S7 - 1 Course Title: Production and Operations Management Course Code: MGT 362 Course Book: Operations Management 10 th Edition. By Jay Heizer & Barry Render

S7 - 2 Chapter 7S: Capacity and Constraint Management

S7 - 3 Summary  Capacity  Design and Effective Capacity  Capacity and Strategy  Capacity Considerations  Managing Demand  Demand and Capacity Management in the Service Sector

S7 - 4 Summary– Continued  Bottleneck Analysis and Theory of Constraints  Process Times for Stations, Systems, and Cycles  Theory of Constraints  Bottleneck Management  Break-Even Analysis  Single-Product Case  Multiproduct Case

S7 - 5 Outline  Reducing Risk with Incremental Changes  Applying Expected Monetary Value to Capacity Decisions  Applying Investment Analysis to Strategy-Driven Investments  Investment, Variable Cost, and Cash Flow  Net Present Value

S7 - 6 Profit corridor Loss corridor Break-Even Analysis Total revenue line Total cost line Variable cost Fixed cost Break-even point Total cost = Total revenue – 900 – 800 – 700 – 600 – 500 – 400 – 300 – 200 – 100 – – |||||||||||| Cost in dollars Volume (units per period) Figure S7.5

S7 - 7 Break-Even Analysis BEP x =break-even point in units BEP $ =break-even point in dollars P=price per unit (after all discounts) x=number of units produced TR=total revenue = Px F=fixed costs V=variable cost per unit TC=total costs = F + Vx TR = TC or Px = F + Vx Break-even point occurs when BEP x = F P - V

S7 - 8 Break-Even Analysis BEP x =break-even point in units BEP $ =break-even point in dollars P=price per unit (after all discounts) x=number of units produced TR=total revenue = Px F=fixed costs V=variable cost per unit TC=total costs = F + Vx BEP $ = BEP x P = P = F (P - V)/P F P - V F 1 - V/P Profit= TR - TC = Px - (F + Vx) = Px - F - Vx = (P - V)x - F

S7 - 9 Break-Even Example Fixed costs = $10,000 Material = $.75/unit Direct labor = $1.50/unit Selling price = $4.00 per unit BEP $ = = F 1 - (V/P) $10, [( )/(4.00)]

S Break-Even Example Fixed costs = $10,000 Material = $.75/unit Direct labor = $1.50/unit Selling price = $4.00 per unit BEP $ = = F 1 - (V/P) $10, [( )/(4.00)] = = $22, $10, BEP x = = = 5,714 F P - V $10, ( )

S Break-Even Example 50,000 – 40,000 – 30,000 – 20,000 – 10,000 – – |||||| 02,0004,0006,0008,00010,000 Dollars Units Fixed costs Total costs Revenue Break-even point

S Break-Even Example BEP $ = F ∑ 1 - x (W i ) ViPiViPi Multiproduct Case whereV= variable cost per unit P= price per unit F= fixed costs W= percent each product is of total dollar sales i= each product

S Multiproduct Example Annual Forecasted ItemPriceCostSales Units Sandwich$5.00$3.009,000 Drink ,000 Baked potato ,000 Fixed costs = $3,000 per month

S Multiproduct Example Annual Forecasted ItemPriceCostSales Units Sandwich$5.00$3.009,000 Drink ,000 Baked potato ,000 Fixed costs = $3,000 per month Sandwich$5.00$ $45, Drinks , Baked , potato $72, AnnualWeighted SellingVariableForecasted% ofContribution Item (i)Price (P)Cost (V)(V/P)1 - (V/P)Sales $Sales(col 5 x col 7)

S Multiproduct Example Annual Forecasted ItemPriceCostSales Units Sandwich$5.00$3.009,000 Drink ,000 Baked potato ,000 Fixed costs = $3,000 per month Sandwich$5.00$ $45, Drinks , Baked , potato $72, AnnualWeighted SellingVariableForecasted% ofContribution Item (i)Price (P)Cost (V)(V/P)1 - (V/P)Sales $Sales(col 5 x col 7) BEP $ = F ∑ 1 - x (W i ) ViPiViPi = = $76,759 $3,000 x Daily sales = = $ $76, days.621 x $ $5.00 = 30.6  31 sandwiches per day

