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N N 7 . N N ! $ 7 * 1 1 1 1 2 2 2 2 5 ( $ 5 ( 4 1 25& 1 24 5& 1 24 !5 -! 1 2 4$1 2 5 - 5! 7 7 7 7 $ 5 &5 5& &55 -!5 -! 5- -5- 7 . ; N ! N / 0 . $ 7 * 5 $( $ 5 ( 5 9 < 1 6 ( )6 2 1 ( 2 ; ; # ! * 1 5 ( 5 ( 5 2 5 & 1 5 6 ( $ 5 )6 2 , ! = $ 4 4 - 4 -7 4 - 4 4 5 5 5 4 /( ( ; 5 $( 5 ( $ 5 -( ( ( ≥- 5 5- 5 5 = # -(--- 4 4 4 '(--- ≥ ' ≥ ≥ ! ≤ ' = = = ? ≥- ' ; . ; / 7 - / 0 = - . ; . - 7 = ; / 3 . ? = - - - # ; ; ; * 1 !( -2K 1 5 '( 5G . ( -2K 1'( -2K 1'( &2 5 & $(--- & D# = 7 < '-J -J 5 5 ' ---- < E D 6 $ . . E $-J -J # - &! 5 5 - '- 1- &!2 4 - $- 1- &!2 - $- 4 - ! 5 5 - - 1- &!2 4 - - 1- &!2 - ! 4- - ) 7 - - 3 ; . , , , 5 , 4, # 5 - 1- $- 4 - ! 2 4 - 1- ! 5 -& 4-! 445 -) 4 - )$ . ; , , < . 4- - 5 1- ) 5 4 - )$ 2 4 1- 4 $ 5 4 , )! 5 < 5 1 )! 5 -' = # -' 7 ' 5 - - 4- - 2 4- - 2 ; , 4 - 2 1 4 - 4 4 $ 2 - - $- ! 4 4 (E ≥ - - ! - - * 5 )$--( E 5 ! --( ' ---- ≤ ≤ = D# 5 - ' 1)$--2 4 7 - 1! --2 5 G 5 5 = & !- 4 ) -- - - -$ 4 - 4 4 ≥ - ( ≥ ≤ 5 $ - 1 1 1 - - -J -J 2 - 2 2 ( '- / 0 . 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P 9 < , P ' , $ 5 9 + $ 7 5 ( 8 ' - 55 9 7 * 5 ( 5 -( 5 / ; + $ 5 5 . , I I * 4 -- ≤ ≥ ≥ $ 7 3 O / 0 = -- 7 . 3 7 -- = O D# . '- # / '- K '- I - I / # -- -- + K - I # # '- I & , & . $-- !-; -- -/ 1 !( !2 -- -- 1 !-( --2 / ; 1 5 ; -- -# # , 5 1 , 5 1 !-2 4 1 --2 5 -!- !( (1 5 !2 4 1 1 5 !-( 5 !2 !( 5 1 5 !( 5 !2 1 5 !-( !2 5 -!- 5 --2 1 !2 4 1 !2 -- -- $-- @ !2 5 &! &! 5 ' ! ( 5 --2 / 0 . ' $-- !-- 9 I !-- -- -9 7 -- -- . / -- -- < 5 = 5 ; !( !( 5 !- 5 &! ! .5 ;5 1 !2 4 1 &!2 5 $ ! # --( 5 --( 5 -- I I ; ; ! #$' $ % $-- &! I < ) -- -- * "! #$% = 7 * & #$% ( ) , ) ) *"+ + , *"!+ '$ ! %$ -. ; I )G! -- ; . ; '/% + I '! &! " )$'! 1 2 !- ( # $% = < = # D = G! -- . . -- 4 !- ! -' 4 4 ! ≤ ≤ ≥ - *7 & -- , & -- 7 - J I J 9 # * -J ≥- 1 ≥- 1 4 2 ≥4-' ≥- 2 - & ' ! ! -1'!2 4 -1 &!2 5 G!- 5 5 + I &! ; ; ' $- - ; G! !- 3 L 0 $% I / 0 . !-7 -J< I 7 1$! !( $ ' 2 -- -- , -- . - -- -- ; 5 $! ! !-1 $ ' 2 5 G !(' ' 0 5 G -(--- -- -- 5 $ ' K 5 --1$! !2 4 ( 0 # K 5 $! 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'( ) $ %%%%%%%%%%%%%%% * $ %%%%%%%%%%%%%%%%%% %+ %+ %+, + & & , 012# &/ 2 # 2 # 2 ! 32 4 5 %%%%%%%%%%%%%%% ++ %%%%%%%%%%%% # %%%%%%%%%%%%%%% & + + 4 5 &/-, / & && 4 5 + + + ! 36 6 * *2 ! # $ %%%%%%%%%%%% ,+ , . , + ,+ ., , /, +.- / . .& ./ 9'9 9'44@ + 0 0 0 0 0 0 + +, )) 5 %%%%%%%%%%%%%%% )) 5 )) 5 +, /- , )) 5 & + & )) 5 32 4 5 %%%%%%%%%%%%%%% + & & - # %%%%%%%%%%%%%%% , & " " " " " " " 1 ? 1 = )) 5 %%%%%%%%%%%%%%% / / -,-/- & + -,, , ., . + ,, & )) 5 ; % 1 3 3 3 2 59 ;222 3 3 2 59 0 9'222 ? 0 5222 D E D E 3 0 5'922 1 D E 3 0 2'222 = 9'222 0 3 D E 3 D9E ? % 1 % = 3 ≤ '222 ? D4E ≤ '222 1 D;E ≤ 4'222 = D 2E ≤ 5'222 ? D E ≤ 5'222 1 D E ≤ 5'222 = !% > > - > > % > % > % , , , % % % 0 <5' 92 ? > - > 02 D@E , > 02 0 D5E % 0 9'222 0 ) 3 2 59 1 '222 2 '222 5'222 '222 '222 2 '922 = 5'922 '922 2 2 > = 0 0 0 0 0 0 0 0 0 0 I = 6 ? '" 6' 22 % > 1 # # H =' 1 '? ! - > 1 # 3 922 3 ' 4 + + ! 222 3 422 3 222 3 422 3 ! 422 3 922 422 3 922 22 3 22 22 3 22 3 0 922 3 0 922 3 4 3 3 0 22 3 3 4 0 22 0 922 3 3 4 0 922 > 3 3 4 3 3 > 3 4 22 0 22 3 22 22 ? 3 3 22 3 3 4 3 0 22 = ≤ 22 ≤ 22 ≤ 22 ≤ 22 9 % > - # ≤ ≤ , - ! < 2 2 0 2 94 0 2 = 6 , F , ! > > , > N 22 22 N 9 9 4: 22 2: N @ 94 0 0 0 0 0 9 50 @0 40 2 3 2 3 3 3 " 3 2 3 3 3 2 N !22 !22 !22 !22 !22 !22 9!22 5!22 ) 1 > 3 2 3 3 3 3 2 3 3 3 5 3 2 9 3 2 9 9 3 9 3 9 3 I≥2 5 5 3 5 3 5 3 0 ' ' 4 ? 5 <@ 52 D @ 3 2 @ @ 3 @ 3 @ 3 E 0 < 2 2 > 4 4 4 4 4 ) ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ 4 4 @ 9 2 2 5 5 $ . !22 !22 !22 !22 !22 !22 9!22 5!22 @!22 4!22 ;!22 > = 1 # '? = 1 " , - %%%%%%%%%%%%%% 9 9 9+ 9& 9 9, 9. 9- %%%%%%%%%%%%%%% - # $ %%%%%%%%%%%%%% '( ) $ %%%%%%%%%%%%%%% ! " " # $ $ %%%%%%%%%%%%%%%%%% & , + & * $ %%%%%%%%%%%%%%%%%% % - & . % - & , . / % - & & $ 4 !22 !22 9!22 5!22 5 $ $ 0 <524 @ !22 !22 4!22 ;!22 % " 0 0 0 0 90 50 !22 !22 !22 !22 !22 !22 @ % @0 40 ;0 9!22 5!22 @!22 ) " . " I E 0 < ; +∑ 2 2 = 42 = ) " ? ' ≥4 3 ) $ " 405 " 04 0 90 @0 ! * $ 42 ! 4 ∑ <@ 52 D * , 0 4 5 <92 52 , 2 , 04 0 90 @0 ' 4 $ 1. Learn about applications of linear programming that are solved in practice. 2. Develop an appreciation for the diversity of problems that can be modeled as linear programs. 3. Obtain practice and experience in formulating realistic linear programming models. 4. Understand linear programming applications such as: data envelopment analysis revenue management portfolio selection game theory 5. Know what is meant by a two person, zero sum game. 6. Be able to identify a pure strategy for a two person, zero sum game. 7. Be able to use linear programming to identify a mixed strategy and compute optimal probabilities for the mixed strategy games. 8. Understand the following terms: game theory two person, zero sum game saddle point pure strategy mixed strategy The application problems of Chapter 5 are designed to give the student an understanding and appreciation of the broad range of problems that can be approached by linear programming. While the problems are indicative of the many linear programming applications, they have been kept relatively small in order to ease the student's formulation and solution effort. Each problem will give the student an opportunity to practice formulating a linear programming model. However, the solution and the interpretation of the solution will require the use of a software package such as , 's Solver or LINGO. 5 1 Chapter 5 1. a. Min s.t. 285.2 123.80 106.72 + + 34.62 + 27.11 + 148 + 27 + 162.3 + 128.70 + 64.21 48.14 43.10 253 41 + 285.2 + 1123.80 + 106.72 , 2. + + 36.72 + 45.98 + 175 + 23 + 275.7 + 348.50 + 104.10 , , + + + + + + + + 33.16 56.46 160 84 210.4 154.10 104.04 = ≥ ≥ ≥ ≥ ≤ ≤ ≤ 1 48.14 43.10 253 41 0 0 0 ≥0 b. Since = 1.0, the solution does not indicate General Hospital is relatively inefficient. c. The composite hospital is General Hospital. For any hospital that is not relatively inefficient, the composite hospital will be that hospital because the model is unable to find a weighted average of the other hospitals that is better. a. Min s.t. 55.31 49.52 281 47 250 +310 316 +134.6 94.4 +116 + + + + + + + + 37.64 55.63 156 3 278.5 114.3 106.8 + + + + + + + + , b. + + + + + + + + 32.91 25.77 141 26 165.6 131.3 65.52 , + + + + + + + + 33.53 41.99 160 21 250 316 94.4 , , , = 0.924 = 0.074 = 0.436 = 0.489 All other weights are zero. c. is relatively inefficient Composite requires 92.4 of 's resources. 5 2 32.48 55.30 157 82 206.4 151.2 102.1 , ≥0 + + + + + + + + 48.78 81.92 285 92 384 217 153.7 + + + + + + + + 58.41 119.70 111 89 530.1 770.8 215 = 1 ≥ 33.53 ≥ 41.99 ≥ 160 21 ≥ 0 ≤ 0 ≤ 0 ≤ Advanced Linear Programming Applications d. 34.37 patient days (65 or older) 41.99 patient days (under 65) 3. e. Hospitals A, C, and E. a. Make the following changes to the model in problem 27. New Right Hand Side Values for Constraint 2 32.48 Constraint 3 55.30 Constraint 4 157 Constraint 5 82 New Coefficients for Constraint 6 Constraint 7 Constraint 8 b. 4. = 1; in 206.4 151.2 102.1 = 1; all other weights = 0 c. No; = 1 indicates that all the resources used by Hospital Hospital . are required to produce the outputs of d. Hospital is the only hospital in the composite. If a hospital is not relatively inefficient, the hospital will make up the composite hospital with weight equal to 1. a. Min s.t. 110 22 1400 3800 25 8 + 96 + 16 + 850 + + 4600 + 32 + 8.5 + 110 + 22 + 1400 , , + + 4400 + 35 + 8 + 100 + 18 + 1200 , , 5 3 ≥0 + + + + + + + 6500 30 10 125 25 1500 + + + + + + + 6000 28 9 120 24 1600 = ≥ ≥ ≥ ≤ ≤ ≤ 1 4600 32 8.5 0 0 0 Chapter 5 b. $$$$$$$$$$$$$$ % &' &" &, & & $$$$$$$$$$$$$$$ " # $$$$$$$$$$$$$$ ( 3 + * /0 1 # $$$$$$$$$$$$$$$ ) ! ! " # # $$$$$$$$$$$$$$$$$$ ()* + *)* -* .* 2 # $$$$$$$$$$$$$$$$$$ - -$ $ (- + (-3 ( )( (+ c. Yes; = 0.960 indicates a composite restaurant can produce Clarksville's output with 96% of Clarksville's available resources. d. More Output (Constraint 2 Surplus) $220 more profit per week. Less Input Hours of Operation 110 = 105.6 hours FTE Staff 22 1.71 (Constraint 6 Slack) = 19.41 Supply Expense 1400 129.614 (Constraint 7 Slack) = $1214.39 The composite restaurant uses 4.4 hours less operation time, 2.6 less employees and $185.61 less supplies expense when compared to the Clarksville restaurant. 5. e. = 0.175, restaurants. = 0.575, and = 0.250. Consider the Bardstown, Jeffersonville, and New Albany a. If the larger plane is based in Pittsburgh, the total revenue increases to $107,849. If the larger plane is based in Newark, the total revenue increases to $108,542. Thus, it would be better to locate the larger plane in Newark. Note: The optimal solution to the original Leisure Air problem resulted in a total revenue of $103,103. The difference between the total revenue for the original problem and the problem that has a larger plane based in Newark is $108,542 $103,103 = $5,439. In order to make the decision to change to a larger plane based in Newark, management must determine if the $5,439 increase in revenue is sufficient to cover the cost associated with changing to the larger plane. 