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Unformatted text preview: BUS! 410 — Homework #1 Answers
1. The linear programming formulation is: Decision Variabies:
C = number of dozen chocolate chip cookies produced R = number of dozen oatmeal raisin cookies produced Objective Function
Maximize profit 3 2.50 C + 3.10 R Constraints:
C s 100 (demand for 00)
R s 50 (demand for OR)
8 C + to R s 1200 (Kristen & her partner's time)
C, R a 0 (nonnegativity constraints)
The constraint for Kristen’s and her partner's time is obtained as follows:
total time available = 20 hours * 60 minslhr = 1200 mins;
time required for each dozen of choc chip cookies = 4 mins for mixing &
dishing + 1 min for setting the oven timer + 2 mins for packaging + 1 min
for receiving payment 3 8 mine. The time for oatmeal raisin cookies is
determined similarly. Solution (see the first page of the Answer & Sensitivity Reports): Make 100
dozen chocolate chip cookies and 40 dozen oatmeal raisin cookies. Doing so
will use up all 20 hours of Kristen and her friend's time, and generate $374 in profit. 2. This problem is like the Piedmont Power & Light problem, LP Formulation
Exercise #3 in the coursepeck. Decision Variables:
X(i, j) a shipment from plant i to distribution center 3'. Since there are three plants
and four distribution centers, we have 12 decision variables. For example, X(1,1)
for shipments from Cleveland to Dallas, X(1,2) for shipments {tom Cleveland to Atianta, and so on. Objective function:
Minimize 8 X(t,1) + 6 X(1,2) + 10 X(1,3) + 7 X(t,4) + 9 X(2,1) + 12 X(2,2)
+ 13 X(2,3) + 5 X(2,4) + 10 X(3,i) + 13 X(3,2) + 14 X(3,3) + 7 X(3,4) Subject to:
Supply constraints:
X(1,1) + X(1,2) + X(‘t,3) + X(1,4) s 35
Total shipment from Cleveland must be less than the availability at Cleveland.
Etc. for shipments out of Chicago and Boston
Demand constraints:
Total shipment to Atlanta must be equal to or greater than it’s requirement.
X(1,2) + X(2,2) + X(3,2) ..>. 20
Etc. for shipments into Dallas, San Francisco and Philadelphia.
X(i,j) 2 O for all plants i and distribution centers 3‘. Note: In order for this problem to be feasible, the total availability must be at least as
large as the total requirement. ln our case, note that the total availability is equal to
the total demand, both being 125 units. ' Optimal Answers (see the second and third pages of the attached Answer/Sensitivity
Reports): There is more than one optimal soiution to this problem. One is: Dallas hie
Cleveland 0 20
Chicao 5 O
0 A second is: Cleveland Chica _ 0
Boston There may be others as well. All have the same total cost = $1060. 3. The firm wants to select a production plan (how many of each kind of product to
manufacture) in order to maximize contribution. Decision variables:
C = the number of chairs to produce
8 a the number of benches to produce
T a the number of tables to produce Objective function:
Max Contribution: $3.00 C + $3.00 13 + $5.00 T Constraints:
1.2 C + 1.7 B + 1.2 T s 1000 (Bending time capacity)
0.8 C + 2.3 T s 1200 (Welding time capacity)
2 C + 3 B + 4.5 T s 2000 (Tubing availabiiity) Non—negativity Soiution: From the Answer Report we see we should produce 700 chairs, no
benches, and 133 1/3 tables. The total contribution is $2,766.67. From the
Sensitivity Report we see that the value of one more unit of tube~bending time
is $1 .1667, and the vatue of one more unit of welding time is $0. Also,
Outdoors could buy up to 555.56 additional pounds of tubing at 79 cents per
pound. Finally, the reduced cost for benches is ($13833), so adding one
bench to the production mix wouid decrease contribution by $13833. in
order to determine whether that reduced cost applies for 100 benches, the
problem must be resoived with a new constraint: B a 100. When i did this I
found that the same reduced cost appties to all 100 benches, so the cost to
the company of this requirement is $138.33. 4. For this problem the formulation is: Decision variahtes:
P1 m the number of cameras produced in month 1
11 m the ending inventory of cameras in month 1
Etc. for P2, P3, P4, P5, P6, 12, i3, l4, l5, & l6
Objective function:
Min Cost 3 $110 P1 + $41+ $115 P2 + $4 l2 + $120 P3 + $4 i3
+ $125 P4 + $4 i4 + $130 P5 + $4 l5 + $130 P6 + $4 l6 Constraints:
P1 5 300 (Month 1 production capacity)
Etc. for the other months 11 s 200 (Month 1 storage capacity)
Etc. for the other months Ending inventory = Beg. inventory + Production — Demand
it = ID + P1 —— Demand (Month 1 unit balance)
Etc. for the other months Nonnegativity Soiution: From the Answer Report we shouid produce 250, 150, 300, 300, 200, and 300 in months 1 through 6. The resuiting ending inventories wiil be
200, 200, 200, 150, 100, and 0. The minimum cost wiii be $186,650. 5. Using the template we provided, the cells/rows are defined/calculated as follows: Decision variables: The loan amounts — ceil 821 (the long—term ioan) and row
22 (the short term loans). “Data” (inputs): The interest rates  ceils 69:811. the minimum ending cash
balance ~ ceil [314, the initial cash carried over — celi B15, and the cash
inﬂow/outflow values m row 28. The calculations: Beginning balance — row 19  the previous ending balance
plus the interest earned on this money, Longterm interest/payback —— row 24
—— the interest paid each month on the long term ioan, except cell N24, which
aiso includes the iong term loan payback, Short~term interest/payback w row
25 — the amount of the previous month’s short term loan pius its interest,
Baiance after loan activities ~— row 26 — the net money available for the month,
Baiance at end of month m row 29 ~— the balance after loan activities plus or
minus the cash inflow/outflow, iVlin total interest paid w cell B34  the totai
interest paid on ail loans, and Max final balance — celi 635 m the Jan balance
(ceii N26) after ail loans are paid off. (There is one more row — row 31  which is iust a copy of the value in cell
BM). There are'two basic constraints for each month in this problem:
Cash inflow = Cash Outflow Cash balance at the end of the month 5 $5,000 The optima! solution is given on the Answer/Sensitivity Report sheets. A
rather large (aimost $40,000) long term loan is taken, then this is augmented
with short term loans in the middle of the year. The maximum ending cash
position for the year is $19,286. 6. The decision variables are:
W; = the number of drivers working in month i
H; = the number of new drivers hired at the beginning of month i
Fi = the number of drivers fired at the beginning of month i The objective is to minimize iabor costs. in mathematicai terms: Min $1000 W1 + $400 H: + $200 F1 + $1000 W2 4 $400 H2 + $200 F2
+ . .. + $1000 W5 + $400 H5 + $200 F5 There are two types of constraints: W; = WM + H, —~ Li for each month i (Driver baiance constraints)
W1 2 60 etc. for each month (Driver requirements) The Part B Soiution is shown on the Part 8 Solution sheet. in words, the plan is
to hire and fire to exactiy meet each month's needs. 80, an additional 10 drivers
are hired in February to bring the incoming 50 up to February’s needs of 60.
Then an additional 10 drivers are hired in March, 20 are iaid off in April, etc. The
total cost for Tarheei Limo is $342,000. For comparison purposes with Parts C &
D, the total driver compensation is $320,000. The Part C solution is shown on the Part 0 Solution sheet. Here, it is more cost
effective to retain 15 extra drivers in the low month, Aprii, so they wiil be available
to meet the driver demand in May. Tarheel Limc’s totai cost is $350,000, and the
totai driver compensation is $337,000. So, the drivers are better off under the
labor organization’s plan, aithough Tarheel Lirno is worse off. The Part D soiution is shown on the Part D Solution sheet. The hiring/firing plan
is the same as in Part B, i.e., hire when more drivers are needed and fire any
excess drivers. Because the cost of hiring has gone down, the totai cost for
Tarheei Limo has also gone down  to $328,500. Unfortunateiy, the drivers are
no better off, at $320,000, using this plan than they were under Plan B. Tarheel
Limo benefits from the economists proposai, but not the drivers. Microsoft Excel 12.0 Answer Report
Worksheet: [Kristens Cookies.xls]Mod Cookies
Report Created: 2/ 3/ 2030 12:20:05 PM Target I () . ..
