Answers:

1.The energy modifications did not generate the expected savings; labor and capital productivity decreased.

Productivity / Last Year / This Year / Change
Labor (hrs) / 11.429 / 10.667 / -6.67%
Capital ($) / 0.267 / 0.222 / -16.67%
Energy (kWh) / 1.333 / 1.538 / 15.38%
Multifactor productivity / 0.196 / 0.170 / -13.34%

2.

Service call: / Lawn Mover / Refrigerator / Washer
No. of service calls attended / 10 / 2 / 3
Price per service call / 50 / 200 / 120
Output value in dollars / 50 x 10 = 500 / 2 x 200 = 400 / 3 x 120 = 360

Output in dollars = 500 + 400 + 360 = $1260

(a)Labor Input = 3 workers x 8hours per day = 24 hours

Labor productivity = 1260/24 = $52.50 per hour of labor

(b)Input = Material + Overhead + Labor

i.e. = 200 + 50 + 6 workers x 8 hours per day x $12 per hour = $826

Multifactor productivity = 1260/826 = $1.525 per dollar of input

3. From the following graph the break-even is at x = 80000.

Algebraic solution:

Revenue = 10 x

Cost = F + V xwhere, x = volume, F = fixed cost, V = variable cost

It is known for x = 40, cost = 600. i.e. 600 = F + V(40)

Also known for x = 60, cost = 700. i.e. 700 = F + V(60)

Solving these two equations, we get V = 5 and F = 400

Therefore, Cost = 400 + 5 x

At break-even volume, Revenue = Cost, i.e. 10x = 400 + 5x.

Solving for x, we get x = 80. i.e. Break-even volume is 80,000 visits.

#4 / 0.25

/ Weak
-25
-25 / -25
0.3
Small / Fair
25
0 / 46.25 / 25 / 25
0.45
Strong
100
100 / 100
2
167.5 / 0.25
Weak
-200
-200 / -200
0.3
Expand / Fair
-25
0 / 167.5 / -25 / -25
0.45
Strong
500
500 / 500

#5

Alternative / 1 / 2 / 3 / 4
Criteria / Weight
A / 9 / 8 / 4 / 3 / 40
B / 6 / 7 / 5 / 10 / 30
C / 9 / 5 / 8 / 6 / 20
D / 2 / 5 / 9 / 8 / 10
Weghted score: / 740 / 680 / 560 / 620

Alternative 1, with a weighted score of 740.

#6

Alternative / Fixed Cost / Variable cost
A / 450000 / 12.5 / A is dominated by B
B / 326000 / 9 / Cost = 326000 + 9x
C / 425000 / 7.2 / Cost = 425000 + 7.2x
D / 400000 / 8 / Cost = 400000 + 8x

Graphical solution:

The only break-even point occurs at the intersection of B and C.

Then, 326000 + 9x = 425000 + 7.2x, Then, x = 55,000

Therefore, for 0 x 55,000 B is best; for x 55000 C is best. At x = 55,000 both B and C are best.

#7

Demand

Probability

/ 0.2 / 0.35 / 0.45
Low / Medium / High / Optimistic / Pessimistic / Laplace / EV
Decision Alternatives / Small / 400 / 400 / 400 / 400 / 400* / 400 / 400
Medium / 100 / 600 / 600 / 600 / 100 / 433.333* / 500*
Large / -300 / 300 / 900 / 900* / -300 / 300 / 450
Best outcome = / 400 / 600 / 900

Regret Matrix:

Low / Medium / High / Maximum Regret
Small / 0 / 200 / 500 / 500
Medium / 300 / 0 / 300 / 300*
Large / 700 / 300 / 0 / 700

(b) Medium with EV = 500 from the last column in the table above.

(c)

Expected value with perfect information = 400(.2) + 600(.35) + 900 (.45) = 695

Expected value of perfect information = EVPI = 695 – 500 = 195

#8

Probability / 0.2 / 0.5 / 0.3
Decline / Same / Improve / Optimistic / Pessimistic / Laplace / EV
South Korea / 21.7 / 19.1 / 15.2 / 15.2 / 21.7 / 18.667 / 18.45
China / 19 / 18.5 / 17.6 / 17.6 / 19* / 18.367 / 18.33
Taiwan / 19.2 / 17.1 / 14.9 / 14.9 / 19.2 / 17.067* / 16.86*
Philippines / 22.5 / 16.8 / 13.8 / 13.8 / 22.5 / 17.700 / 17.04
Mexico / 25 / 21.2 / 12.5 / 12.5* / 25 / 19.567 / 19.35
Best outcome / 19 / 16.8 / 12.5

(b) Taiwan with EV = 16.86 from the last column in the table above.