5 4 Advanced Linear Programming Applications b. Using a larger plane based in Newark, the optimal allocations are: PCQ = 33 PCY = 16 NCQ = 26 NCY = 15 = 32 = 46 = 23 PMY= 6 = 56 NMY = 7 CMY = 8 COY = 10 = 43 POY = 11 NOQ = 39 NOY = 9 The differences between the new allocations above and the allocations for the original Leisure Air problem involve the five ODIFs that are boldfaced in the solution shown above. c. Using a larger plane based in Pittsburgh and a larger plane based in Newark, the optimal allocations are: PCQ = 33 PCY = 16 NCQ = 26 NCY = 15 = 37 = 44 PMQ = 44 PMY= 6 = 56 NMY = 7 CMY = 8 COY = 10 = 45 POY = 11 NOQ = 39 NOY = 9 The differences between the new allocations above and the allocations for the original Leisure Air problem involve the four ODIFs that are boldfaced in the solution shown above. The total revenue associated with the new optimal solution is $115,073, which is a difference of $115,073 $103,103 = $11,970. 6. d. In part (b), the ODIF that has the largest bid price is COY, with a bid price of $443. The bid price tells us that if one more Y class seat were available from Charlotte to Myrtle Beach that revenue would increase by $443. In other words, if all 10 seats allocated to this ODIF had been sold, accepting another reservation will provide additional revenue of $443. a. The calculation of the number of seats still available on each flight leg is shown below: !" !" 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 PCQ PMQ POQ PCY PMY POY NCQ NMQ NOQ NCY NMY NOY CMQ CMY COQ COY 33 44 22 16 6 11 26 36 39 15 7 9 31 8 41 10 25 44 18 12 5 9 20 33 37 11 5 8 27 6 35 7 5 5 8 0 4 4 1 2 6 3 2 4 2 1 4 2 6 3 Chapter 5 Flight Leg 1: 8 + 0 + 4 + 4 + 1 + 2 = 19 Flight Leg 2: 6 + 3 + 2 + 4 + 2 + 1 = 18 Flight Leg 3: 0 + 1 + 3 + 2 + 4 + 2 = 12 Flight Leg 4: 4 + 2 + 2 + 1 + 6 + 3 = 18 Note: See the demand constraints for the ODIFs that make up each flight leg. b. The calculation of the remaining demand for each ODIF is shown below: !" !" 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 c. PCQ PMQ POQ PCY PMY POY NCQ NMQ NOQ NCY NMY NOY CMQ CMY COQ COY 33 44 45 16 6 11 26 56 39 15 7 9 64 8 46 10 25 44 18 12 5 9 20 33 37 11 5 8 27 6 35 7 8 0 27 4 1 2 6 23 2 4 2 1 37 2 11 3 The LP model and solution are shown below: 4 (). "56- . 56--. 563. "76+* 76* 76( 3.* "76+++ 76*. 76() " 563. " 76--+" 56*.-" 7 (8 -8 38 +8 *8 8 )8 .8 8 ( 8 ((8 (-8 (38 (+8 (*8 ( 8 ()8 (.8 ( 8 - 8 ( "56( ( "56( ( 56( ( 56( ( "59. ( 59( ( 59-) ( "79+ ( 79( ( 79( "59 ( 59-3 ( 59( "79+ ( 79( 79( (" 593) (" 79(" 59(( (" 793 56( 56( 76( 76( 56( "76( 56( "76( 56( 76(" 56( 76(" 76( 76( 56(" 56(" 5 6 79( 79(. 79(79(. "56-+ 563+ 56 Advanced Linear Programming Applications (*)3 $$$$$$$$$$$$$$ "5 5 5 "7 7 7 "5 5 5 "7 7 7 " 5 " 7 " 5 " 7 $$$$$$$$$$$$$$$ . ( 3 + ( - ! ! " # # $$$$$$$$$$$$$$$$$$ 3 + ( 3 ) 3 Note: The values shown above provide the allocations for the remaining seats available. The bid prices for each ODIF are provided by the deal prices in the following output. " # $$$$$$$$$$$$$$ ( 3 + * ) . ( (( ((3 (+ (* ( () (. ( 7. a. /0 1 # $$$$$$$$$$$$$$$ 2 # $$$$$$$$$$$$$$$$$$ + ) () --+ ()+ .* -+ 3) -)3 33(** 3(* ( * -. 3+ - ( + 3*. Let CT = number of convention two night rooms CF = number of convention Friday only rooms CS = number of convention Saturday only rooms RT = number of regular two night rooms RF = number of regular Friday only rooms RS = number of regular Saturday only room 5 7 Chapter 5 b./c.The formulation and output obtained using % : : ' % 4 --*" 6(-3" 6(3 " 6- * (8 (" 9+ -8 (" 938 (" 9(* +8 ( 9*8 ( 93 8 ( 9-* )8 (" 6(" .8 (" 6(" 8 (" 6(" ( 8 (" 6(" is shown below. 6(+ ;+. ;+. 6( 6( 6( 6( 6(*- 9 9 -*3(+ $$$$$$$$$$$$$$ " " " $$$$$$$$$$$$$$$ 3 ((* -. -* " # $$$$$$$$$$$$$$ ( 3 + * /0 1 # $$$$$$$$$$$$$$$ + . ! ! " # # $$$$$$$$$$$$$$$$$$ 2 # $$$$$$$$$$$$$$$$$$ -. +) * $-3 ) . 3 (+ ( - ( ',%" % " % $$$$$$$$$$$$ " " " " % :% < = $$$$$$$$$$$$$$$ (-3 * ( -+. (-3 ( - " $$$$$$$$$$$$$$$ --* (-3 (3 - * (+ (*- 5 8 11 = $$$$$$$$$$$$$$$ -*3 (+ 11 = 11 = ( 3 11 = Advanced Linear Programming Applications :> > 2 " # $$$$$$$$$$$$ ( 3 + * ) . ( d. 8. 2% :% < = $$$$$$$$$$$$$$$ 3 ((( (. -. -( + < = . 3 " $$$$$$$$$$$$$$$ + (* 3 -* +. +. 11 = $$$$$$$$$$$$$$$ 11 = 11 = -3 -3 11 = -. * *( . ( The dual price for constraint 10 shows an added profit of $50 if this additional reservation is accepted. To determine the percentage of the portfolio that will be invested in each of the mutual funds we use the following decision variables: = proportion of portfolio invested in a foreign stock mutual fund = proportion of portfolio invested in an intermediate term bond fund = proportion of portfolio invested in a large cap growth fund = proportion of portfolio invested in a large cap value fund = proportion of portfolio invested in a small cap growth fund = proportion of portfolio invested in a small cap value fund a. A portfolio model for investors willing to risk a return as low as 0% involves 6 variables and 6 constraints. Max 12.03 + 6.89 ! 20.52 + 13.52 + 21.27 + 13.18 s.t. 10.06 13.12 13.47 45.42 21.93 + + + + + 17.64 3.25 7.51 1.33 7.36 + + + + 32.41 18.71 33.28 41.46 23.26 + + + + + 32.36 20.61 12.93 7.06 5.37 + + + + + " " " + " 5 9 " ≥0 33.44 19.40 3.85 58.68 9.02 + + + + + 24.56 25.32 6.70 5.43 17.31 ≥ ≥ ≥ ≥ ≥ = 0 0 0 0 0 1 Chapter 5 b. The solution obtained using is shown. (. + $$$$$$$$$$$$$$ ! ! " # # $$$$$$$$$$$$$$$$$$ (3 - ) 3+) * (-* - $$$$$$$$$$$$$$$ ' : : " # $$$$$$$$$$$$$$ ( 3 + * *) 3+3 /0 1 # $$$$$$$$$$$$$$$ 3 3 . -( +-. -3 + +3. 2 # $$$$$$$$$$$$$$$$$$ $ 3 ) (. + The recommended allocation is to invest 65.7% of the portfolio in a small cap growth fund and 34.3% of the portfolio in a small cap value fund. The expected return for this portfolio is 18.499%. c. One constraint must be added to the model in part a. It is ≥ .10 The solution found using is given. () (). $$$$$$$$$$$$$$ $$$$$$$$$$$$$$$ ( ' : : " # $$$$$$$$$$$$$$ ( 3 + * ! ! " # # $$$$$$$$$$$$$$$$$$ 3+) * (-* - * . 3 /0 1 # $$$$$$$$$$$$$$$ -) -* -( .( 3 * ( 2 # $$$$$$$$$$$$$$$$$$ $ 3 ) (. + $(3 - ) ) The recommended allocation is to invest 10% of the portfolio in the foreign stock fund, 50.8% of the portfolio in the small cap growth fund, and 39.2 percent of the portfolio in the small cap value fund. The expected return for this portfolio is 17.178%. The expected return for this portfolio is 1.321% less than for the portfolio that does not require any allocation to the foreign stock fund. 5 10 Advanced Linear Programming Applications 9. To determine the percentage of the portfolio that will be invested in each of the mutual funds we use the following decision variables: = proportion of portfolio invested in a large cap stock mutual fund = proportion of portfolio invested in a mid cap stock fund = proportion of portfolio invested in a small cap growth fund = proportion of portfolio invested in an energy sector fund # = proportion of portfolio invested in a health sector fund = proportion of portfolio invested in a technology sector fund $ = proportion of portfolio invested in a real estate sector fund a. A portfolio model for investors willing to risk a return as low as 0% involves 7 variables and 6 constraints. Max 9.%& + 5.91 ! 15.20 32.3 23.2 0.9 49.3 22.8 + 20.8 + 22.5 + 6.0 + 33.3 + 6.1 + + 11.74 + 7.34# + 16.97 + 15.44$ s.t. 35.3 20.0 28.3 10.4 9.3 + + + + + + + + + 49.1# + 5.5# + 29.7# + 77.7# 24.9# + # + " b. 25.3 33.9 20.5 20.9 2.5 " " "# " The solution obtained using "$ + + + + + 46.2 21.7 45.7 93.1 20.1 + 20.5$ + 44.0$ 21.1$ + 2.6$ + 5.1$ + $ ≥0 is shown. (* *3 $$$$$$$$$$$$$$ $$$$$$$$$$$$$$$ ! ! " # # $$$$$$$$$$$$$$$$$$ * ) (( *+. * % > + -) ( * (+3 3*) " # $$$$$$$$$$$$$$ ( 3 + * /0 1 # $$$$$$$$$$$$$$$ -- 3--. * 2 # $$$$$$$$$$$$$$$$$$ $ -. ..) $ )) (* ) + 5 11 ≥ ≥ ≥ ≥ ≥ = 2 2 2 2 2 1 Chapter 5 The recommended allocation is to invest 50% of the portfolio in the small cap stock fund, 14.3% of the portfolio in the technology sector fund, and 35.7% of the portfolio in the real estate sector fund. The expected portfolio return is 15.539%. c. The portfolio model is modified by changing the right hand side of the first 5 constraints from 2 to 0. d. The solution obtained using is shown. (* ) + $$$$$$$$$$$$$$ ! ! " # # $$$$$$$$$$$$$$$$$$ * ) (( *+. $$$$$$$$$$$$$$$ -** % > + -) ( * -(*33 " # $$$$$$$$$$$$$$ ( 3 + * /0 1 # $$$$$$$$$$$$$$$ 3. 33 ).3 2 # $$$$$$$$$$$$$$$$$$ $ - *3 $ )) (* ) + The recommended allocation is to invest 25.5% of the portfolio is the small cap stock fund, 21.2% of the portfolio in the technology sector fund, and 53.3% of the portfolio in the real estate sector fund. The expected portfolio return is 15.704%. This is an increase of .165% over the portfolio that limits risk to a return of at least 2%. Most investors would conclude that the small increase in the portfolio return is not enough to justify the increased risk. 10. # 1 # $ 2 3 Minimum 1 8 5 7 5 2 2 4 10 2 8 5 7 Maximum The maximum of the row minimums is 5 and the minimum of the column maximums is 5. The game has a pure strategy. Player A should take strategy 1 and Player B should take strategy 2 . The value of the game is 5. 11. By definition, a pure strategy solution means each player selects a single strategy with probability 1. Thus, if a linear programming solution has a probability decision variable equal to 1, the game has a pure strategy solution. 5 12 Advanced Linear Programming Applications 12. The expected payoffs for Company A are as follows: Strategy 1 : Expected Payoff = 0.5(3) +0.5(2) = 2.5 Strategy 2 : Expected Payoff = 0.5(4) +0.5(1) = 2.5 Strategy 3 : Expected Payoff = 0.5( 2)+0.5(5) = 1.5 2 with If Company B were to implement strategy 0.50, Company A can select strategy 1 or in market share. 13. 