S Reducing Risk with Incremental Changes (a)Leading demand with incremental expansion Demand Expected demand New capacity (c)Attempts to have an average capacity with incremental expansion Demand New capacity Expected demand (b)Capacity lags demand with incremental expansion Demand New capacity Expected demand Figure S7.6

S Reducing Risk with Incremental Changes (a)Leading demand with incremental expansion Expected demand Figure S7.6 New capacity Demand Time (years) 123

S Reducing Risk with Incremental Changes (b)Capacity lags demand with incremental expansion Expected demand Figure S7.6 Demand Time (years) 123 New capacity

S Reducing Risk with Incremental Changes (c)Attempts to have an average capacity with incremental expansion Expected demand Figure S7.6 New capacity Demand Time (years) 123

S Expected Monetary Value (EMV) and Capacity Decisions  Determine states of nature  Future demand  Market favorability  Analyzed using decision trees

S Example SHS a company that makes hospital gowns is considering capacity expansion. SHS major alternatives are to do nothing, build a small plant, build a medium plant, or build a large plant. Cost of capacity Large 100,000 $ (Favorable) ; -90,000 $ (unfavorable) Medium 60,000 $ (F) ; 10,000 $ (uf) Small 40,000 $ (F) ; 5,000 $ (uf) Probability Favorable Market 40 % ; Unfavorable Market 60%

S Expected Monetary Value (EMV) and Capacity Decisions -$90,000 Market unfavorable (.6) Market favorable (.4) $100,000 Large plant Market favorable (.4) Market unfavorable (.6) $60,000 -$10,000 Medium plant Market favorable (.4) Market unfavorable (.6) $40,000 -$5,000 Small plant $0 Do nothing

S Expected Monetary Value (EMV) and Capacity Decisions -$90,000 Market unfavorable (.6) Market favorable (.4) $100,000 Large plant Market favorable (.4) Market unfavorable (.6) $60,000 -$10,000 Medium plant Market favorable (.4) Market unfavorable (.6) $40,000 -$5,000 Small plant $0 Do nothing EMV =(.4)($100,000) + (.6)(-$90,000) Large Plant EMV = -$14,000

S Expected Monetary Value (EMV) and Capacity Decisions -$90,000 Market unfavorable (.6) Market favorable (.4) $100,000 Large plant Market favorable (.4) Market unfavorable (.6) $60,000 -$10,000 Medium plant Market favorable (.4) Market unfavorable (.6) $40,000 -$5,000 Small plant $0 Do nothing -$14,000 $13,000$18,000

S Strategy-Driven Investment  Operations may be responsible for return-on-investment (ROI)  Analyzing capacity alternatives should include capital investment, variable cost, cash flows, and net present value

S Net Present Value (NPV) whereF= future value P= present value i= interest rate N= number of years P = F (1 + i) N F = P(1 + i) N In general: Solving for P:

S Net Present Value (NPV) whereF= future value P= present value i= interest rate N= number of years P = F (1 + i) N F = P(1 + i) N In general: Solving for P: While this works fine, it is cumbersome for larger values of N

S NPV Using Factors P = = FX F (1 + i) N whereX=a factor from Table S7.1 defined as = 1/(1 + i) N and F = future value Portion of Table S7.1 Year6%8%10%12%14%

S Limitations 1.Investments with the same NPV may have different projected lives and salvage values 2.Investments with the same NPV may have different cash flows 3.Assumes we know future interest rates 4.Payments are not always made at the end of a period

S Summary  Capacity  Design and Effective Capacity  Capacity and Strategy  Capacity Considerations  Managing Demand  Demand and Capacity Management in the Service Sector

S Summary  Bottleneck Analysis and Theory of Constraints  Process Times for Stations, Systems, and Cycles  Theory of Constraints  Bottleneck Management  Break-Even Analysis  Single-Product Case  Multiproduct Case

S Summary  Reducing Risk with Incremental Changes  Applying Expected Monetary Value to Capacity Decisions  Applying Investment Analysis to Strategy-Driven Investments  Investment, Variable Cost, and Cash Flow  Net Present Value