經(jīng)濟定量分析

1、練習(xí)：1 All-Star Bat Manufacturing, Inc., supplies baseball bats to major and minor league baseball teams. After an initial order in January, demand over the 6-month baseball season is approximately constant at 2000 bats per month. Assuming that the bat production process can handle up to 5000 bats per

2、 month, holding costs have a monthly rate of 1%, the bat production setup costs are $50 per setup, and that the production cost is $10 per bat. What production lot size would you recommend to meet the demand during the baseball season?A=2000bats, B=5000bats，k=$50, h=$0.01, c=$10(bats )The optimal pr

3、oduction quantity is 1826 bats.Feasible regionOptimal solution2x1+3x2=0x1+3x2=20x1+x2=10-2x1-x2=-16The optimal solution is (5,5). And the corresponding value is 25.The optimal solution is (15/4,3/4). And the corresponding value is 33/4.WPW1W2W3W4CapacityP141468P2226110P337514Demand6763Getting starte

4、d： The Vogels approximation rule WarehousePlantW1W2W3W4CapacityP14174168P226264110P33751134Demand6763Finding the entering cell :Row and column numbers 0140WarehousePlantW1W2W3W4CapacityP104417-41+668P22262+-164-1-110P31327551134Demand6763Finding the new solution :The closed-loop path and finding the

5、 entering cell: Row and column numbers 1140WarehousePlantW1W2W3W4CapacityP10431345668P212624611010P31317551134Demand6763All the improvement difference is not positive shown in this table, so the solution is the optimal solution to the transportation problem. The total cost of this shipment schedule

6、is computed to be 51. WarehousePlantW1W2W3W4CapacityP141241116P22103910P38511622P400002Demand1014121450Getting started： The Vogels approximation rule WarehousePlantW1W2W3W4CapacityP141241211416P22101030910P38514116822P4000022Demand1014121450Finding the entering cell :Row and column numbers 310411War

7、ehousePlantW1W2W3W4CapacityP1041122412+114-16P2-121010130-9+-110P3-581051411126822P4-11080107022Demand1014121450Finding the new solution :The closed-loop path and finding the entering cell: Row and column numbers 410411WarehousePlantW1W2W3W4CapacityP104012241211416P2-2210102319010P3-58951411126822P4

8、-11070107022Demand1014121450All the improvement difference is positive shown in this table, so the solution is the optimal solution to the transportation problem. The total cost of this shipment schedule is computed to be 230. Show out the earliest possible event times and the critical pathin the fo

9、llowing network16105616192132298100Letting TE: the earliest possible event times that can be expected TL: the latest allowable event times. EST: event slack timeTE(1)=0;TE(2)=TE(1)+t12=0+5=5;TE(3)=TE(1)+t13=0+8=8;TE(6)=TE(1)+t12=0+6=6;TE(7)=TE(4)+t47=10+6=16;TE(8)=TE(6)+t68=6+10=16;TL(12)=32;TL(11)=

10、TL(12)- t1112=32-3=29;TL(10)= TL(11)- t1011=29-8=21;TL(9)= TL(11)- t911=29-5=24;TL(8)= TL(9)- t89=24-0=24;TL(6)= TL(8)- t68=24-10=14;TL(5)= TL(9)- t59=24-9=15;TL(3)= TL(4)- t34=10-2=8;TL(2)= TL(4)- t24=10-4=6;EST(1)=0-0=0; EST(2)=6-5=1; EST(3)=8-8=0; EST(4)=10-10=0; EST(5)=15-10=5; EST(6)=14-6=8; ES

11、T(7)=16-16=0; EST(8)=24-16=8; EST(9)=24-19=5; EST(10)=21-21=0; EST(11)=29-29=0; EST(12)=32-32=0;So, the critical path is 1à3à4à7à10à11à12.EVENT (probability)ACTA1A2A3A4S1 (0.2)19151210S2 (0.5)21263822S3 (0.3)25303640Act 3 is the optimal one.Present the structure of the following decision problem in decision tree and c