2 probability 0.50 and strategy 3 with probability to obtain an expected payoff providing a 2.5% increase The row minimums are 5, 6, and 4. Station A prefers the maximin strategy 2 to obtain a gain of at least 6. The column maximums are 10, 8 and 7. Station B prefers the minimax strategy 3 to limit its maximum loss to no more than 7. However, because the maximum of the row minimums is not equal to the minimum of the row maximums, the game does not have a pure strategy. A mixed strategy solution with a value of the game between 6 and 7 exists. The linear programming formulation and solution for Station A follows. Max ' (' s.t. 10)'1 + 8)'2 + 4)'3 ' (' > 0 $ (Strategy 1) # 5)'1 + 7)'2 + 8)'3 ' (' > 0 (Strategy 2 ) 3)'1 + 6)'2 + 7)'3 ' (' > 0 (Strategy 3 ) )'1 + )'2 + )'3 = 1 )'1, )'2, )'3 > 0 + $$$$$$$$$$$$$$ ( 3 : $$$$$$$$$$$$$$$ " # $$$$$$$$$$$$$$ ( 3 + /0 1 # $$$$$$$$$$$$$$$ ! ! " # # $$$$$$$$$$$$$$$$$$ - + + 2 # $$$$$$$$$$$$$$$$$$ $ - ( $ The optimal strategy is for Station A to implement strategy 2 . + with probability 0.6 and strategy 3 with probability 0.4. Using the absolute value of the dual prices, we see that it is optimal for Station B to implement strategy 1 with probability 0.2 and strategy 3 with probability 0.8. The expected value of the game is 6.4. This is an expected increase of 6400 viewers for Station A. 5 13 Chapter 5 14. The row minimums are 15, 10, 25, and 10. The Republican candidate prefers the maximin strategy 4 to obtain a gain of at least 10. The column maximums are 30, 20, 10, and 20. Station B prefers the minimax strategy 3 to limit its maximum loss to no more than 10. The maximum of the row minimums is equal to the minimum of the row maximums. The game has a pure strategy. The Republican candidate goes to South Bend and the Democratic candidate goes to Fort Wayne. The value of the game shows a 10,000 voter increase for the Republican candidate. 15. a. The row minimums are 1, 3, and 4. Player A prefers the maximin strategy Red to obtain a payoff of at least 1. The column maximums are 5, 4 and 2. Player B prefers the minimax strategy Blue to limit its maximum loss to no more than 2. However, because the maximum of the row minimums is not equal to the minimum of the column maximums, the game does not have a pure strategy. A mixed strategy solution with a value of the game between 1 and 2 exists. The linear programming formulation and solution for Player A is as follows. Max ' (' s.t. )'1 2)'1 )'1 + + 5)'2 4)'2 3)'2 )'2 + + + 2)'3 3)'3 4)'3 )'3 ' (' ' (' ' (' > > > = 0 0 0 1 # $ # (Red Chip) (White Chip) (Blue Chip) )'1, )'2, )'3 > 0 * $$$$$$$$$$$$$$ ( 3 : $$$$$$$$$$$$$$$ ) 3 " # $$$$$$$$$$$$$$ ( 3 + /0 1 # $$$$$$$$$$$$$$$ ( ! ! " # # $$$$$$$$$$$$$$$$$$ ( * 2 # $$$$$$$$$$$$$$$$$$ $ $ * * * The optimal mixed strategy for Player A is to select a red chip with a 0.7 probability and a white chip with a 0.3 probability. Using the absolute value of the dual prices, the optimal mixed strategy for Player B is to select a white chip with a 0.5 probability and a blue chip with a 0.5 probability. b. The value of the game is 0.5. This is an expected gain of 50 cents for Player A. c. Player A is the preferred player. Over the long run, Player A average winning 50 cents per game. To make the value of the game 0 and thus, fair for both players, Player A should pay Player B to play the game. The game would be considered fair if Player A pays Player B 50 cents per game. 5 14 Advanced Linear Programming Applications 16. The row minimums are 0, 2, 2 and 2. Company A prefers the maximin strategy 3 to obtain a payoff of at least 2. The column maximums are 4, 6, 5 and 6. Player B prefers the minimax strategy 1 to limit its maximum loss to no more than 4. However, because the maximum of the row minimums is not equal to the minimum of the row maximums, the game does not have a pure strategy. A mixed strategy solution with a value of the game between 2 and 4 exists. The linear programming formulation and solution for Player A is as follows. Max * s.t. 3) 1 2) 1 + 2) 3 2) 2 + 1) 3 + 5) 3 4) 1 + 2) 2 2) 1 + 6) 2 ) 1 + ) 2 + + 4) 4 6) 4 1) 3 + ) 3 + * ≤ 0 # (Strategy 1) * ≤ 0 (Strategy 2 ) * ≤ 0 (Strategy 3 ) * ≤ 0 (Strategy 4 ) ) 4 # = 1 ) 1, ) 2, ) 3, ) 4 > 0 - . $$$$$$$$$$$$$$ '( ''3 '+ ' $$$$$$$$$$$$$$$ + " # $$$$$$$$$$$$$$ ( 3 + * /0 1 # $$$$$$$$$$$$$$$ ( 3 - ! ! " # # $$$$$$$$$$$$$$$$$$ ( - - . 2 # $$$$$$$$$$$$$$$$$$ . $- . The optimal mixed strategy solution for Company B is to select strategy strategy 2 with probability 0.4 and with probability 0.6. Using the dual prices, the optimal mixed strategy for Company A is to select strategy 3 with a probability 0.8 and a strategy gain for Company A is 2.8%. 17. 1 4 with a probability 0.2. The expected The payoff table is as follows: %$ $ & Run Defense Pass Defense Minimum Run 2 6 2 Pass 11 1 1 Maximum 11 6 5 15 Chapter 5 The Bears prefer the maximin strategy run to obtain a payoff of at least 2 yards. The Packers prefer the minimax strategy pass defense to limit its maximum loss to no more than 6 yards. However, because the maximum of the row minimums is not equal to the minimum of the row maximums, the game does not have a pure strategy. A mixed strategy solution with a value of the game between 2 and 6 exists. The linear programming formulation and solution for the Bears is as follows. Max '( '$ s.t. % 2)'1 6)'1 )'1 + 11)'2 1)'2 + )'2 ' (' ' (' ≥ 0 ≥ 0 = 1 $ # # (Run Defense) (Pass Defense) + -* $$$$$$$$$$$$$$ ( : '% $$$$$$$$$$$$$$$ )* -* + -* " # $$$$$$$$$$$$$$ ( 3 /0 1 # $$$$$$$$$$$$$$$ ! ! " # # $$$$$$$$$$$$$$$$$$ 2 # $$$$$$$$$$$$$$$$$$ $ +3) $ * 3 + -* The optimal mixed strategy is for the Bears to run with a 0.75 probability and pass with a 0.25 probability. Using the absolute value of the dual prices, the optimal mixed strategy for the Packers is to use a run defense with a 0.437 probability and a pass defense with a 0.563 probability. The expected value of the game shows that with the mixed strategy solution, the Bears average 4.25 yards per play. 5 16 1. Be able to identify the special features of the transportation problem. 2. Become familiar with the types of problems that can be solved by applying a transportation model. 3. Be able to develop network and linear programming models of the transportation problem. 4. Know how to handle the cases of (1) unequal supply and demand, (2) unacceptable routes, and (3) maximization objective for a transportation problem. 5. Be able to identify the special features of the assignment problem. 6. Become familiar with the types of problems that can be solved by applying an assignment model. 7. Be able to develop network and linear programming models of the assignment problem. 8. Be familiar with the special features of the transshipment problem. 9. Become familiar with the types of problems that can be solved by applying a transshipment model. 10 Be able to develop network and linear programming models of the transshipment problem. 11. Know the basic characteristics of the shortest route problem. 12. Be able to develop a linear programming model and solve the shortest route problem. 13. Know the basic characteristics of the maximal flow problem. 14. Be able to develop a linear programming model and solve the maximal flow problem. 15. Know how to structure and solve a production and inventory problem as a transshipment problem. 16. Understand the following terms: network flow problem transportation problem origin destination capacitated transportation problem assignment problem transshipment problem capacitated transshipment problem shortest route maximal flow source node sink node arc flow capacities 6 1 Chapter 6 1. The network model is shown. Atlanta 1400 Dallas 3200 Columbus 2000 Boston 1400 2 6 Phila. 5000 6 2 1 2 New Orleans 3000 5 7 2. a. Let 11 12 : : • • • Min 14 11 s.t. 11 Amount shipped from Jefferson City to Des Moines Amount shipped from Jefferson City to Kansas City + 9 12 + 7 13 + 12 + 13 + + 11 8 21 + 21 21 12 13 10 22 + + 22 + 23 + 22 + 23 11, 12, 13, 21, 22, 23, ≥ 0 b. Optimal Solution: Amount 5 15 10 20 Jefferson City Des Moines Jefferson City Kansas City Jefferson City St. Louis Omaha Des Moines Total 6 2 Cost 70 135 70 160 435 5 23 ≤ ≤ = = = 30 20 25 15 10 Distribution and Network Models 3. a. Hamilton 1 400 Butler 2 200 Clermont 3 300 10 Southern 1 500 20 15 12 400 b. Let Min s.t. 15 Northwest 2 18 = amount shipped from supply node to demand node . 10 11 + 20 12 + 15 13 11 + 12 + 13 + + 11 12 12 21 21 21 + 15 22 + 22 + 22 13 + 18 23 + 23 + 23 ≤ ≤ = = = 500 400 400 200 300 ≥ 0 for all , c. Optimal Solution Amount 200 300 200 200 Southern Hamilton Southern Clermont Northwest Hamilton Northwest Butler Total Cost d. Cost $ 2000 4500 2400 3000 $11,900 To answer this question the simplest approach is to increase the Butler County demand to 300 and to increase the supply by 100 at both Southern Gas and Northwest Gas. The new optimal solution is: Amount 300 300 100 300 Southern Hamilton Southern Clermont Northwest Hamilton Northwest Butler Total Cost Cost $ 3000 4500 1200 4500 $13,200 From the new solution we see that Tri County should contract with Southern Gas for the additional 100 units. 6 3 Chapter 6 4. a. 1 Pittsburg 3000 10 20 9000 2 1 5 Seattle Mobile 5000 9 10 2 10 2 3 8 4000 Columbus Denver 4000 30 6 1 3 8000 New York 20 7 4 10 Los Angeles 6000 4 5 Washington b. 3000 The linear programming formulation and optimal solution as printed by The Management Scientist are shown. The first two letters of the variable name identify the “from” node and the second two letters identify the “to” node. Also, The Management Scientist prints “<” for “≤.” 