Audio products inc produces two AM/FM/CD players for cars. The radio/CD units are identical but the mounting hardware and finish trim differ. The standard model fits intermediate and full size cars, and the sports model fits small sports cars...
(Complete problem found in attached files)© BrainMass Inc. brainmass.com October 24, 2018, 6:39 pm ad1c9bdddf
This posting provides solution to several problems of operations management including Inventory Control Models and Linear Programming Models
Lila Battle has determined that the annual demand for number 6 screws is 100,000 screws. Lila, who works in her brother's hardware store, is in charge of purchasing. She estimates that it costs $10 every time an order is placed. This cost includes her wages, the cost of the forms used in placing the order, and so on. Furthermore, she estimates that the cost of carrying one screw in inventory for a year is one-half of 1 cent. Assume that the demand is constant throughout the year.
(a) How many number 6 screws should Lila order at a time if she wishes to minimize total inventory cost?
(b) How many orders per year would be placed? What would the annual ordering cost be?
(c) What would the average inventory be? What would the annual holding cost be?
It takes approximately 8 working days for an order of number 6 screws to arrive once the order has been placed. (Refer to problem 6-17). The demand for number 6 screws is fairly constant, and on the average, Lila has observed that her brother's hardware store sells 500 of these screws each day. Because the demand is fairly constant, Lila believes that she can avoid stock-outs completely if she only orders number 6 screws at the correct time. What is the ROP?
Jan Gentry is the owner of a small company that produces electric scissors used to cut fabric. The annual demand is for 8,000 scissors, and Jan produces the scissors in batches. On the average, Jan can produce 150 scissors per day, and during the production process, demand for scissors has been about 40 scissors per day. The cost to set up the production process is $100, and it costs Jan 30 cents to carry one pair of scissors for one year. How many scissors should Jan produce in each batch?
The demand for barbeque grills has been fairly large in the past several years, and Home Supplies, Inc., usually orders new barbeque grills five times a year. It is estimated that the ordering cost is $60 per order. The carrying cost is $10 per grill per year. Furthermore, Home Supplies, Inc., has estimated that the stockout cost is $50 per unit. The ROP is 650 units. Although the demand each year is high, it varies considerably. The demand during the lead time is shown in the following table:
DEMAND DURING LEAD TIME PROBABILITY
The lead time is 12 working days. How much safety stock should Home Supplies, Inc., maintain?
Georgia Products offer the following discount schedule for its 4- by 8-foot sheets of good-quality plywood:
ORDER UNIT COST ($)
9 sheets or less 18.00
10 to 50 sheets 17.50
More than 50 sheets 17.25
Home Sweet Home Company orders plywood from Georgia Products. Home Sweet Home has an ordering cost of $45. The carrying cost is 20%, and the annual demand is 100 sheets. What do you recommend?
The demand for product S is 100 units. Each unit of S requires 1 unit of T and ½ unit of U. Each unit of T requires 1 unit of V, 2 units of W, and 1 unit of X. Finally, each unit of U requires ½ unit of Y and 3 units of Z. All items are manufactured by the same firm. It takes two weeks to make S, one week to make T, two weeks to make U, two weeks to make V, three weeks to make W, one week to make X, two weeks to make Y, and one week to make Z.
(a) Construct a material structure tree and a gross material requirements plan for the dependent inventory items.
(b) Identify all levels, parents, and components.
(c) Construct a net material requirements plan using the following on-hand inventory data:
ITEM S T U V W X Y Z
On-Hand Inventory 20 20 10 30 30 25 15 10
The Electrocomp Corporation manufactures two electrical products: air conditioners and large fans. The assembly process for each is similar in that both require a certain amount of wiring and drilling. Each air conditioner takes 3 hours of wiring and 2 hours of drilling. Each fan must go through 2 hours of wiring and 1 hour of drilling. During the next production period, 240 hours of wiring time are available and up to 140 hours of drilling time may be used. Each air conditioner sold yields a profit of $25. Each fan assembled may be sold for a $15 profit. Formulate and solve this LP production mix situation to find the best combination of air conditioners and fans that yields the highest profit. Use the corner point graphical approach.