(c)

Expected value with perfect information = 19(.2) + 16.8(.5) + 12.5 (.3) = 15.95

Expected value of perfect information = 16.86 – 15.95 = .91

9. Max 10 x1 + 9 x2

s.t. 7/10 x1 + x2 630Cutting and dyeing

1/2 x1 + 5/6 x2 600Sewing

x1 + 2/3 x2 708 Finishing

1/10 x1 + 1/4 x2 135 Inspection & Packaging

x1 , x2 0

Constraint / Slope
Cutting and dyeing / -0.7
Finishing / -1.5
Inspection and packaging / -0.4

Slope of the objective function is –10/9 = -1.11. Comparing this slope with the slopes of the constraints, the slope of the objective function lines falls between Cutting & Dyeing and Finishing. Therefore the optimal solution is an intersection of these two lines, which is point C. Now, solve the equation of these two lines:

3/2 x Finishing:3/2 x1 + x2 = 1062

Cutting & Dyeing:7/10 x1 + x2 = 630

Subtract:8/10 x1= 432, i.e. x1= 540

Substitute x1= 540 into Cutting & Dyeing, (7/10)540 + x2 = 630, i.e. x2= 252

Then, z = 10(540) + 9(252) = 7668

Z = / 10 / 9 / 7668
x1 / x2
Decision Variable: / 540 / 252 / LHS / RHS
Cutting and dyeing / 0.7 / 1 / 630 / 630
Sewing / 0.5 / 0.833333 / 480 / 600
Finishing / 1 / 0.666667 / 708 / 708
Inspection & Packaging / 0.1 / 0.25 / 117 / 135

(c)

Constraint / Status / Slack
Cutting and dyeing / Binding / 0
Sewing / Not Binding / 120
Finishing / Binding / 0
Inspection and packaging / Not Binding / 18

(d) Sensitivity range for objective function coefficients:

For x1:6.300 to 13.500

For x2:6.667 to 14.286

Sensitivity range for RHS of constraints:

Cutting and dyeing495.6 to 682.364

Sewing480 to 

Finishing580 to 900

Inspection and packaging117 to 

Constraint / RHS range / Shadow price
Cutting and dyeing / 495.6 to 682.364 / 4.375 / Binding constraint. For each hour of labor added to Cutting and Dyeing, the profit will increase by $4.375. All other parameters are constant.
Sewing / 480 to  / 0 / Non-binding constraint. Available hours are not fully used; 120 hours remain unused. So adding more hours to Sewing will not increase the profit. All other parameters are constant.
Finishing / 580 to 900 / 6.9375 / Binding constraint. For each hour of labor added to Finishing, the profit will increase by $6.9375. All other parameters are constant.
Inspection & packaging / 117 to  / 0 / Non-binding. Available hours are not fully used; 18 hours remain unused. So adding more hours to Inspection & Packaging will not increase the profit. All other parameters are constant.

(f) Add 100 hours to Finishing. Increase in net profit will be $193.75.

10) Min .06 x1 + .05 x2

s.t. 0.30 x1 + 0.20 x2 5Protein

0.15 x1 + 0.30 x2 3Fat

x1 , x2 0

Constraint / Slope
Protein / -1.5
Fat / -0.5

Slope of the objective function is –.06/.06= -1.2. Comparing this slope with the slopes of the constraints, the slope of the objective function lines falls between slopes of the two constraints. Therefore the optimal solution is an intersection of these two lines, which is point B. Now, solve the equation of these two lines:

2 x Fat:.3x1 +.6x2 = 6

Protein:.3x1 + .2x2 = 5

Subtract:.4 x2= 1, i.e x2= 2.5

Substitute x2= 2.5 into Protein, .3x21 + 0.2(2.5) = 5, i.e. x1= 15

Then, z = .06(15) + .05(2.5) = 1.025

(c)

Constraint / Status / Surplus
Protein / Binding / 0
Fat / Binding / 0

Both constraints are binding.