6 4 Distribution and Network Models ! ! ! ! ! " #$%&'()% *+,&'(-, ./0+% ! ./1(/#0% 44444444444444 c. ./0+% 444444444444444 %2+&%2 -3'3 444444444444444444 The new optimal solution actually shows a decrease of $9000 in shipping cost. It is summarized. Optimal Solution Units Seattle Denver Seattle Los Angeles Columbus Mobile New York Pittsburgh New York Los Angeles New York Washington 4000 5000 5000 4000 1000 3000 6 5 Cost $ 20,000 45,000 50,000 4,000 10,000 12,000 Total: $141,000 Chapter 6 5. a. 100 180 B 2 75 C 3 100 D 4 85 125 Avery 1 160 A 1 115 100 120 135 Baker 2 160 115 120 155 150 140 Campbell 3 140 130 b. Let = number of hours from consultant assigned to client . Max 100 11 + 125 12 + 115 13 + 100 + 120 24 + 155 31 + 150 s.t. 11 + 13 + 12 + 21 + + 11 21 + 12 13 14 + 120 21 + 135 32 + 140 33 + 130 14 22 + 23 + 31 + + 31 + 22 + 23 + 14 ≥ 0 for all , 6 6 22 + 115 23 34 24 32 + 33 + 32 + 24 33 + ≤ ≤ 34 ≤ = = = 34 = 160 160 140 180 75 100 85 Distribution and Network Models Optimal Solution Hours Assigned 40 100 40 35 85 140 Avery Client B Avery Client C Baker Client A Baker Client B Baker Client D Campbell Client A Total Billing c. New Optimal Solution Hours Assigned 40 100 75 85 140 Avery Client A Avery Client C Baker Client B Baker Client D Campbell Client A Total Billing 6. Billing $ 5,000 11,500 4,800 4,725 10,200 21,700 $57,925 Billing $ 4,000 11,500 10,125 10,200 21,700 $57,525 The network model, the linear programming formulation, and the optimal solution are shown. Note that the third constraint corresponds to the dummy origin. The variables 31, 32, 33, and 34 are the amounts shipped out of the dummy origin; they do not appear in the objective function since they are given a coefficient of zero. 6 7 Chapter 6 Demand Supply D1 2000 D2 5000 D3 3000 D4 2000 32 5000 34 C.S. 32 40 34 30 3000 D. 28 38 0 0 4000 0 Dum 0 Note: Dummy origin has supply of 4000. Max 32 11 + 34 12 + 32 13 + 40 14 + 34 21 + 30 22 + 28 23 + 38 24 s.t. 11 + 12 + 13 + 31 + + 11 14 32 + + 21 + 12 21 + 22 + 33 + 34 + + 24 31 22 13 23 + 32 23 + + 14 ≥ 0 for all 6 8 24 33 + 34 ≤ 5000 ≤ 3000 ≤ 4000 = 2000 = 5000 = 3000 = 2000 Dummy Distribution and Network Models Optimal Solution Units Cost Clifton Springs D2 Clifton Springs D4 Danville D1 Danville D4 4000 $136,000 1000 40,000 2000 68,000 1000 38,000 Total Cost: $282,000 Customer 2 demand has a shortfall of 1000 Customer 3 demand of 3000 is not satisfied. 7. a. 1 Boston 50 7 1 100 11 Denver 8 13 2 20 Dallas 70 3 Los Angeles 60 17 100 2 Atlanta 12 10 8 18 3 150 13 Chicago 16 4 St. Paul b. There are alternative optimal solutions. Solution #1 Solution # 2 Denver to St. Paul: 10 Atlanta to Boston: 50 Atlanta to Dallas: 50 Chicago to Dallas: 20 Chicago to Los Angeles: 60 Chicago to St. Paul: 70 Denver to St. Paul: 10 Atlanta to Boston: 50 Atlanta to Los Angeles: 50 Chicago to Dallas: 70 Chicago to Los Angeles: 10 Chicago to St. Paul: 70 6 9 80 Chapter 6 Total Profit: $4240 If solution #1 is used, Forbelt should produce 10 motors at Denver, 100 motors at Atlanta, and 150 motors at Chicago. There will be idle capacity for 90 motors at Denver. If solution #2 is used, Forbelt should adopt the same production schedule but a modified shipping schedule. 8. The linear programming formulation and optimal solution are shown. Let Min 1A 1B • • • = = Units of product A on machine 1 Units of product B on machine 1 3C = Units of product C on machine 3 1A + 1.2 1B + 0.9 1C + 1.3 2A + 1.4 2B + 1.2 2C + 1.1 3A + 3B + 1.2 3C s.t. 1A + 1B + 1C 2A + 2B + 2C 3A + + 1A + 2A + 1B + 1C 3A + 2B 1 1 2 3 3 A C A A B Units Cost 300 1200 1200 500 500 $ 300 1080 1560 550 500 Total: $3990 Note: There is an unused capacity of 300 units on machine 2. 6 10 3B + 2C ≥ 0 for all Optimal Solution 3B + ≤ 1500 ≤ 1500 3C ≤ 1000 = 2000 = 500 3C = 1200 Distribution and Network Models 9. a. 10 1 1 1 16 Jackson Client 1 1 32 14 2 1 2 22 Client 2 Ellis 1 40 22 24 3 1 Smith 3 34 Client 3 1 b. Min 10 11 + 16 12 + 32 13 + 14 21 + 22 22 + 40 23 + 22 31 + 24 32 + 34 33 s.t. ≤ 1 11 + 12 + 13 ≤ 1 21 + 22 + 23 31 + 32 + 33 ≤ 1 + + = 1 11 21 31 + 22 + 32 = 1 12 + 23 + 33 = 1 13 ij ≥ 0 for all , Solution 12 = 1, 21 = 1, 33 = 1 Total completion time = 64 6 11 Chapter 6 10. a. 30 44 1 Red 1 31 25 38 47 1 1 1 2 W hite 2 1 1 3 Blue 3 1 1 4 Green 4 1 5 1 34 44 26 43 5 Brown 1 28 b. Min s.t. 30 11 + 44 12 + 38 13 + 47 14 + 31 15 11 + 12 + 13 + 14 + 15 21 + 22 + 23 + 24 + 25 31 + 32 + 33 + 34 + 35 41 + 42 + 43 + 44 + 45 53 + 54 + 21 + 31 + + + 22 + 32 13 + 23 + 14 + 24 15 + 51 11 + 12 ≥ 0, + 25 21 + ⋯ 41 + + + 42 + 52 33 + 43 + 53 + 34 + 44 + 54 25 + 35 + 45 + 52 = 1, 2,.., 5; 55 51 55 + 28 55 ≤ ≤ ≤ ≤ ≤ = = = = = 1 1 1 1 1 1 1 1 1 1 = 1, 2,.., 5 Optimal Solution: Green to Job 1 Brown to Job 2 Red to Job 3 Blue to Job 4 White to Job 5 $26 34 38 39 25 $162 Since the data is in hundreds of dollars, the total installation cost for the 5 contracts is $16,200. 6 12 Distribution and Network Models 11. This can be formulated as a linear program with a maximization objective function. There are 24 variables, one for each program/time slot combination. There are 10 constraints, 6 for the potential programs and 4 for the time slots. Optimal Solution: NASCAR Live Hollywood Briefings World News Ramundo & Son 5:00 – 5:30 p.m. 5:30 – 6:00 p.m. 7:00 – 7:30 p.m. 8:00 – 8:30 p.m. Total expected advertising revenue = $30,500 12. a. This is the variation of the assignment problem in which multiple assignments are possible. Each distribution center may be assigned up to 3 customer zones. The linear programming model of this problem has 40 variables (one for each combination of distribution center and customer zone). It has 13 constraints. There are 5 supply (≤ 3) constraints and 8 demand (= 1) constraints. The problem can also be solved using the Transportation module of optimal solution is given below. Plano: Flagstaff: Springfield: Boulder: 13. Assignments Kansas City, Dallas Los Angeles Chicago, Columbus, Atlanta Newark, Denver Total Cost . The Cost ($1000s) 34 15 70 97 $216 b. The Nashville distribution center is not used. c. All the distribution centers are used. Columbus is switched from Springfield to Nashville. Total cost increases by $11,000 to $227,000. A linear programming formulation and the optimal solution are given. For the decision variables, , we let the first subscript correspond to the supplier and the second subscript correspond to the distribution hub. Thus, = 1 if supplier is awarded the bid to supply hub and = 0 if supplier is not awarded the bid to supply hub . 6 13 Chapter 6 Min 190 11 +175 12 + 125 13 + 230 14 + 150 21 + 235 22 + 155 + 135 33 +260 34 + 170 41 + 185 42 + 190 43 + 280 + 240 54 + 270 61 + 200 62 + 130 63 + 260 64 23 44 + 220 + 220 24 51 + 210 + 190 31 52 + 225 + 140 32 53 s.t. + 11 + + 12 + 21 + 31 + 13 22 + 23 + 42 + 32 41 + 51 14 + 21 + 31 12 + + + + 41 22 51 32 13 + + + 24 + 43 33 52 + 61 11 + + 34 + 53 62 + 44 + 63 + 54 + 64 + 46 61 42 23 + + 52 33 14 + + + 62 43 24 + + 53 34 + + 63 44 + 45 ≤ ≤ ≤ ≤ ≤ ≤ = = = = 1 1 1 1 1 1 1 1 1 1 ≥ 0 for all , 14. Optimal Solution Bid Martin – Hub 2 Schmidt Materials – Hub 4 D&J Burns – Hub 1 Lawler Depot – Hub 3 175 220 170 130 695 A linear programming formulation of this problem can be developed as follows. Let the first letter of each variable name represent the professor and the second two the course. Note that a DPH variable is not created because the assignment is unacceptable. Max 2.8AUG + 2.2AMB + 3.3AMS + 3.0APH + + + AMS BMB CUG + + + CUG BMB AMS + + + + 3.2BUG + OOO + 2.5DMS s.t. AUG AUG + AMB BUG BUG AMB + + APH BMS CMB DUG DUG CMB BMS APH + + + BPH CMS DMB + + CPH DMS + + + DMB CMS BPH + + DMS CPH All Variables ≥ 0 Optimal Solution: A to MS course B to Ph.D. course C to MBA course D to Undergraduate course Max Total Rating 6 14 Rating 3.3 3.6 3.2 3.2 13.3 ≤ ≤ ≤ ≤ = = = = 1 1 1 1 1 1 1 1 Distribution and Network Models 15. a. Min 150 11 + + 210 12 170 21 + + + 270 13 230 22 180 31 + + + 220 23 230 32 + 160 41 + 225 33 240 42 + 230 43 s.t. 11 + 12 21 + + + 21 11 13 22 31 + + + 31 + 22 12 13 + 32 + 23 23 32 + 41 + + 41 + 42 + 33 33 42 + + 43 43 ≤ ≤ ≤ ≤ = = = 1 1 1 1 1 1 1 ≥ for all , Optimal Solution: 12 = 1, 23 = 1, 41 = 1 Total hours required: 590 Note: statistician 3 is not assigned. b. The solution will not change, but the total hours required will increase by 5. This is the extra time required for statistician 4 to complete the job for client A. c. The solution will not change, but the total time required will decrease by 20 hours. d. The solution will not change; statistician 3 will not be assigned. Note that this occurs because increasing the time for statistician 3 makes statistician 3 an even less attractive candidate for assignment. 16. a. The total cost is the sum of the purchase cost and the transportation cost. We show the calculation for Division 1 Supplier 1 and present the result for the other Division Supplier combinations. Division 1 Supplier 1 Purchase cost (40,000 x $12.60) Transportation Cost (40,000 x $2.75) Total Cost: 6 15 $504,000 110,000 $614,000 Chapter 6 Cost Matrix ($1,000 ) Supplier 1 2 3 4 5 6 1 614 660 534 680 590 630 2 603 639 702 693 693 630 3 865 830 775 850 900 930 4 532 553 511 581 595 553 5 720 648 684 693 657 747 Division b. Optimal Solution: Supplier 1 Supplier 2 Supplier 3 Supplier 5 Supplier 6 17. a. Division 2 Division 5 Division 3 Division 1 Division 4 $ 603 648 775 590 553 Total $3,169 Network Model Demand Supply 450 1 P1 600 4 4 W1 8 300 8 C3 300 9 C4 400 4 8 5 3 5 W2 5 380 7 C2 6 6 3 P3 300 4 7 2 P2 6 C1 7 7 6 6 16 Distribution and Network Models b. & c. The linear programming formulation and solution as printed by The Management Scientist is shown. 