The dean of the Western College of Business must plan the school's course offerings for the fall semester. Student demands make it necessary to offer at least 30 undergraduate and 20 graduate courses in the term. Faculty contracts also dictate that at least 60 courses be offered in total. Each undergraduate course taught costs the college an average of $2,500 in faculty wages, and each graduate course costs $3,000. How many undergraduate and graduate courses should be taught in the fall so that total faculty salaries are kept to a minimum?
MSA Computer Corporation manufactures two models of minicomputers, the Alpha 4 and the Beta 5. The firm employs five technicians, working 160 hours each per month, on its assembly line. Management insists that full employment (i.e., all 160 hours of time) be maintained for each worker during next month's operations. It requires 20 labor hours to assemble each Alpha 4 computer and 25 labor hours to assemble each Beta 5 model. MSA wants to see at least 10 Alpha 4s and at least 15 Beta 5s produced during the production period. Alpha 4s generate $1,200 profit per unit, and Beta 5s yield $1,800 each. Determine the most profitable number of each model of minicomputer to produce during the coming month.
Solve the following LP problem using the corner point graphical method:
maximum profit = 4X + 4Y
subject to: 3X + 5Y < 150
X - 2Y < 10
5X + 3Y < 150
X, Y > 0
Consider the LP formulation:
minimize cost = $X + 2Y
subject to: X + 3Y > 90
8X + 2Y > 160
3X + 2Y > 120
Y < 70
X, Y > 0
Graphically illustrate the feasible region and apply the isocost line procedure to indicate which corner point produces the optimal solution. What is the cost of this solution?
Graph the following LP problem and indicate the optimal solution point
maximum profit = $3X + 2Y
subject to: 2X + Y < 150
2X + 3Y < 300
(a) Does the optimal solution change if the profit per unit of X changes to $4.50?
(b) What happens if the profit function should have been $3X + $3Y?
Chapter 6 Hints
ABCO example in the book - Table 6.6: The re-order point they have determined to use is 50 (see page 212). After the table is set up as in Table 6.6 and the EMV is calculated, the lowest EMV = $110 and that's for ROP = 70 (confusing terminology). Since the ROP they use is 50, 70 - 50 = 20. In order for them to have the lowest EMV, they'll reorder when they get down to 70--that means SS = 20.
Q number pieces to order
EOQ = Q* optimal number of pieces to order
? b) Nr of orders =
Annual ordering cost =
? c) Average inventory level =
Annual holding cost =
? .5 ½ of 1 cent = ½ of .01 = .005
6-31 & 6-32
? Need to develop a table like Table 6.6 on page 213. How to calculate values:
Along the diagonal:
If ROP = demand over lead-time, the total cost equals $0
Above the diagonal are the cases where the demand is greater than ROP:
If ROP < demand over lead-time
TC = number of units short x stockout cost per unit x number or orders per year
If: ROP = 650
Demand During Lead Time = 70
Stockout Cost per Unit = $50
Number of orders per year = 5
TC = (700 - 650) x $50 x 5 = $12,500
Below the diagonal are the cases where ROP is greater than the demand over the lead time:
If ROP > demand over lead-time
TC = number of surplus units x the carrying cost
If: ROP = 650
Demand During Lead Time = 600
Carrying Cost = $10
TC = (650 - 600) x $10 = $500
? Then you must calculate EMV = sum of each probability times the related cost. In the case of a ROP of 650, the EMV is xcalculated as such: ipi (X=TC and p=probability)
EMV = (500 x 0.3) + (0 x 0.2) + (12,500 x 0.1) + (25,000 x 0.1) + (37,500 x 0.05)
+ (50,000 x 0.05) + (62,500 x 0.05) + (75,000 x 0.05) + (87,500 x 0.05) +
(100,000 x 0.03) + (112,500 x 0.02) = $33,375
? Calculate Q* based on D, , C (cost per unit) and = IC
? Then calculate Total Cost using equation 6-14 on page 206:
? See Figures 6.11, 6.12, and 6.13 on pages 219-221.
