Solutions to Assignment 10

• Problem 4-46
• Problem 4-50 thru 4-57
• Problem 4-50 thru 4-58

• 4-46


• a) H=20; F=25
• b) OV= $1450.
• c) Nothing, the constraint is not binding
• d) H=20; F=20; OV=$1200, Machine A's capacity is binding.

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• 4-50
  
a) No effect the short dovel constraint is not binding and has 360 slack units;
b)nothing, again 360 extra short dovels can absorb this shortfall;
c) 1240 and up ( allowable increase is infinity allowable decrease 360, the slack.


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• 4-51
• a) This is a binding constraint and the 10 unit reduction is within the allowable decrease thus : SP* chnage in RHS= Change in OV  or . 6* (-10) = -60
• b) Can be tightened by 90, the amount of surplus, i .e, can tighten the constraint to x1 + x2 >= 190.
• c) Shadow price of legs is  $6.0 and  is valid for 50 aditional legs , hence OV will increase by $6 * 50 = $300

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• 4-52
Optimal solution is degenerate, in general when the allowable increase or decrease of a RHS is zero the solution is degenerate.
Also if the allowable increase or decrease of an objective function coefficient is zero then we know there are alternative optima.

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• 4-53
Degenerate, because there are 6 constraints and only 5 non-zero variables: Qty of Captains and  Mates, Surplus variable of chair production constraint, Slacks of short dowels and light seats..


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• 4-54
• a) Since the reduction of $10 is within the allowable decrease of $12 the solution (130,60) remains the same.
• b) However OV will be reduced by Change in coeff * value of the variable = (-10)*60 = -$600; i.e., they will be making $10 less for each of 60 Mates they are making and selling.

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• 4-55
• a) Since the increase equals the max allowable increase, the solution will remain optimal albeit an alternative optimum will exist.
• b) OV increases by  change in coeff * value of the variable = $24 * 130 = 3120 i.e.,  they will be making $24 more for each of the 130 captains  they are making and selling.
  

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• 4-56
• a) T4=0; reduced cost is $91.11, thus it must become cheaper by at least 91.11 before it becomes attractive enough to use ore 4.
• b) No change in solution (allowable increase is 120), however the OV will be reduced by $80*0.259=$20.72.
• c) No change in solution ( allowable increase is 223.64), however the OV increases by $100 * 0.259 = $25.90.

• 4-57
• a) No change in solution ( allowable increase is 85.71), however the OV will increase by change in coeff * value of the variable = $50 * 0.037 =$1.85/per ton.
  
• b) Since the change is exactly equal to the allowable decrease, the solution will remain optimal, however there will be an alternative optimum that uses more T3 and less of the other ores. 
• c) The new OV is less by $118.269 * 0.037 = $4.38.

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• 4-58.
• a)We need to pay close attention to how the problem variables are measured:
• W -> Wheat raised (acres)
• AR -> Alfalfa Raised (tons)
• B -> Beef raised (tons)
• AB -> Alfalfa bought (tons)
• AS -> Alfalfa sold (tons)

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The objective function must express the net contribution of each activity (in the units they are measured) in dollar terms. e.g., since wheat is measured in acres we must calculate the contribution of wheat in dollars per acre, etc.
• W (wheat): 70 bushels per acre * $2 per bushel - $20 cost per acre = $120 per acre.
• AR (alfalfa raised): $28 per acre/4 tons per acre = -$7 per ton (cost).
• B (Beef raised): $500 per ton - $50 per ton = $450 per ton.
• AB (Alfalfa bought): -$28 per ton (ton)
• AS (Alfalfa sold): $22 per ton
First constraint is acreage constraint: so we must calculate acres needed per ton of alfalfa raised.
4 tons/acre => 1/4=.25 acres per ton.

 The second constraint is in acre-feet: so we must calculate acre-feet per ton of AR.
.25 acres/ton * 3 acre-feet per acre = .75 acre-feet per ton.

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• b) Since the slack is 600, a total of 400 acre-feet of water is being used ( we see the same under "final value" column).

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• c) 8000 tons.

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• d) Buys 40,000 tons of alfalfa. Note that it will never be optimal to have AB and AS both positive. 

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• e) None. The shadow price is zero, or the water constraint is inactive. He does not use all of his allotment.

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• f) An additional acre of land will add $3,100 to the profits. Notice also that this return for additional land holds for up to 1200 acres.

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• g) If the price of wheat tripled it will be 140*3 = $420. The objective function coefficient will become $420 - $20 =$400. Since the allowable increase in the coefficient of wheat is $2980 it is still too cheap to allocate any resources to wheat. Also, in general, the reduced cost of a variable whose value is currently zero (such as wheat), measures the amount by which the variable must become more attractive before it can assume a positive value. Here the reduced cost says wheat must become more attractive (high value) by at least $2,980 before it can become a viable crop. Hence W is still zero and OV remains at $2480000.

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• h)$2,480,000

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• i) Since the decrease of $1 is within the allowable range, the optimal variable values will stay the same but the OV will decrease by a dollar for every ton of Alfalfa bought (40,000 tons). Thus the OV changes by  40,000 ($-1) = -$40,000. Or the new OV will be $2,440,000.
  

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• j) Notice that as the cost of buying alfalfa decreases the OFC of this variable increases. Since the allowable increase in the OFC of AB is $6, the cost must become $22 or better before the solution changes. If that happens, Sod will probably buy more AB and increase his profits.
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