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 4 5 4 5 4 5 4 5 4 5 4 5 ! ! ! ! ! ! " #$%&'()% *+,&'(-, ./0+% ! ./1(/#0% 44444444444444 5 5 5 5 5 5 5 5 5 5 5 5 5 5 ./0+% 444444444444444 There is an excess capacity of 130 units at plant 3. 6 17 %2+&%2 -3'3 444444444444444444 5 5 Chapter 6 18. a. Three arcs must be added to the network model in problem 23a. The new network is shown. Demand Supply 1 P1 450 300 7 C2 300 8 C3 300 9 C4 400 4 6 7 4 4 W1 8 4 8 2 P2 600 6 C1 2 5 2 3 6 5 W2 5 7 6 3 P3 380 7 7 b.&c. The linear programming formulation and optimal solution as printed by The management Scientist follow: 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 ! ! ! 5 ! 6 18 5 5 5 5 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 4 5 ! ! Distribution and Network Models " #$%&'()% *+,&'(-, ./0+% ! ./1(/#0% 44444444444444 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 ./0+% 444444444444444 %2+&%2 -3'3 444444444444444444 The value of the solution here is $630 less than the value of the solution for problem 23. The new shipping route from plant 3 to customer 4 has helped ( 39 = 380). There is now excess capacity of 130 units at plant 1. 19. a. 300 1 Augusta 7 100 150 6 NewYork 100 7 Philadelphia 150 8 3 Albany 5 5 2 Tupper Lake 5 Boston 7 5 3 4 4 Portsmouth 6 10 6 19 Chapter 6 b. Min s.t. 7 13 + 5 14 + 3 23 + 4 24 + 8 35 + 5 36 + 7 37 + 5 45 + 6 46 + 10 47 13 + 14 23 23 13 + 14 24 + 35 + 36 + 37 + + 24 35 + 36 37 ≥ 0 for all and c. Optimal Solution: Variable 13 14 23 24 35 36 37 45 46 47 Value 50 250 100 0 0 0 150 150 100 0 Objective Function: 4300 6 20 45 45 + 46 + 46 + 47 + 47 ≤ ≤ = = = = = 300 100 0 0 150 100 150 Distribution and Network Models 20. 3 1 Muncie 34 4 Louisville 34 4 8 Concord 3 9 Chatham 3 32 3 8 57 35 5 Cincinnati 28 9 5 7 Greenwood 44 2 Brazil 3 Xenia 2 8 6 6 6 Macon 24 3 A linear programming model is Min 8 14 + 6 15 + 3 24 + 8 25 + 9 34 + 3 35 + 44 46 + 34 47 + 34 48 + 32 49 + 57 56 + 35 57 + 28 58 + 24 59 s.t. 14 + ≤ 3 15 24 + ≤ 6 25 34 + 24 14 15 + 34 25 ≤ 5 35 46 + 47 + 48 + 35 46 56 + + 56 57 + 49 58 + 59 = 0 = 2 = 4 57 + 48 6 21 + + 47 ≥ 0 for all = 0 49 = 3 58 + 59 = 3 Chapter 6 Optimal Solution Units Shipped Muncie to Cincinnati 1 Cincinnati to Concord 3 Brazil to Louisville 6 Louisville to Macon 2 Louisville to Greenwood 4 Xenia to Cincinnati 5 Cincinnati to Chatham 3 Cost 6 84 18 88 136 15 72 419 Two rail cars must be held at Muncie until a buyer is found. 21. +3 2 10 12 6 +5 11 1 3 11 5 4 8 9 12 4 +2 The positive numbers by nodes indicate the amount of supply at that node. The negative numbers by nodes indicate the amount of demand at the node. 22. a. Min 20 12 + 25 15 + 30 25 + 45 27 + 20 31 + 30 42 + 25 53 + 15 54 + 28 56 s.t. 31 12 31 + = 8 15 25 + 27 36 53 54 53 + 54 + 56 15 12 27 67 ≥ 0 for all , 6 22 = 5 42 = 3 + 74 = 3 42 = 2 25 36 74 + 35 36 + 12 67 + 27 74 + 56 67 = 5 = 6 Distribution and Network Models b. 12 15 25 27 31 36 42 = = = = = = = 0 0 8 0 8 0 3 = = = = = = 53 54 56 67 74 56 5 0 5 0 6 5 Total cost of redistributing cars = $917 23. Origin – Node 1 Transshipment Nodes 2 to 5 Destination – Node 7 The linear program will have 14 variables for the arcs and 7 constraints for the nodes. 1 if the arc from node to node is on the shortest route  = 0 otherwise  Let Min 7 +6 12 +9 57 + 2 + 18 14 65 + 3 67 13 +3 23 +5 25 +3 32 +4 35 +3 46 +5 52 +4 53 +2 56 s.t. Flow Out Node 1 12 Node 2 23 Node 3 32 + + + Node 4 46 Node 5 52 + Node 6 65 + 13 + Flow In =1 14 − 25 12 − 35 53 + 56 + 57 13 − − 23 − − 52 =0 53 =0 =0 25 − 46 + Node 7 − 32 14 − 67 − 57 − + 35 − 65 =0 56 =0 67 =1 > 0 for all and Optimal Solution: 12 =1, 25 =1, 56 = 1 , and 67 =1 Shortest Route 1 2 5 6 7 Length = 17 24. The linear program has 13 variables for the arcs and 6 constraints for the nodes. Use same six constraints for the Gorman shortest route problem as shown in the text. The objective function changes to travel time as follows. 6 23 Chapter 6 Min 40 + 36 12 +8 45 +6 13 +8 23 + 11 54 Optimal Solution: 12 +6 46 =1, 32 + 23 24 + 12 24 + 12 46 =1 42 + 25 26 + 15 35 + 15 53 56 = 1 , and Shortest Route 1 2 4 6 Total Time = 63 minutes 25. a. Origin – Node 1 Transshipment Nodes 2 to 5 Destination – Node 6 The linear program will have 13 variables for the arcs and 6 constraints for the nodes. 1 if the arc from node to node is on the shortest route  = 0 otherwise  Let Min 35 +16 12 + 30 46 13 +15 + 12 53 + 18 23 +12 54 +30 24 + 39 26 + 12 32 + 15 35 + 18 42 + 12 45 56 s.t. Flow Out + Node 1 12 Node 2 23 Node 3 32 Node 4 Node 5 =1 13 + + 42 + Flow In + 53 + 24 + − 26 − 35 45 + − − 46 54 + 56 + Node 6 12 13 − − − − 35 32 23 − 42 =0 53 =0 54 =0 45 =0 24 26 − + 46 + 56 =1 > 0 for all and b. Optimal Solution: 12 =1, 24 = 1 , and 46 =1 Shortest Route 1 2 4 6 Total time = 69 minutes c. Allowing 8 minutes to get to node 1 and 69 minutes to go from node 1 to node 6, we expect to make the delivery in 77 minutes. With a 20% safety margin, we can guarantee a delivery in 1.2(77) = 92 minutes. It is 1:00 p.m. now. Guarantee delivery by 2:32 p.m. 6 24 Distribution and Network Models 26. Origin – Node 1 Transshipment Nodes 2 to 5 and node 7 Destination – Node 6 The linear program will have 18 variables for the arcs and 7 constraints for the nodes. Let 1 if the arc from node to node is on the shortest route  = 0 otherwise  Min 35 +9 12 + 30 13 + 20 43 + 15 47 + 12 14 52 +8 + 10 23 53 + 12 +5 25 +8 + 20 56 32 +9 57 34 + 15 + 10 74 35 + 20 + 20 75 +5 36 76 s.t. Flow Out Node 1 12 + Node 3 + 32 + Node 4 43 + Node 5 52 Node 2 13 23 + + Flow In =1 14 25 34 + 35 + 36 47 53 + 56 + 57 74 + 75 + 12 − 14 − − + − Node 6 Node 7 − − 76 − − 13 − − 23 32 34 − − 25 + 47 − 36 35 56 − + 52 =0 43 − =0 53 74 =0 75 =0 76 =1 57 =0 > 0 for all and Optimal Solution: 14 =1, 47 = 1 , and 76 =1 Shortest Route 1 4 7 6 Total Distance = 40 miles 27. Origin – Node 1 Transshipment Nodes 2 to 9 Destination – Node 10 (Identified by the subscript 0) The linear program will have 29 variables for the arcs and 10 constraints for the nodes. Let 1 if the arc from node to node is on the shortest route  = 0 otherwise  6 25 Chapter 6 Min 8 12 + 13 13 + 15 14 + 10 15 + 5 23 + 15 27 + 5 32 + 5 36 + 2 43 + 4 +3 46 + 4 54 + 12 59 + 5 63 + 3 64 + 4 67 + 2 68 + 5 69 + 15 72 + 4 +2 78 + 4 70 + 2 86 + 5 89 + 7 80 + 12 95 + 5 96 + 5 98 + 5 90 45 76 s.t. Flow Out + Node 1 12 Node 2 23 + Node 3 32 Node 4 + 43 + Node 5 54 + Node 7 63 + 72 + Node 8 86 Node 9 95 Node 6 13 + + + Flow In 14 + =1 15 27 36 45 + 46 64 + 76 + 96 12 − − − 13 − 59 89 − + + 67 + 78 + 68 + − − 69 70 − 80 98 + − 90 + Node 10 − 14 − 15 − 36 − 27 − 68 59 − − 70 + 32 − − 54 − 23 45 − 46 − =0 72 43 − =0 63 =0 64 =0 95 76 − 86 − 96 =0 67 78 69 =0 − − 80 + =0 98 89 =0 90 =1 > 0 for all and Optimal Solution: 15 =1, 54 = 1, 46 = 1, 67 = 1 , and 70 =1 Shortest Route 1 5 4 6 7 10 Total Time = 25 minutes 28. Origin – Node 0 Transshipment Nodes 1 to 3 Destination – Node 4 The linear program will have 10 variables for the arcs and 5 constraints for the nodes. Let 1 if the arc from node to node is on the minimum cost route  = 0 otherwise  6 26 Distribution and Network Models Min 600 + 1000 01 +800 02 + 2000 + 1600 23 24 03 + 2800 + 700 04 + 500 12 + 1400 34 s.t. Flow Out Node 0 01 + + Node 1 12 Node 2 23 + Node 3 34 + 02 13 + Flow In 03 + =1 04 − 14 24 Node 4 =0 01 − 02 − − − 03 − 04 − =0 12 − 14 − 13 =0 23 24 − 34 =1 > 0 for all and Optimal Solution: 02 =1, 23 = 1 , and 34 =1 Shortest Route 0 2 3 4 Total Cost = $2500 29. The capacitated transshipment problem to solve is given: Max s.t. 61 + + 34 + 42 + 54 + 12 13 24 25 61 ≤2 ≤1 34 ≤ 3 42 ≤ 1 54 ≤ 1 + 36 43 61 12 42 13 + 56 36 =0 =0 43 = 0 14 + 45 + 46 25 45 46 56 ≤6 ≤4 36 ≤ 2 43 ≤ 3 56 ≤ 6 12 13 24 25 14 24 14 ≤3 45 ≤1 34 46 54 ≤3 ≥ 0 for all , 6 27 =0 =0 =0 13 + 2100 14 Chapter 6 3 2 2 5 1 3 3 1 4 1 4 6 2 4 Maximum Flow 9,000 Vehicles Per Hour 2 3 The system cannot accommodate a flow of 10,000 vehicles per hour. 30. 4 2 3 5 1 3 3 1 4 6 3 5 6 2 11,000 2 3 31. The maximum number of messages that may be sent is 10,000. 32. a. 10,000 gallons per hour or 10 hours b. 33. Flow reduced to 9,000 gallons per hour; 11.1 hours. Current Max Flow = 6,000 vehicles/hour. With arc 3 4 at a 3,000 unit/hour flow capacity, total system flow is increased to 8,000 vehicles/hour. Increasing arc 3 4 to 2,000 units/hour will also increase system to 8,000 vehicles/hour. Thus a 2,000 unit/hour capacity is recommended for this arc. 34. Maximal Flow = 23 gallons / minute. Five gallons will flow from node 3 to node 5. 35. a. Modify the problem by adding two nodes and two arcs. Let node 0 be a beginning inventory node with a supply of 50 and an arc connecting it to node 5 (period 1 demand). Let node 9 be an ending inventory node with a demand of 100 and an arc connecting node 8 (period 4 demand to it). 6 28 Distribution and Network Models b. + 2 15 Min s.t. + 5 26 + 3 37 + 3 48 + 0.25 56 + 0.25 67 + 0.25 78 + 0.25 89 = 05 15 26 37 48 05 + 15 + 26 56 56 + 37 + 48 ≥ 0 for all and Optimal Solution: 05 = 50 15 = 600 26 = 250 37 = 500 48 = 400 67 67 56 = 250 67 = 0 78 = 100 89 = 100 Total Cost = $5262.50 6 29 78 78 89 89 ≤ ≤ ≤ ≤ = = = = = 50 600 300 500 400 400 500 400 400 100 Chapter 6 36. a. Let R1, R2, R3 O1, O2, O3 D1, D2, D3 represent regular time production in months 1, 2, 3 represent overtime production in months 1, 2, 3 represent demand in months 1, 2, 3 Using these 9 nodes, a network model is shown. 