? For the Gross and Net Material Requirements: Work backward from S. You must make sure T and U are ready for S and that V, W & X are ready for T and Y & Z are ready for U. It takes 2 weeks to make S so T & U have to be ready two weeks before S is needed ("the end"). T has a 1 week lead time so V, W & X have to be ready 1 week before T is needed. When you work backward, you'll determine the number of weeks required.
? Gross Materials Requirements Plan
o But S takes 2 weeks so the Order Release would be in Week 5. Everything that goes into S must be completed no later than Week 5.
o T must be completed no later than Week 5. T has a 1 week lead time. Therefore the Required Week is 5 and the Order Release is Week 4
o Continuing, everything that goes into T must be completed no later than Week 4.
? Net Materials Requirements Plan
? You must take into account the number on hand when you calculate the number that are needed.
o For S, 80 must be ordered since 20 are on-hand.
o Each S is comprised of 1 T and ½ U
o 80 T are required but 20 are on hand so 60 are ordered
o 40 U are required and 10 are on hand so 30 are ordered
o Each T is comprised of 1 V, 2 W, 1 X and 3 Z
o Since 60 T are ordered, 60 V, 120 W , 60 X are required
o 60 V are required but 30 are on hand so 30 are ordered
o 120 W are required but 30 are on hand so 90 are ordered
o 60 X are required but 25 are on hand so 35 are ordered.
o Each U is comprised of ½ Y and 3 Z
o 30 U are ordered so 15 Y and 90 Z are required.
o 15 Y are required but there are 15 on hand so none need to be ordered.
o 90 Z are required but there are 10 on hand so 80 need to be ordered.
? You must not only take into account the number on hand when you calculate the number that are needed, but the number of components to which it belongs to be acquired.
Chapter 7 Hints
For each one of the problems in Chapter 7:
1. State the objective function to be maximized or minimized.
2. State the constraints. Must state constraints correctly with , , or =.
3. Plot the constraint lines on a graph.
4. Identify feasible region
5. Identify corner points of feasible region.
6. If two lines cross, and where they cross is a corner point, solve simultaneous solutions to find the point of intersection.
7. Test all corner points of in objective function.
8. Identify the point that meets all the criteria and maximizes (or minimizes) the objective function.
When the goal is to maximize the object function, the feasible region is below the "highest" constraint line and sometimes below all the constraint lines, but never above all the constraint lines. When the goal is to minimize the object function, the feasible region is above the constraint lines.
To solve simultaneous equations:
5X + 3Y <= 150
3X + 5Y <= 150
Multiply the first equation by 3:
3(5X + 3Y <= 150)
and the 2nd equation by -5
-5(3X + 5Y <= 150)
Which results in these two equations
15X + 9Y <=450
-15X -25Y <= -750
Then add them:
-16Y <= - 300
Y <= 18.75
Now plug this in either equation:
5X + 3(18.75) <= 150
5X + 56.25 <= 150
5X <= 93.75
X <= 18.75
So C = (18.75, 18.75)
The Corner Point Method calls for testing all endpoints. See Table 7.3.
? Minimize cost: $2,500U + $3,000G
? Subject to 30U + 20G < 60
? Must offer at least 30 U and 20 G courses. This is written: U >= 30 and G >= 20
? This is what we have:
Minimize cost: $2,500U + $3,000G
U + G >= 60
U >= 30
G >= 20
? Subject to 20A4 + 25B5 = 800 hrs (note the = sign)
? Must have at least 10 A4s and 15 B5s. Constraint is written: A4 >= 10 and B5 >= 15
? This is what we have:
Maximize profit = $1,200A4 + $1,800B5
20A4 + 25B5 = 800
A4 >= 10
B5 >= 15
? Need to draw X - 2Y<= 10 above the x axis. Since two points determine a line, simply extend the line until it crosses the other lines. This line is one of the constraint lines and calculating the intersection provides you another corner point of the feasible region.