《运筹作业.11doc1.doc》由会员分享，可在线阅读，更多相关《运筹作业.11doc1.doc（8页珍藏版）》请在金锄头文库上搜索。
1、 Logistics operation2.1-1 (P. 15)Farmer Jones must determine how many acres of corn and wheat to plant this year. An acre of wheat yields 25 bushels of wheat and requires 10 hours of labor per week. An acre of corn yields 10 bushels of corn and requires 4 hours of labor per week. All wheat can be so
2、ld at $4 a bushel, and all corn can be sold at $3 a bushel. Seven acres of land and 40 hours per week of labor are available. Government regulations require that at least 30 bushels of corn be produced during the current year. Let x1=number of acres of corn plated, x2=number of acres of wheat plante
3、d. Using these decision variables, formulate an LP whose solution will tell Farmer Jones how to maximize the total revenue from wheat and corn.Solution O.f. mix=30X1+100X2 S.T. X1+X2=7 4X1+100X2=30 X1 、X2=0 By constraints and objective function, by use of excel find out optimal value, if the maximal
4、 profit and qualified: when clamps its X1 = 3 X2 = 2.8 yields Z biggest, is 370.2.2-2 (P. 23)Leary Chemical manufactures three chemicals: A, B, and C. These chemicals are produced via 2 production processes: 1 and 2. Running process for 1 h costs $4, and yield 3 units of A, 1 of B, and 1 of C. Runni
5、ng process 2 for 1 hour costs $1 and produces 1 unit of A and 1 of B. To meet customer demands, at least 10 units of A, 5 of B, and 3 of C must be produced daily. Graphically determine a daily production plan that minimizes the cost of meeting Leary Chemicals daily demands.Solution：X1=hours of prece
6、ss 1 X2=hours of precess 2O.F. Min Z=4X1+X2S.T. 3X1+X2=10 X1+X2=5 X1=3 X20Make clamps its x1 x2 seperately process 1, 2 production process of the amount of time. By TiYi know there is nothing beats A, five of tenve of B, and 3 of C daily. Wish all demoting listed constraint conditions, by use of exc
7、el ask out clamps its X1, X2 respectively 3, 2, the least expensive, is 14.Problem 2.4-1There are 3 factories on the Momiss River. Each emits 2 types of pollutants into the river. If the waste from each factory is processed, the pollution in the river can be reduced. It costs ($15, $10, $20) to proc
8、ess a ton of (factory 1, factory 2, factory 3) waste , and each ton processed reduces the amount of pollutant 1 and 2 by(0.10, 0.45),(0.20,0.25),(0.40,0.30) ton. The sate wants to reduce pollutants at least (30,40) tons. Formulate an LP that will minimize the total cost.Solution: x1=reduced pollutio
9、n of the factory 1 x2=reduced pollution of the factory 2 x3=reduced pollution of the factory 3O.F. Min z=15x1+10x2+20x3S.T. 0.1x1+0.20x2+0.40x3=30 0.45x1+0.25x2+0.30x3=30 x1 x2 x30To make the minimum, according to the regulation pollutants cost constraint, finally proper clamps its x1 =7.692308472,
10、x2 = 146.1538458, spend a minimum, for 1576.923085.2.5-4 (P 35) *Suppose the post office has 25 full-time employees and was not allowed to hire or fire any employees. Formulate an LP that could be used to schedule the employees in order to maximize the number of weekend days off received by the empl
11、oyees.Employees RequiredDay 1=Monday17Day 2=Tuesday13Day 3=Wednesday15Day 4=Thursday19Day 5=Friday14Day 6= Saturday16Day 7= Sunday11 Solution:Defining xi to the number of employees working on day i.O.F. max z=x5+2x6+x7 S.T. x1+x2+x3+x4+x5+x6+x7=25x1+x4+x5+x6+x7=15x1+x2+x5+x6+x7=19x1+x2+x3+x6+x7=14x1
12、+x2+x3+x4+x7=16x1+x2+x3+x4+x5=11x2+x3+x4+x5+x6=17x3+x4+x5+x6+x7=13To make the number of weekend days off received by the employees is the biggest, so only two forms, Friday, Saturday rest and on Saturday and Sunday rest, the objective function is Max z = x5 + 2x6 + x7, according to the topic request
13、 list constraint, finally come to a maximum of z is 23. 2.6 2 (P. 39) *Two investments with varying cash flows ( in thousands of dollars) are available, as shown in Table 7. At time 0, $10,000 is available for investment, and at time 1, $7,000 is available. Assuming that r = 0.10, set up and LP whos
14、e solution maximizes the NPV obtained from these investments. Graphically find the optimal solution to the LP. (Assume that any fraction of an investment may be purchased.)Cash flow (in thousands)Time 0123Investment 1-6-579Investment 2-8-397Solution:x1=fraction oftheinvestment1purchased,x2=fraction
15、oftheinvestment2 purchased, O.F. Z=12.547x1+12.697x2 S.T. 6x1+8x2=10 5x1+3x20 Z is with net present value formula calculate out of,0 period shall not exceed 10 thousand, pxix can not be more than seven thousand, but do not need to be a net present value2.8 1 *You have decided to enter the candy business. You are considering producing two types of candi