275 100 O1 200 R2 50 O2 100 R3 50 b. R1 D1 150 D2 250 D3 300 O3 Use the following notation to define the variables: first two letters designates the "from node" and the second two letters designates the "to node" of the arc. For instance, R1D1 is amount of regular time production available to satisfy demand in month 1, O1D1 is amount of overtime production in month 1 available to satisfy demand in month 1, D1D2 is the amount of inventory carried over from month 1 to month 2, and so on. 6 30 Distribution and Network Models ≤ ≤ ≤ ≤ ≤ ≤ 4 ! 4 ! ! c. Optimal Solution: ./1(/#0% 44444444444444 ./0+% 444444444444444 Value = $46,750 Note: Slack variable for constraint 2 = 75. d. The values of the slack variables for constraints 1 through 6 represent unused capacity. The only nonzero slack variable is for constraint 2; its value is 75. Thus, there are 75 units of unused overtime capacity in month 1. 6 31 ! # " " $ " " % &'() * () + , " - . ' () / & 0 1 # # # 1 1 1 !00 #0 #0 ≥0 " . ≤ ≤ ≤ ' () / 2 , 3 3 % 5 () 6 ! "! 78 # ! # 1 4 9 ! ! 0 0 . ! () / # % 9 ! " ! ! / 9 " 9! ' 4 9! 7 2 ' 5 % ' ( $ 60"#8 # ! # 1 4 9 !0 0 0 . 9 0" ! 9# ' # % !0 . ' 4 2 % # 1 () 6 ! 8 9 # 6!" 8 0 0 ! # % 4 ) - . () / 9 !" 9 ' # . 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() / # % 9 4 7%" 9 # !4 ' !! . !! / 9 % 7% 4 7%" 9# % 7% . ' 4 5 6 7" %8 % # ! 1 9 # % 7 0 0 ! . &'() . 9 1 " 9 7" 9# ' 1 #1 %9 #90 1 !9 !≤ !≤ !≤ !≤ !≥ 9% ' 1 7 4 7 ( ) - 4  = 0 ( !000 1 %000 1 0#00 1 !000 ! 1 4000 # 1 000 % 000 000 !000 1 1 1 #00 #00 #00 1 1 1 %000 !000 #000 1 1 1 000 ! #00 ! 400 ! 1 #000 # 1 000 # 1 !000 # 1 1 1 000 % #00 % 700 % " 5 " " !" #" % 9 0" $ ('?@5 9 ! 9 % 9 A 9 "#00 . 1 ≤ ? . , !90 7 ? ! ≤ 4000 ! % ≤ %000 % ! ≤ 4000 ! % ≤ %000 % !1 % 9 & # ! 1 4 % 1 000 ! 1 #00 % 0 ≤ ≤ ≤ 0"#00 "000 4" #0 ' 0 ( 9 ( ) "0 ; 1 1 - 1 ; 1 1 - @ 1 @ 1 @ 1 @ 1 C 1 1 1 ; 1 B 1 - 1 B B 1 1 B 1 C C 1 C 1 1 ; 1 - D D ≥ ≥ ≥ ≥ D ≥ D ≥ D ≥ 6 6 6 6 6 6 6 8 8 8 8 8 8 8 ! # % B ( 9 & # 1 # 1 1 1 # 4 1 # 1 1 1 " 0 ≤ ≤ ≤ # # !#0 #0 #0 ≥ 0 " . 9 %0 9 40 9 %0 A 9 ##!0 . E##!0 $ 6 8 " E ##0 9 E!00 1 E##0 1 E%00 $ 6 8" ) > 9 E##!0 0 ##0 9 E 770 , > , ≤ " # ≤ #0 ≤ !0 " F + - & # 1 # 1 1 1 # 4 1 # 1 1 1 0 !00 ##0 %00 # # # #0 !0 " . . ≥ 0= " & " " 9 0" $ 9 00 9 00 9 0 9 9 9 0 . A 9 ! #0 E! #0 . E %0 6 8 ≤ #1 # ) ≤ #1 # @ ≤ #1 # G )1 @1 G ≤ > , 00 ) & 1 4# @ 1 #0 1 1 1 ) %0 @ 1 G 2 9 ≤ ≤ ≤ ≤ %0 #1 # ) #1 # @ #1 # G " " ≥ 0 )" @" G 9 0" 5 $ + 9 #" 9 #" 9 0 ) 9 0" @ 9 0" G 9 A 9 #0 ≤ ≤ ≤ ≤ ≤ ≤ !#0 #0 #0 0 0 0 ' 5 ) 2 1 ? ( 9 5 A $ + 9 " 9 " 9 0" 9 0" # 9 0" # 9 0 9 7!0 $ $ ( " 9 0" ! 1 !≤ ? 5 A ! $ + ! 9 " ! 9 0" # 9 0 9 4%0 ( & ? ? ! # H : H : 5 - ' . ) . ! √ √ √ √ # B - √ 00 00 700 %00 00 000 00 !00 00 700 √ √ √ √ √ √ √ √ √ √ √ √ √ & √ √ ! # : - C 00 700 !00 00 00 00 000 00 %00 00 I ? H H ? H H ? H ? H - %0 %# 0 # # %0 -  = 0 ( # & 00 #00 5 # $ + 9 # 1 # 1 1 !00 1 400 !0 ! 1 # # 1 %00 ! 1 700 ! 1 1 00 # 00 # 1 1 400 1 !00 ≥ 700 . ≥ 700 B : 9  = 0  = 0 ( . 2 7# 1•••1 1 7% # . = - + + 1 1 , 1 1 1 ! • • • - + 1 1 % • • • 1 • • • 1 5 ≥ ≥ 1 1 ≥ 1 , 1 1•••1 9 . 7"000 ; 7"000 - . 1 1•••1 . %"000 5 - 9 0 %"000 ! # 0 ' ( ' ) = "0#4"000 F " %! 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" E707 7 9# 9# 9 . . = # # 8 % ! 4 0 7 % ! % % - ( 9 9 9 !9 #9 %9 9 )) )) )) )) )) )) )) 1 & 1 1 1 ( F / ; > & "0 "0 "0 "0 "0 "0 "0 1 1 ! 1 1 1 1 1 1 1 ! ! ! ! ! ! ! 1 1 1 1 1 1 % 1 % 1 1 1 % % % 1 ≥ ≥ ≥ ≥ ≥ ≥ ≥ 1 1 ; ; ; - 5 ! # % 4 1 ) 1%1 1 1 #1 %1 #141 1 % 01 01 !1 7 41%1 01# 1 1 01 71 1 #1 71!1 %1 0 7 # 9 9 9 9 9 9 9 9 ; # !7 % 4 7 0 7 #7 6( 6F 6/ 6; 6> 6& 6 8 8 8 8 8 8 8 . $ C K K "0 $ 1 1 1 1 1 1 1 1 1 1 5 K ) 9 9 ( & # # # 1 1 ( 4 # 1 # 0 4 7 7 ! 1 1 1 1 1 1 1 1 1 # % # 4 0 % ! 1 % 1 1 1 1 1 1 1 1 1 1 ! 7 % ! 1 1 1 1 1 1 1 1 "0 4 % % 0 % 1 1 1 1 1 1 1 1 % ! 0 0 # % 1 1 1 1 1 1 1 1 % ! 7 # # 1 1 !1 !1 !1 !1 0 !1 % !1 1 1 !1 !1 !1 !1 !1 !1 ! ! ! ! % 7 # 0 4 0 7 0 # !7 % 4 7 0 7 #7 ! ! ! ! ! 0 7 ! ! ! 1 1 ! % 1 ! 1 ! ! # % 4 ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ 9 9 9 9 ' . 9 " : 9 9 9 > . ! 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"#$%&'($ )*+%&',+ -/'-".$ -.*$ -.*$ 7 6%8 $A-/) 0J - ,+0&/-'+& .-%12 */3.*0 . $A-/; " 000 A-/-& A-/-& # ! "#$%&'($ )*+%&',+ -/'-".$ ,+0&/-'+& -.*$ -.*$ .-%12 */3.*0 0 % 6%8 . 6 8 . 2 ' ! A + " 5 ) " + 5 ( #+ # 1 1 ≤ 1 # 1 # 1 #! 1 #! 1 #! 1 ## 1 ## 1 #% ≤ 5 > : + > : + 1 # 1 1 1 ! 1 1 ! 1 1 1 6 8 1 6? H :8 1 6) 8 6 8 6> 8 6/ 8 6/ 8 6C 8 6$ 8 6G : 8 6. 8 6& 8 9 0" 1 1 1 1 1 1 +) 1 # 1 # 1 ! 1 # 1 % 1 1 1 1 1 1 1 ! ! ! ! ! ! ? +? + 1 1 1 1 1 1 # # # # # # 1 1 1 1 1 1 "$ + ! ? ≤ 1 1 # 1 % 1 % 1 % ≤  =  0 ( ? 1 H :" % % % % % % % 1 1 1 1 1 1 1 4 1 7 1 1 1 1 1 4 4 4 4 4 7 7 7 7 7 1 1 1 1 0 0 1 0 1 0 1 . H :" / . "$ + 1 1 1 1 0 1 ") . "C . 1 1 1 ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ! ' "#$ % & ( * + + . & & . . 2 ) , ! / ,& , / 0 1 ! . 1 3 4 5 % ! 6 ! ! 0 Management Scientist ! ! , , 4 "#$ Microsoft Excel5 & The 1 ( "#$ % & 4 X7 4Y7 "#$ % & 4 X7 4 8 ( . X: Y 8 ≤ 8 4Y789 4 ! ; ! " ! 8 9 ! ! 9 ! ; 8 * 7 T 7 '<= . 8 > '8? 7 '< 8 7 ' ' * 7 P/T 7 '8< ' 7 88 * 7 N 7 P/T N = P<= '<= ( "#$ q 7 9'8 > q 7 '88 : 8 > 2?? = = % & T7@ = 888? 1 $ & ! 2 N7 8! '8 '=B888? 7 88 ! 2 ! '= 888? : A=B888? 7 1 $ 8 > P)P< ' 7 P – = < '?P 7 c 7 84888 : '88q 7 84888 : '88= '8? 7 4 '4888 7c 7 84888 : 888q 7 84888 : 888= 88? 7 84888 7 G 7 p q + p q −c −c 7 888= '8? : B888= 88? > 4 7 4 '4888 '4888 > 84888 7 G 7 p q + p q −c −c q q ! ! ! G G = p =9'8 − ' p + A p ? + p = '88 + p − ' p ? − = 84 888 + '88=9'8 − ' p + A p ?? − = 84 888 + 888= '88 + p − ' p ?? = − ' p − ' p + p p + 888 p + '8 p − 4 B'4 888 p p P 7 98 . # ( "#$ q 7 % & p 7C A ' ! 8 . p 7CB A' . G 7C 49 . - 0 + = ? . q 7 ' ! 4 A' B + p q + p q −c −c st c = 8888 + '88q c = 8888 + 888q q = 9'8 − ' p + A p q = '88 + p − ' p . 0 4 3 . ! . ! . = ? . = ? ' 3 C 888 C 888 S = − R − 8M − RM + R+ = − = ? − 8= ? − = ?= ? + ! M = ?+ =? = & C 4888 ! - ! 3 - − R − 8 M − RM + R+ M R+M ≤ ( B "#$ 6 "#$ + 7 & 4! . "#$ ! 8 8 ! C 4'88 C A4888 # D 8 % C'88 ! 4 ! "#$ A% 8 . X7 9A Y 1 $ "#$ E "#$ ! ? X78 # A )% . E$ & $ Y7 , & B B $ & 3 4 ; & ! =! B '' - "#$ - + ' L ' C A' 'L + A'C ≤ A'888 L4 C ≥ 8 ( "#$ % & ; ! . 4 A'8 " % E8 L 7 A'8 C 7 A'8 ! & 1E8 - '8L + 88C st 8 L8 8 C 8 A8 = '8888 L4 C ≥ 8 ( B "#$ % & 4 L7 C A 48 B 9 " % ! 9 E8 1E8 . OT − x + x − x + C7 ! & - x + A88 − 'OT st x + Bx ≤ + OT x 4 x 4 OT ≥ 8 ( 8 "#$ % x 7 BBBBA " X & 4 OT 7 x 7 BBBBA 88888 A ! , Y 6 4 - X − X +'+Y + Y + st X +Y = X 4Y ≥ 8 ( "#$ ; % & A' 4 F7 A' G7 ' ! , # - D C'88 " ! ! ! ! " ' " #$ #$ & #$ #$ #$ !% &" " ! ! . " #$ #% !$ . #% #% ' #% " #% " #% 4 "/ 7 '88 "#$ !$ !$ & !$ & & !$ ' !$ !% !% !% !% !% ' " " ( H , I . "#$ / ) #)*+, )-./0+, 1),2./)3 4)235 6789*./:9 :+,29; <).+, 1),:9= /.9=+./)31; )59, </.,9; >6? <@ %+=/+7,9 "& %+,29 ' "& ' "' " #$ #% !$ !% & 952*95 A)1. ' "' % & ' " & & ! "#$ KDH%% J " ' " '& & "& " ! ' & " + 1 / ! " % & / & α . t: 4 1 1 Ft + = α Yt + = − α ? Ft ! t " 1 % B 0 . 4! , ! & 4 9 'B α "#$ 4 ' / " & # B # B # B "#$ " "#$ 3 , 4G 7 A 4 " & ' " * 0 D B B B B B B B B " 8 A 6 G # # # # # # # # " & ! % B . " & ' t ' " & B & ' ' # - + " ' & " & ' & & "#$ * 26* 7 8 A 6 G , ! ! # $ L 9A L 9 ' L 99 8 8 L 88 '9 L 8 8 L 8 L 8 L 8 L 8' L 8B & 3 # B " $ ! # $ A 'A 8 A 9 B B 8' A ' 8 8 8 BBB A B 8' 8 B 8 6 #; < <# C B 8 BA 9 'A )<<!!! >6? <@ % A @ 6 <B 6 < 6# 6 ' " & C !A ' C !A & C !A ' C !A &" C !A ' 6 <D # " 6 " 6 A# <D # " 6 & 6 A# '' 6 A# <D # 6 <D # 6 6 A# <D # " 6 A# A 8 '8B 8 99'B 8 8 ' A8 8' 9 8 B 8 8 ) 1&" ) A 8 8 ' A AA A' ' 8 8A 8 B ' 8 '' A 8 A' 8 " 8 89B 9 B ' " 89 9 B " " 8 8 B 8 8B 8 88B' 8 9A 9 8 89 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# C !A ' C !A &" C !A 6 C !A 6 '' 6 6 A# <D # 6 A# 6 <D # " # " C!A " ' C D!< # C !A 6 A# <D 6 " # 6 & # 6 ' # C!A + <D # C!A - ' 6 A# <D 6 A# " <D 6 A# & <D '" ' ' D## % !< 6 A# 6 <D ! <B 6 A# ' D<D # D ! $ < % " & ' #)*+, )-./0+, 1),2./)3 4)235 6789*./:9 :+,29; <).+, 1),:9= /.9=+./)31; )59, </.,9; <AB %+=/+7,9 $ !E 6 < AB <D %+,29 " & & " '" " " ' " '"" & " & & & """ " " " & ' # 6 A# B & ! 952*95 A)1. " ! ? D1P . ! ! D 4 = 4 ! ! D1N D P − DIN = R − RBAR 2 # 0 . 0 4 ; . 9 4! "#$ - 1 < <# C C @ >6? <@ ? <B ! <B ! % <B A< ! C A A ! #$ & " #$ " " & #$ #$ ' ! " #$ % !< < # % #! !D <6 #$ #% !$ !% A< <D ! ! ! ! C % C <D " <B #% #% ' #% #% " #% " !$ !$ & !$ & & !$ ' !$ !% !% ' " !% !% !% " D ( C C !D > 6! < % <B <D $ < % % % # ! D < 6 !< A< #)*+, )-./0+, 1),2./)3 4)235 6789*./:9 :+,29; <).+, 1),:9= /.9=+./)31; )59, </.,9; 6 " & >6? <@ ? <B ! %+=/+7,9 % A %+,29 952*95 A)1. & ' " &" ! ' ' & &" #$ #% !$ !% & & ' & " &'" " & &" & " & ' " & ' " "" 8 " # . B9 M ! 4 - BM + 4B M 4 ! ' 9M A % * & . ; ! 4 BA ! - N H , + , " σ 99M − σ − − σ σ * ! ! 99M ! * + H , , N - - "#$ # % & 4 N B H , & " " " #$ , ! ! ! 99M #% !$ #$ #$ & #$ #$ #$ !% " #% #% ' #% #% " #% !$ !$ & !$ & & !$ ' !$ !% !% ' " " D %+ ! D ? 4 !< 6 6 8 ! " ! σ . − M . K2&#= O? 7 8 8 4 O7 = H , > σ − %+ ! ! " ! ! ' ! ! M N 4 "#$ # H+& "&.= O? 7 8 8 . M . @ @ !< %+ <G9 1),2./)3 /1 #)*+, )-./0+, 1),2./)3 4)235 6789*./:9 :+,29; <).+, 1),:9= /.9=+./)31; & ' !% !% !% 1 )59, </.,9; >6? <@ %+=/+7,9 % ! %+,29 & ' & " "' #$ #% !$ !% " & ' 952*95 A)1. ' '& & N H + H ,4 ; - & '& ' ' " & " &'' !< ' " D 6 @ # 4 N - + " , H , " ! σ =+ − ; - N σ σ− ! 4σ + " ! + σ =+ − D - . / - ,& % + σ =− + σ ! C A A ! 6 < 6 C <B ! C % C ! = − ++ σ− σ σ & + 6 A ## 6 < 6 @ < ! @ ! < C # A> !AB6# ! D# < ! C!< > 6 A ! A &" CAD < !<6A> A C # % A C # !< % < 6 ' C # !> < <B 6 <B < ## <6 <D < ! #66> < <B A#6! 6 D$D!< D < # A#6! 6 ! < ! 6 C< C? < C E$ ? ># %6# < # < C 6 # @ !< @ % < #6$ ! ! % < 6 < ! # ?% % ! - + "' " ?% % @ C C !> %+=/+7,9 A ! @ %+,29 " ' &" " "' ' " < ! "&& ' " "' " % ?% . 9 ! . ,L & 8< D 8 , " , ! AB < > D!< & !! $ C< > C< > C< > , C F"L 7 , % . C % AB #6 $ . 6A> 0 , 1 , . 