⽯家庄经济学院第⼗届校内数学建模竞赛试题
⽯家庄经济学院第⼗届校内数学建模竞赛试题
⼤⼀年级组选做A题
A题：交通管理中的黄灯问题
在⼗字路⼝的交通管理中，亮红灯之前，要亮⼀段时间的黄灯，这是为了让那些正⾏驶在⼗字路⼝的⼈注意，告诉他们红灯即将亮起，假如你能够停住，应当马上刹车，以免冲红灯违反交通规则。黄灯时间的设定与该路⼝的汽车速度、司机的反应时间、汽车的制动距离、路⼝宽度、汽车长度等因素有关。假设某⼀路⼝宽度为40m，该路⼝限速标志为40km/h。请研究下列问题:
（1）汽车的刹车距离由反应距离和制动距离组成，驾驶⼿册规定具有良好刹车性能的汽车在以80km/h的速率⾏驶时，可以在56m的距离内刹住；在以48km/h 的速率⾏驶时可以在24m的距离被刹住。我们随机选择了该路⼝的⼏辆家⽤轿车做了⼀个刹车实验，当汽车速度为20km/h时，汽车的平均制动距离(从制动器开始制动到汽车完全停⽌的距离)为 6.36m，利⽤这些信息和所学的知识建⽴汽车刹车距离与车速之间关系的数学模型。
（2）建⽴数学模型分析该路⼝黄灯亮多久才⽐较合适？
⼤⼆及⼤⼆以上年级组选做B题
B题：客房预定的价格和数量问题
某著名的旅游景区中的宾馆主要提供举办会议和游客使⽤。客房通过电话或互联⽹预定，这种预定具有很⼤的不确定性，客户很可能由于各种原因取消预定。宾馆为了争取更⼤的利润，⼀⽅⾯要争取客户，另⼀⽅⾯要降低客户取消预定遭受的损失。为此，宾馆采⽤⼀些措施。⾸先，要求客房提供信⽤卡号，预付第⼀天房租作为定⾦。如果客户在前⼀天中午以前取消预定，定⾦将如数退还，否则定⾦将被没收。其次，宾馆采⽤变动价格，根据市场需求情况调整价格，⼀般来说旅游旺季价格⽐较⾼，淡季价格略低。
（1）请建⽴客房预定价格的数学模型，并对以下实例作分析。表1给出了某宾馆2005年10⽉～2010年3⽉期间，每⽉标准间平均价格(单位：元)，⽤你的模型说明价格变动的规律，并据此估计未来⼀年内的标准房参考价格。你还可以收集更多的数据来佐证你模型的价值（要求注明出处）。
（2）在旅游旺季，宾馆往往可以预定出超过实际套数的客房数，以减低客户取消预定时宾馆的损失。当然这样做可能会带来新的风险，因为万⼀届时有超出客房数的客户出现，宾馆要通过升级客房档次或赔款来解决纠纷，为此宾馆还会承担信誉风险。某宾馆有总统套房20套，豪华套房100套，标准间500套。试为该宾馆制定合理的预定策略，并论证你的理由。
单位：元)
h=[x=[10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3];
x=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,4
7,48,49,50,51,52,53,54];
y=[328,263,251,241,249,316,344,360,320,344,384,368,401,363,336,366,331,390,401,439,397,463,509,474,508,458,412,369,403,436,44 7,483,439,514,550,489,534,498,402,397,416,451,486,507,458,493,562,474,528,436,398,442,404,428];
cftool
Smoothing spline:
f(x) = piecewise polynomial computed from p
Smoothing parameter:
p = 0.99997739
Goodness of fit:
SSE: 0.008141
R-square: 1
Adjusted R-square: 0.9999
RMSE: 0.6978
Linear interpolant:
f(x) = piecewise polynomial computed from p
where x is normalized by mean 27.5 and std 15.73
Coefficients:
p = coefficient structure
Goodness of fit:
SSE: 0
R-square: 1
Adjusted R-square: NaN石家庄汽车网
RMSE: NaN
Linear model Poly6:
f(x) = p1*x^6 + p2*x^5 + p3*x^4 + p4*x^3 + p5*x^2 +
p6*x + p7
Coefficients (with 95% confidence bounds):
p1 = 0.01878 (-0.001125, 0.03868)
p2 = -0.7598 (-1.538, 0.01812)
p3 = 11.83 (0.005115, 23.66)
p4 = -88.76 (-176.9, -0.6539)
p5 = 325.2 (-6.245, 656.6)
p6 = -498.6 (-1076, 78.56)
p7 = 491.5 (146.2, 836.8)
Goodness of fit:
SSE: 1441
R-square: 0.9451
Adjusted R-square: 0.8792
RMSE: 16.97
Linear model Poly6:
f(x) = p1*x^6 + p2*x^5 + p3*x^4 + p4*x^3 + p5*x^2 +
p6*x + p7
Coefficients (with 95% confidence bounds):
p1 = 0.01668 (-0.0177, 0.05106)
p2 = -0.6666 (-2.01, 0.6773)
p3 = 10.29 (-10.14, 30.72)
p4 = -77.57 (-229.8, 74.63)
p5 = 295.4 (-277.2, 867.9)
p6 = -502.8 (-1500, 494.3)
p7 = 639.4 (42.92, 1236)
Goodness of fit:
SSE: 4300
R-square: 0.8734
Adjusted R-square: 0.7214
RMSE: 29.32
Linear model Poly7:
f(x) = p1*x^7 + p2*x^6 + p3*x^5 + p4*x^4 + p5*x^3 +
p6*x^2 + p7*x + p8
Coefficients (with 95% confidence bounds):
p1 = -0.0003383 (-0.02308, 0.02241)
p2 = 0.03221 (-1.004, 1.069)
p3 = -0.9379 (-20.05, 18.18)
p4 = 12.47 (-170.4, 195.4)
p5 = -84.41 (-1051, 882.5)
p6 = 287.4 (-2480, 3055)
p7 = -417.9 (-4277, 3442)
p8 = 601.1 (-1339, 2541)
Goodness of fit:
SSE: 7701
R-square: 0.7344
Adjusted R-square: 0.2695
RMSE: 43.88
Linear model Poly1:
f(x) = p1*x + p2
Coefficients (with 95% confidence bounds):
p1 = 42.46 (3.685, 81.23)
p2 = 21.94 (-293.1, 336.9)
Goodness of fit:
SSE: 4453
R-square: 0.7971
Adjusted R-square: 0.7295
RMSE: 38.53
Linear model Poly1:
f(x) = 42.46 *x + 21.94 Coefficients (with 95% confidence bounds): p1 = (3.685, 81.23) p2 = (-293.1, 336.9)
Goodness of fit:
SSE: 4453
R-square: 0.7971
Adjusted R-square: 0.7295
RMSE: 38.53
Linear model Poly1:
f(x) = 39.5*11+ 44.6
Coefficients (with 95% confidence bounds): p1 = (0.6155, 78.38)
p2 = (-271.3, 360.5)
Goodness of fit:
SSE: 4479
R-square: 0.777
Adjusted R-square: 0.7026
RMSE: 38.64
f = 39.59 *11 +44.21
Coefficients (with 95% confidence bounds): p1 = (-2.735, 81.92)
p2 = (-299.7, 388.1)