Ce] ell Constraints I I
$E$9 Choc. Chip Demand Total 100 $E$9<=SF$9 Binding 0
SESlO Oatmeal Raising Demand Total 40 SESlO<=$F$10 Not Binding 10 Microsoft Excel 12.0 Answer Report
Worksheet: {Compressonxlsﬁolution
Report Created: 2/3/2010 12:58:52 PM TartCeii (Min) al Va_ ‘ ' 0 '
Cleveland Atianta 0 20
Cleveland San Francisco 0 15
Cleveland Philadelphia 0 0
Chicago Dallas 0 5
Chicago Atlanta 0 0
Chicago San Francisco 0 15
Chicago Philadelphia 0 30
Boston Sallas O 40
Boston Adante O 0
Boston San Rancisco O 0
Boston Philadelphia 0 O Constraints Received Dallas B$1>$B$1 Bn $C$16 Received Atlanta 20 $C$16>=$C$17 Binding 0
SD$16 Received San Francisco 30 SD516>=$D$17_ Binding 0
SES16 Received Philadelphia 30 SE516>=$E$17 Binding 0
$F$13 Cieveiand Shipped 35 SF$13<2$G$13 Binding 0
$F$l4 Chicago Shipped 50 $F$14<m$6514 Binding 0
$F$1S oston hiped I H I ‘ N ‘ 40 $F$15<2$G$15 Binding 0 Microsoft Excel 12.0 Answer Report
Worksheet: [Compressor.xls]$oiution2
Report Created: 2/3/2018 3:05:36 PM iargeCe' _
Cell Ad'e”5
eli e Final Value " Ce Daiias ' 0 0
$C$13 Cleveland Atlanta 0 20
$9613 Cieveland San Francisco 0 15
$ES13 Cleveiand Philadelphia 0 0
$B$14 Chicago Dallas 0 20
$C$14 Chicago Atlanta 0 O
SD$14 Chicago San Francisco 0 0
SES14 Chicago Philadelphia 0 30
SBSlS Boston Dallas 0 25
$C$15 Boston Atlanta 0 0
$0515 Boston San Francisco 0 15
$615 Boston Piladelphia O 0 Constraints _ a“ i .. .c .... 081g ,,
SCS16 Received Atianta 20 $C$16=$C$17 Not Binding 0
$D$16 Received San Francisco 30 $D$16=$DS17 Not Binding 0
$5516 Received Philadelphia 30 SE$16a$E$17 Not Binding 0
$F$13 Cleveland Shipped 35 $F$13=SG$13 Not Binding 0
$F$14 Chicago Shipped SD $F$14=$G$14 Not Binding 0
SFES Bston Shipped 0 F315:SGI No Binding 0 Microsoft Excel 12.0 Answer Report
Worksheet: [BookZiSolution
Report Created: 2/3/2010 2:00:00 $3M Tart Cell (Max) ,_ r s ‘ 24. _ 2,765.67 $BS11 Production Flan: Chair SCSI: Production Pian: Bench
D$1 Production Pian: Table Costraints 257 Overall 7<zs Bing SESS WeldéngTimezOveralE 867 $E$8<2$F$8 NotBinding 3333333333 Mmmmmmmmwm:mmwwwmwwm..rm¢mm .‘ . ...,.._:_M.&wwo“Emmwm $3an .‘mwmw mmmmmmm.mmm om+mm QQNH ooood now 2820 “map 3225 mmmm mmmmmmmmmﬁ 8N
Emmmhumo mmmwuuu: m3m26=< o5m25=d uEmbmcou =m$>06§hm$nmmm Nwmm . Hpﬂmmou .. mama? ummmmmam omit” mmmmmmmmmé m mmmmmmmmﬂw o ﬁcmm Ema :25:on ammuw , m
wmwwhut mungﬁw‘n— mamas: $332.4 $33.30 “.833. .‘ “moo w2m> .mcm “Ego “cm—m comwusvogm Hammw 2:3. 55 853 QEQQN 6386 38mm gonzomﬁxooa "$23.25 roam”. >wm>wmmcmm 0.3 Euxm tawohumﬁ Microsoft Excel 12.0 Answer Report Worksheet: [Kodak.xis]$olution Regort Created: 2/3/2010 3:13:14 PM “’9‘ C'M‘“) .. Adjutab'l 55‘ . . Name 3514 ‘ odctioniVio 1 ' . i 3.50 25
$C$14 Production Mo 2 150 150
$13314 Production Mo 3 300 300
$15514 Production Mo 4 300 300
$FSl4 Production Mo 5 300 200
sos'm Production Mo 6 300 300
$BSlS Ending Inventory Mo 1 100 200
$0515 Ending Inventory Mo 2 100 200
$D$15 Ending inventory Mo 3 100 200
$£$15 Ending Inventory Mo 4 50 150
$F$15 Ending Inventory Mo 5 100 100
$6515 Ending Inventory Mo 6 0 H Costrai ‘ H ‘ shipped " 10 SBS1=B$1 0
'SC$16 Cameras Shipped Mo 2 150 $C$162$C$£1 Not Binding 0
$0516 Cameras Shipped Mo 3 300 SD$16¢$D$11 Not Binding 0
$E$15 Cameras Shipped Mo 4 350 SE316=SE$11 Not Binding 0
$F$16 Cameras Shipped Mo 5 250 $F$16=$F$11 Not Binding 0
$G$16 Cameras Shipped Mo 6 400 SG$16=$G$11 Not Binding 0
SBS14 Production Mo 1 250 $B$14<=SB$5 Not Binding _5_0
$CSl4 Production Mo 2 150 $C$14<=$BSS Not Binding 150
SD$14 Production Mo 3 300 $D$14<ESBSS Binding 0
SE$14 Production Mo 4 300 $E$la<e$8$5 Binding 0
$F$14 Production Mo 5 200 $F$14<¢$B$5 Not Binding 100
$6514 Produetion Mo 6 300 SGS14<5$B$5 Binding 0
SBSIS Ending inventory Mo 1 200 $8515<=$B$6 Binding 0
SC$15 Ending inventory Mo 2 200 $C$15<m$3$6 Binding 0
$D$15 Ending Inventory Mo 3 200 SD$15<mSB$6 Binding 0
$E$15 Ending Inventory Mo 4 150 SESlS<mSBS6 Not Binding 50
$F$15 Ending Inventory Mo 5 100 SF515<2$B$6 Not Binding 100
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