8 ; 6A> , , D!< !! $ C 6A> C 6A> C 6A> < , > , ,4 . B 6 ! 4 , 1 4 4 ! , , . 8 4! ! 88P 8PF PF & = 88P 8? , ! ; , * 4 88 88 F 7 , F ; & & <G9 1),2./)3 .) .G/1 0)59, /1 $,)7+, )-./0+, 1),2./)3 4)235 6789*./:9 :+,29; H.93595 1),:9= 1.9-1; <).+, 1),:9= /.9=+./)31; . 7 . ; & & , , & & & 1 %+=/+7,9 . %+,29 , , 4 4% "#$ 8 . ; & 952*95 A)1. , ! - , , , , . & * ; ! , - + - , 0 8 . ) + =A'Q + 'Q + 88? + = 'Q + 'Q + '8? Q +Q = 8 Q 4Q ≥ 8 . * "#$ C '8 8 & & ; ! ) & ' C A #AD# < " & ' F F F F F F F F F C A #AD# < ! ! ! ! ! ! <B A A A A A A A A" A& F A' F A F A 6 A !< A A A A A A A A" A& A' A A <B 6 A !< AB 6 '8 . ' - " 8 C ; . % & 6 AB 6 C '88 # " & ' " & ' - + !" !& !' ! ! ! C ! # ! ! " !" < ' ! !& C AD D# < % A A A '' A ' C !< ( C " ' ! &" !' ! # ! " ' A" < ? <B 6 " C ##6? ?6 6 K<C $6 F ! ! ' " ! ! ' ! ' ! < A " A& < # A !< & & " ' A A' ' " ' " A A & ' !< 6 <6 & 6 & ! 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( ; ; ? ? , ( " $ ; . , 6 4 7 8 9 5 ! ." ? @ .! $ A ; * B 3 # + 3 0 6 4 % # * C " ( 1 / . . , % % , " . ; " , ;; ;; 1 ;; 1 1& / / . ;; /, ; / . / 20 64 " " 2 " ;; D 1 1& D . ;; < & 1& " 2; D; ,D; , < ; (/ C ; ; ; (( ; , ; , ; 3 4 8 5 9 0 6 7 * 3 # + 3 0 " 4 7 8 9 5 ; ;; ; ;; ( 1& ( 1& ( ;; ; 1& & ./ & ./ ./ 6 4 7 8 9 5 ; ;; ; 1& , ./ ( 1& / ./ ./ & 1& & ./ ./ 4 ( 1& ( ;; ; 1& & ./ 1 ;; 1& ,; ./ , ./ 23689 % @ A < 3D 6D 8D 9 < ( 1& D 1 1& D 1 D ( < , .. + σ < σ3 D σ6 D σ8 D σ9 < ; , D ; , D ; (( D ; " = ,− @ A σ = % , − , .. ;, C ( 1& ( ;; . ;; 1 1& ( ;; ;; / 1& 1 ;; ( ;; 6 3 # + 3 0 < 4 ;, ) = −; ., . ; , ; (( ; ; , ; (( ; ;. ; (( ; 4 ( 1& ( 1& ./ & ./ ./ & ./ ./ ./ , ./ 5 $ ; ;; ; 1& ; 1& ; ;; & ./ ; 1& ; / ; ;; ; ;; 3 # + ? ? ? ? 5 4 "3 *) % % + , < ; ., + 2 ; ( < ; ,;;; ; ( < ; ,/1 & * 3 # + 3 0 " , & / . / & ; 1 6 4 7 8 4 " C ; ; ;& ; ; (( ; (( ; ; (( /1 % " .) 3 6 4 8 @ A < ,D.D&D1 < σ < ; < D ; (( D ; " @ A σ < D /1 < (( " (( 4 "3 " < < ; .( @ * 3 ; &1 ! $ A < ; ,;;; ; &1 < ; . % " < <; @ " < , <D @ 3 ; (/ . ! $ A < ; ,;;; D ; (/ . < ; / . ( 3 # + 3 0 6 4 7 8 * " 1 / , ; , / / 3 ; ;;;; ! $ A < ; ,;;; C /1 ; ( ;; ; (( ( ;; ( ;; /1 /1 3 # + 3 0 5 6 4 7 8 23 % 5 ; 1 1 1 ; ; &; (, @ A < 3D <1D 4 4 ; ( 1 , ; 1 , ; & &; &; (, (, , 5 1 $ 3 # + ? ; . ; / ; ; ; &; &; (, , ? ? ; ; ? ? 78 D D 7D 8 D ;D ,D/<, ! $ @ + A σ = σ +σ +σ +σ +σ < /1 D ( ;; D ( ;; D " = (( − @ A = % (( − , σ =− + < ,/ − @ A = # + $ % < ; ,;;; ; (1 . < ; ; /( % ,/ − , ( &( 4 "3 % % ; (1 . (( ! $ σ 9 ! ) *) " = /1 < ( &( ( &( 4 "3 % % /1 D ) = & *) + < " & ; (;.. ,/ ! $ < ; ,;;; ; (;.. < ; ; &( $ + " % &" " " 3 0 " )6 # + ;" 0 " $ + ! , 4 5 9 6 7 1 8 4 5 2 % 3 # + 3 0 * " C ; ;& ; (( ; ; ;& ; ;& ; ; (( ; ; ;& & 6 4 7 8 9 3 # + 3 0 ( ( 5 5 ; 4 4 , , & , ; 6 & ; , . , 4 7 8 9 5 ; ( . 1 & , & & , & & , 230 789 < D&D(D(D < ,! $ C σ < ; ;& D ; (( D ; (( D ; 4 "3 . *) ! @ A < σ < ,D && < D ; ;& < ; % % , < ;, ;, < ;, + < D && && /( ! $ 367: @ A <.D,D&D < . σ < /1 D /1 D ; , D ; 3 <& 4: @ A <.D&D D < & σ < /1 D ; D ; ;& D ; < ;& $ ; ; , , ; ; ; 3 # + ? ? ? ? ? 089: @ A < D(D D < ; σ < ; (( D ; . D ; ;& D ; % < / 367: < 3 ; & . < ; 3 < ; (/1& D ; ,;;; < ; /1& 4: < ; & < ( ;& 3 *" + ;;;; 3 *" + ;;;; 089: < ; ; < 1 1/ / +! " 9 " +% # # ! # % $) # + " " + % % % % + " " # + # % % + " " # # = - 0 % * 6 306 " <(,D1;D.;< 1,! $ < E&;;; D E,;;; < E1;;; 0 @B " " A&D,D D( < < E1;;; D (@E(;;A < E )1;; - @ " " A1D D.D < E1;;; D &/@E(;;A < E )1;; 0 ,; < E1;;; D 1 ,@E(;;A < E ,)(;; % % + " ! * * ; ( + < &/ " + " 4 %+ " " # " ) ) " 3 % " / 5 % % + + % 5 2 σ = & (/ = 1. 0 < E .)1;; < E1);;; D 6 + @E(;;A (;; 6 + < .)1;; 1;;; < 1)1;; 6 + < " % % % " + + < @ F A − 1, = 11 1. = 4 "3 *) 3 < ,;;; (. < ;&; 1 7 6 5 3 8 0 9 4 4 % * 3 # + 3 0 " 6 4 7 8 9 3 # + 3 0 6 4 7 8 9 * 5 ; ;; / / / / / / / / / & / 230 6789 " C / . ;; ( ;; ;; & ;; ;; ;; ;; ;; 5 ; ;; / & / / / ; / / / / & / " < ; ;& ; (( ; (( ; ; ; ; ; ; ;; 4 4 / , ; & & ( / / / / / / / / / /D.D / / & & & ( D D 5 $ ; ;; ; ;; ;; ; ;; & ;; ; ;; ; ;; ; ;; ; ;; / / / / / / / / / D < ( /! $ 3 # + ? ? ? ? ? ? " % % + " &! $ σ < σ3 D σ0 D σ6 D σ7D σ8 D σ9 < ; ;& D ; (( D ; & < @ A σ D; & < D; ( / < ; 1; D ; ;; < ; 1; & 3 @ & ! $ A < ; ,;;; ; (;( < ; ; , ; (;( - ! % % +) " * 3 # + 3 0 4 7 8 9 : ; . 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" & , ; G ../ G .;; &,; (,; &.; G E )( / ; 6 & & 4 7 8 3 # + 3 0 6 4 7 8 3 1 & ; 5 5 4 ; ; & , 1 ; ; & , 1 , & , & 4 5 & , , 1 & , , 1 & & . & & . $ ; ; ; ; ; ; ; ; 3 # + ? ? ? ? ? ? ? ? # 3 # + 3 0 5 5 ; ; & 6 4 7 1 1 ; 4 4 ; & & & 1 ; 1 / ( ; 5 & & 1 1 ( ( ( $ ; 3 # + ? ; ? ; ? 5 23 " % " < 3D D <&D,D.< ( + < E1)(;; * 3 # + 3 0 6 + 6 4 7 3 0 .;; 3 D ≥ & 3 D 0 3 D 6 0 D D D 4 7 D (;; D (;; 6 D ,;; D (;; ≥ , ≥ , ≥ . D 4 0 ≥ D 6 D /;; 6 + .;; /;; (;; (;; ,;; (;; ,;; 6 ≥ 4 4 6 7 4 D ≥ ≥ ≥ ; 49> 49> ≥ . 7 ≥ ; 49> ≤ ≤ 3 0 ≤ ≤ 6 ≤ ≤ ≤ 4 7 3 ) ≤ ≥ ; % 3 # + + + E(;; ,;; E ;; , 4 D (;; 7 2 <> " D < E1)(;; D E ;; < E )&;; & % " " # # " ! $ 5 " ) % "" 9) # " " ! # + 9) " " 2 9 " ! " " # " 9! 3 # + 3 3 ≥ τ3 0 ≥ τ0 3 ≥ τ 0 6 6 3 ≥ τ6 3 ≥ τ 4 4 7 7 ≥ τ4 6 ≥ τ7 4 ≥ τ7 7 8 0 ≥ τ8 ≥ τ8 8 8 9 7 ≥ τ9 8 ≥ τ9 9 9 3 ≥ ; ( 0 3 5 4 4 6 % 3 # + 3 0 6 4 5 5 ; ; 1 ; / 1 4 4 ; ; 1 5 ; 1 1 / / 1 1 & . ; 1 1 1 1 & $ ; ; ; ; 5 3 # 3) 0) ) 4 * 3 # 3 0 > "% " E (; &; ; ; , , E 5 4 3 5 ; 1 ( 1 ( & 6 & ! $ - $ 6 3 # + 3 0 " 2 4 ; 1 ( 1 ( & 4 1 ( & ( & . 5 1 ( & ( & . $ ; ; ; ; ; ; # E ),;; @ A , < < < < < D 3 3 3 D D 0 3 ≥ τ3 0 ≥ τ0 D 6 6 D D 4 7 8 9 : : 4 D D D 49> 0 6 6 D ≥ τ6 ≥ τ ≥ τ4 ≥ τ7 7 D D ≥ τ6 0 D D 9 3 6 D D ≥ τ 0 D 6 0 0 # + ! $ *" " " ! $ # + > "% ! $ # + " " # + > " " " # + ≥ τ8 8 9 6 ≥ τ9 9 7 ≥ τ9 : 4 ≥ τ: : 8 ≥ τ: 9 ≥ ; / D 4 4 D 7 7 D 8 8 D 9 9 D : : 49> : ≥ ; 49> ≥ τ: # % " " # % " +3 ≤ 3 4 ≤ 0 0 7 9 " & 6 ≤ 4 7 ≤ 8 8 ≤ 6 ≤ 9 9 ≤ : : 6 2 *" " ≤ ≤ ≤ % @5 " # +@ # +@ A " " @ A 1 A % " A !" # $ % & ( ' ) ' ! * + + , & - $ ' ' ' ' . / ' & ' ' ' ' ' ' 0 1 ' !" ' ' ' ' 2 ' r = dm = T= #, DCo = Ch Q3 = #, ( (" = - ( Q3 ( %#. ." = = # %# #, D TC = QCh + 5 6 QCh = D Co = Q 2 ' 8 4# . ,% D ( = ( ( ( "# " =% = (" 8 r ≤ Q3+ ' r 8 dm 8 9 " = 4# . ,# " = ,% # DCO = Ch r 8 dm 8 9 #, %#. . 2 Q3 = d= (" #" + (" #" = 4 ,% # %#. ." D #, Co = %#. . Q # %#. ." 2 5 7 " " = %#. . ( #" (" 8 # r ≤ Q3+ ' r 8 dm 8 # (" 8 ( : r 8 dm 8 ( 8( % 8 / 8/ 7 % "8 Q3 8 % 7 %(" 8 / : Q3 8 % 7 : ' + DCo = Ch Q3 = % r = dm = /( %(" + QCh = 6 ( 2 < Q8 = . /( %(" + /( %( D Co = Q 7 /( %( (" = % ( ( Q3 ( = D T= " (" = ( " " " ( " = 4 -. ., (" = -# ., 8 4(%- - TC 8 = " " 7 ( "> + = " (" 8 ( > # Q8 4(( ' r = dm = , ' + ( ( "= % D 8 ) 8 Q3 = DCo = Ch " % % "- " = ( /( ", " QCh = 6 0 ( /(" " , " = 4 + (# D % - " = 4 + (# Co = Q ( /( 7 2 D=Q 8 T= 8 4 + , % = ( /( 8 ( % ( Q3 ( % " = = ,, ( /( D # 8 4(( 4(%- - 8 4 . 7 ; 2 DCo = Ch Q3 = - @ I? 7 DCo = I ? C DCo = IC Q? = 5 . "=- A 2 2 6 Q3 = D 8 (= 8<) 2 8 4 +, = " 84 " , " + " = /+ " DCo = Ch 0 "8, + " 8 4 /+ -# * D = Q3 8 , = 8 ( 8 QCh = A " /+ 8 D = Q "CO = , = 2 7 84 Q3 = ' I I? = I 3 Q I? ∴Q ? = / DCo I ?C Q? Q? Q ?= Q3 = . DCo IC DCo = Ch r = dm = ( r = dm = ( 9 ( ( "= + ( (" ". " =% " "= % #(" = - + ' 7 ' - 7 7 * ' 6 % 8# + ' ' + (+ Q3 8 % r8- " " = 4 (+ ) +, : + 7 ' % : T= " ( " DCo = − D = P "Ch Q3 = % % Q = D B   −%  " " = % % % = =- - Q3 8 Q3 8 Q3 8 Q3 8 P 8 .+ P8 + P8# + P8 + !; -./ ### ./% . - 1 ,% DCo = Ch Q3 = " ,% " "  ,%   − P    DCo = − D = P "C h Q3 = = % % = . . Q P= 8    ; " " =. 9 * Q3 B 5 ! Q3 Q3 + P D+ C + Ch +' ! Q3 Q3 = DCo = − D = P "Ch 2 TC = B TC = 9 # = ( ∴;   −.  "# " =   ,  Q8( D  D  − P  QCh +  Q  Co =     7   −.   (  Q3 8 2   −.  1    , "+ , "+ 1 # =, 4# = 8 C Q3 8 ( +, =4 ( # =# + =4 ( 2 ' D - DCo = − D = P "Ch Q3 = # ( Q ( -. = D - B " = -. ." % ( " = -. 8 , ,. = #- %# -. Q = = P= ( ( = ( 8 ; ) -. :  D   − P Q =  − (    6    " = -,- , -. 2  D  − P  QCh =   = 7 2   − (     D Co = Q -. 8 4 + C8 C?8 ." % ( " = 4 " ( "=4 -% -% (" = %# #C DCo = − D = P "Ch Q3 = DC =( − D = P " IC Q? 8 * DCo − D = P" IC ? Q? = DCo − D = P " IC ? Q? = Q3 Q? 8 -. # %.  D  r = dm =  m = (  (  %    8 D = Q3 8 - ' T= " ( "   − (  DCo − D = P" IC / - Q3" 8 / = C? = C = C? C = #C C -( " 8 %( / , # = / - : DCo  Ch + Cb   = Ch  Cb  Q3 = ( '  Ch  S3 = Q 3  =  Ch + Cb  T= ( Q3 ( = D 6 %. / " & 1 2 1 4( = # /% % !  r = dm =   ' % 4(%- - 5  ( = %  ( ' ' r 8 dm S 8 % % %( 8 7 - %% %, S Cb = # -% Q 7 @ % %( D Co = , Q 1 , %. / Q − S" Ch = 4 #- #. Q 1 7 (   %. /  =  ( +( 8 Q3 S 3 8 ; ) " ("  ( + (  =  ( (   #( (( ' ' !; 7 . DCo = Ch Q3 = = 2 QCh + " ( " = . .% # D .  . .%  C =  # + . .% ( " = 4.%. (# Q   B 9 ; Q3 = DCo  Ch + Cb   = Ch  Cb  . " ( "  #+  #  -   =# ##  ' ; 2  Ch S3 = Q 3   Ch + Cb 7 2 E 1 S = Q 8 #/ (, = # # # #C 8 # ' ' ' E 8 S = d 8 #/ (, = . & .  1 r = dm =    (  ; % 7 2 1 = DCo Ch Q3 = = QCh + D C = Q  Ch + Cb   =  Cb  - .( + - .( = 4% ( - , " + %(   = #/ %(  %. " ("  " , "   =.  Q − S" Ch + ' & 4% ( - % %% %. " (" = #% ,% ", "  Ch S3 = Q 3   Ch + Cb 2 = ,% 1 r 8 dm S 8 DCo = Ch Q3 = 7 = ( " 8 1 !; ' ( ( 8 4(- . , -(C" & / + #/( ,# + ( , = 4-/ & 4.%. (# -/ .   = #/ (,  Q − S" D S Ch + Co + Cb = #%% Q Q Q = 2   # =# ##   #+  D S Co + Cb = %( . + .% ,. + #. .. = 4#,/ #, Q Q = S . = d %. = # =( # ) #,/ #, 8 4%, #% ( # C" . :  %. !1 r = dm =  # & 1 r 8 dm  , = / ,  S 8 #/ DCo Ch Q= Q = " " = (# (# " Q = ( Q = " " = ( /, ( . (" Q =( " Q# = " ( -" = , ,- ' Q# = (C ' C 6 0 2 2 ! ( .( - # Q8 B 2 # 7 ; ( 2 /# -( -. # ##- ( /, %. % 7 2 2 #, #% # % 4#+-./ -( 4#+,%, # 4#+, ( ' C Q= DCo Ch Q = ( "% " = % % " Q = ( "% " = %# (/ / -" 9 Q 7 2 // 1 = QCh + ' +Q D Co + DC = #/ . + #/ . + %+.( Q D 8% ( "8 + Co 8 4# I 8 C 84 . / ' = 4(+ #" 0 . (, Q 8 %# (/ 2 5 1 Q8( C 8 4 ." TC 8 = Q" Ch" > D=Q"Co > DC 8 = ( " " ." > =( " # " > 8 % > > (,+ 8 (-+( ! A DCo Ch Q3 = ! // // // " " " " # # ' E Q # ' = 4 .