F5, B10:C10Changing cells:B8:C8Target cell:F8Output cells:D5, F8 e)This is a linear programming model because the decisions are represented by changing cells that can have any value that satisfy the constraints.

48、Each constraint has an output cell on the left, a mathematical sign in the middle, and a data cell on the right. The overall level of performance is represented by the target cell and the objective is to maximize that cell. Also, the Excel equation for each output cell is expressed as a SUMPRODUCT f

49、unction where each term in the sum is the product of a data cell and a changing cell.f)Maximize P = 60W + 30Asubject to6W + 8A 48W 6A 4andW 0, A 0.g)Algebraic Versiondecision variables:W, Afunctional constraints:6W + 8A 48W 6A 4objective function:Maximize P = 60W + 30Aparameters:all of the numbers i

50、n the above algebraic modelnonnegativity constraints:W 0, A 0Spreadsheet Versiondecision variables:B8:C8functional constraints:D8:F8, B8:C10objective function:F8parameters:B2:C2, B5:C5, F5, B10:C10nonnegativity constraints:“Assume nonnegativity in the Options of the Solverh)Optimal Solution: (W, A)

51、= (x1, x2) = (6, 1.5) and P = $405.i)Solution unchanged when profit per wood-framed window = $40, with P = $285.Optimal Solution = (W, A) = (2.667, 4) when the profit per wood-framed window = $20, with P = $173.33.j)Optimal Solution = (W, A) = (5, 2.25) if Doug can only make 5 wood frames per day, w

52、ith P = $32a)b)Let x1 = number of 27 TV sets to be produced per monthLet x2 = number of 20 TV sets to be produced per monthMaximize P = $120 x1 + $80 x2,subject to20 x1 + 10 x2 500 x1 40 x2 10andx1 0, x2 0.c)Optimal Solution: (x1, x2) = (20, 10) and P = $3200.2.13a)If x2 = 0 then x1 = 2. If

53、 x1 = 0 then x2 = 4.b)c)slope = 2d)x2 = -2x1 + 4, slope = 2, x2-intercept = 42.14a)If x2 = 0 then x1 = 5. If x1 = 0 then x2 = 2.b)c)slope = 0.4d)x2 = 0.4x1 + 2, slope = 0.4, x2 intercept = 22.15a)If x2 = 0 then x1 = 6. If x1 = 0 then x2 = 4.b)c)slope = 0.667d)x2 = 0.667x1 4, slope = 0.667, x2 interc

54、ept = 42.16a)b)c)d)2.17a)b)c)d)2.18a)b)c)d)2.19a)The decisions to be made are how many of each light fixture to produce. The constraints are the amounts of frame parts and electrical components available, and the maximum number of product 2 that can be sold (60 units). In addition, negative producti

55、on levels are not possible. The overall measure of performance for the decisions is the profit to be made.b)frame parts:1 (# product 1) + 3 (# product 2) 200electrical components:2 (# product 1) + 2 (# product 2) 300product 2 max.:1 (# product 2) 60Nonnegativity:(# product 1) 0, (# product 2) 0Profi

56、t = $1 (# product 1) + $2 (# product 2)c)d)Let x1 = number of units of product 1 to producex2 = number of units of product 2 to produceMaximize P = $1x1 + $2x2,subject tox1 + 3x2 2002x1 + 2x2 300 x2 60andx1 0, x2 0.e)Optimal Solution: (x1, x2) = (125, 25) and P = $175.2.20a)The decisions to be made

57、are what quotas to establish for the two product lines. The constraints are the amounts of work hours available in underwriting, administration, and claims. In addition, negative levels are not possible. The overall measure of performance for the decisions is the profit to be made.b)underwriting:3 (

58、# special risk) + 2 (# mortgage) 2400administration:1 (# mortgage) 800claims:2 (# special risk) 1200Nonnegativity:(# special risk) 0, (# mortgage) 0Profit = $5 (# special risk) + $2 (# mortgage)c)d)Let S = units of special risk insuranceM = units of mortgagesMaximize P = $5S + $2M,subject to3S + 2M

59、2,400M 8002S 1,200andS 0, M 0.e)Optimal Solution: (S, M) = (x1, x2) = (600, 300) and P = $3600.2.21a)P=6P=12P=18b)slope-intercept formslopex2 interceptP=6x2 = 0.667x1 + 20.6672P=12x2 = 0.667x1 + 40.6674P=18x2 = 0.667x1 + 60.66762.22a)P=100P=200P=300b)slope-intercept formslopex2 interceptP=100 x2 = -

60、2.5x1 + 102.510P=200 x2 = -2.5x1 + 202.520P=300 x2 = -2.5x1 + 302.5302.23a)Cost=100200300b)slope-intercept formslopex2 interceptC=300 x2 = 5x1 3005300C=200 x2 = 5x1 2005200C=100 x2 = 5x1 100 x251002.24a)x2 = 0.5x1 + 10b)slope = 0.5, x2-intercept = 10c)2.25x2 = 1.6x1 + 82.26a)x2 = 2x1 + 4b)x2 = 0.667