(d) shadow prices: .175, .05; (i) the cost of increasing the minimum requirement of 5 ounces of protein per day is $0.175 per ounce. All other parameters are constant; (ii) the cost of increasing the minimum requirement of 3 ounces of fat per day is $0.05 per ounce. All other parameters are constant.

(e) decrease minimum protein requirement by 2 ounces.

11) Min550x1 + 425x2 + 350x3

s.t.67x1 + 55x2 + 46x3 600Budget

(15x5) x1 + (20x2.5)x1 + (25x1)x3 550Demand

x1 + x2 + x3 15No more than 15

x3 3At least 3 econo

x1 .5 (x1 + x2 + x3) , or 0.5x1 - 0.5 x2 – 0.5 x3 0No more than 50% Super

x1 , x2 , x3 0

Monthly operating cost / 550 / 425 / 350 / 4650
Super / Regular / Econo
Number of tankers / 5 / 2 / 3 / LHS / RHS
Purchase cost / 67 / 55 / 46 / 583 / 600
Demand / 75 / 50 / 25 / 550 / 550
No more than 15 new trucks / 1 / 1 / 1 / 10 / 15
At least 3 Econo / 1 / 3 / 3
No more than 1/2 Super / 0.5 / -0.5 / -0.5 / 0 / 0

X1=5, X2=2, X3=3; Z = 4650

Constraint / Status / Slack
Purchase cost LHS / Not Binding / 17
Demand LHS / Binding / 0
No more than 15 new trucks LHS / Not Binding / 5
At least 3 Econo LHS / Binding / 0
No more than 1/2 Super LHS / Binding / 0

Sensitivity for objective function coefficients

Decision Variable / Lower Limit / Upper Limit
Super / -50 / 637.5 / The optimal solution of X1=5, X2=2, X3=3 will remain the same if the operating cost for Super tankers is within 0 to 637.5. All other parameters are constant.
Regular / 366.6667 / 575 / The optimal solution of X1=5, X2=2, X3=3 will remain the same if the operating cost for Regular tankers is within 366.67 to 575. All other parameters are constant.
Econo / 230 / 1E+30 / The optimal solution of X1=5, X2=2, X3=3 will remain the same if the operating cost for Econo tankers is within 230 to . All other parameters are constant.
Constraint / Shadow Price / Lower limit / Upper limit
Purchase cost / 0 / 583 /  / The shadow price of $0 will be valid for budget of at least $583,000. All other parameters constant.
Demand / 7.8 / 300 / 567.42 / The shadow price of $7.80 will be valid for monthly demand in the range of 300,000 to 567,420 gallons. All other parameters constant.
No more than 15 new trucks / 0 / 10 /  / The shadow price of $0 will be valid for the required number of new tankers of at least 10. All other parameters constant.
At least 3 Econo / 120 / - / 4.10 / The shadow price of $120 will be valid for the econo-tankers requirement of 0 to 4. All other parameters constant.
No more than 1/2 Super / -70 / -1.37 / 1.67 / Not interpretable

Shadow price:

Constraint / Shadow price / Interpretation
Budget / 0 / Constraint is not binding. Increasing the budget will have no effect on operating cost. All other parameters are constant.
No more than half super-tankers / -70 / Not interpretable
No more than 15 vehicles / 0 / Constraint not binding.
550000 monthyl demand / 7.8 / Each additional gallon of demand, will increase monthly operating cost by $7.8 All other parameters are constant.
At least 3 Econo-tankers / 120 / Increasing the minimum number of econo-tankers to be bought by a unit will increase the monthly operating cost by $120. All other parameters are constant.

Review sheet problem #12

13.

Item / A / B
ROP / 50 / 50
Q / 100 / 100
Service / 0.87 / 0.55
No. of orders / 22.00 / 29.00
Ordering cost / 11000 / 14500
Average Inventory / 44.25 / 26.92
Holding cost / 11947.51 / 7268.07
Orders quantity satisfied / 2186.50 / 2850.00
Total cost / 22,948 / 21,768
Profit before inventory costs / 109,325 / 85,500
Net profit / 86,377 / 63,732

Average inventory = Mean inventory (int. buff.) of lead time WS + Mean inventory of ROP buffer