= TC = QIC + D Co + DC Q ' Q A Q* / #% %, Ch #," 8 # "8,% # "8, ." 8 ( , 32 5 ." TC 3 ## Q3 G // Q TC 5 " I+ D+ 4%. H C "+ ' C C TC Q 7 < F ,%+.-, , +/ (-+ % + TC ' + ) + TC : % co 8 ( cu 8 # ( 8 P D ≤ Q3" = ( cu ( = = ( cu + co ( + ( ( ' co 8 ( 8 P D ≤ Q3" = < ( cu ( = = -( ( + ( cu + co (+ z 8 Q8( > co 8 . ( 8# cu 8 ( ,- ," 8 (. % ( . 8 %( P D ≤ Q3" = cu %( = = , cu + co %( + # B A ≤ !3" 8 σ 8. , !3 < Q3 8 ' , P9 Q3+ z 8 , ' > ( ( ." 8 5 " 8 P D ≥ Q3" 8 , 8 % co 8 4 ( ' 7 6 cu 8 4 % 4 ( + 4 % 4 ( 8 4/ ' ; 2 ≤ Q3" = PA z8 # + zσ 8 ( ∴ Q3 = E # " 8 %, . %"8 P - cu / = = #-( cu + co / + ( #-( 8 co 8 / 8 / cu 8 ,( /8 %, P D ≤ Q3" = cu = cu + co , ( %, = - -%, + / B A ≤ !3" 8 σ 8# - -- ( !3 < - -- Q3 8 ( > ' / . cu = cu + co +z8 %( # " 8 #, ( 7 ' .8 / # %, = # ., %, + /% Q3 8 ( cu 8 - -- 8 /% # ., co 8 . ( 8 # (( (( # " 8 ,, ( (8 P D ≤ Q3" = < ' " 8 P D ≥ Q3" 8 P9 co 8 Q3+ z 8 %( : P D ≤ Q3" = Q3 ' cu = cu + co % C +# ' - Q3 8 > P " 8 P D ≥ Q3" 8 % , P D ≤ Q3" 8 Q3 8 > . " 8 %% .( P 9 .( , = % % 8 "8 , ( "8- g8 cu 8 > P D ≤ Q3" = 8 ." > g 8 >g cu = .( cu + co 9 cu 8 cu 8 > g 8 g8 ( / ' r 8 dm 8 = ( " (8 D=Q 8 = ( 8 . = 7 P9 = " 8 =. 8 ( σ 8 ( B 9 r < z= r8 8 r− ( > = ( (+ z 8 ( ( (" 8 % .-( 0 ( # "8 ( ; 2 # 9 9 8# 5 2 P9 8 # 4(" 8 4 (= = " 8 =. 8 ( σ 8 ( B 9 < z= r8 8 r− ( > (+ z 8 "8 ,- = ,- ,- (" 8 # ,-( 0 % ' σ 8 ( B 9 r < 8 r8 > 9 9 9 9 ## +z8 (" 8 - . ( "8 % '" 8 . 8 8, ## 0 . 2 2 % 8 4(" 8 4 8 , 4(" 8 4# "8 ( : " ( (" = -/ .% . DCo = Ch Q3 = # ' σ 8( B 9 ( < 8 r8 (> +z8 ( (" 8 #( # 9 9 8 #( 9 9 2 "8 r ( 0 #( (8 8 " 4." 8 4. = σ 8( B 9 # ( z 8 r ( ( 8 # 5 # ( 8 z8 P9 D ( = ' Q3 = "8I ( "8 #% # 8 (.- (.- 9 = DCo = Ch # 8 (.- D ' " 8 " (" =# , (" " ( (.- D=Q 8 ; 2 ' D = Q3 8 / %/ "= P9 D = Q3 = . σ 8, B 9 8 r8 (> ## ' 9 8 9 9 2 ' 0 (8( 8( > zσ 8 , > M8 . (" "84 ( / M 8 #( > #% . ," 8 / ., 9 =( 8 .+ z 8 - " 8 .( /. ." .( #(" 8 .% P9 "8 r8 > zσ 8 ( > z8 9 5 9 2 M8 > zσ 8 %( > 9 5 9 2 8 %# 8 /# ## ## % " 8 %# ( 8 /# " /(" 8 4(% .## - " 8 , # 8 , # %( 8 ,# 8 ,# " /(" 8 4/, - 7 ) 4/, - 4(% .-" 8 4% # ' 0 0 2 . r ( < "8 8 4 /( 5 A 8 4%+ # ) , 6 J + : #( ' z= %− . = , < z ' +P 9 "8 ( < (C+ z 8 /, M 8 > zσ 8 . > /, ," 8 / -, 0 M8# 7 Q8# #, 8@ >@ 8 ,> ,>.8% ' #% # 8 σ 8K @ ">K @ 8 (> (> (8, ( σ = , ( =-/ .8 > 7 ">K 7 , P9 "8 = ,8 #.( z 8 -M 8 > zσ 8 % > -- - /" 8 (% (% (.- . 8 #, - " ; ! " # $ % & % & % ! ! ' ' ! ' ' ( ) $ * * + * , λ - (/. "0 x λ P /x0 - λ e x1 x e x1 x . P(x) . ( . 3.3 . 3.3 . 4." P/x ≤ 0 - 0 - P/x 6 0 - P/5 . 4(3 - . " 7 -.) P/ ≤ 0- e /. )0 - . "( P/ ≤ 0- e /. )0 - . )744 P/ 6 0- . )744 - . . λ - ." - . .) /. "0 λ / λ0 . ) /. ) . "0 P. Lq - L - Lq 2 λ Wq - Lq - λ 2 ." .) - ." W - Wq 2 - 2 - ( .) Pw - λ - . " - . )))3 .) " Pn - λ n P/n 6 0 ( λ / n /. P/n ≤ 0 - λ - P. - Lq - P. - . " .) λ0 0 n Pn . . . . "4 . .744 . 4. " - . 73) . - . ))3 - . / .0 - " ))3 8 Lq Wq - - . " )3 λ W - Wq 2 Pw = ) λ P. = − Pw = = − / − λ0 Lq λ λ / . 0 =.4 λ Lq = Wq = λ 0 - ( . = / ( = ( = . 3( ( / − (0 = ." 3 ." 3 =.4 ( = ( = /(. =.) ( 9 3 (. λ Lq = / − λ0 L = Lq + Wq = Lq λ λ λ = W = Wq + Pw = 0 = = / (0 = . (... (/( − (0 = . (... + ( = ( . (... =. . ( = . .+ ( = . (. ( ( = . ". / 0 / " 0 # & , 4 λ - - P. = − λ / − λ0 L = Lq + Wq = Lq λ λ λ = ( = = = W = Wq + Pw = = − λ Lq = ( =. . (/. (0 + ( = =" = = ( + ( =" = . 4. : - (' ; 7 P. = − λ = − ( = . () λ P =   P. = /. ()0 = . ")" (   λ   P =   P. =   /. ()0 = . .4" (     λ   P =   P. =   /. ()0 = . ."33 (     0 - P/& 0 /P. 2 P 2 P 2 P 0 . 7) ( - . . 3( P/& Lq = Wq = λ / − λ0 Lq λ = = . (3 (/( − 0 = . "(3 /7 " 0 " ' 8 . λ- - 9 9 9 9 /Lq0 /L0 /Wq0 /W0 /Pw0 .... . )))3 .... . )))3 < - = ./L0 2 = " - ./ 0 2 " - =3" :! - = ./L0 2 = . - ./ 0 2 . - =(. ∴> λ - -" . (... .... . (.. . (... . (... ! ( - ).? . - ) λ Lq - / λ0 - ( ) /) (0 - . 73) L - Lq 2 λ - . 3 " Wq - Lq - . λ W - Wq 2 Pw - λ - 7. /3 " 0 - . 4(3 ( - . " )3 ) < @ Wq - 3 " A / 0 - ).?4 - 3 ( λ Lq Wq - / Lq λ λ0 - ( 3 ( /3 ( - . .))3 (0 /" - . ))3 0 ; ( # & , λ - P. - λ Lq - ( - . ( . ( ./ . - λ0 / (0 - ( L - L 2λ Wq - Lq - . ( λ W - Wq 2 Pw - λ 8 /7 0 - . . / 0 ( - . 3( . Wq - 7 ' + 9 $( B + λ - ( - λ Lq Wq - λ0 / Lq . - ( ./ . - .. λ (0 - . (. / 0 ; λ - ( - A ' P. - . "("( ; /λ ? 0 λ P. - / ( ? .0 / (0 / .0 /. "("(0 - . 1 / / .0 (0 /". (0 Lq Wq - . Lq - . ..4 λ C /. "7 0 + ; ; " = ' ). =4 3( P. = − λ = − 3 ( = . 3(. 4 ) Wq - 8 λ Lq = Wq = Pw = / − λ0 Lq λ λ = ( = ." 3 4/4 − (0 ." 3 = . .4 ( = / ( 0 ( = . ) (. 4 = ) (D ( ( k - ' λ - (' -4 * Lq = Wq = P .' P . - . ( 4 λ? λ P. = . .)3) 1/ k − λ 0 Lq λ = P. = . ( . .)3) = .. ( ( P= 4 Pw - P/n ≥ 0 - λ? 1 /. 4 ( P. = /. ( 4 P/n ≤ 0 .( 4 . 0 40 = . 3" - . "44 ; " 44D .4 ) Wq = λ / − λ0 Lq λ E ' A 0 ( - . (. . = ( = . (. ./ . − (0 =. W = Wq + 3 /"7 λ - P. - Lq = 3" /) =. 0 / 0 Wq - ) ; ' P. - . ). L q - /λ ? 0 λ P. - . . 1 /k λ0 3 B # & , Wq - Lq λ - . ..)3 W - Wq 2 / " 0 - . .)3 /) " 0 ; " ).D 4 9 $λ-(" % $ P. ' λ ? - Lq = k (λ ? ) λ /k − 01/k − λ 0 L = Lq + Wq = Lq λ λ = P. = 4' k - ' P. - . .( ) / 40 /( "0/ 0 /. .( )0 = 3 )3 / − 01/) − ( "0 = 3 )3 + 4 = 7 "3 3 )3 = " (" W = Wq + = " +. = 3( k Pw = λ  k  / 0  ( "   . .( ) = . 4( )     P. =    k 1   k − λ  −( " 1 / 0   ; 3 )3 ; F ; 7 3( A k 6 λ' ' 8 λ - ( "' - ; * P. ' λ ? - ' Lq 3' k - ' P. - . . ( ( / λ? - 40 Lq = k (λ ? ) λ /k − 01/k − λ 0 P. = > k k- / 30 /( "0/ 0 /. . ( (0 = 3 "( 1/) − ( "0 F . 4 λ? - ) 8 ; Lq 3 "( = (" = λ W = Wq + = 4 4 + . ( = 44 $; ! P. ' ! P. G $ P. - . . 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'-$ < & 7 ? : '- 6'- - - - - '6 - '6 , 4-- 4--- * - 7 & 5 % 0 % 0 ! 0 0 % 0 ! 0 0 0 4 $ $ ≥- $ 7 ≤ ≤ ≤ ≥ 4$--$--'$--$--- *AG 2 > 8 B # 7 5 % 0 % 0 ! 0 - 0 4 0 % 0 ! $ ≤ 4$--- *AG 0 ≤ 4$--- 2 0 ≤ ≥ :$--$--- >A8 $ $ ≥- 2 , ---$ 8 , -$ , %- , @!--- 7 B , ---$ > 8 # - , --$ ,- --- 9 8 ! " $ & # % % ' ! ( ) * + * + ) , * - ! % ≤ ≥ , ! ≤ ≤ ≤ ≤ , + , - ( ! ≤ $ $ . ≤ . ≤ * , / 0 "1 ", / "0 "1 ≤ , / 0 "1 ≥ , ≥ " * , / 0 "1 ", /"0 "1 ≤ , / 0 "1 ≤ , ≤ ( "≤ 2 ≤ * , !, / 0 "1 $!0" . . / 0 "1 ", / "0 "1 ≤ , . / 0 "1 - , ≤ , ≤ "0 / !0"1 3 &,0 3 &" . . 4 * , !, / 0 "1 !0" . ( . / 0 "1 ", /"0 "1 ≤ , . / 0 "1 ,≤, ≥ / "0 1 / $0"1 3 ≤ 2 ! ≤ . ,0 3 ! &" $0"≤, . . ≤ $0" 6 3 " 1 - 8", / ,1 3 8 9 , 8 "/ : 5 ; 3 : 5 ; 3 : 5 ; 3 , 3 , - "/ 1 3 &, 3 , - "/ 1 3 " 3 , - "/ 1 3 " 7 3 ,- "/ 1 3 3 , ,/ 1 3 3 , ,/ 1 3 !, 3 , ,/ 1 3 7 3 , 7 " ,, , ,/ 1 3 , ! , , , - / 1 / 1 / 1 ≤ , ≤ ,, ≤ , , , ≤ - ≥ , → ≥ , → ≥ , → ≤ ≤ ≥ !, , , , ", /&1 / 1 / 1 ≥ ≥ ≥ , , , → → → ≥ ≥ ≥ ,0& , , 5 ' , ≤ ≤ !, ≤ , ≤ $, , , , ≤ ≤ ≥ , → ≥ , ≥ , / 1 /,1 /,1 ≤ , , , " - , /,1 / 1 /,1 , , , ≥ < ( ≥ 2 / 0 "1 ≥ / 0 "1 ≥ /"0 "1 ≥ - , $$ 0 ≤ ( ≥ ≥ ≥ 0 ≤ 2 , , , ,, ≤ ≥ $$ 0 ", ≥ → → → ≤ ,, ≤ !,, < $0" , + ; $ 3 , / $0"1 3 ,! < ! 9 ! , , , , , , , , , ,0 $ , , , "0 $ , ,0 $ , , 0 , , 0 $ "0 9 / &, , 10 $ , / , , / , , / &,10 $ ! ,0 $ , " , , , "!, 90$! 910 $ , 9 , 10 $ , $ -"!, . &,1 0 $ ≤ , → ≤ , 10 $≤, → ≥$ / , / 9 ( 2$ ≤ ≤ . 4 5 " " * / ,, , 10 $≤, → ≥$ 0 / ,,1 0 $ ≤ , → ≤ ! 0& 2$ 0 ≤ ( ≤ ! 0& < + "!, / ,1 - " /&1 3 & $! 8$ 0 : 8 ! 0& = < 3 ,,+ > 3! , 7 2 89 ,, & " - , "!, !!≤ 2 !9" $ ≤ ( ≥ ≥ ≥ ≥ / ,0 $1 / "0 $1 / ,0 $1 / 0 1 ≤$ ≤" , , , , → → → → ≥ ≤ ≤ ≤ $ $ ! ,≤ ≤ 9,, " ,≤ &≤ ! < " !! ! " $ ' ! , 9 , , , , , , ! , , 6 + 9 9 , , , , ,0 $ , 0 $ , " 0 "0 , & ,0 , , "0 $ , ,0 $ , ,0 $ , 0 , 90$! 9 &,0 $ , 0 $ , , &,0 $ , 0 $ , " , $+ $ 0 3 & 9& 0 ! + .  &,   $ ,=   ,, $ : . ,0 $,, !, 3 "$, + "&, ) ! ,,  &,   $  $, = $   , < = + 6 < , , , , , , , , , ,0 , ,09 , 0 , "0 , ,, /"0 , "0 !09 , , 1 ,, , ? " 0 ,,0 ,, ≤ , " 09 , ≤, $ " "09 ,, ,, - " 9 " 9 $,, - " . . 4 5 ) " ≤ ,, 0 ≤ $, ) " ≥ ,,0 09 ≥ /90 "1 / ,,0 1 ≥ ! < !≤ ≤ $, / ,, , 10 ≤, → ≥ , / , ,,,1 0 9 ≤ , → ≤ ", 4 ≤ ", 2 ,≤ ( ' 8 ", ? + 8 , / "1 - 8 , / ,1 3 * 8 @ (@' , " - ≤ $≤ ( 2 , ( !≤ - " " ≤ , , , → → → ≥ ≤ ≤ $ 0 " 0 0 /,1 / 1 /,1 ≥ ≥ ≥ , , , → → → ≥ ≥! 2 , ≥ ≥ ≥ / ,0 1 / 0 1 / "0 1 8!! !! - / ,091 /!091 / "091 ≥ ≥ ≥ , , , → → → ≤ ≥ ≥ & 9 90! 9 : ' 0! ≤ ( ≤9 0! ≤ 2 ≤ , 5 2 !,,09 7 0! ≤ ≤ , < ! " ( "0 , , " , , 0 "0 90 " , &0 , "0 &0$ " 0 0 , 0 "0 90 , , , , ≤ 2 . + < 3 , ! 3 "0 3 !,0 /. 1 ) . 90 ≤ , ( . ! 2 ! !,,0 ≤ + ≤ ,, &" ≤ &"0 ≤ ≤! 0 ≤$ " , 0 , ! , , ,, . " < 4 2 " " , , , " , 3 !+ 3 0 , , , , , > , 5 "0 : ≤ ≤ $, " < * ! 0 0 - / 1 / 0!1 / 0!1 !≤ 2,≤ ( ≥ ≥ ≥ , , , → → → ≥ ≤ ≤ ≤ ≤$ * ! 0 0 - /,1 / 0 1 / 0 1 ≤ 2 ≤ ≥ ≥ ≥ , , , → → → , 0 0 A ≤ ≤ ! 0! 2 "0 + "+ , " , , 0! 2 &" 0 , 0 ' < , 0! , !"0 , , 0! &" ( . ≥ −1 ≤3 ≤ ≤$ 9 ! 0 "0 " , "0 " , &" 5 ' * - ! 0 0 ≥ ≥ ≥ /,1 /,1 / 1 , , , → → → ≥ 0 0 ≤ 2! 0 ≤ ( < $ 4 % / 1 , , , , , 0$, 0$ , 0&" 0 , !+,,, 0$, 0$ , ,+,,, &0 ,, 0$ , $ +,,, &0 ,, 0$ , , , < , , &+,,, ( A 3 $ +,,, < 0$ 3 ) &,,, !,,, ,+,,, / 0$1 / 0 1 / 0$1 +,,, ≤ . + ! +,,,≤ $& . + A ≥ ≥ ≥ , , , 8 $& → → → ≤ ≥ ≤ ! +,,, +,,, $,+,,, ≤ ! +,,, ≤ , +,,, < 3 3 3 7 . &,,, !,,, ,+,,, ",,,/ 0$1 ",,,/ 0 1 ",,,/ 0$1 3 $ +,,, ",,,/ 3 3 3 &,,,0$ &+,,,0 ""+,,,0$ 3 3 3 $+ $$ $$& "+$$$ $$& 9+ $$ $& 0$1 3 ! &+,,,0$ 3 & + + + , & + . . 4 5 < A < + + 2 9/!+,,,1 - / ,+,,,1 3 $$+,,, & < 2 @ "", - &,, " - ! ! - + > 2 3 0 ,+ 3 ,+ + ,, ≥ ≥ ≥ ≥ + ≥, ! $ 3 "!0 , 3 ,+ < 3 "+ !3, . @ 3 0 ,+ ' < 3 ,+ 3 "!0 ,+ 4 2 @ " - " - " " - ! - + ≥, ! ,, ≤ ≤ $,,, ,, ≤ + 9 ' / B ,1 "!0 , < @ - - - - " - ! ! + + + !+ ! ≤ ≤ ≤ ≤ ≥, < @ , C , CC C CC C CC C , " - - CC C + CC + + ≥, " ≥ ≥ ≥ " ≥ , , " " 3 "+ ' , @ , - , - , ≤ ≤ + ≥, + < , , , 0 , , , < = , 0 0 0 0 "" , 0 , " , , " , " , , " , A 3 , 3 , 3 " 0 + @ 8 ", " - ," + + , - , ≤ ! ≤ ≤ $ ≥, < " " , , , , , , , " , * "+ , , , , , ! , 0 0! 0! 0 0! 0! 0 , !"0 , , "0 0 "0 " , "0 " , &" 3 "0 + 3 3 , < ) + !/ "0 1 - / "1 3 &" &" . 3 "0 3 " . @ , - . 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